West Bengal Madhyamik Result 2020: The West Bengal Board of Secondary Education will release the results of West Bengal Madhyamik Pariksha on July 15, 2020, from 10 AM onwards. The examination was held in February. The results will be declared online on the official website wbresults.nic.in. Read on to find out all the details about the results.

Direct link to Check West Bengal Madhyamik Result 2020

State Board Results

West Bengal Madhyamik Pariksha Results 2020

Steps to Check West Bengal Madhyamik Pariksha Results

The results can be checked through two ways

1. Online Mode
2. Offline Mode/SMS Mode

Online Mode

The results can be checked online by following the steps below

1. Go to the official website wbresults.nic.in
2. Click on “Result for Madhyamik Pariksha (S.E)”
3. Enter your roll number and date of birth
4. Click on submit to generate the result.

Offline Mode/ SMS Mode

The results can also be checked via SMS. The candidates have type the following according to the operator to obtain the results via SMS

Details Mentioned on West Bengal Madhyamik Scorecard

Scorecard of West Bengal Madhyamik Pariksha Results consist of following details

• Name of the Candidate
• Roll Number of the Candidate
• Student’s Date of Birth
• Parents’ Name
• Category
• School Code
• Marks Obtained in each Subject
• Grade Obtained
• Total Marks
• Qualifying Status

Previous Year Statistics of West Bengal Madhyamik Pariksha Results

We hope the detailed article on West Bengal Madhyamik Result 2020 is helpful. If you have any doubt regarding this article or West Bengal Madhyamik Result 2020, drop your comments in the comment section below and we will get back to you as soon as possible.

The post West Bengal Madhyamik Result Released | West Bengal Board Class 10 Result @ wbresults.nic.in appeared first on Learn CBSE.

CBSE Class 10 Result 2020: The much awaited CBSE 10th result 2020 has been declared on the official website on 15th July 2020 at 1.13 PM. The officials of Central Board of Secondary Education released CBSE Class 10th Results on the official website of cbse.nic.in.

Check CBSE 10th Result 2020 – Direct Link 

State Board Results

Students who have appeared for the CBSE Class 10 board exam can visit the official website of cbse.nic.in to check their Class 10 CBSE Results in 2020. In order to check their CBSE Class 10 Result 2020, candidates must have their registration number & D.O.B which is as mentioned in CBSE Class 10 hall ticket 2020. In this article, we will provide you with all the necessary information regarding CBSE Class 10 Result 2020. Read on to find out CBSE 10th Class Result 2020.

CBSE Class 10 Result 2020 Date and Time

Before getting into the details of CBSE Class 10 Result 2020, let’s have an overview of CBSE Class 10 Result date & time:

How To Check CBSE Class 10 Result 2020?

Follow the steps as listed below to check your CBSE Class 10 Results 2020:

• Step 1: Visit the official website: cbse.nic.in.
• Step 2: Now click on the link “Click for CBSE Results”
• Step 3: The page will be directed.
• Step 4: Now look into the “News & Events” box which is displayed on the right side of the screen.
• Step 5: Click on the link CBSE Class 10 Result.
• Step 6: Now enter your Registration Number & D.O.B.
• Step 7: Click on Submit.
• Step 8: The Results will be displayed on the screen.
• Step 9: Now download your CBSE Class 10 Results safely for future reference.

Details Mentioned On CBSE Class 10 Results 2020

The following details will be mentioned on CBSE 10th class result 2020:

1. Candidates Name
2. Candidates Roll No
3. Students Date of Birth
4. Candidates Parents Name
5. Gender
6. Candidates Photograph
7. Marks Obtained In Each Subject
8. Total Marks Obtained
9. Grade Obtained
10. Qualifying Status

Ways To Check CBSE Class 10 Results

Candidates can CBSE Class 10 Results through the following ways:

If a candidate is not satisfied with their class 10 results then they have the provision to apply for re-evaluation of their answer sheets. Students who need to verify their marks can do so by making a payment of Rs 500 per subject. The links for CBSE Class 10 Result will be activated day after the declaration of the result on the official website, cbse.nic.in. Candidates can also obtain a photocopy of their board exam answer booklet by  Rs 550 for class 10 exams, per subject. For re-evaluation purposes, the link will only be activated for one day. Students will have to pay Rs 100 per question for the re-evaluation process.

Highlights of CBSE Class 10 Grade Verification

• Candidates must pay Fee of Rs.500/- per subject through payment gateway (Credit/Debit card) or through e-challan.
•  Candidates cannot process the fee payment through offline mode.

•  If the candidate is paying through offline mode then they have a facility to pay through DD.
• Only those candidates who have applied for the verification of grades will be able to get their photocopy of evaluated answer books.
• The CBSE Class 10 grade verification application will be accepted only online through the official website of CBSE: cbse.nic.in
• If candidates come across any mistakes in their answer book then they must write a formal letter will also be sent by speed post.
• If any of the answers are not evaluated, then candidates must report to their respective Regional Offices of CBSE not later than seven days of receipt with the photocopy of the answer book.

CBSE Class 10 Result – Grading System

The CBSE Class 10 grading system is tabulated below:

CBSE Class 10 Result – How to calculate your CGPA?

For example, if your grade points for the five main subjects are:

• Subject 1: 8;
• Subject 2: 9;
• Subject 3: 7;
• Subject 4: 9;
• Subject 5: 9;

Now, Add the grade points: 8+9+7+9+9 = 42 and Divide it by 5

i.e., 42/5 = 8.4

So, your CGPA is 8.4.

CBSE Class 10 Result – Previous Years Statistics

The CBSE Class 10 Result Previous years analysis are given below.

Now that you are provided all the necessary information regarding CBSE Class 10 Result 2020. If you have any query regarding CBSE 10th Class Results 2020, leave your comments in the comment section below and we will get back to you as soon as possible.

The post CBSE Class 10 Result 2020 (Released) | Check CBSE Class 10 Result @ cbseresults.nic.in appeared first on Learn CBSE.

UP Scholarship 2020-21: The scholarships in UP are intended for students who are a member of the learning system of the northern state which claims of more than 2 lakh schools and around 60 universities. Strived at contributing financial backing to students at various levels of education in the state, the UP scholarship carries multiple pre-matric and post-matric plans for SC/ST/OBC/Minority/General category students.

यूपी छात्रवृत्ति 2020-21: यूपी में छात्रवृत्ति का उद्देश्य उन छात्रों के लिए है जो उत्तरी राज्य के शिक्षण प्रणाली के सदस्य हैं जो 2 लाख से अधिक स्कूलों और लगभग 60 विश्वविद्यालयों का दावा करते हैं। राज्य में शिक्षा के विभिन्न स्तरों पर छात्रों को वित्तीय सहायता प्रदान करने के प्रयास में, यूपी छात्रवृत्ति अनुसूचित जाति / अनुसूचित जनजाति / अन्य पिछड़ा वर्ग / अल्पसंख्यक / सामान्य श्रेणी के छात्रों के लिए कई प्री-मैट्रिक और पोस्ट-मैट्रिक योजनाएं चलाती है।

Apart from, the key characteristic of the scholarship application, the Scholarship UP gateway also allows the candidates to examine the status of the applied scholarship. Whether you are a new candidate or applying for a comeback, you can pursue your application request online by its dedicated online arrangement. While the fresh candidates can simply log in to the dashboards to check current application status and the existing users can seek their previous year’s application through the “Status” section.

You can also find more Scholarship Articles for 12th passed, 10th passed Students and many more.

If Students forgot registration number of up scholarship 2020 they can visit the official website to rest their new login id and password. We will discuss here the importance of each UP scholarship checking including the application procedure, eligibility criteria, rewards, etc. Candidates can check here and seek their dream academic profession with these UP accomplishments.

इसके अलावा, छात्रवृत्ति आवेदन की प्रमुख विशेषता, छात्रवृत्ति यूपी प्रवेश द्वार भी उम्मीदवारों को लागू छात्रवृत्ति की स्थिति की जांच करने की अनुमति देता है। चाहे आप एक नए उम्मीदवार हैं या वापसी के लिए आवेदन कर रहे हैं, तो आप अपनी समर्पित ऑनलाइन व्यवस्था द्वारा अपना आवेदन अनुरोध ऑनलाइन कर सकते हैं। जबकि ताजा उम्मीदवार वर्तमान आवेदन स्थिति की जांच करने के लिए केवल डैशबोर्ड में प्रवेश कर सकते हैं और मौजूदा उपयोगकर्ता “स्थिति” खंड के माध्यम से अपने पिछले वर्ष के आवेदन की तलाश कर सकते हैं।

हम यहां प्रत्येक यूपी छात्रवृत्ति के महत्व पर चर्चा करेंगे जिसमें आवेदन प्रक्रिया, पात्रता मानदंड, पुरस्कार आदि शामिल हैं। उम्मीदवार यहां देख सकते हैं और इन यूपी उपलब्धियों के साथ अपने सपनों के अकादमिक पेशे की तलाश कर सकते हैं।

UP Scholarship Time Table has been released

• UP Post-Matric Scholarship Time Table PDF
• UP Pre-Matric Scholarship Time Table PDF

UP Scholarship Pre matric correction window started on October 16, 2019. Candidates can make the corrections on the application form till October 25, 2019, on the official site at http://scholarship.up.nic.in

• Apply Online (Fresh/Renewal)
• Login for Complete Form (Fresh/ Renewal)
• Download Notification Pre-Matric | Post Matric

UP Scholarship Check Status | यूपी छात्रवृत्ति स्थिति

Candidates, who have successfully applied for the UP Scholarship and filled the application form along with the needed documents can check the status of the application for the scholarship provided by Uttar Pradesh Board, from time to time, on the university website.

With this, they will be able to track the advancement of their application and can get an approach if they will get the scholarship or not. All they have to do is, use their registration ID and password and submit the details to check the UP Scholarship status 2020. Check the table below for detailed information;

उम्मीदवार, जिन्होंने यूपी छात्रवृत्ति के लिए सफलतापूर्वक आवेदन किया है और आवश्यक दस्तावेजों के साथ आवेदन पत्र भरा है, उत्तर प्रदेश बोर्ड द्वारा प्रदान की गई छात्रवृत्ति, समय-समय पर, विश्वविद्यालय की वेबसाइट पर आवेदन की स्थिति की जांच कर सकते हैं।

इसके साथ, वे अपने आवेदन की उन्नति को ट्रैक करने में सक्षम होंगे और एक दृष्टिकोण प्राप्त कर सकते हैं कि क्या उन्हें छात्रवृत्ति मिलेगी या नहीं। उन्हें बस इतना करना है कि अपनी पंजीकरण आईडी और पासवर्ड का उपयोग करें और विवरण जमा करने के लिए यूपी छात्रवृत्ति स्थिति 2019 की जांच करें। विस्तृत जानकारी के लिए नीचे दी गई तालिका देखें;

UP Scholarship 2020-21

Every year UP scholarship schemes is announced for the students who want to seek the accomplishment for their higher education. The dates to apply for the scholarship are mentioned in the below table. The dates specified are applicable only for the scholastic year 2020-21. It may vary every year on the responsibility of the scholarship organizing committee. UP scholarship, online process and UP Scholarship registration start from 1st August 2020. UP Scholarship last date to apply is 12th November 2020.

हर साल यूपी छात्रवृत्ति योजनाओं की घोषणा उन छात्रों के लिए की जाती है जो अपनी उच्च शिक्षा के लिए उपलब्धि चाहते हैं। छात्रवृत्ति के लिए आवेदन करने की तारीखें नीचे दी गई तालिका में उल्लिखित हैं। निर्दिष्ट तिथियां केवल शैक्षिक वर्ष 2020-21 के लिए लागू हैं। यह छात्रवृत्ति आयोजन समिति की जिम्मेदारी पर हर साल भिन्न हो सकता है। यूपी छात्रवृत्ति, ऑनलाइन प्रक्रिया और यूपी छात्रवृत्ति पंजीकरण 1 अगस्त 2020 से शुरू होता है। यूपी छात्रवृत्ति आवेदन करने की अंतिम तिथि 12 नवंबर 2020 है।

UP Scholarship Eligibility Criteria | यूपी छात्रवृत्ति पात्रता मापदंड

Now, as you are informed of all the list UP scholarships, let us know further the respective eligibility criteria. Every scholarship has its own eligibility criteria but there is one criterion, which each scholarship has stated, which is, candidates need to be a permanent resident of UP state and should have a domicile certificate along with them.

Now let us check the academic qualifications and family income for all the category students required as per the eligibility criteria defined by the conducting organization.

अब, जैसा कि आपको सभी सूची यूपी छात्रवृत्ति के बारे में बताया गया है, आइए आगे संबंधित पात्रता मानदंड के बारे में जानें। प्रत्येक छात्रवृत्ति की अपनी पात्रता मानदंड है लेकिन एक मानदंड है, जिसे प्रत्येक छात्रवृत्ति ने कहा है, जो, उम्मीदवारों को यूपी राज्य का स्थायी निवासी होना चाहिए और उनके साथ एक अधिवास प्रमाण पत्र होना चाहिए।

अब आइए हम सभी योग्य छात्रों के लिए शैक्षणिक योग्यता और परिवार की आय की जाँच करें जो कि आयोजन संगठन द्वारा परिभाषित पात्रता मानदंड के अनुसार आवश्यक हैं।

• Domicile in Uttar Pradesh.
• Enrolled with School, College, Institute or University.
• Pre Matric Class 9: Passed Class 8 Exam and Enrolled in Class 9.
• Pre Matric Class 10: Passed Class 9 Exam and Enrolled in Class 10.
• Post Matric 11: Passed Class 10 Exam and Enrolled in Class 11.
• Post Matric 12: Passed Class 11 Exam and Enrolled in Class 12.
• Dashmottar: Enrolled in any UG, PG, Diploma Certificate Examination.

Pre-matric Scholarship Uttar Pradesh for ST/SC/General Category

• Students of class 9th and 10th who belong to SC/ST/General category are eligible for this scholarship
• The annual salary of the house should not be beyond Rs. 1 Lakh.

Pre-Matric Scholarship Uttar Pradesh for Minorities

• Students who are belonging to minority category and studying at graduation, postgraduation, Ph.D. or higher level, are eligible for this scholarship.
• The annual salary of the house should not be beyond Rs. 2 Lakh from all sources.

Pre-Matric Scholarship Uttar Pradesh for OBC Students

• Students belonging to the OBC category and studying in class IXth and Xth are eligible for this scholarship.
• The annual salary of the house should not be beyond Rs. 1 Lakh from all sources.

Post matric Intermediate Scholarship, Uttar Pradesh for ST/SC/General Category

• Students belonging to SC/ST/General category and are studying in class 11 and 12 are eligible for this scholarship.
• The annual salary of the house should not be beyond Rs. 2 Lakh (for General candidates) and Rs. 2.5 Lakh (for SC/ST candidates) from all sources.

Post matric Other State Scholarship, Uttar Pradesh for ST, SC, General Category

• Students belonging to SC/ST/General category and are studying in class 11th or above, are eligible for this scholarship.
• The annual salary of the house should not be beyond Rs. 2 Lakh (for General candidates) and Rs. 2.5 Lakh (for SC/ST candidates) from all sources.

Post matric Intermediate Scholarship, Uttar Pradesh for Minorities

• Students belonging to the minority community and are studying in class 11th and 12th, are eligible for the scholarship.
• The annual salary of the house should not be beyond Rs. 2 Lakh from all sources.

Post matric Intermediate Scholarship, Uttar Pradesh for OBC Students

• Students belonging to the OBC category and studying in class 11thand 12th are eligible for this scholarship.
• The annual salary of the house should not be beyond Rs. 2 Lakh from all sources.

Post matric (Other than Intermediate) Scholarship, Uttar Pradesh for ST, SC, General Category

• Students belonging to SC/ST/General category and are studying at graduation, post-graduation, Ph.D. or higher level, are eligible for this scholarship.
• The annual salary of the house should not be beyond Rs. 2 Lakh (for General candidates) and Rs. 2.5 Lakh (for SC/ST candidates) from all sources.

Post matric (Other Than Intermediate) Scholarship, Uttar Pradesh for OBC

• Students belonging to the OBC category and studying at graduation, postgraduation, Ph.D. or higher level are eligible for the scholarship.
• The annual salary of the house should not be beyond Rs.2 Lakh from all sources.

Post-Matric (Other Than Intermediate) Scholarship, Uttar Pradesh for Minorities

• Students belonging to the minority community and are studying at graduation, post-graduation, Ph.D. or higher level, are eligible for this scholarship.
• The annual salary of the house should not be beyond Rs. 2 Lakh from all sources.

TCS Internship is hosted by Tata Consultancy Services for some of the brightest graduate, post-graduate students.

UP Scholarship Application Procedure – How To Apply?

जैसा कि आपने पहले ही छात्रवृत्ति की सूची और उनके पात्रता मानदंड की जानकारी दे दी है, अब इसकी आवेदन प्रक्रिया से संबंधित जानकारी एकत्र करने का समय है। यूपी छात्रवृत्ति फॉर्म केवल ऑनलाइन उपलब्ध होगा। आइए अब नीचे दिए गए इन चरणों का पालन करके यूपी छात्रवृत्ति के लिए आवेदन करने का तरीका देखें।

As you have already given the information on the list of scholarship and their eligibility criteria, now its time to collect the information regarding its application process. UP Scholarship Form will be available online only. Let us now check how to apply for UP scholarships by following these steps given below.

• Visit the official website http://scholarship.up.nic.in/
• Click on the “Student” section and choose “New Registration”.

• Pick the scholarship for which you need to apply.
• Fill in all the necessary details.
• Click on the “Submit” button to register.
• Take the print out of the registration slip for later reference.
• Once registered, click on the “Student” section and choose the “Fresh Login” option for new applications and “Renewal Login” choice for renewal applications as per the scholarship.
• Log in with your registration number, date of birth and password.
• You will be directed to a webpage comprising important instructions about form filling.
• Read all instructions thoroughly and tick the box given at the end of the page and click “Proceed”.
• The minute you click on the “Proceed” button, you will be directed to the user dashboard.
• Click on the “Fill in application form” part.
• Fill in additional details in the scholarship application form.
• Click on the submit button.
• Once you click on submit after finishing the application form, you need to upload your photograph and other supporting documents.
• Before heading towards the final submission of the application form, the candidates are suggested to go through every detail filled carefully to avoid any sort of error afterward. Also, there is no requirement of making changes to the information filled by the candidate, once finally submit the application form.
• After the final submission of the application form, the candidates are asked to take the print out of the form and submit it along with other supporting documents to their respective educational institutions.

Documents Required for Application Process

For Fresh Candidates

• Last Qualifying Exam Mark Sheet
• Category Certificate
• Family Income Certificate
• Bank Passbook
• Fee Receipt Number
• Annual Non Refundable Amount
• Enrollment Number
• Student ID Proof
• Domicile Certificate
• Aadhar Card Number
• Latest Passport Size Photograph

For Renewal Candidates: Kindly Use Last Year Registration Number to log in to Renewal Section and Enter the Fresh Details.

List of UP Scholarships

Let us discuss here, the list of UP Scholarships, who provide these scholarships, when to apply for it, etc. in this section. Check the table below for further details.

Important Instructions for UP Scholarship

• UP scholarship is fairly available for all the candidates belonging to any of the categories such as General, OBC, SC, ST.
• Applicants from those institutes, which are blacklisted by the government, are not eligible to apply for the scholarship. These barred colleges include Chaudhary Babulal Institute of Technology and Management, Hathras, Excellence College of Professional Studies, Meerut and Shri Krishna College of Engineering and Technology, Baghpat.
• It is necessary for students to attach a hardcopy of the photocopy of the bank passbook with the application form.
• Registrants must submit the application form in the department concerned, in person and must receive the receipt.
• Candidates must provide their own effective email ID and mobile phone number.
• All the details and documents produced by the candidate must be authentic and real. If the given information is obtained wrong, then the application and data collected will be rejected.
• Suspect applications will not be sent for scholarship and will be discarded at the same time.
• Failed students are not eligible and should not apply for the scholarship.
• Candidates who have applied for the scholarship or willing to apply are suggested to not share their individual details, bank account number, IFSC code, class 10th/12th roll number, password, and other delicate information to anyone.
• Candidates can review their application status from time to time on the official website.
• Candidates are required their ID and password produced at the time of application to see the scholarship status.
• Candidates, those who are already enrolled with the portal have to renew the account by adding the updates. No new registration is required.
• UP Scholarship renewal status can check on through online by visiting the web portal of UP Scholarship login
• All updates regarding the scholarship are updated on the official portal in a regular interval of time. Candidates are requested to stay in contact for regular updates.
• Candidates are suggested to check the scholarship updates on the portal on a regular basis. They must obey all the instructions timely and carefully so that they cannot miss the chance to get the scholarship.

UP Scholarship – Rewards | यूपी छात्रवृत्ति पुरस्कार

The amount of financial assistance that a student can receive under UP scholarship and the expenses that can be covered with the help of these scholarships are mentioned in this section. Even, the conditions on the basis of which the awards are decided are covered in this section. The financial service provided to students majorly depends on your individual financial need. Also, there are some scholarships that recognize educational merit for the disbursal of scholarships. Moreover, the amount of scholarship that the students will receive is given in the table below.

एक छात्र को यूपी छात्रवृत्ति के तहत मिलने वाली वित्तीय सहायता की राशि और इन छात्रवृत्ति की मदद से खर्च किए जा सकने वाले खर्च का उल्लेख इस खंड में किया गया है। यहां तक कि, जिन स्थितियों के आधार पर पुरस्कार तय किए जाते हैं, वे इस खंड में शामिल हैं। छात्रों को प्रमुख रूप से प्रदान की जाने वाली वित्तीय सेवा आपकी व्यक्तिगत वित्तीय आवश्यकता पर निर्भर करती है। इसके अलावा, कुछ छात्रवृत्तियाँ हैं जो छात्रवृत्ति के संवितरण के लिए शैक्षिक योग्यता को मान्यता देती हैं। इसके अलावा, छात्रों को मिलने वाली छात्रवृत्ति की मात्रा नीचे दी गई तालिका में दी गई है।

UP Scholarship – Statistics for Last Year Session

Here is the UP scholarship statistics as of 12th December 2019. It shows the number of students who have enrolled in a particular category, finally submitted application and number of applications forwarded by the institution.

UP Scholarship Contact Details

• UP Scholarship Customer Care Phone Number: 0522-2209270, 0522-2288861, 0522-2286199
• UP Scholarship Toll-free Number: 18001805131 (Backward Class Welfare), 18001805229 (Minority Welfare)

FAQ’s on UP Scholarship

Question 1.
What is the UP scholarship?

Answer:
UP scholarship promotes the education of students who are day scholars in Uttar Pradesh (UP). Both pre-matric and post-matric scholarships are accessible for students belonging to SC/ ST/ OBC/ General categories. The scholarships covered under this parasol are offered by the Minority Welfare Department, Backward Classes Welfare Department, Tribal Welfare Department, and Samaj Kalyan Vibhag.

प्रश्न 1
यूपी छात्रवृत्ति क्या है?

उत्तर:
यूपी छात्रवृत्ति उत्तर प्रदेश (यूपी) में छात्रों की शिक्षा को बढ़ावा देती है। प्री-मैट्रिक और पोस्ट-मैट्रिक दोनों छात्रवृत्ति अनुसूचित जाति / अनुसूचित जनजाति / अन्य पिछड़ा वर्ग / सामान्य श्रेणियों के छात्रों के लिए सुलभ हैं। इस छत्र के अंतर्गत आने वाली छात्रवृत्ति अल्पसंख्यक कल्याण विभाग, पिछड़ा वर्ग कल्याण विभाग, आदिम जाति कल्याण विभाग और समाज कल्याण विभागा द्वारा दी जाती है।

Question 2.
How much amount of financial support is given to the students under the UP scholarship scheme?

Answer:
The UP government does not define the amount of scholarship it will provide to scholars. The scholarship amount changes based on the class and category to which the learners belong.

प्रश्न 2
यूपी छात्रवृत्ति योजना के तहत छात्रों को कितनी वित्तीय सहायता दी जाती है?

उत्तर:
यूपी सरकार छात्रवृत्ति प्रदान करने वाली छात्रवृत्ति की मात्रा को परिभाषित नहीं करती है। छात्रवृत्ति राशि वर्ग और श्रेणी के आधार पर बदल जाती है जिससे शिक्षार्थी संबंधित होते हैं।

Question 3.
How can I Apply for UP Scholarship?

Answer:
The applications for all UP scholarships can be filled online through its Scholarship and Fee Reimbursement Online System named ‘Saksham’.

प्रश्न 3
यूपी छात्रवृत्ति के लिए आवेदन कैसे करें?

उत्तर:
सभी यूपी छात्रवृत्ति के लिए आवेदन पत्र अपनी छात्रवृत्ति और शुल्क प्रतिपूर्ति ऑनलाइन प्रणाली के माध्यम से ऑनलाइन भरे जा सकते हैं, जिसका नाम। साक्षाम ’है।

Question 4.
Is UP scholarship available for only UP board students?

Answer:
Candidates studying at any recognized school/college/institution in UP can apply for the scholarships.

प्रश्न 4
UP Scholarship Status 2020 कैसे चेक करें?

उत्तर:
यूपी स्कालरशिप 2020-21 स्टेटस चेक करने के लिये छात्रवृत्ति एवं शुल्क प्रतिपूर्ति ऑनलाइन प्रणाली की वेबसाइट पर लॉगिन करना होगा।

The post UP Scholarship Status 2020 | Registration, How To Apply?, Eligibility Criteria appeared first on Learn CBSE.

Factoring Calculator: If you ever need assistance with finding factors of a given number take the help of the online tool provided. Just enter the input integer and you will get the factors in the blink of an eye. Scroll down to find the detailed procedure on how to evaluate Factoring using different methods. To help you out, we even jotted down the solved examples showing step by step.

How to Factor Numbers?

Use the simple procedure on how to factor numbers using the trial division method. Following it you will arrive at the solution easily. They are along the lines

• Find the square root of an integer number n and round it to the closest number. Let the number be s
• Start with 1 and find the corresponding factor pair n ÷ 1 = n. Thus 1, n are factor pairs as the division results in a whole number leaving a remainder zero.
• Continue the process with number 2 and test all the integers (n ÷ 2, n ÷ 3, n ÷ 4… n ÷ s) through the square root rounded to s. List out the factor pairs wherein the remainder is zero.
• By the time you have reached n ÷ s and noted all the factor pairs, you are done with the successful factoring of n.

Get Top Notch Calculators of Maths included with a variety of concepts like area, perimeter, factor tree, percentages, algebra, trigonometric functions, etc. all at one place Onlinecalculator.guru

Example

What are the Factors of 14

Solution: 

The Square root of 14 is 3.7416 and rounding it we get 4

Testing the integer value 1 through 4 for the division into 14 and checking for the factors that leave a remainder zero.

The numbers that leave a remainder zero when dividing the number 14 are 1, 2, 7, 14

Factor Pairs are (1,14)(2,7)(7,2)(14,1)

Take the help of Factoring Calculator to evaluate the Factors of Number 25 i.e. 1, 5, 25 the integers that divide the number 25 leaving a remainder zero.

Factor Tree of 25 to evaluate the factors

• Split the number 25 into two factors that can result in the same product after multiplication. That means 5*5 = 25
• 5 is a Prime number and the procedure ends here as the numbers can’t be expressed as a product anymore. Factor Tree of 25 looks as under

Here are some samples of factoring calculations.

• Factors of 128
• Factors of 160
• Factors of 162
• Factors of 176
• Factors of 180
• Factors of 189
• Factors of 192
• Factors of 196

How to find Factors of 25?

To find factors of 25 check for the numbers that divide 25 exactly and leaves a remainder zero. Check for all the integers that divide 25 by all the numbers up to 25. Make a note of numbers that results in the remainder zero.

1, 5, 25 are factors that divide 25 leaving a remainder zero.

Factoring Calculator | Table of Factors from 1 to 100

Example: All the factors of 12

• 2 × 6 = 12,
• but also 3 × 4 = 12,
• and of course 1 × 12 = 12.

So 1, 2, 3, 4, 6 and 12 are factors of 12.

And also -1,-2,-3,-4,-6 and -12, because you get a positive number when you multiply two negatives, such as (-2)×(-6) = 12

Answer: 1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12

Factoring Calculator – How Can I Do It Myself?

Work from the outside in!

Example: All the factors of 20.

Start at 1: 1×20=20, so put 1 at the start, and put its “partner” 20 at the other end:

Then try 2. 2×10=20 works, so put in 2 and 10:

Then try 3. 3 doesn’t work (3×6=18 is too low, 3×7=21 is too high).

Then try 4. 4×5=20 works, so put them in:

There is no whole number between 4 and 5 so you are done! (Don’t forget the negative ones).

FAQs on Factoring Numbers | Factors of a Number Calculator

1. What is a factor?

A factor is a number that divides into another number exactly and without leaving a remainder.

2. How to find factoring of numbers easily?

Take the help of the Factoring Calculator to find factors of given numbers in the blink of an eye.

3. Where do I get a detailed procedure on Factoring Numbers?

You can get a detailed procedure on factoring numbers on our page.

4. Where can I find Solved Examples on Factoring?

You can find Solved Examples on Factoring provided with detailed steps on our page.

5.  What are the Factors of 25?

Factors of 25 are 1, 5, 25.

6. How to find Factors of 25?

You can find Factors of 25 by simply dividing with all the integers up to 25. Factors are those numbers that leave a remainder zero.

7. Where do I get the detailed procedures on finding factors of 25?

You can get a detailed procedure on finding factors of 25 on our page.

You can Also Check:

• Maths Formulas
• Trigonometry Formulas

The post Factoring Calculator | Find Factors of a Number appeared first on Learn CBSE.

NCERT Solutions for Class 6 English are solved by experts of LearnCBSE.in in order to help students to obtain excellent marks in their annual examination. All the questions and answers that are present in the CBSE NCERT Books has been included in this page. We have provided all the Class 6 English NCERT Solutions (Honeysuckle, A Pact With the Sun) with a detailed explanation i.e., we have solved all the question with step by step solutions in understandable language. So students having great knowledge over NCERT Solutions Class 6 English can easily make a grade in their board exams.

NCERT Solutions for Class 6 English – Honeysuckle

NCERT Solutions for Class 6 English Honeysuckle Prose

• Chapter 1 Who Did Patrick’s Homework
• Chapter 2 How the Dog Found Himself
• Chapter 3 Taros Reward
• Chapter 4 An Indian American Woman in Space
• Chapter 5 A Different Kind of School
• Chapter 6 Who I Am
• Chapter 7 Fair Play
• Chapter 8 A Game of Chance
• Chapter 9 Desert Animals
• Chapter 10 The Banyan Tree

NCERT Solutions for Class 6 English Honeysuckle Poetry

1. A House, A Home
2. The Kite
3. The Quarrel
4. Beauty
5. Where Do All the Teachers Go?
6. The Wonderful Words
7. Vocation
8. Whatif

NCERT Solutions for Class 6 English(Download PDF)

NCERT Solutions for Class 6 English – A Pact with the Sun

NCERT Solutions for Class 6 English A Pact With the Sun – Supplementary Reader

• Chapter 1 A Tale of Two Birds
• Chapter 2 The Friendly Mongoose
• Chapter 3 The Shepherd’s Treasure
• Chapter 4 The Old-Clock Shop
• Chapter 5 Tansen
• Chapter 6 The Monkey and the Crocodile
• Chapter 7 The Wonder Called Sleep
• Chapter 8 A Pact with the Sun
• Chapter 9 What Happened to the Reptiles
• Chapter 10 A Strange Wrestling march

CBSE Class 6 English Reading

• Unseen Passages Type I
• Unseen Passages Type II

CBSE Class 6 English Writing

• Message Writing
• Notice Writing
• Postcard Writing
• Telegram Writing
• Paragraph Writing
• Composition Based on Verbal Input
• Composition Based on Visual Input
• Letter Writing
• Application Writing

CBSE Class 6 English Grammar

• Articles
• Noun
• Pronouns and Possessive Adjectives
• Adjectives
• Agreement of Verb and Subject
• Preposition
• Verb
• Tenses
• Active and Passive Voice
• Reported Speech
• The Sentence
• The Parts of Speech
• Kinds of Noun
• Uses of Articles (A, An and The)
• Degrees of Comparison
• Modals
• Correct Use of Verbs
• Correct Uses of Prepositions
• Editing Task (Omissions)
• Editing Task (Error Correction)
• Rearrangement of Words (Jumbled Words)
• Word Power

NCERT Solutions for Class 6 English

In this page, each and every question originate with a step-wise solution. Working on NCERT Solutions for Class VI will help students to get an idea about how to solve the problems. With the help of these NCERT Solutions for Class 6 English you can easily grasp basic concepts better and faster. Moreover, it is a perfect guide to help you to score good marks in CBSE board examination. Just click on the chapter wise links given below to practice the NCERT Solutions for the respective chapter.

Advantages of Solving NCERT Solutions for Class 6 English From LearnCBSE.in

•  All the Class 6 English NCERT Solutions provided in this page are clear and concise in nature.
• NCERT Solutions for Class 6 English are solved in easily understandable language to help students to grasp everything on the go.
• Accessible to everyone at any time anywhere without any difficulty.
• All the questions are solved strictly based on the NCERT (CBSE) Syllabus and Books. So mastering these solutions will definitely help students to score good marks in the examination.
• NCERT Solutions for Class 6 English Honeysuckle, A Pact with the sun Supplementary Reader, Reading,writing, unseen passages, Grammar given in this page are of free of cost.

We hope the NCERT Solutions for Class 6 English provided in this page helps in your board exam preparation. If you have any questions, ping us through the comment section below and we will get back to you as soon as possible.

The post NCERT Solutions for Class 6 English appeared first on Learn CBSE.

NCERT Solutions for Class 6 are solved by experts of LearnCBSE.in in order to help students to obtain excellent marks in their board examination. All the questions and answers that are present in the CBSE NCERT Books has been included in this page. We have provided all the Class 6 NCERT Solutions with a detailed explanation i.e., we have solved all the questions with step by step solutions in understandable language. So students having great knowledge over NCERT Solutions Class 6 can easily make a grade in their board exams. Read on to find out more about NCERT Solutions for Class 6 Maths, Class 6 Science, Class 6 Social Science Geography, History, Civics, Class 6 Honeysuckle Prose and Poetry, Class 6 English A Pact with the Sun, Class 6 English Reading, Class 6 English Writing, Class 6 English Grammar, Class 6 Hindi, Class 6 Hindi Unseen Passages, Class 6 Hindi Grammar, Class 6 Hindi लेखन कौशल, Class 6 Sanskrit.

NCERT Solutions for Class 6

Here we have provided Updated NCERT Solutions for Class 6 for 2020 Exams. NCERT Solutions clear your concepts to the core ensuring to stay with you in the long run. These are created by experts in accordance with the CBSE curriculum. You can find 100% accurate step-by-step solutions for every question in all the latest NCERT books. Practice with these daily to perform better in boards.

NCERT Solutions for Class 6 Maths

• Chapter 1 Knowing Our Numbers
• Chapter 2 Whole Numbers
• Chapter 3 Playing With Numbers
• Chapter 4 Basic Geometrical Ideas
• Chapter 5 Understanding Elementary Shape
• Chapter 6 Integers
• Chapter 7 Fractions
• Chapter 8 Decimals
• Chapter 9 Data Handling
• Chapter 10 Mensuration
• Chapter 11 Algebra
• Chapter 12 Ratio and Proportion
• Chapter 13 Symmetry
• Chapter 14 Practical Geometry

NCERT Solutions for Class 6 Science

• Chapter 1 Food Where Does It Come From
• Chapter 2 Components of Food
• Chapter 3 Fibre to Fabric
• Chapter 4 Sorting Materials Into Groups
• Chapter 5 Separation of Substances
• Chapter 6 Changes Around Us
• Chapter 7 Getting to Know Plants
• Chapter 8 Body Movements
• Chapter 9 The Living Organisms and Their Surroundings
• Chapter 10 Motion and Measurement of Distances
• Chapter 11 Light Shadows and Reflection
• Chapter 12 Electricity and Circuits
• Chapter 13 Fun with Magnets
• Chapter 14 Water
• Chapter 15 Air Around Us
• Chapter 16 Garbage In Garbage Out

NCERT Solutions for Class 6 Social Science Geography

• Chapter 1 The Earth in the Solar System
• Chapter 2 Globe Latitudes and Longitudes
• Chapter 3 Motions of the Earth
• Chapter 4 Maps
• Chapter 5 Major Domains of the Earth
• Chapter 6 Major Landforms of the Earth
• Chapter 7 Our Country India
• Chapter 8 India Climate Vegetation and Wildlife

NCERT Solutions for Class 6 Social Science – History

• Chapter 1 What, Where, How and When?
• Chapter 2 On The Trial of the Earliest People
• Chapter 3 From Gathering to Growing Food
• Chapter 4 In the Earliest Cities
• Chapter 5 What Books and Burials Tell Us
• Chapter 6 Kingdoms, Kings and an Early Republic
• Chapter 7 New Questions and Ideas
• Chapter 8 Ashoka, The Emperor Who Gave Up War
• Chapter 9 Vital Villages, Thriving Towns
• Chapter 10 Traders, Kings and Pilgrims
• Chapter 11 New Empires and Kingdoms
• Chapter 12 Buildings, Paintings, and Books

NCERT Solutions for Class 6 Social Science – Civics

• Chapter 1 Understanding Diversity
• Chapter 2 Diversity and Discrimination
• Chapter 3 What is Government
• Chapter 4 Key Elements of a Democratic Government
• Chapter 5 Panchayati raj
• Chapter 6 Rural Administration
• Chapter 7 Urban Administration
• Chapter 8 Rural Livelihoods
• Chapter 9 Urban Livelihoods

NCERT Solutions for Class 6 English Honeysuckle Prose

• Chapter 1 Who Did Patrick’s Homework
• Chapter 2 How the Dog Found Himself
• Chapter 3 Taros Reward
• Chapter 4 An Indian American Woman in Space
• Chapter 5 A Different Kind of School
• Chapter 6 Who I Am
• Chapter 7 Fair Play
• Chapter 8 A Game of Chance
• Chapter 9 Desert Animals
• Chapter 10 The Banyan Tree

NCERT Solutions for Class 6 English Honeysuckle Poetry

1. A House, A Home
2. The Kite
3. The Quarrel
4. Beauty
5. Where Do All the Teachers Go?
6. The Wonderful Words
7. Vocation
8. Whatif

NCERT Solutions for Class 6 English(Download PDF)

NCERT Solutions for Class 6 English – A Pact with the Sun

NCERT Solutions for Class 6 English A Pact With the Sun – Supplementary Reader

• Chapter 1 A Tale of Two Birds
• Chapter 2 The Friendly Mongoose
• Chapter 3 The Shepherd’s Treasure
• Chapter 4 The Old-Clock Shop
• Chapter 5 Tansen
• Chapter 6 The Monkey and the Crocodile
• Chapter 7 The Wonder Called Sleep
• Chapter 8 A Pact with the Sun
• Chapter 9 What Happened to the Reptiles
• Chapter 10 A Strange Wrestling march

CBSE Class 6 English Reading

• Unseen Passages Type I
• Unseen Passages Type II

CBSE Class 6 English Writing

• Message Writing
• Notice Writing
• Postcard Writing
• Telegram Writing
• Paragraph Writing
• Composition Based on Verbal Input
• Composition Based on Visual Input
• Letter Writing
• Application Writing

CBSE Class 6 English Grammar

• Articles
• Noun
• Pronouns and Possessive Adjectives
• Adjectives
• Agreement of Verb and Subject
• Preposition
• Verb
• Tenses
• Active and Passive Voice
• Reported Speech
• The Sentence
• The Parts of Speech
• Kinds of Noun
• Uses of Articles (A, An and The)
• Degrees of Comparison
• Modals
• Correct Use of Verbs
• Correct Uses of Prepositions
• Editing Task (Omissions)
• Editing Task (Error Correction)
• Rearrangement of Words (Jumbled Words)
• Word Power

NCERT Solutions for Class 6 Hindi

• Chapter 1 वह चिड़िया जो
• Chapter 2 बचपन
• Chapter 3 नादान दोस्त
• Chapter 4 चाँद से थोड़ी सी गप्पे
• Chapter 5 अक्षरों का महत्व
• Chapter 6 पार नज़र के
• Chapter 7 साथी हाथ बढ़ाना
• Chapter 8 ऐसे ऐसे
• Chapter 9 टिकट अलबम
• Chapter 10 झाँसी की रानी
• Chapter 11 जो देखकर भी नहीं देखते
• Chapter 12 संसार पुस्तक है
• Chapter 13 मैं सबसे छोटी होऊं
• Chapter 14 लोकगीत
• Chapter 15 नौकर
• Chapter 16 वन के मार्ग में
• Chapter 17 साँस-साँस में बाँस

NCERT Solutions for Class 6 Hindi Bal Ramkatha बाल रामकथा

CBSE Class 6 Hindi Unseen Passages अपठित बोध

• अपठित गद्यांश
• अपठित पद्यांश

CBSE Class 6 Hindi Grammar व्याकरण

• भाषा, लिपि और व्याकरण
• वर्ण-विचार
• शब्द-विचार
• संज्ञा
• संज्ञा के विकार
• वचन
• कारक
• सर्वनाम
• विशेषण
• क्रिया
• काल
• वाच्य
• अव्यय या अविकारी शब्द
• संधि
• समास
• उपसर्ग
• प्रत्यय
• वाक्य-विचार
• विराम-चिह्न
• अशुद्ध वाक्यों का संशोधन
• मुहावरे और लोकोक्तियाँ
• शब्द-भंडार

CBSE Class 6 Hindi लेखन कौशल

• अनुच्छेद-लेखन
• पत्र-लेखन
• निबंध-लेखन

NCERT Solutions for Class 6 Sanskrit

• Chapter 1 – अकारान्त – पुल्लिङ्ग:
• Chapter 2 – आकारान्त – स्त्रीलिंग
• Chapter 3 – अकारान्त – नपुंसकलिंग
• Chapter 4 – क्रीडास्पर्धा
• Chapter 5 – वृक्षाः
• Chapter 6 संम्रुद्रतटः
• Chapter 7 बकस्य प्रतिकार
• Chapter 8 सूक्तिस्तवकः
• Chapter 9 अडगुलीयकं प्राप्तम्
• Chapter 10 कृषिकाः कर्मवीरः
• Chapter 11 पुष्पोत्सवः
• Chapter 12  दशमः त्वम् असि
• Chapter 13 लोकमड्गलम्
• Chapter 14 अहह आः च
• Chapter 15 मातुलचन्द्र 

Class 6 NCERT Solutions All Subjects

• NCERT Solutions for Class 6 Maths
• NCERT Solutions for Class 6 Science
• NCERT Solutions for Class 6 Social Science
• NCERT Solutions for Class 6 English
• NCERT Solutions for Class 6 Hindi
• NCERT Solutions for Class 6 Sanskrit

Advantages of Solving NCERT Solutions for Class 6 from LearnCBSE.in

•  All the Class 6 NCERT Textbook Solutions provided in this page are clear and concise in nature.
• NCERT Solutions for Class 6 Books are solved in easily understandable language to help students to grasp everything on the go.
• Accessible to everyone at any time anywhere without any difficulty.
• All the questions are solved strictly based on the NCERT (CBSE) Syllabus and Books. So mastering these solutions will definitely help students to score good marks in the examination.
• NCERT Solutions for Class VI  given in this page are of free of cost.

CBSE class 6 study materials are provided for students in downloadable PDF format. These materials are free and students can access them as per their requirement. LearnCBSE provides here NCERT solutions, revision notes, syllabus, important questions, sample papers and previous year question papers for students of 6th standard, so that they can prepare for exams thoroughly and score good marks in the final examinations. All these materials are designed keeping into consideration the syllabus as per the NCERT curriculum by our subject experts and teachers. 

Class 6 is the next level for students after they have crossed their primary schooling. The solutions provided by us covers all the subjects chapters-wise and exercise-wise, so that students can get the best answers for all the questions here. Also, they can prepare for the exam by solving the question papers and sample papers to understand the paper pattern and weightage of each chapter, subject-wise. Important questions covers all the important topics which will be asked necessarily in the final exam. To have a quick revision, students can prefer to our notes where each topic is described briefly. Below are the links given for each subject material. Click on the links and prepare well.

We hope the NCERT Solutions for Class 6 provided in this page helps in your board exam preparation.

The post NCERT Solutions for Class 6 appeared first on Learn CBSE.

CBSE Class 6 English Letter Writing

Question 1.
To friend inviting to spend vacation
You are Ramesh living at ABC Janakpuri, Delhi. Write a letter to your friend inviting him to come to Delhi during the vacation. Mention the few attractions that you want both of you to visit.
Answer:
ABC Janakpuri
Delhi
2 April, 2007
Dear Rohit
The holidays have started. I wish to pass some time with you.
Please come to Delhi during the holidays. We shall be visiting historical places in and around Delhi. We will go to the restaurants, cinema houses etc. We will go for picnic on the river bank. We will shop in Connaught Place. I shall take you to the British Council Library and the American Study Centre.
I am sure you shall come soon.
With love
Yours lovingly
Ramesh

Question 2.
Describing how you celebrated your last birthday
You are Suresh of 25, Nehru Nagar, Hyderabad. Write a letter to your friend Vinaya describing your birthday party.
Answer:
25, Nehru Nagar
Hyderabad
10 June, 2007
My dear Vinaya
Thank you for the good-wishes card on my birthday. The card was really beautiful. However, I missed you.
Any way, my parents held a party on the occasion. All my relatives and friends came to wish me a happy birthday.
Everybody liked the birthday cake. There was a special programme of dance and music. We enjoyed ourselves till late in the evening.
My uncle from England gave us a surprise visit. My parents were very glad at it. My uncle brought with him a special gift from London. I will show you the same when you come here.
With love
Yours affectionately
Suresh

Question 3.
To father for replacement of watch
Your father has sent you a watch as a present on your birthday, but it is not keeping correct time. Write a letter informing him of this fact and re-questing him to get it replaced. You are Sudha/Sudhir.
Answer:
Miranda Hostel
ABC School
New Delhi
1 March, 2007
Dear Father
Thank you very much for sending the beautiful wrist watch on my birth¬day. However, the watch has some defect. It loses about five minutes in eveiy twenty-four hours. It has one year’s guarantee. So kindly get it replaced. Kindly tell me whom to send it to.
I shall wait for your letter. I am quite well here. I hope that everything is fine there at home.
With regards
Yours lovingly
Sudhir

Question 4.
To uncle thanking him for a nice birthday gift
Your Uncle has presented a book to you on your birthday. It is related with the art of living. Pointing out the usefulness of the book in life, write a letter in about 100 words thanking your uncle for sending this valuable gift. You are Navin/Nalini living at 250, Rajdhani Enclave Baroda.
Answer:
250 Rajdhani Enclave
Baroda
2 March, 2007
My dear Uncle
It is always a pleasure to hear from you. Last Friday was my 11th birth¬day. The same day I received your gift of a book along with your blessings.
Thank you very much for this valuable gift. I have not read the entire book. But the little that I have read is highly useful and entertaining. I feel that these are the things that matter most. I am sure the book will remain with me all my life.
1 thank you once again for the kind gift.
Yours affectionately
Navin

Question 5.
To a friend inviting him to the Book Fair
Your friend, Kunal is fond of reading good books. Write a letter to him inviting him to visit the Book Fair which is going to be held in your city in December. You are Rajiv Mittal, living at H-15, Ballabh Colony, New Delhi
Answer:
H-15, Ballabh Colony
New Delhi
2 December, 2007
My dear Kunal
Thank you very much for your letter. I am glad that everybody is fine. Our city is holding a Book Fair in the third week of December. All the major publishers of the country and from abroad are taking part. It is an opportunity for book-lovers. You are very fond of reading good books. So I invite you to visit this Book Fair in our city. When are you coming to us for this purpose?
Yours sincerely
Rajiv

NCERT Solutions for Class 6 English

The post CBSE Class 6 English Letter Writing appeared first on Learn CBSE.

NCERT Solutions For Class 10 Science Chapter 6 Life Processes: In this article, you will find all the necessary information regarding NCERT Solutions For Class 10 Science Chapter 6 Life Processes.  Students who are planning to build their career stream in the field of medicine can refer to this article as biology plays a major role in the medical field. Candidates having strong command over life processes class 10 NCERT solutions will be able to easily crack the competitive exams like NEET, AIIMS, JIPMER, etc.,

Did you check the Life Process Important Questions which were given in the previous Board Papers?

In this article, we have covered all the important topics in the exercises and each answer comes with a detailed explanation to help the class 10 students to understand concepts better. Read on to find out everything about NCERT class 10 science book activities solutions chapter 6.

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

Before getting into the details of NCERT Solutions For Class 10 Science Chapter 6 Life Processes, let’s have an overview of topics and subtopics under class 10 science chapter 6 activities solutions:

1. Life Processes
2. What Are Life Processes?
3. Nutrition
4. Respiration
5. Transportation
6. Excretion

Free download NCERT Solutions for Class 10 Science Chapter 6 Life Processes PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

• जैव-प्रक्रम कक्षा 10 विज्ञान हिंदी में
• Life Processes Class 10 Notes
• Life Processes NCERT Exemplar Solutions
• Life Processes Class 10 Extra Questions
• Class 10 Life Processes Mind Map

NCERT Solutions for Class 10 Science Chapter 6 Intext Questions

Page Number: 95

Question 1
Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans ?
Answer:
In multicellular organisms like humans, all the body cells are not in direct contact with the surrounding environment. Therefore, every cell of the body will not get oxygen as per need by the process of diffusion from the environment. Therefore diffusion is insufficient to meet the oxygen requirements of multicellular organisms.

Question 2
What criteria do we use to decide whether something is alive ?
Answer:
The main criteria used to decide whether something is alive are breathing and respiration. However, living beings also show growth and movement.

Question 3
What are outside raw materials used by an organism ?
Answer:
Any organism uses organic molecules as raw material. Heteroptrophs use food and autotrophs use carbon dioxide, minerals, water and all organisms use oxygen (for respiration) as raw materials.

Question 4
What processes would you consider essential for maintaining life ?
Answer:
Processes essential for maintaining life are :
(i) Nutrition
(ii) Respiration
(iii) Transportation
(iv) Excretion

Page Number: 101

Question 1
What are the differences between autotrophic nutrition and heterotrophic nutrition ?
Answer:

Question 2
Where do plants get each of the raw materials required for photosynthesis ?
Answer:
(i) Carbon dioxide : Plants get carbon dioxide from the environment/atmosphere through stomata.
(ii) Water : Plants absorb water from the soil through roots and transport to leaves.
(iii) Sunlight : Plants get sunlight from the sun.
(iv) Chlorophyll : It is present in chloroplast found in green leaves and green parts of plants.

Question 3
What is the role of the acid in our stomach ?
Answer:
Role of acid in our stomach is :
(i) To make acidic medium which is necessary for the activation of the enzyme pepsin.
(ii) To kill bacteria which the food may contain.

Question 4
What is the function of digestive enzymes ?
Answer:
The food we eat is complex in nature, i.e., it contains complex molecules. Digestive enzymes break down these complex molecules into smaller simpler molecules so that they can be absorbed by the walls of the intestine.

Question 5
How is the small intestine designed to absorb digested food ?
Answer:
The small intestine is designed to provide maximum area for absorption of digested food and its transfer into the blood for its circulation into the body. For this the inner lining of the small intestine has numerous finger-like projections called villi. The villi are richly supplied with blood vessels which take the absorbed food to each and every cell of the body.

Page Number: 105

Question 1
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration ?
Answer:
Aquatic organisms use oxygen dissolved in surrounding water. Since air dissolved in water has fairly low concentration of oxygen, the aquatic organisms have much faster rate of breathing.
Terrestrial organisms take oxygen from the oxygen-rich atmosphere through respiratory organs. Hence, they have much less breathing rate than aquatic organisms.

Question 2
What are the different ways in which glucose is oxidised to provide energy in various organisms ?
Answer:
First step of breakdown of glucose (6 carbon molecules) takes place in the cytoplasm of cells of all organisms. This process yields a three carbon molecule compound called pyruvate.
Further break down of pyruvate takes place in different ways in different organisms.

(i) Anaerobic respiration : The anaerobic respiration in plants (like yeast) produces ethanol and carbon dioxide as end products.
(ii) Aerobic respiration : In aerobic respiration break down of pyruvate takes place in presence of oxygen to give rise three molecules of carbon dioxide and water. The release of energy in aerobic respiration is much more than in anaerobic respiration.
(iii) Lack of oxygen : Sometimes, when there is lack of oxygen especially during physical exercise, in our muscles, pyruvate is converted into lactic acid (3 carbon molecule compound). Formation of lactic acid in muscles causes cramp.

Question 3
How is oxygen and carbon dioxide transported in human beings ?
Answer:
(i) Transport of oxygen : Haemoglobin present in the blood takes up the oxygen from the air in the lungs. It carries the oxygen to tissues which are deficient in oxygen before releasing it.
(ii) Transport of carbon dioxide : Carbon dioxide is more soluble in water. Therefore, it is mostly transported from body tissues in the dissolved form in our blood plasma to lungs. Here it diffuses from blood to air in the lungs.

Question 4
How are the lungs designed in human beings to maximise the area for exchange of gases ?
Answer:
Within the lungs, the air passage divides into smaller and smaller tubes, called bronchi which in turn form bronchioles. The bronchioles terminate in balloon-like structures, called alveoli. The alveoli present in the lungs provide maximum surface for exchange of gases. The alveoli have vary thin walls and contain an extensive network of blood vessels to facilitate exchange of gases.

Page Number: 110

Question 1
What are the components of the transport system in human beings ? What are the functions of these components ?
Answer:
The transport system (circulatory system) in human beings mainly consists of heart, blood and blood vessels.

(i) Function of heart : The heart receives deoxygenated blood from the body parts and pumps it to lungs for enriching with oxygen. It receives purified blood from lungs and pumps it around the body.
(ii) Function of blood : Blood transports oxygen, carbon dioxide, digested food, hormones and nitrogeneous waste like urea. It also protects the body from diseases and regulates the body temperature.
(iii) Function of blood vessels : The blood pushed by the heart flows through the blood vessels (arteries, veins and capillaries) and also comes back to the heart through them.

Question 2
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds ?
Answer:
Separation of oxygenated and deoxygenated blood allows good supply of oxygen to the body. This system is useful in animals that have high energy requirement. Mammals and birds constantly need oxygen to get energy to maintain their body temperature constant.

Question 3
What are the components of the transport system in highly organised plants?
Answer:
In highly organised plants there are two conducting tissues xylem and phloem.
Xylem consists of vessels, tracheids and other xylem tissues. The interconnected vessels and tracheids form a continuous system of water conducting channels reaching all parts of the plant. Xylem carries water and minerals.
Phloem conducts soluble products of photosynthesis from leaves to different parts of the plant body.

Question 4
How are water and minerals transport in plants ? [AICBSE 2015]
Answer:
The roots of a plant have hair called root hair.
The root hair are directly in contact with the film of water in between the soil particles. Water and dissolved minerals get into the root hair by the process of diffusion. The water and minerals absorbed by the root hair from the soil pass from cell to cell by osmosis through the epidermis, root cortex, endodermis and reach the root xylem.

The xylem vessels of the root of the plant are connected to the xylem vessels of its stem.
Therefore the water containing dissolved minerals enters the root xylem vessels into stem xylem vessels. The xylem vessels of the stem branch into the leaves of the plants. So, the water and minerals carried by the xylem vessels in the stem reach the leaves through the branched xylem vessels which enter from the petiole (stalk of the leaf) into each and every part of the leaf. Thus the water and minerals from the soil reach through the root and stem to the leaves of the plants. Evaporation of water molecules from the cells of a leaf creates a suction which pulls water from the xylem cells of roots. The loss of water in the form of vapour from the aerial parts of the plant is known as transpiration.

Question 5
How is food transported in plants ?
Answer:
The movement of food in phloem (or translocation) takes place by utilizing energy. The sugar (food) made in leaves is loaded into the sieve tubes of phloem tissue by using energy from ATR Water now enters the sieve tubes containing sugar by the process of osmosis due to which the pressure in the phloem tissue rises. This high pressure produced in the phloem tissue moves the food to all parts of the plant having less pressure in their tissues. This allows the phloem to transport food according to the needs of the plant.

Page Number: 112

Question 1
Describe the structure and functions of nephrons.
Answer:
Structure of nephron : Each nephron is composed of two parts. First one is a cup-shaped bag at its upper end which is called Bowman’s capsule.
The Bowman’s capsule contains a bundle of blood capillaries which is called glomerulus. One end of the glomerulus is attached to the renal artery which brings the impure blood containing the urea waste into it. These impurities are filtered. The other part of the nephron is coiled. In this part, the substances like sugar (glucose), amino acid, ions and excess water which are required by the body, are reabsorbed. The substance remained in the nephron is mainly urine containing dissolved urea in water which is expelled from the body through urethra from time to time.

Functions of nephron : Filtration of blood takes place in Bowman’s capsule from the capillaries of glomerulus. The filtrate passes into the tubular part of the nephron. This filtrate contains glucose, amino acids, urea, uric acid, salts and water.
Reabsorption : As the filtrate flows along the tubule, useful substances such as glucose, amino acids, salts and water are selectively reabsorbed into the blood by capillaries surrounding the nephron tubule.
Urine : The filtrate which remained after reabsorption is called urine. Urine contains dissolved nitrogenous waste like urea and uric acid, excess salts and water. Urine is collected from nephrons to carry it to the ureter from where it passes into urinary bladder.

Question 2
What are the methods used by plants to get rid of excretory products ?
Answer:
(i) The plants get rid of gaseous products-through stomata in leaves and lenticels in stems.
(ii) The plants get rid of stored solid and liquid waste by the shedding off leaves, peeling off bark and felling off fruits.
(iii) The plants get rid of wastes by secreting them in the form of gums and resins.
(iv) Plants also excrete some waste substances into the soil around them.

Question 3
How is the amount of urine produced regulated ?
Answer:
The amount of urine is regulated by kidney. It depends on the quantity of excess water and wastes dissolved in water.

(i) Quantity of water : When water is abundant in the body tissues, large quantities of dilute urine is excreted out. When water is less in quantity in the body tissues, a small quantity of concentrate urine is excreted.
(ii) Quantity of dissolved wastes : Dissolved wastes, especially nitrogenous wastes, like urea and uric acid and salts are excreted from the body. When there is more quantity of dissolved wastes in the body, more quantity of water is required to excrete them. Therefore, the amount of urine produced increases.
(iii) Hormones : The amount of urine produced is also regulated by certain hormones which control the movement of water and Na+ ions in and out of the nephrons.

NCERT Solutions for Class 10 Science Chapter 6 Textbook Chapter End Questions

Question 1
The kidneys in human beings are a part of the system for
(i) nutrition
(ii) respiration
(iii) excretion
(iv) transportation
Answer:
(iii) Excretion

Question 2
The xylem in plants are responsible for
(i) transport of water
(ii) transport of food
(iii) transport of amino acids
(iv) transport of oxygen
Answer:
(i) Transport of water

Question 3
The autotrophic mode of nutrition requires
(i) carbon dioxide and water
(ii) chlorophyll
(iii) sunlight
(iv) all of the above
Answer:
(iv) All of the above

Question 4
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in
(i) cytoplasm
(ii) mitochondria
(iii) chloroplast
(iv) nucleus
Answer:
(ii) Mitochondria

Question 5
How are fats digested in our bodies ? Where does this process take place ?
Answer:
Digestion of fats takes place in the small intestine.
Bile juice secreted by the liver poured in the intestine along with pancreatic juice. The bile salts present in the bile juice emulsify fhe large globules of fats. Therefore, by enulsification large globules break down into fine globules to provide larger surface area to act upon by the enzymes.
Lipase enzyme present in the pancreatic juice causes break down of emulsified fats. Glands present in the wall of small intestine secrete intestinal juice which contains lipase enzyme that converts fats into fatty acids and glycerol.

Question 6
What is the role of saliva in the digestion of food ?
Answer:
Saliva contains salivary amylase enzyme that breaks down starch into sugars like maltose.

Saliva keeps the mouth cavity clean and moistens the food that help in chewing and breaking down the big pieces of food into smaller ones.

Question 7
What are the necessary conditions (or autotrophic nutrition and what are its by-products ?
Answer:
Necessary conditions for autotrophic nutrition :
(i) Presence of chlorophyll in the living cells.
(if) Provision of supply of water to green plants or cells of the plant.
(iii) Sufficient sunlight.
(iv) Sufficient supply of carbon dioxide.
By-product of auto tropic nutrition is oxygen.

Question 8
What are the differences between aerobic and anaerobic respiration ? Name some organisms that use the anaerobic mode of respiration.
Answer:

Some organisms which use anaerobic respiration are yeast, bacteria etc.

Question 9
How are the alveoli designed to maximise the exchange of gases ?
Answer:
(i) The alveoli are thin walled and richly supplied with a network of blood vessels to facilitate exchange of gases between blood and the air filled in alveoli.
(ii) Alveoli have balloon-like structure. Hence, provide maximum surface for exchange of gases.

Question 10
What would be the consequences of a deficiency of haemoglobin in our bodies?
Answer:
Due to the deficiency of haemoglobin in blood, its oxygen carrying capacity decreases. As a result the production of energy by oxidation will become slower. Therefore, one would fall sick and would feel fatigue most of the time.

Question 11
Describe double circulation in human beings. Why is it necessary ?
Answer:
In our heart blood enters twice and also pumped out twice from the heart. The deoxygenated blood from the body is brought to the right atrium through vena cava from where it is sent to right ventricle. From right ventricle, the blood is pumped to the lungs for oxygenation through pulmonary artery. The oxygenated blood from lungs again enters the left atrium of the heart through pulmonary veins. From left atrium it is send to left ventricle, from where this oxygenated blood is pumped to different parts of body through the arteries. In this way the blood flows through the heart twice, that’s why it is called ‘double circulation’.

Necessity of double circulation: The right side and the left side of the human heart are useful to keep deoxygenated and oxygenated blood from mixing. This type of separation of oxygenated and deoxygenated blood ensures a highly efficient supply of oxygen to the body. This is useful in case of humans who constantly need energy to maintain their body temperature.

Question 12
What are the differences between the transport of materials in xylem and phloem ?
Answer:

Question 13
Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.
Answer:

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

Life processes: ‘Living being’. Basic concept of nutrition, respiration, transport and excretion in plants and animals.

Question 1
How are fats digested in our bodies? Where does this process take place?
Solution:
The small intestine is the site of the complete digestion of carbohydrates, proteins and fats. It receives the secretions of the liver and pancreas for this purpose. The food coming from the stomach is acidic and has to be made alkaline for the pancreatic enzymes to act. Bile juice from the liver accomplishes this in addition to acting on fats. Fats are present in the intestine in the form of large globules, which make it difficult for enzymes to act on them. Bile salts break them down into smaller globules increasing the efficiency of enzyme action. The pancreas secretes pancreatic juice, which contains enzymes like trypsin for digesting proteins and lipase for breaking down emulsified fats. The walls of the small intestine contain glands, which secrete intestinal juice. The enzymes present in it finally convert the proteins to amino acids, complex carbohydrates into glucose and fats into fatty acids and glycerol.

Question 2
What is the role of saliva in the digestion of food?
Solution:
When we eat something we like, our mouth ‘waters’. This is actually not only water, but also a fluid called saliva secreted by the salivary glands. Another aspect of the food we ingest is its complex nature. If it is to be absorbed from the alimentary canal, it has to be broken into smaller molecules. This is done with the help of biological catalysts called enzymes. The saliva contains an enzyme called salivary amylase that breaks down starch, which is a complex molecule to give sugar. The food is mixed thoroughly with saliva and moved around the mouth while chewing by the muscular tongue.

More Resources for CBSE Class 10

• NCERT Solutions
• NCERT Solutions for Class 10 Science

Question 3
What are the necessary conditions for autotrophic nutrition and what are its byproducts?
Solution:
Carbon and energy requirements of the autotrophic organism are fulfilled by photosynthesis. It is the process by which autotrophs take in substances from the outside and convert them into stored forms of energy. This material is taken in the form of carbon dioxide and water, which is converted into carbohydrates in the presence of sunlight and chlorophyll. Carbohydrates are utilised for providing energy to the plant. The carbohydrates, which are not used immediately, are stored in the form of starch, which serves as the internal energy reserve to be used as and when required by the plant.

Download NCERT Solutions for Class 10 Science Chapter 6 Life Processes PDF

Question 4
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Solution:
The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. Diverse organisms do this in different ways – some use oxygen to breakdown glucose completely into carbon dioxide and water; some use other pathways that do not involve oxygen. In all cases, the first step is the breakdown of glucose, a six-carbon molecule, into a three-carbon molecule called pyruvate. This process takes place in the cytoplasm. Further, the pyruvate may be converted into ethanol and carbon dioxide. This process takes place in yeast during fermentation. Since this process takes place in the absence of air (oxygen), it is called anaerobic respiration. Breakdown of pyruvate using oxygen takes place in the mitochondria. This process breaks up the three-carbon pyruvate molecule to give three molecules of carbon dioxide. The other product is water. Since this process takes place in the presence of air (oxygen), it is called aerobic respiration. The release of energy in this aerobic process is a lot greater than in the anaerobic process.

Question 5
How are the alveoli designed to maximise the exchange of gases?
Solution:
Within the lungs, the passage divides into smaller and smaller tubes, which finally terminate in balloon-like structures, which are called alveoli. The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood vessels. As we have seen in earlier years, when we breathe in, we lift our ribs and flatten our diaphragm, and the chest cavity becomes larger as a result. Because of this, air is sucked into the lungs and fills the expanded alveoli. The blood brings carbon dioxide from the rest of the body for release into the alveoli, and the oxygen in the alveolar air is taken up by blood in the alveolar blood vessels to be transported to all the cells in the body. During the breathing cycle, when air is taken in and let out, the lungs always contain a residual volume of air so that there is sufficient time for oxygen to be absorbed and for the carbon dioxide to be released.

Question 6
Describe double circulation in human beings. Why is it necessary?
Solution:
The double circulatory system of blood flow refers to the separate systems of pulmonary circulation and the systemic circulation.
The adult human heart consists of two separated pumps, the right side with the right atrium and ventricle which pumps deoxygenated blood into the pulmonary circulation.
The oxygenated blood re-enters the left side of the heart through the pulmonary vein into the left atrium and passes to the left ventricle where it is pumped to the rest of the body. This part of the circulation is called as systemic circulation. This type of circulation is called double circulation. The advantage of a double circulatory system is that blood can be pumped to the rest of the body at a higher pressure.

Multiple Choice Questions (MCQs) [1 Mark each]

Question 1.
Yeast respires anaerobically using sugar as a substrate. Out of the options given below, choose the correct combination of condition and product?

Answer:
(c) Under an aerobic condition, yeast respires and converts glucose to alcohol and CO₂.

Question 2.
The table shows the percentage composition of four samples of air. Which sample could have been breathed out by a person after vigorous exercise?

Answer:
(b) This is because rapid aerobic respiration occurs – during vigorous exercise in order to obtain more energy.

Question 3.
Cramps caused during sudden activities are due to the formation of
(a) lactic acid
(b) acetic acid
(c) excess of water
(d) ethanol
Answer:
(a) Lactic acid is formed by the breakdown of pyruvate when oxygen is insufficient in muscles instead of forming C02 and water. Accumulation of excess lactic acid in the muscles causes cramps.

Question 4.
Which of the following plays nose like function in plants?
(a) Flower
(b) Phloem
(c) Stomata
(d) Chlorophyll
Answer:
(c) Stomata are pores which help in the passage of air in the plants.

Question 5.
Which changes occur when a person breathe in deeply?

Answer:
(a) When a person breathes deeply the external intercostal muscles contract causing the rib cage to swing up and out. Also, the diaphragm contracts and flattens causing the thoracic cavity to increase in volume and decrease in pressure.

Question 6.

The diagram given above shows part of the lining of the human trachea. What is the function of X?
(a) Gaseous exchange
(b) Mucus removal
(c) Phagocytosis
(d) Secretion of mucus
Answer:
(b) The cilia (X) of the cells lining the air passages move in a sweeping motion to keep the air passages clean. The constant action of these cilia carry mucus and debris upward into the pharynx where they are swallowed.

Question 7.
The table given below shows the percentage composition of a gas in inspired and in expired air.

What is the gas?
(a) Carbon dioxide
(b) Nitrogen
(c) Oxygen
(d) Water vapour
Answer:
(c) The gas is oxygen as atmospheric air has approximately 21% of oxygen

NCERT Solutions for Class 10 Science Chapter 6 Life Processes (Hindi Medium)

Class 10 Science Life Processes Mind Map

Nutrition
Nutrition is the process by which source of energy (food) is transferred from outside the body of the organism to the inside. Most of the food sources are also carbon-based on Earth and depending on the complexity of these carbon sources different organisms use different kinds of nutritional processes.
Autotrophic Nutrition: Carbon and energy requirements of the autotrophic organism are fulfilled by photosynthesis.

• It is the process by which autotrophs convert carbon dioxide & water into carbohydrate in the presence of sunlight and chlorophyll. Oxygen is the byproduct.
• The following events occur during this process:
• Absorption of light energy by chlorophyll.
• Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
• Reduction of carbon dioxide to carbohydrates.

Heterotrophic Nutrition: Heterotrophs depend on other organisms for their nutrition.

• Saprophytes: They break-down the food material outside the body and then absorb it, also termed as extra-cellular digestion. E.g. fungi like bread moulds, yeast, mushrooms etc.
• Parasites: Derive nutrition from plants or animals without killing them. E.g. cuscuta (amar-bel), ticks, lice, leeches, tape-worms etc.
• Holozoic nutrition: These organisms take in whole material & break it down inside their bodies. E.g. cow, deer, lion, tiger, humans etc. What can be taken in and broken down depends on body design and functioning

Respiration
It is the process by which organism uses the food material to produce energy. Diverse organisms do this in different ways:

Energy released during cellular respiration is immediately used to synthesise ATP which is used to fuel all other activities in the cell. Aerobic organisms need to ensure that there is sufficient intake of oxygen:
• Plants: Exchange of gases takes place through stomata by simple diffusion. Large inter-cellular spaces ensure that all cells are in contact with air. Direction of diffusion depends upon the environmental conditions and the requirements of the plant. For e.g. CO₂ elimination majorly takes place at night while oxygen release is the major event of the day time.
• Aquatic animals such as fishes take in water through their mouths & force it past the gills where the dissolved oxygen is taken up by blood.
• In human beings, the passage of air can be written as nostril → trachea → bronchi → bronchioles → alveolar sac. The alveoli provide a surface where the exchange of gases can take place. Blood releases the dissolved CO₂ into the alveoli & carries O₂ from alveolar air. Haemoglobin in RBC of blood transport O₂ from lungs to various tissues of the body.

Life Process
The processes which maintain the body functions and are required for the survival of living being are called life processes. Some of the important life processes are nutrition, respiration, transportation, excretion etc.

Nutrition In Human Beings
The alimentary canal is a long tube extending from the mouth to the anus. The nutrition in human being is divided into five steps:
• Ingestion: Intake of food from outside source. Teeth & saliva crush the food to generate the particles of same size & texture. The food is then passed to stomach via oesophagus. The peristaltic movements occur all along the gut which helps in pushing the food forward.
• Digestion: In mouth, salivary amylase helps in carbohydrates digestion. In stomach, pepsin helps in protein digestion. However, small intestine is the main site of complete digestion of carbohydrates, proteins & fats. It receives pancreas and liver secretions. Bile juice emulsifies fats and pancreatic enzymes, trypsin & lipases digest proteins & emulsified fats. It finally converts proteins to amino acids, complex carbohydrates into glucose & fats into fatty acids & glycerol.
• Absorption: The digested food is taken up by the walls of the intestine. The inner lining of the small intestine has numerous finger-like projections called villi which increase the surface area for absorption. Large intestine absorbs water from the unabsorbed food.
• Assimilation: The villi are richly supplied with blood vessels which take the absorbed food to each & every cell of the body, where it is required either for energy, build up or repair.
• Excretion: The waste material is removed from the body via anus which is regulated by anal sphincter.

Transportation
Transportation in Human Beings

• Blood consists of fluid medium called plasma in which
  the cells are suspended. Plasma transports food, CO₂ & nitrogenous wastes in dissolved form. Oxygen is carried by RBC.
• Heart: Heart is the muscular organ made up of cardiac muscles and is as big as our fist. It is composed of four chambers (2 atria & 2 ventricles) to prevent the mixing of oxygenated & deoxygenated blood.
• Ventricles are thick wailed as they have to pump the blood to various organs of the body. In addition, valves are also present in heart and veins to prevent the backflow of the blood.

Circulation of blood: Oxygenated blood is carried out from lungs to the left atrium with the help of pulmonary’ veins.

• Left atrium contracts to release blood into the left ventricle which relaxes while collecting it. It then pumped out the blood to whole body via aorta.
  a Deoxygenated blood from whole body then enters the right atrium via vena cava vein.
• Right atrium contracts to pump the blood in right ventricle. It then pumps the blood towards lungs via pulmonary’ artery for oxygenation.

Oxygenation of blood: Invertebrates such as birds, mammals etc which constantly use energy to maintain their body temperature, blood goes through heart twice during each cycle which is known as double circulation.

• In contrast, animals like amphibians or many reptiles have three-chambered hearts as they can tolerate some mixing of the oxygenated & de-oxygenated blood streams. They do not use energy for thermoregulation and body temperature depends on the temperature in the environment.
• Fishes, on other hand, have only two chambered heart. Blood is pumped to the gills for oxygenation and passes directly to the rest of the body.

Transportation In Plants
There are two main pathways present in plants: xylem pathway- moves water & minerals from the soil & phloem transports products of photosynthesis from leaves (where they are synthesized) to other parts of the plant.
Transport of Water

• In xylem tissue, vessels and tracheids of roots, stems & leaves are interconnected to form a continuous system of water-conducting channels reaching all parts of the plant.
• At root site, cells actively take up ions from soil which creates concentration gradient. Water then diffuses into the root cells in order to eliminate this gradient.
• It provides steady movement of water into root xylem, creating a column of water that is steadily pushed upwards.
• However, it is not efficient enough to push water over the heights of tall plants.
• So, plants use other method which is known as transpiration to push water upwards. The loss of water in the form of vapour from aerial parts of plant is known as transpiration.
• Evaporation of water molecules from the cells of a leaf creates a suction which pulls water from the xylem cells of roots. It also aids in thermoregulation.
• Transport of food and other substances
• Transport of soluble products of photosynthesis is called translocation.
• The translocation takes place in sieve tubes with the help of adjacent companion cells both in upward & downward directions.
• It utilizes energy (ATP) in contrast to xylem transport.
• Material like sucrose is transferred into phloem tissue using energy from ATP.
• It increases osmotic pressure of tissue causing water to move into it.
• This pressure moves the material in phloem to tissues which have less pressure.
• It allows phloem to move material according to plant’s needs.

Excretion
The biological process involved in removal of harmful metabolic wastes from body is called excretion.
Many unicellular organisms remove these wastes by simple diffusion from body surface into surrounding water. However, complex multi-cellular organisms use specialised organs to perform this function.
Excretion in Human Beings: The excretory system includes pair of kidneys, pair of ureters, urinary bladder & urethra.

• Nephrons are the functional units of kidneys. They are the clusters of thin-walled capillaries. Each cluster is associated with cup-shaped end (Bowmans capsule) of a tube that collects the filtered urine.
• Substance such as glucose, amino acids, salts & a major amount of water are selectively re-absorbed as the urine flows along the tube. The amount of water depends up on amount of excess water & dissolved waste in the body.
• The urine formed in each kidney is carried to urinary bladder by ureter. Urine is stored in urinary bladder until the pressure of the expanded bladder leads to the urge to pass it out through the urethra.
• Excretion in Plants: They get rid of excess water by transpiration.
• Many plant waste products are stored in cellular vacuoles.
• Waste products may be stored in leaves that fall off.
• In addition, some waste products are stored as resins & gums, especially in old xylem.
• Lastly, plants excrete some waste substances into the soil around them.

We hope the detailed information regarding  NCERT Solutions For Class 10 Science Chapter 6 Life Processes is helpful. For any query related to  NCERT Solutions For Class 10 Science Chapter 6 Life Processes, kindly drop your questions in the comment box below and we will get back to you as soon as possible.

The post NCERT Solutions for Class 10 Science Chapter 6 Life Processes appeared first on Learn CBSE.

CBSE Class 10 Science Notes Chapter 6 Life Processes Biology Pdf free download is part of Class 10 Science Notes for Quick Revision. Here we have given NCERT Class 10 Science Notes Chapter 6 Life Processes. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Science pdf Carries 20 Marks.

You can aslo read Life Processes Class 10 NCERT Solutions to score more marks in your examinations.

Life Process Class 10 Science Notes

Nutrition in Plants and Animals – Life Processes Class 10 Notes

• Nutrition: The process by which an organism takes food and utilizes it, is called nutrition.
• Need for Nutrition: Organisms need the energy to perform various activities. The energy is supplied by the nutrients. Organisms need various raw materials for growth and repair. These raw materials are provided by nutrients.
• Nutrients: Materials which provide nutrition to organisms are called nutrients. Carbohydrates, proteins and fats are the main nutrients and are called macronutrients. Minerals and vitamins are required in small amounts and hence are called micronutrients.
• Modes of Nutrition
  1. Autotrophic Nutrition.
  2. Heterotrophic Nutrition.

Autotrophic Nutrition – Life Processes Class 10 Notes

The mode of nutrition in which an organism prepares its own food is called autotrophic nutrition. Green plants and blue-green algae follow the autotrophic mode of nutrition.

• The organisms which carry out autotrophic nutrition are called autotrophs (green plants).
  
• Autotrophic nutrition is fulfilled by the process, by which autotrophs intake CO₂ and H₂O, and convert these into carbohydrates in the presence of chlorophyll, sunlight is called photosynthesis.
• Equation
  

Nutrition in Plants: Green plants prepare their own food. They make food in the presence of sunlight. Sunlight provides energy’, carbon dioxide and water are the raw materials and chloroplast is the site where food is made.

What is Photosynthesis in biology class 10?

Photosynthesis: The process by which green plants prepare food is called photosynthesis.

• During this process, the solar energy is converted into chemical energy and carbohydrates are formed.
• Green leaves are the main site of photosynthesis.
• The green portion of the plant contains a pigment chloroplast, chlorophyll (green pigment).
• The whole process of photosynthesis can be shown by the following equation:
  

Raw Materials for Photosynthesis:

• Sunlight
• Chlorophyll: Sunlight absorbed by chloroplast
• CO₂: Enters through stomata, and oxygen (O₂) is released as a byproduct through stomata on the leaf.
• Water: Water + dissolved minerals like nitrogen, phosphorous etc., are taken up by the roots from the soil.

How do raw materials for photosynthesis become available to the plant?

• Water comes from the soil, through the xylem tissue in roots and stems.
• Carbon dioxide comes in the leaves through stomata.

Site of Photosynthesis: Chloroplast in the leaf. Chloroplast contains chlorophyll (green pigment)

Main Events of Photosynthesis:

• Absorption of light energy by chlorophyll.
• Conversion of light energy into chemical energy + splitting (breaking) of water into hydrogen and oxygen.
• Reduction of CO₂ to carbohydrates.
• Sunlight activates chlorophyll, which leads to splitting of the water molecule.
• The hydrogen, released by the splitting of a water molecule is utilized for the reduction of carbon dioxide to produce carbohydrates.
• Oxygen is the by-product of photosynthesis.
• Carbohydrate is subsequently converted into starch and is stored in leaves and other storage parts.
• The splitting of water molecules is a part of the light reaction.

Other steps are part of the dark reaction during photosynthesis.

Stomata – Life Processes Class 10 Notes

• Stomata: These are tiny pores present in the epidermis of leaf or stem through which gaseous exchange and transpiration occur.

Functions of stomata

• Exchange of gases, O₂ and CO₂.
• Loses a large amount of water (water vapour) during transpiration.
  

Opening and closing of stomatal pores:

• The opening and closing of stomatal pores are controlled by the turgidity of guard cells.
• When guard cells uptake water from surrounding cells, they swell to become a turgid body, which enlarges the pore in between (Stomatal Opening).
• While, when water is released, they become flaccid shrinking to close the pore (Stomatal Closing).
  

Significance of Photosynthesis:

• Photosynthesis is the main way through which solar energy is made available for different living beings.
• Green plants are the main producers of food in the ecosystem. All other organisms directly or indirectly depend on green plants for food.
• The process of photosynthesis also helps in maintaining the balance of carbon dioxide and oxygen in the air.

Heterotrophic Nutrition – Life Processes Class 10 Notes

The mode of nutrition in which an organism takes food from another organism is called heterotrophic nutrition. Organisms, other than green plants and blue-green algae follow the heterotrophic mode of nutrition. Heterotrophic nutrition can be further divided into three types, viz. saprophytic nutrition, holozoic nutrition, and parasitic.

• Saprophytic Nutrition: In saprophytic nutrition, the organism secretes the digestive juices on the food. The food is digested while it is still to be ingested. The digested food is then ingested by the organism. All the decomposers follow saprophytic nutrition. Some insects, like houseflies, also follow this mode of nutrition.
• Holozoic Nutrition: In holozoic nutrition, the digestion happens inside the body of the organism. i.e., after the food is ingested. Most of the animals follow this mode of nutrition.
• Parasitic Nutrition: The organism which lives inside or outside another organism (host) and derives nutrition from it is known as parasites and this type of mode of nutrition is called parasitic nutrition. For example Cuscuta, tick etc.

Nutrition in Amoeba

• Amoeba is a unicellular animal which follows the holozoic mode of nutrition.
• In holozoic nutrition, the digestion of food follows after the ingestion of food. Thus, digestion takes place inside the body of the organism.
• Holozoic nutrition happens in five steps, viz. ingestion, digestion, absorption, assimilation and egestion.
  

Steps of Holozoic Nutrition:

• Ingestion: The process of taking in the food is called ingestion.
• Digestion: The process of breaking complex food substances into simple molecules is called digestion. Simple molecules, thus obtained, can be absorbed by the body.
• Absorption: The process of absorption of digested food is called absorption.
• Assimilation: The process of utilization of digested food, for energy and for growth and repair is called assimilation.
• Egestion: The process of removing undigested food from the body is called egestion.

Amoeba is a unicellular animal which follows the holozoic mode of nutrition. The cell membrane of amoeba keeps on protruding into pseudopodia. Amoeba surrounds a food particle with pseudopodia and makes a food vacuole. The food vacuole contains food particle and water. Digestive enzymes are secreted in the food vacuole and digestion takes place. After that, digested food is absorbed from the food vacuole. Finally, the food vacuole moves near the cell membrane and undigested food is expelled out.

Nutrition in Human Beings – Life Processes Class 10 Notes

Human beings are complex animals, which have a complex digestive system. The human digestive system is composed of an alimentary canal and some accessory glands. The alimentary canal is divided into several parts, like oesophagus, stomach, small intestine, large intestine, rectum and anus. Salivary gland, liver and pancreas are the accessory glands which lie outside the alimentary canal.

Structure of the Human Digestive System:
The human digestive system comprises of the alimentary canal and associated digestive glands.

• Alimentary Canal: It comprises of mouth, oesophagus, stomach, small intestine and large intestine.
• Associated Glands: Main associated glands are
  • Salivary gland
  • Gastric Glands
  • Liver
  • Pancreas

Mouth or Buccal Cavity:

• The mouth has teeth and tongue. Salivary glands are also present in the mouth.
• The tongue has gustatory receptors which perceive the sense of taste.
• The tongue helps in turning over the food so that saliva can be properly mixed in it.
• Teeth help in breaking down the food into smaller particles so that, swallowing of food becomes easier.
• There are four types of teeth in human beings. The incisor teeth are used for cutting the food.
• The canine teeth are used for tearing the food and for cracking hard substances.
• The premolars are used for the coarse grinding of food. The molars are used for fine grinding of food.
  

Salivary glands secrete saliva: Saliva makes the food slippery which makes it easy to swallow the food. Saliva also contains the enzyme salivary amylase or ptyalin. Salivary amylase digests starch and converts it into sucrose, (maltose).

Oesophagus: Taking food from mouth to stomach by Peristaltic movement.

Peristaltic movement: Rhythmic contraction of muscles of the lining of the alimentary canal to push the food forward.

Stomach

• Stomach is a bag-like organ. Highly muscular walls of the stomach help in churning the food.
• The walls of the stomach secrete hydrochloric acid. Hydrochloric acid kills the germs which may be present in food.
• Moreover, it makes the medium inside the stomach as acidic. The acidic medium is necessary for gastric enzymes to work.
• The enzyme pepsin, secreted in the stomach, does partial digestion of protein.
• The mucus, secreted by the walls of the stomach saves the inner lining of the stomach from getting damaged from hydrochloric acid.
  

Small Intestine: It is a highly coiled tube-like structure. The small intestine is longer than the large intestine but its lumen is smaller than that of the large intestine. The small intestine is divided into three parts, like duodenum, jejunum and ileum.

Liver: Liver is the largest organ in the human body. The liver manufactures bile, which gets stored in the gall bladder. From the gall bladder, bile is released as and when required.

Pancreas: Pancreas is situated below the stomach. It secretes pancreatic juice which contains many digestive enzymes.

Bile and pancreatic juice go to the duodenum through a hepatopancreatic duct. Bile breaks down fat into smaller particles. This process is called emulsification of fat. After that, the enzyme lipase digests fat into fatty acids and glycerol. Trypsin and chymotrypsin are enzymes which digest protein into amino acids. Complex carbohydrates are digested into glucose. The major part of digestion takes place in the duodenum.

No digestion takes place in the jejunum: The inner wall in the ileum is projected into numerous finger-like structures, called villi. Villi increase the surface area inside the ileum so that optimum absorption can take place. Moreover, villi also reduce the lumen of the ileum so that food can stay for a longer duration in it, for optimum absorption. Digested food is absorbed by villi.

Large Intestine:

• Large intestine is smaller than the small intestine.
• Undigested food goes into the large intestine.
• Some water and salt are absorbed by the walls of the large intestine. After that, the undigested food goes to the rectum, from where it is expelled out through the anus.
• Large Intestine bbsorb excess of water. The rest of the material is removed from the body via the anus. (Egestion).

Respiration – Life Processes Class 10 Notes

Types of respiration, aerobic and anaerobic respiration, human respiratory system, respiration in plants.

Respiration: The process by which a living being utilises the food to get energy, is called respiration. Respiration is an oxidation reaction in which carbohydrate is oxidized to produce energy. Mitochondria is the site of respiration and the energy released is stored in the form of ATP (adenosine triphosphate). ATP is stored in mitochondria and is released as per need.

Steps of respiration:

• Breaking down of glucose into pyruvate: This step happens in the cytoplasm. Glucose molecule is broken down into pyruvic acid. Glucose molecule is composed of 6 carbon atoms, while pyruvic acid is composed of 3 carbon atoms.
• Fate of Pyruvic Acid: Further breaking down of pyruvic acid takes place in mitochondria and the molecules formed depend on the type of respiration in a particular organism. Respiration is of two types, viz. aerobic respiration and anaerobic respiration.
• Respiration involves
  • Gaseous exchange: Intake of oxygen from the atmosphere and release of CO₂ → Breathing.
  • Breakdown of simple food in order to release energy inside the cell → Cellular respiration

Types of Respiration – Life Processes Class 10 Notes

• Aerobic respiration: This type of respiration happens in the presence of oxygen. Pyruvic acid is converted into carbon dioxide. Energy is released and water molecule is also formed at the end of this process.
• Anaerobic respiration: This type of respiration happens in the absence of oxygen. Pyruvic acid is either converted into ethyl alcohol or lactic acid. Ethyl alcohol is usually formed in case of anaerobic respiration in microbes, like yeast or bacteria. Lactic acid is formed in some microbes as well as in the muscle cells.
  • Glucose (6 carbon molecule) → Pyruvate (3 carbon molecules) + Energy
  • Pyruvate (In yeast, lack of O₂) → Ethyl alcohol + Carbon dioxide + Energy
  • Pyruvate (In muscles, lack of O₂) → Lactic Acid + Energy
  • Pyruvate (In mitochondria; the presence of O₂) → Carbon dioxide + Water + Energy

The equations for the above reactions can be written as follows:

Pain in leg muscles while running:

• When someone runs too fast, he may experience throbbing pain in the leg muscles. This happens because of anaerobic respiration taking place in the muscles.
• During running, the energy demand from the muscle cells increases. This is compensated by anaerobic respiration and lactic acid is formed in the process.
• The deposition of lactic acid causes pain in the leg muscles. The pain subsides after taking rest for some time.

Exchange of gases:

• For aerobic respiration, organisms need a continuous supply of oxygen, and carbon dioxide produced during the process needs to be removed from the body.
• Different organisms use different methods for the intake of oxygen and expulsion of carbon dioxide.
• Diffusion is the method which is utilized by unicellular and some simple organisms for this purpose.
• In plants also, diffusion is utilized for exchange of gases.
• In complex animals, respiratory system does the job of exchange of gases.
• Gills are the respiratory organs for fishes. Fishes take in oxygen which is dissolved in water through gills.
• Since, availability of oxygen is less in the aquatic environment, so the breathing rate of aquatic organisms is faster.
• Insects have a system of spiracles and trachease which is used for taking in oxygen.
• Terrestrial organisms have developed lungs for exchange of gases.
• Availability of oxygen is not a problem in the terrestrial environment so breathing rate is slower as compared to what it is in fishes.

Terrestrial organisms: Use atmospheric oxygen for respiration.
Aquatic organisms: Use dissolve oxygen for respiration.

Human respiratory system – Life Processes Class 10 Notes

The human respiratory system is composed of a pair of lungs. These are attached to a system of tubes which open on the outside through the nostrils.
Following are the main structures in the human respiratory system:

1. Nostrils: There are two nostrils which converge to form a nasal passage. The inner lining of the nostrils is lined by hair and remains wet due to mucus secretion. The mucus and the hair help in filtering the dust particles out from inhaled air. Further, air is warmed up when it enters the nasal passage.
2. Pharynx: It is a tube-like structure which continues after the nasal passage.
3. Larynx: This part comes after the pharynx. This is also called voice box.
4. Trachea: This is composed of rings of cartilage. Cartilaginous rings prevent the collapse of trachea in the absence of air.
5. Bronchi: A pair of bronchi comes out from the trachea, with one bronchus going to each lung.
6. Bronchioles: A bronchus divides into branches and sub-branches inside the lung.
7. Alveoli: These are air sacs at the end of bronchioles. The alveolus is composed of a very thin membrane and is the place where blood capillaries open. This is alveolus, where the oxygen mixes with the blood and carbon dioxide exits from the blood. The exchange of gases, in alveoli, takes place due to the pressure differential.

Passage of air through the respiratory system in human beings:

Breathing Mechanism

• The breathing mechanism of lungs is controlled by the diaphragm and the intercostalis muscles.
• The diaphragm is a membrane which separates the thoracic chamber from the abdominal cavity.
• When the diaphragm moves down, the lungs expand and the air is inhaled.
• When the diaphragm moves up, the lungs contract and air are exhaled.

Transportation – Life Processes Class 10 Notes

Circulatory system of human being, transportation in plants. Human beings like other multicellular organism need a regular supply of foods, oxygen etc. This function is performed by a circulatory system or transport system.

Transportation in Human Beings: The circulatory system is responsible for transport of various substances in human beings. It is composed of the heart, arteries, veins and blood capillaries. Blood plays the role of the carrier of substances.

1. Heart: Heart is a muscular organ, which is composed of cardiac muscles.

• It is so small that, it can fit inside an adult’s wrist. The heart is a pumping organ which pumps the blood.
• The human heart is composed of four chambers, viz. right atrium, right ventricle, left ventricle and left atrium.
• Systole: Contraction of cardiac muscles is called systole.
• Diastole: Relaxation of cardiac muscles is called diastole.

2. Arteries:

• These are thick-walled blood vessels which carry oxygenated blood from the heart to different organs.
• Pulmonary arteries are exceptions because they carry deoxygenated blood from the heart to lungs, where oxygenation of blood takes place.

3. Veins:

• These are thin-walled blood vessels which carry deoxygenated blood from different organs to the heart, pulmonary veins are exceptions because they carry oxygenated blood from lungs to the heart.
• Valves are present in veins to prevent back flow of blood.

4. Capillaries: These are the blood vessels which have single-celled walls.

Blood: Blood is a connective tissue which plays the role of the carrier for various substances in the body. Blood is composed of 1. Plasma 2. Blood cells 3. Platelets.

• Blood plasma: Blood plasma is a pale coloured liquid which is mostly composed of water. Blood plasma forms the matrix of blood.
• Bloods cells: There are two types of blood cells, viz. Red Blood Cells (RBCs) and White Blood Cells (WBCs).
  (a) Red Blood Corpuscles (RBCs): These are of red colour because of the presence of haemoglobin which is a pigment. Haemoglobin readily combines with oxygen and carbon dioxide. The transport of oxygen happens through haemoglobin. Some part of carbon dioxide is also transported through haemoglobin.
  (b) White Blood Corpuscles (WBCs): These are of pale white colour. They play important role in the immunity.
• Platelets: Platelets are responsible for blood coagulation. Blood coagulation is a defense mechanism which prevents excess loss of blood, in case of an injury.

Lymph:

• Lymph is similar to blood but RBCs are absent in lymph.
• Lymph is formed from the fluid which leaks from blood capillaries and goes to the intercellular space in the tissues. This fluid is collected through lymph vessels and finally return to the blood capillaries.
• Lymph also plays an important role in the immune system.
• Lymph a yellowish fluids escape from the blood capillaries into the intercellular spaces contain less proteins than blood.
• Lymph flows from the tissues to the heart assisting in transportation and destroying germs.

Double circulation: In the human heart, blood passes through the heart twice in one cardiac cycle. This type of circulation is called double circulation. One complete heartbeat in which all the chambers of the heart contract and relax once is called cardiac cycle. The heart beats about 72 times per minute in a normal adult. In one cardiac cycle, the heart pumps out 70 mL blood and thus, about 4900 mL blood in a minute. Double circulation ensures complete segregation of oxygenated and deoxygenated blood which is necessary for optimum energy production in warm-blooded animals.

Transportation in plants: Plants have specialized vascular tissues for transportation of substances. There are two types of vascular tissues in plants.

• Xylem: Xylem is responsible for transportation of water and minerals. It is composed of trachids, xylem vessels, xylem parenchyma and xylem fibre. Tracheids and xylem vessels are the conducting elements. The xylem makes a continuous tube in plants which runs from roots to stem and right up to the veins of leaves.
• Carry water and minerals from the leaves to the other part of the plant.
• Phloem: Phloem is responsible for transportation of food. Phloem is composed of sieve tubes, companion cells, phloem parenchyma and bast fibers. Sieve tubes are the conducting elements in phloem.
• Carries product of photosynthesis from roots to other part of the plant.

Transportation in plants

Ascent of sap: The upward movement of water and minerals from roots to different plant parts is called ascent of sap. Many factors are at play in ascent of sap and it takes place in many steps. They are explained as follows :

• Root pressure: The walls of cells of root hairs are very thin. Water from soil enters the root hairs because of osmosis. Root pressure is responsible for movement of water up to the base of the stem.
• Capillary action: A very fine tube is called capillaiy, water, or any liquid, rises in the capillary because of physical forces and this phenomenon is called capillary action. Water, in stem, rises up to some height because of capillaiy action.
• Adhesion-cohesion of water molecules: Water molecules make a continuous column in the xylem because of forces of adhesion and cohesion among the molecules.
• Transpiration pull: Loss of water vapour through stomata and lenticels, in plants, is called transpiration. Transpiration through stomata creates vacuum which creates a suction, called transpiration pull. The transpiration pull sucks the water column from the xylem tubes and thus, water is able to rise to great heights in even the tallest plants.
• Transport of food: Transport of food in plants happens because of utilization of energy. Thus, unlike the transport through xylem, it is a form of active transport. Moreover, the flow of substances through phloem takes place in both directions, i.e., it is a two-way traffic in phloem.
  Transpiration is the process of loss of water as vapour from aerial parts of the plant.

Functions

• Absorption and upward movement of water and minerals by creating pull.
• Helps in temperature regulation in plant.

Transport of food from leaves (food factory) to different parts of the plant is called Translocation.

Excretion – Life Processes Class 10 Notes

Human excretory system, excretion in plants.

Excretion in human beings:

• Removal of harmful waste from the body is called excretion.
• Many wastes are produced during various metabolic activities.
• These need to be removed in time because their accumulation in the body can be harmful and even lethal for an organism.

Human Excretory System:

• The human excretory system is composed of a pair of kidneys.
• A tube, called ureter, comes out of each kidney and goes to the urinary bladder.
• Urine is collected in the urinary bladder, from where it is expelled out through urethra as and when required.

Excretory system of human beings includes :

• A pair of kidneys.
• A urinary bladder.
• A pair of the ureter.
• A urethra.

Kidney:

• Kidney is a bean-shaped organ which lies near the vertebral column in the abdominal cavity.
• The kidney is composed of many filtering units, called nephrons.
• Nephron is called the functional unit of kidney.

Nephron

• It is composed of a tangled mess of tubes and a filtering part, called glomerulus.
• The glomerulus is a network of blood capillaries to which renal artery is attached.
• The artery which takes blood to the glomerulus is called afferent arteriole and the one receiving blood from the glomerulus is called efferent arteriole.
• The glomerulus is enclosed in a capsule like portion, called bowman’s capsule. The bowman’s capsule extends into a fine tube which is highly coiled.
• Tubes from various nephrons converge into collecting duct, which finally goes to the ureter.

Urine formation in the kidney: The urine formation involves three steps:

• Glomerular filtration: Nitrogenous wastes, glucose, water, amino acid filter from the blood into bowman’s capsule of the nephron.
• Tubular reabsorption: Now, useful substances from the filtrate are reabsorbed back by capillaries surrounding the nephron.
• Secretion: Extra water, salts are secreted into the tubule which opens up into the collecting duct and then into the ureter.

Urine produced in the kidneys passes through the ureters into the urinary bladder where it is stored until it is released through the urethra.
The purpose of making urine is to filter out waste product from the blood i.e., urea which is produced in the liver.

Haemodialysis: The process of purifying blood by an artificial kidney. It is meant for kidney failure patient.

Excretion in Plants

• Other wastes may be stored in leaves, bark etc. which fall off from the plant.
• Plants excrete some waste into the soil around them.
• Gums, resin → In old xylem
• Some metabolic wastes in the form of crystals of calcium oxalates in the leaves of colocasia and stem of Zamikand.

Nutrition in Plants and Animals

Nutrition: Process of obtaining and utilizing of food is known as nutrition.

Mode of nutrition:

• Autotrophic Nutrition (All green plants)
• Heterotrophic Nutrition (Animals, Man, Non-green plants)
  • Saprotrophic nutrition
  • Parasitic nutrition
  • Holozoic nutrition

Autotrophs: It is a mode of nutrition in which organisms can make their own food from simple raw material. Example, all green plants.

Heterotrophs: It is a mode of nutrition in which organisms cannot prepare their food on their own and depend on others. Example, animals.

Saprotrophic Nutrition: It is the process by which the organism feeds on dead and decaying matter. Example, Rhizopus, Mucor, yeast.

Photosynthesis: It is the process by which green plants prepare their own food.

Raw materials for photosynthesis:

• Water and Minerals: These are absorbed by the roots from the soil.
• Carbon dioxide: Carbon dioxide enters the leaves through tiny pores called stomata.
• Sunlight: Energy from the sun is called solar energy.
• Chlorophyll: Chlorophyll pigment helps leaves to capture solar energy.

Products of Photosynthesis: Carbohydrate-glucose- It is converted to starch.

Symbiotic relationship: Two organisms live in a close association and develop a relationship that is beneficial to both this is called a symbiotic relationship.
Example, Lichen is a living partnership between a fungus an alga. Fungus absorbs water and provides shelter and alga prepare food by photosynthesis

Insectivores: Plants feed on insects for their nitrogen requirements.

Holozoic nutrition: It means feeding on solid food. Organism takes complex organic food into the body. Example, man, amoeba, dog, etc.

• Herbivores: Animals which feed on plants only. Example, deer, cow.
• Carnivores: Animals which feed on flesh or meat. Example, tiger.
• Omnivores: Animals which feed on both plant and flesh. Example, man, dog.

Steps of Holozoic nutrition:

• Ingestion: Taking food into the mouth.
• Digestion: Break down of large insoluble food into small water-soluble molecules by enzymes.
• Absorption: Digested food absorbed through the intestinal wall into the blood.
• Assimilation: Absorbed food is taken by body cells for releasing energy, growth and repair.
• Egestion: Eliminating undigested food from the body.

Digestive organs of human beings: Mouth, oesophagus, stomach, small intestine and large intestine with glands like salivary gland, liver, pancreas.

Teeth: An organ which breaks down the complex food and helps in chewing the food.

• Milk teeth: The first set of 20 small teeth when the baby is 6-7 months old.
• Permanent teeth: The second set of 32 larger teeth, when a child is 6-7 years old and comes by replacing milk teeth.

Enamel: A white, strong, shining, protective material covering on teeth.

Tongue: A muscular organ attached to the floor of the buccal cavity which helps in tasting and mixing the food with saliva for digestion.

Transportation in Plants and Animals

• Vascular tissue: A plant tissue which helps in transportation.
• Xylem tissue: It helps in transporting water and minerals in plants.
• Phloem: It helps in transporting food in plants.
• Translocation: The process of transporting food from leaves to other parts of plants.
• Transpiration: A loss of water from stomata in leaves.
• Blood: A red colour fluid which circulates in the body of animals.
• Plasma: Fluid part of the blood which consists of nutrients, hormones, and waste products.
• Blood vessel: Tube-like structure present in the body for carrying blood inside the body.
• Artery: It carries oxygenated blood from the heart to body parts.
• Vein: It carries deoxygenated blood from body parts to the heart.
• Capillary: A thin-walled narrow tube which connects artery and vein.
• Heart: A muscular organ present in the thoracic cavity and helps in pumping blood in the body.
• Double circulation: A circulatory system in which blood travels twice through the heart in one complete cycle.
• Heartbeat: One complete contraction and relaxation of the heart (72 times in a minute).
• Stethoscope: Instrument which measures heartbeat.
• Systolic pressure: Maximum pressure at which blood flows during contraction of the heart. (120 mm Hg)
• Diastolic pressure: Minimum pressure at which blood flows during relaxation of the heart. (80 mm Hg)
• Sphygmomanometer: Instrument which measures blood pressure.
• Lymph: A light yellow liquid flowing from body tissue to the blood circulatory system and provides immunity.

Excretion in Plants and Animals

• Excretion: It is the process of removing waste products from the body.
• Excretory products of plants: CO₂, O₂, water vapour, peel of bark, fruits, leaves, gum, raisin, etc.
• Excretory products of humans: Carbon dioxide, urea, etc.
• Kidney: Organ which removes the toxic substance urea from blood and filters it.
• Urine: A yellowish liquid which contains water and urea.
• Dialysis: The procedure used for cleaning the blood of a person in case of kidney failure.
• Nephron: Functional unit of excretory system present in the kidney for filtering blood.
• Renal Artery: Blood vessels which bring blood from heart to kidney.
• Renal Vein: Blood vessel which brings blood from kidney to heart.

We hope the given CBSE Class 10 Science Notes Chapter 6 Life Processes Pdf free download will help you. If you have any query regarding NCERT Class 10 Science Notes Chapter 6 Life Processes, drop a comment below and we will get back to you at the earliest.

The post Life Processes Class 10 Notes Science Chapter 6 appeared first on Learn CBSE.

NCERT Solutions for Class 10 Social Science History Chapter 3 Nationalism in India

Q.1. Explain:
(a) Why is the growth of nationalism in the colonies linked to an anti-colonial movement ?
(b) How did the First World War help in the growth of the National Movement in India?    [CBSE2014]
Or
Explain any four facts to show how did the First World War help in the growth of the National Movement in India. [CBSE March 2011]
(a) Why were Indians outraged by the Rowlatt Act ?
(b) Why did Gandhiji decide to withdraw the Non-Cooperation Movement?
Ans. (a)

1. In India, as in Vietnam and many other colonies, the growth of modem nationalism is intimately connected to the Anti-colonial Movement. People began discovering their unity in the process of their struggle with colonialism. The sense of being oppressed under colonialism provided a shared bond that tied many different groups together.
2. The European powers considered their culture more civilised, modern and superior. They forcefully started imposing their culture on the colonies. This also aroused the feeling of nationalism.
3. Gandhiji used ‘Satyagraha’ against the Britishers. This also promoted the spirit of nationalism among the people.
4. The anti-colonial movement was a united struggle by the people against the foreigners. The united struggle was responsible for arousing the spirit of nationalism.

(b) The War created a new economic and political situation :

1. It led to a huge increase in defence expenditure which was financed by war loans and increasing taxes, customs duties were raised, and income tax introduced.
2. Through the war years, prices increased – doubling between 1913 and 1918 – leading to extreme hardships for the common people.
3. Villagers were called upon to supply soldiers, and the forced recruitment in rural areas caused widespread anger.

(c) 

1. Rowlatt Act was passed through the Imperial Legislative Council on a report of the Sedition Committee, headed by Justice Rowlatt.
2. It was the black act which gave the government and the police to repress political activities, and allowed detention of political prisoners without tried for two years.
3. The Act was passed despite the united opposition of the Indian members of the Council.
   This Act became one of the factors due to which Gandhiji launched Non-Cooperation Movement.

(d) In February 1922, Gandhiji decided to withdraw the Non-Cooperation Movement due to the following reasons:

1. The movement was turning violent. At Chauri-Chaura in Gorakhpur, a peaceful demonstration in a bazar turned into a violent clash in which more than 20 policemen were killed.
2. Gandhiji felt that the Safyagrahis needed to be properly trained before they would be ready for mass struggle.
3. Within the Congress, some leaders were tired of mass struggles and wanted to participate in elections to the provincial councils, which were set up under the Government of India Act, 1919.
4. Industrialists, workers, peasants etc. interpreted the term ‘Swaraj’ in their own way. At many places like that of Andhra Pradesh, leaders like Alluri Sitaram Raju asserted that India could be liberated only by the use of force. But there values were not approved by the Congress.

Q.2. What is meant by the idea of Satyagraha ?
Or
Explain the idea of Satyagraha according to Gandhiji.        [CBSE 2014 (D)]
Ans.

• It was a non-violent method of mass agitation against the Oppressor.
• It emphasised the power of truth and the need to search the truth.
• It suggested that if the cause was true, if the struggle was against injustice, there is no need for physical force to fight the oppressor.
• People-including the oppressors had to be persuaded to see the truth instead of being forced to accept truth through the use of violence.
• By this struggle, truth was bound to be victorious.

More Resources for CBSE Class 10

• NCERT Solutions
• NCERT Solutions for Class 10 Science
• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Social
• NCERT Solutions for Class 10 English
• NCERT Solutions for Class 10 Hindi
• NCERT Solutions for Class 10 Sanskrit
• NCERT Solutions for Class 10 Foundation of IT
• RD Sharma Class 10 Solutions

Q.3. Write a newspaper report on :
(a) The Jallianwala Bagh massacre
(b) The Simon Commission
Ans. (a) The Jallianwala Bagh massacre : A public meeting was announced for the 13th April, 1919, at Jallianwala Bagh, Amritsar to protest against the Rowlatt Act. The people were allowed to assemble there. After they had gathered there in thousands, General Dyer marched there with armoured cars and troops. Without giving any warning to the people to disperse, he ordered firing on the unarmed, and the peaceful people. The casualties among the Indians were very heavy. Dyer’s purpose in doing so was to ‘produce a moral effect’, to create in the minds of Satyagrahis, a feeling of terror and awe. This massacre of innocent people in thousands converted Mahatma Gandhi into a non-cooperator.

(b)

(i) The Indian members of the Central Legislative Assembly exposed the drawbacks in the Government of India Act of 1919 A.D. As a result of it, the Simon Commission was appointed in 1927 A.D. to suggest any further constitutional reforms. This commission consisted of seven members and its Chairman was Sir John Simon.

(ii) Why was it boycotted by the Indians?
But Indians boycotted the Simon Commission, because there was no Indian member in this commission. The terms of the commission’s appointment did not give any indication of ‘Swaraj’, while the demand of the Indians was only ‘Swaraj’. Therefore, the Indian National Congress, the Muslim League and other parties decided to oppose the Simon Commission.

(iii) Methods : Indian people organised hartals all over the country. They also held black flag demonstration with the slogan, “Simon go back”, when the Commission reached Bombay (Mumbai). Such demonstrations were held everywhere it went.

Q.4. List all the different social groups which joined the Non-Cooperation Movement of 1921. Choose any three, and write about their hopes and struggles to show why they joined the movement.
Ans. Social Groups who took part in the Non­Cooperation Movement. In the Non- Cooperaton Movement (1920-1922), the following social groups took part.
(I) Middle-class people in the towns.

1. The movement in the cities : The Movement started with middle-class participation in the cities. Thousands of students left government-controlled schools and colleges, headmasters and teachers resigned, and lawyers gave up their legal practices.
2. Boycott of council elections : The Council elections were boycotted in most provinces except Madras (Chennai), where the Justice Party, the party of the non­Brahmans, felt that entering the council was one way of gaining some power, something that usually only Brahmans had an access to.
3. Swadeshi : The Non-Cooperation Movement had a great impact on the Indian textile industry. Swadeshi goods, especially cloth got a great impetus. Foreign goods were boycotted, liquor shops picketed, and foreign cloth burnt in huge bonfires.
4. Impact on industry : In many places, merchants and traders refused to trade in foreign goods or finance foreign trade. Due to this, the demand of Indian textile mills and handlooms went up. The increase in demand provided a big relief to the vanishing textile industry of India.
5. Movement in the countryside: Though people in the countryside interpreted the idea of ‘Swaraj’ in their own way but they participated in the movement on large scale. In Awadh, peasants launched the movement against the talukdars and landlords. Whereas the plantation workers launched the movement against the tea estate owners.

(II) Peasants in the rural areas.

(i) Participants : In the countryside, the movement was led by the peasants, tribals and the local leaders. For example, in Awadh, it was Baba Ramchandra sanyasi, who had earlier been to Fiji as an indentured labourer.

(ii)  Why the rural people participated ?
The movement here was not against the Britishers but against talukdars and landlords. The problems of the rural people were different from those of the urban people:

• The talukdars and landlords were demanding very high rents and a variety of other taxes.
• Peasants had to do begarand work at the landlord’s farms without any payment.
• The peasants had no security of tenure. They were regularly evicted so that they could acquire no security of tenure.

As the problems of the people were different, their demands were also different. The peasant
movement demanded:

• Reduction of revenue
• Abolition of begar
• Redistribution of land
• Social boycott of oppressive landlords.

(iii) Ways of protests : The Movement in the countryside had a different angle. In many places, Nai-dhobi bandhs were organised by the Panchayats to deprive the landlords of the services of barbers, cobblers, washermen, etc. Even national leaders like Jawaharlal Nehru went to villages in Awadh to know the grievances of the people. By October, the Awadh Kissan Sabhas were set up headed by Jawaharlal Nehru, Baba Ramchandra, and a few others.
When the movement spread in 1921, the houses of talukdars and merchants were attacked. The movement turned violent which was not liked by some of the Congress leaders.

(III) Tribal people.
Most of the tribal people were dependent on forests for their livelihood but under the new Forest Policy, the government had put several restrictions on the people :

• Closing large forest area for the tribal people.
• Forcing the local people to contribute begar.
• Preventing people from entering the forests to graze their cattle, or to collect fuelwood and fruits.

All these steps enraged the hill people. Not only were their livelihoods affected, but they felt that their traditional rights were also being denied. So the people revolted.

(IV)  Plantation workers.
(i) For plantation workers in Assam, freedom meant the right to move freely in and out of the confined space in which they were enclosed, and it meant retaining a link with the village from which they had come.

• The government had passed the Inland Emigration Act of 1859 under which plantation workers were not permitted to leave the tea estates without permission, and in fact, they were rarely given such permission.
• When the plantation workers heard of the Non-Cooperation Movement, thousands of them defied the authorities, left the plantations and headed towards their homes.
• The plantation workers believed that the Gandhi Raj was coming, and everyone would be given land in their own villages.

Formulae Handbook for Class 10 Maths and Science

Q.5. Discuss the Salt March to make clear why it was an effective symbol of resistance against colonialism.  [CBSE 2015 (O)]
Ans. The Salt March was an effective symbol of resistance against colonialism because-

1. It was the first time that Indian leaders decided to violate law. People were now asked not only to refuse cooperation with the British, but also to break colonial laws.
2. Thousands of Indians in different parts of the country broke the salt law, manufactured salt and demonstrated in front of the government salt factories.
3. As the movement spread, foreign cloth was boycotted and liquor shops were picketed. Peasants refused to pay revenue and ‘chaukidari taxes’, village officials resigned, and in many places forest people violated forest laws – going into Reserved Forests to collect wood and graze cattle.
4. Worried by the development, the colonial government began arresting the Congress leaders, one by one. This led to violent clashes in many places. Angry crowd demonstrated in the streets, facing armoured cars and police firing. Many were killed.
5. When Mahatma Gandhi himself was arrested, industrial workers in Sholapur attacked police posts, municipal buildings, law courts and railway stations – all structures that symbolised the British rule.
6. The outcome of the movement was the Gandhi-Irwin Pact which was signed by Gandhiji with Irwin on 5th March, 1931. By this Gandhi-Irwin Pact, Gandhiji consented to participate in a Round Table Conference in London and the government agreed to release the political prisoners.

Q.6. Imagine you are a woman participating in the Civil Disobedience Movement. Explain what the experience meant to your life.
Or
‘Women played a very important role in the Civil Disobedience Movement.’ Explain.
Ans.

1. Women participated in large numbers in the Civil Disobedience Movement.
2. During the movement, thousands of women came out of their homes to listen to Gandhiji.
3. They participated in protest marches, manufactured salt, and picked foreign cloth and liquor shops.
4. Many were put to jail by the police.
5. Moved by Gandhiji’s call, they began to see service to the nation as a sacred duty of women.

Q.7. Why did the political leaders differ sharply over the question of separate electorates ?[CBSE2015]
Ans. By the system of separate electorates, we mean such a system when people of one religion only vote for a candidate of their own religion. Using such a system, was a mischief of the British Government who wanted to divide the people to weaken the national movement. By doing so, the British wanted to prolong their stay in India.
The different political leaders differed over the question of separate electorates because of the following reasons :

(1) The Congress leaders opposed the policy of the British Government in instigating different peoples in demanding separate electorate. They knew well that it was all the mischief of the British Government who encouraged different people to ask for separate electorates because such a policy would weaken the national movement, and prolong Britishers stay in India. The Congress leaders were one and all in favour of joint electorates.

(2) The Muslim leaders, like Muhammed Iqbal and Mr Jinnah asked for separate electorates to safeguard the political interests of the Muslims. In their opinion, the majority of the people were Hindus, and so in case of joint electorates, the Muslims would have little chance of winning the seats. As such, they would always be at the mercy of the Hindus.

(3) The leaders of the Depressed Classes, Dr. B.R. Ambedkar, also asked for separate electorates because in the joint electorates, he feared the dominance of the upper electorates or the upper caste Hindus in the elections. By the Poona Pact he, however, agreed to have joint electorates with the Hindus, provided the seats for the Depressed Classes were fixed or reserved in the Provincial and Central Legislative Councils.
Outcome : Lord Irwin announced in October 1929, a vague, offer of ‘dominion status’ for India.

Multiple Choice Questions

1. Which of the following agreements gave seats to the depressed classes in Provincial and Central Legislative council? [AI 2012]
(a) Lucknow Pact
(b) Gandhi-irwin Pact
(c) Poona Pact
(d) None of these

2. Which one of the following Viceroys announced a vague offer of dominion status for India in October 1929? [AI 2012]
(a) Lord Mount batter
(b) Lord Dalhousie
(c) Lord Irwin
(d) None of these

3. Which one of the following combination of colours was there in the Swaraj flag designed by Gandhiji in 1921? [AI 2012]
(a) Red, Green and White
(b) Red, Green and Yellow
(c) Orange, White and Green
(d) Yellow, White and Green

4. In which of the following region was Dalit participa¬tion limited in the civil disobedience movement? [AI 2012]
(a) Maharashtra and Nagpur
(b) Awadh and Maharashtra
(c) Bengal and Punjab
(d) Kerala and Karnataka

5. Why were the Dalits ignored by the Congress for a longtime? [AI 2012]
(a) Fear of offending the sanatanis
(b) Fear from Dr. B.R. Ambedkar
(c) Fear from socialism
(d) Fear from industralists

6. Which of the following Acts did not permit plantation workers to leave the tea gardens without permission? [AI 2012]
(a) Indian Independence Act
(b) Inland Emigration Act of 1859
(c) Impaired Legislature Council Act
(d) Inland Immigration Act

7. In which of the following year Mahatma Gandhi inspired the peasants of Champaran district of Bihar to struggle against the oppressive plantation system ?
(a) 1916
(b) 1917
(c) 1918
(d) 1919

8. In which one of the following Indian National Congress Session, the idea of Non-Cooperation movement was accepted ?
(a) Lahore Session
(b) Nagpur Session
(c) Calcutta (Kolkata) Session
(d) Madras (Chennai) Session

9. Why did Gandhiji withdraw the Non-Cooperation Movement ? [Delhi 2011]
(a) Gandhiji realised that people were losing interest in the movement.
(b) Gandhiji felt the movement was turning violent in many places.
(c) Some Congress leaders wanted to participate in elections to Provincial Councils.
(d) Some Congress leaders wanted more radical mass agitations.

10. Which one of the following statements is false about Alluri Sitaram Raju? [Delhi 2011]
(a) He claimed he had special powers.
(b) He was inspired by the non-cooperation movement.
(c) He believed in the principle of non-violence advocated by Gandhiji.
(d) He persuaded people to give up drinking.

11. Which one of the following statements is not related to the Gandhi-irwin Pact? [AI 2011]
(a) Gandhiji agreed not to launch any further mass agitations against the British.
(b) Gandhiji agreed to participate in the Round Table Conference.
(c) Gandhiji decided to call off the Civil Disobedience Movement.
(d) The British agreed to release the political prisoners.

12. Why did Nationalists in India tour villages to gather folk songs and legends ? Choose the most appropriate reason from the following : [AI 2011]
(a) Nationalists wanted to study their own culture.
(b) Nationalists wanted to publish it and earn money.
(c) Nationalists did it because it gave a true picture of traditional culture.
(d) Nationalists wanted to keep folk culture intact.

13. Who among the following wrote ‘Vande Mataram’? [Foreign 2011]
(a) Abanindranath Tagore
(b) Rabindranath Tagore
(c) Bankim Chandra Chattopadhyay
(d) Natesa Shastri

14. Which of the following statements is not true about the Simon Commission? [Foreign 2011]
(a) It was appointed by Sir John Simon.
(b) It did not have any Indian member.
(c) It was opposed by all parties in India.
(d) It was set up to look into the Constitutional system in India.

15. The Jallianwalla Bagh incident took place in the city of
(a) Amritsar
(b) Agra
(c) Meerut
(d) Lahore

16. Which of the following was the reason for calling off the Non-Cooperation Movement by Gandhiji ?
(a) His arrest
(b) The Chauri-Chaura incident
(c) Pressure from the British Government
(d) Round Table Conference

17. The resolution of Puma Swaraj was adopted at which session?
(a) Karachi Congress
(b) Haripur Congress
(c) Lahore Congress
(d) Lucknow Congress

18. The Simon Commission was boycotted because
(a) there was no Indian in the commission
(b) it supported the Muslim League
(c) Congress felt that the people deserved Swaraj
(d) there were differences among the members.

19. When did Jallianwalla Bagh incident take place ?
(a) 13th April 1919
(b) 14th April 1920
(c) 13th March 1919
(d) 15th March 1920

20. Justice Party of Madras was a party of
(a) non-muslims
(b) non-brahmins
(c) non-tamils
(d) judges

21. Who led a peasant movement during the Non-Cooperation Movement ?
(a) Jawaharlal Nehru
(b) Gandhiji
(c) Baba Ramchandra
(d) Sardar Patel

22. Baba Ramchandra led a Peasant Movement in Avadh against the
(a) British government
(b) the ruler of Avadh
(c) the Talukdars
(d) the moneylenders

23. By whom was the Swaraj Party formed?
(a) Motilal Nehru and C.R. Das
(b) Subhas Chandra Bose and Sardar Patel
(c) Jawaharlal Nehru and Rajendra Prasad
(d) Motilal Nehru and Rajendra Prasad

24. With which main demand did the Civil Disobedience Movement start ?
(a) Abolition of Untouchability
(b) Abolition of Dowry
(c) Abolition of Salt Law
(d) None of the above

25. At which of the following places did Gandhiji make salt out of sea water to defy the salt law ?
(a) Sabarmati
(b) Wardha
(c) Dandi
(d) Ahmedabad

26. Who among the following led the Civil Disobedience Movement in Peshawar ?
(a) Lala Lajpat Rai
(b) Maulana Abul Kalam Azad
(c) Khan Abdul Gaffar Khan
(d) Jawaharlal Nehru

27. By what name were the dalits referred by Gandhiji ?
(a) Untouchables
(b) Shudras
(c) Harijans
(d) Achhuts

28. When was the Indian Chambers of Commerce and Industries set up ?
(a) 1926
(b) 1927
(c) 1928
(d) 1929

29. Why was Alluri Sitaram Raju well known ?
(a) He led the militant movement of tribal peasants in Andhra Pradesh
(b) He led a peasant movement in Avadh
(c) He led a satyagraha movement in Bardoli
(d) He set up an organisation for the uplifment of the dalits.

30. Who organised the dalits into the Depressed Classes Association in 1930 ?
(a) Sitaram Raju
(b) Dr. B.R. Ambedkar
(c) Mahatma Gandhi
(d) Sardar Patel

31. By whom was the song ‘Vande Mataram’ written ?
(a) Bankim Chandra Chattopadhyay
(b) Rabindranath Tagore
(c) Sarat Chandra
(d) Natesa Sastri

32. Which of the following was the most important feature of Satyagraha Movement advocated by Gandhiji ?
(a) Abolition of untouchability
(b) Social equality
(c) Truth and non-violence
(d) Basic education

33. Why did Gandhiji organise Satyagraha in 1917 in Kheda district of Gujarat ?
(a) To support the plantation workers.
(b) To protest against high revenue demand.
(c) To support the mill workers to fulfil their demand.
(d) To demand loans for the farmers.

34. Gandhiji organised Satyagraha against Rowlatt Act in 1919 because
(a) the Act was unjust, which denied the civil rights to Indians.
(b) the Act was passed by the British.
(c) the Act discriminated against the Muslims.
(d) the Act denied educational rights to Indians.

35. Why was Satyagraha organised in Champaran in 1916 ?
(a) To oppose the British laws.
(b) To oppose the plantation system.
(c) To oppose high land revenue.
(d) To protest against the oppression of the mill workers.

36. Why did the Indians oppose the Rowlatt Act ?
(a) It introduced the Salt Law.
(b) It increased taxes on land.
(c) It gave the British the power to arrest and detain a person without a trial.
(d) It put a ban on the Congress party.

37. Why did Gandhiji organise a Satyagraha in Ahmedabad Mill in 1918 ?
(a) To protest against the poor working conditions in the factory.
(b) To demand for higher wages for workers.
(c) To protest against high revenue demand.
(d) None of the above.

38. Who among the following two leaders led the Khilafat Movement ?
(a) Shaukat Ali and Muhammad AM
(b) Gandhiji and Sardar Patel
(c) Muhammad Ali Jinnah and Abul Kalam Azad
(d) Abul Kalam Azad and Jawaharlal Nehru

39. What is meant by begar ?
(a) Unemployment
(b) Forced labour without payment
(c) Beggary
(d) Working for nominal payment

40. ‘Hind Swaraj’ was written by
(a) Abul Kalam Azad
(b) Mahatma Gandhi
(c) Sardar Patel
(d) Subhas Chandra Bose

41. What does the term Khalifa refer ?
(a) Sultan of a Muslim country
(b) Spiritual leader of the Muslims
(c) Nawab of a Muslim state
(d) Badshah of Mughal period

42. Mark the correct response
Under the Inland Emigration Act of 1859 the peasants were not permitted to
(a) leave their village
(b) settle in the city
(c) leave their plantation without permission
(d) allow the women to leave farmlands without permission

43. Why did Mahatma Gandhi decide to withdraw the Non-Cooperation Movement ?
(a) The leaders failed to organise the movement well
(b) People lacked courage
(c) The movement turned violent
(d) The movement was not widespread enough to continue

44. Why was the Simon Commission sent to India ?
(a) To look into the Indian constitutional matter and suggest reform
(b) To choose members of Indian Council
(c) To settle disputes between the government and the Congress leaders
(d) To set up a government organisation

45. Why was the Round Table Conference held in England ?
(a) To discuss the provisions of future Indian Constitution.
(b) To discuss the steps to be taken to check Indian National Movement.
(c) To give concessions to Indians
(d) To make plans for improvement of agriculture in India.

46. By whom was the first image of Bharatmata painted ?
(a) Rabindranath Tagore
(b) Abanindranath Tagore
(c) Ravi Verma .
(d) Nandalal Bose

47. During which of the following movements did the women participate in large numbers for the first time ?
(a) Swadeshi and Boycott Movement
(b) Non-Cooperation Movement
(c) Civil Disobedience Movement
(d) Quit India Movement

48. What kind of movement was launched by the tribal peasants of Gudem Hills in Andhra Pradesh ?
(a) Satyagraha Movement
(b) Militant Guerrilla Movement
(c) Non-Violent Movement
(d) None of the above.

49. Which of the following is the most important factor for the growth of nationalism in India ?
(a) British administrative reforms.
(b) Introduction of railways.
(c) Social reforms.
(d) Colonial exploitation under the British rule.

50. Why did General Dyer open fire on peaceful crowd in Jallianwalla Bagh?
Mark the most important factor
(a) To punish the Indians.
(b) To take revenge for breaking martial laws.
(c) To create a feeling of terror and awe in the mind of Indians.
(d) To disperse the crowd.

ANSWERS

The post NCERT Solutions for Class 10 Social Science History Chapter 3 Nationalism in India appeared first on Learn CBSE.

Important Question for Class 10 Science Life Processes PDF will help you in scoring more marks.. This consists of 1 mark Questions, 3 Mark Numericals Questions, 5 Marks Numerical Questions and previous year questions from Life Processes Chapter.

Life  Processes Chapter Wise Important Questions Class 10 Science

Short Answer Type Questions[l] [2 Marks] -Year 2015

1.What are enzymes? Name any one enzyme of our digestive system and write its function.
Answer. Enzymes are biological catalysts. Catalysts are proteins that increase the rate of chemical reactions
without being used up. For example: Amylase catalyses the breakdown of starch into sugars in the mouth and small intestine

2.(i) Write the balanced chemical equation for the process of photosynthesis,
(ii) When do the desert plants take up carbon dioxide and perform photosynthesis ?
Answer.
(i) Photosynthesis can be represented using a chemical equation. The overall balanced equation is

(ii) Desert plants open up their stomata during night and take in CO₂. Stomata remains close during the day time to prevent the loss of water by i transpiration. They store the CO₂ in their cells until the sun comes out and they can carry on with photosynthesis during the day time.

CBSE Class 10 Science – More Resources

Short Answer Type Questions[ll] [3 Marks] -Year 2015

3.In single celled organisms diffusion is sufficient to meet all their requirements of food, exchange of gases or removal of wastes but it is not in case of multicellular organisms. Explain the reason for this difference.
Answer. Unicellular organisms can absorb sufficient oxygen because of its complete contact with the atmosphere, but in multicellular organisms the rate of absorption and diffusion becomes very less because all cells are not in direct contact with the atmosphere. Multicellular organisms require greater amount of oxygen to sustain life processes which cannot be fulfilled by the process of diffusion.

4. Draw a diagram of human excretory system and label kidneys, ureters on it.
Answer.

5.Name the acid presents in the following:
(i) Tomato (ii) Vinegar (iii) Tamarind
Answer.
(i) Oxalic acid (ii) citric acid (iii) Tartaric acid.

6.State the role of the following in human digestive system :
(i) Digestive enzymes (ii) Hydrochloric acid (iii) Villi
Answer.
(i) Digestive enzymes – Foods need to be broken into their small or simpler
molecules so that they can be absorbed into the bloodstream. However, the physical breakdown of food is not enough. Enzymes are hence needed for the chemical breakdown of food and speeding up the digestive process. The products of digestion can hence be small enough to be absorbed.
(ii) Hydrochloric acid – Hydro chloric acid helps to kill the germs which might have entered in to the system through food. It creates acidic medium for the pepsin to act on food to breakdown proteins.
(iii) Villi – Villi are finger like projections in the small intestine. They help to increase the surface area for absorption of the digested food. Villi are richly supplied with blood vessel which help to absorb digested food in to the blood stream.

Long Answer Type Question [5 Marks] -Year 2015

7.(a)Explain how does the exchange of gases occur in plants across the surface of stems, roots and leaves.
(b) How are water and minerals transported in plants ?
Answer.
(a) In plants there are tiny pores called stomata on leaves and lenticels in stem which facilitate the exchange of gases. Carbon dioxide is taken in and oxygen given out {during photosynthesis} and vice versa during respiration.
(b) Water and minerals are transported within the plant by the Xylem vessels (mainly in an upward direction); these are part of the vascular system which also includes Phloem vessels.
Phloem transports the products of photosynthesis within the plant, to all parts like the stem, roots, fruits etc. in all directions.

Very Short Answer Type Question [1 Mark] -Year 2014

8. Mention the raw materials required for photosynthesis.
Answer. The following raw materials are required for photosynthesis:
(i) Carbon Dioxide: Plants get CO₂ from atmosphere through stomata.
(ii)Water: Plants absorb water from soil through roots and transport to leaves.
(iii) Sunlight: Sunlight, which is absorbed by the chlorophyll and other green parts of the plant.

Short Answer Type Questions[l] [2 Marks] -Year 2014

9. Why do herbivores have longer, small intestine than carnivores ?
Answer.Digestion of cellulose takes a longer time. Hence, herbivores eating grass need a longer small intestine to allow complete digestion of cellulose. Carnivorous animals cannot digest cellulose due to the absence of enzyme CELLULASE, hence they have a shorter intestine.

10.Write correct sequence of four steps of method for the preparation of temporary mount of a stained leaf peel.
Answer.

1. Take a healthy leaf from the potted plant.
2. Remove a part of the peel from the lower surface of the leaf. You can do this by folding the leaf over and gently pulling the peel apart using forceps. Keeps the peel in a watch glass containing water.
3. Put a few drops of safranin stain in a watch glass.
4. After 2-3 minutes take out the peel and place it on a clean glass slide.
5. Put a drop of glycerin over the peel and place a clean covers lip gently over it with the help of a needle.
6. Remove the excess stain and glycerin with the help of blotting paper.
7.  Observe the slide under magnifications of the compound microscope.

Short Answer Type Questions[ll] [3 Marks] -Year 2014

11.In mammals and birds why is it necessary to separate oxygenated and de-oxygenated blood ?
Answer. Mammals and birds are warm blooded animals. This means they can control their body temperature and do not have to depend on environment for their body temperature regulation. Because of this birds and mammals require optimum oxidization of glucose which would be possible with good supply of oxygen. So it is required to have separate oxygenated and de-oxygenated blood to supply the required amount of oxygen.

12.Draw a neat diagram of excretory system of human beings and label on it:
(i) Left kidney (ii) Urinary bladder
Answer.

13. Draw a diagram of human respiratory system and label on it :
(i) Diaphragm (ii) Larynx
Answer.

14.(a) Name the site of exchange of material between the blood and surrounding cells. (b) Draw a schematic representation of transport and exchange of oxygen and carbon dioxide in human body.
Answer.
(a) Capillaries
(b)

Very Short Answer Type Question [1 Mark] -Year 2013

15.What would be the consequences of deficiency of hemoglobin in your body?
Answer.The deficiency of hemoglobin in our body is called anemia. In anemia, the blood is unable to carry the sufficient amount of oxygen required by the body. So, respiration would be less and less energy will be available to the body. The hemoglobin deficient person will feel weak, pale, lethargic and will be
unable to perform heavy physical work.

Short Answer Type Questions[ll] [3 Marks] -Year 2013

16. List three characteristics of lungs which make it an efficient respiratory surface.
Answer. These features which particularly make our lungs efficient for gas exchange.
1.Thin: the air sac walls are very thin so that gases can quickly diffuse through them. Oxygen is absorbed in to the blood and carbon dioxide is given out in to the lungs to be exhaled out.
2.Moist: the air sacs are moist with mucus so that gases can dissolve before diffusing.
3.Large surface area: the surface area for gases to diffuse through in human lungs is roughly the same as a tennis court. The alveoli help to increase the surface area for absorption of oxygen.
4.Good blood supply: the air sacs or the alveoli have a large capillary network so that large volumes of gases can be exchanged. More the flow of blood more exchange.

17.(a) What is the role of HCl in our stomach ?
(b) What is emulsification of fats ?
(c) Which protein digesting enzyme is present in pancreatic juice ?
Answer.
(a)(i) It sterilises food by killing pathogens and other microbes.
(ii) It has a pH of 2, which is perfect for entyaus such as pepsin to break down proteins as effectively as possible.
(iii) Helps emulsify food (digestion of protein and stimulates the pancreas to produce digestive enzymes and bile) and protects against harmful ‘ bacteria
(b) Breakdown of large gloubule fats into smaller fats droplets is known as emulsification.
(c) Trypsin is the enzyme secreted by the pancreas which aids in digestion of proteins.

18. List in tabular form three differences between arteries and veins.
Answer.

long Answer Type Questions [5 Marks] -Year 2013

19. Draw a diagram of human excretory system and label renal artery and urethra.
State in brief the function of :

1. renal artery
2.  kidney
3.  ureter
4. urinary bladder

Answer.

1. Renal artery: The renal artery carries blood to the kidneys from the abdominal aorta. This blood comes directly from the heart and is sent to the-kidneys to be filtered before it passes through the rest of the body. Up to one-third of the total cardiac output per heartbeat is sent to the renal arteries to be filtered by the kidneys. Each kidney has one renal artery that supplies it with blood. The filtered blood then can exit the renal vein.
2. Kidney: The kidneys perform the essential function of removing waste products from the blood and regulating the water fluid levels. The kidneys regulate the body’s fluid volume, mineral composition and acidity by excreting and reabsorbing water and inorganic electrolytes.
3. Ureter: The ureter is a tube that carries urine from the kidney to the urinary bladder.’ There are two ureters, one attached to each kidney.
4. Urinary bladder: The urinary bladder is an expandable muscular sac that stores urine before it is excreted out of the body through the urethra.

20.(a) Draw a diagram of excretory system in human beings and label the following parts. Aorta, kidney, urinary bladder and urethra.
(b) How is urine produced and eliminated ?
Answer.(a)

(b) Blood from the heart comes into the kidneys afferent and efferent arteriols from the renal arteries where it enters about 2-3 million nephrons per kidney. Then, it goes through the glomerulus a tugt or bunch of blood capillaries and get rid of some of the unwanted substances like urea, uric acid, creatinine in the blood and then continues through the renal tubules. The loop of Henley, reabsorb certain substances such as water (actually if body is dehydrated, body will send anti-diuretic hormone (ADH) to kidneys to prevent extra water from going into urine and thus saving water for body and get rid of anything else that isn’t wanted, then the urine goes through ureters to bladder and then to urethra where it is excreted out of body as urine.

Short Answer Question [I] [2 Marks] -Year 2012

21. Why do the walls of the trachea not collapse’when there is less air in it?
Answer. Rings of cartilages are present in trachea. These rings support the trachea and do not allow the trachea to collapse when there is less air in it.

Short Answer Type Questions [ll] [3 Marks] -Year 2012

22. In human alimentary canal, name the site of complete digestion of various components of food. Explain the process of digestion.
Answer.In small intestine, complete digestion of various components of food take place. The process of digestion of food in mouth, stomach and small intestine in human
body are as follows:
Mouth: Digestion of food begins in the mouth. Saliva present in mouth contains a digestive enzyme, called salivary amylase, maltose and dextrins, which breaks down starch into sugar.
Stomach: Stomach stores and mixes the food received from the oesophagus with gastric juices. The main components of gastric juice are hydrochloric acid, mucus and pepsinogen. Hydrochloric acid dissolves bits of food and creates an acidic medium. In this medium, pepsinogen is converted to pepsin which is a protein-digesting enzyme. Mucus protects the inner lining of the stomach from the action of HC1.
Small Intestine: Small intestine is the site of complete digestion of carbohydrates, proteins and fats. Small intestine produces intestinal juice from the glands present in its wall. The intestinal juice helps in further digestion of food. Small intestine also obtains digestive juices from liver and pancreas. The liver produces bile juice that causes emulsification of fats and the pancreas produces pancreatic juice for digesting proteins and emulsified fats. This digested food is finally absorbed through the intestinal walls.

23. List the three kinds of blood vessels of human circulatory system and write ¦ their functions in tabular form.
Answer. Three types of blood vessels in human circulatory system are: Arteries, Veins and Capillaries.
Their functions are tabulated below:

Lonq Answer Type Questions [5 Marks] -Year 2012

24.(a) Draw a diagram to show open stomatal pore and label on it:
(i) guard cells
(ii) chloroplast
(b) State two functions of stomata.
(c) How do guard cells regulate the opening and closing of stomatal pore?
Answer.(a)

(b) Two functions of stomata are:
(i) Exchange of gases between the plant and the atmosphere takes place through stomata.
(ii)Transpiration in plants takes place through stomata.
(c) Opening and Closing of Stomatal Pore: The opening and closing of the pore is a function of the guard cells. The guard cells swell when water flows into them causing the stomatal pore to open. Similarly, the pore closes if the guard cells shrink. As large amount of water is lost through these stomata, the plant closes these pores when it does not require carbon dioxide for photosynthesis.

25.(a) Draw a diagram of human respiratory system and label the following:
(i) part where air is filtered by fine hair and mucus
(ii) part which terminates in balloon – like structures
(iii) balloon – like structures where exchange of gases takes place. (iv) part which separates chest cavity from abdominal cavity.
(b) Why is the rate of breathing in aquatic organisms much faster than in terrestrial organisms?
Answer.(a)

(b) Quantity of dissolved oxygen is fairly low in water as compared to the amount of oxygen in air. Aquatic organisms therefore have to breath faster than terrestrial organisms to absorb the required amount of oxygen from the water.

26. Draw a neat diagram of excretory system of human beings and label the following:
(i) Kidney
(ii) Ureter
(iii) Urinary Bladder
(iv) Urethra
Answer.

Short Answer Type Questions[l] [2 Marks] -Year 2011

27.What are the final products after digestion of carbohydrates and proteins?
Answer.The final product produced after digestion of carbohydrates is glucose and of proteins is amino acids.

28.What is saliva? State its role in the digestion of food.
Answer. Saliva is a watery fluid secreted by the salivary glands in the mouth. The digestive functions of saliva include moistening food, and helping to create a food bolus, so it can be swallowed easily. Saliva contains the enzyme amylase that breaks some starches down into maltose and dextrin.

29.Explain the process of nutrition in Amoeba.
Answer. Amoeba is an important protozoa found in fresh water. It feeds on microscopic plants and animals present in water. The mode of nutrition in amoeba is Holozoic. And the process of obtaining food by amoeba is called phagocytosis. The different processes involved in the nutrition of amoeba are:

1. Ingestion: Ingestion is the process of taking food in the body. Amoeba is a unicellular animal, so it doesn’t have a mouth for ingestion of food. Amoeba ingests the food by encircling it by forming pseudopodia. When the food is completely encircled , the food is engulfed in the form of a bag called food vacuole.
2. Digestion: Digestion is the process of breaking the large and insoluble molecules in small and water soluble molecules. In amoeba, several digestive enzymes react on the food present in the food vacuoles and break it down into simple and soluble molecules.
3. Absorption: The food digested by digestive enzymes is then absorbed in the cytoplasm by the process of diffusion. While the undigested food remains in the food vacuole. If a large amount of food is absorbed by amoeba, the excess food is stored in the cytoplasm ih the form of glycogen and lipids.
4. Assimilation: During this step the food absorbed by the cytoplasm is used to obtain energy, growth and repair. This process of utilizing absorbed food for obtaining energy, repair and growth is called assimilation.
5. Egestion: When a sufficient amount of undigested food gets collected in the food vacuole, it is thrown out of the body by rupturing cell membrane. The process of removal of undigested food from the body is called egestion.

30. State two differences between arteries and veins.
Answer. Arteries

1. Arteries carry oxygenated blood, away from the heart except pulmonary artery.
2. These are thick-walled, highly muscular except arteries of cranium and vertebral column.
3. Valves are absent.

Blood in arteries moves with pressure.Veins

1. Veins carry deoxygenated blood, towards the heart except pulmonary veins.
2. These are thin-walled.
3. Valves are present which provide unidirectional flow of blood.
4.  Blood in veins moves under very low pressure.

31. How are the alveoli designed to maximise the exchange of gases?
Answer.Alveoli are small pouches or sacs like structure. They are surrounded by blood capillaries. Thus a large amount of air is brought in contact with the air in the lungs. More than millions of alveoli are present in the lungs. The presence of millions of alveoli in the lungs provides a very large surface area for the exchange of gases. The availability of large surface area maximises the exchange of gases.

32. Name two excretory products other than 02 and CO₂ in plants.
Answer. The two excretory products other than 02 and CO₂ in plants are resins and gums.

Short Answer Type Questions[ll] [3 Marks] -Year 2011

33.(a) “The breathing cycle is rhythmic whereas exchange of gases is a continuous process”. Justify this statement.
(b) What happens if conducting tubes of circulatory system develops a leak? State in brief, how could this be avoided?
(c) How opening and closing of stomata takes place?
Answer. (a) The breathing cycle involves inhalation and exhalation of air due to alternate expansion and contraction of thoracic cavity. Thus it is a rhythmic process. But exchange of gases is a continuous process as it takes place between the blood and each and every cell, by diffusion.
(b) The circulatory system will become inefficient if it develops a leak. This could be avoided by maintaining a normal blood pressure.
(c) When water flows into the guard cells, the guard cells swell and the stomatal pore opens up. When water moves out the guard cells shrinks and the stomatal pore closes.

34. Draw a diagram of the front view of human heart and label any six parts including at least two, that are concerned with arterial blood supply to the heart muscles.
Answer.

Long Answer Type Questions [5 Marks] -Year 2011

35.(a) Draw a schematic representation of transport and exchange of oxygen and carbon dioxide during transportation of blood in human beings and label on it:
Lung capillaries, Pulmonary artery to lungs, Aorta to body, Pulmonary veins from lungs.
(b) What is the advantage of separate channels in mammals and birds for oxygenated and deoxygenated blood?
Answer.(a) A schematic representation of transportation and exchange of oxygen and
carbon dioxide during transportation of blood in human beings

(b) It is necessary to separate oxygenated and deoxygenated blood in mammals and birds because they need high energy and large amount of oxygen. The separation of oxygenated and deoxygenated blood provides high oxygen supply to the organs.

36.(a) Draw a diagram depicting Human Alimentary Canal and label on it: Gall bladder, Liver and Pancreas.
(b) State the roles of Liver and Pancreas.
(c) Name the organ which performs the following functions in humans:
(i) Absorption of digested food
(ii) Absorption of water.
Answer.(a)

(b) Liver: It synthesize and store bile juice secreted by gall bladder which breaks down fats into fat globules.
(c) The organ which performs the following functions in humans are as follows:
Pancreas: It secretes pancreatic juice which contains protein-digesting and starch- digesting enzymes. *
(i) Absorption of digested food – Small intestine.
(ii)Absorption of water – Large intestine.

37.(a) Draw a sectional view of the human heart and label on it – Aorta, Right ventricle and Pulmonary veins.
(b) State the functions of the following components of transport system:
(i) Blood (ii) Lymph
Answer.(a)

(b) The functions of blood and lymph are as follows:
(i) Blood

• Oxygen is transported by the blood to the tissues of the body for the breakdown of digested food.
• Carbon dioxide is transported to the lungs by the blood plasma.
• The digested and absorbed nutrients are transported by blood to the tissues. Nitrogenous wastes are transported to the kidneys.
• It regulates the body temperature and maintains the pH of the body tissues.
• It transports various hormones from one region to another and bring about the coordination.
• It maintains water balance to constant level.
• The lymphocytes produce antibodies against the invading antigens and protect from diseases.
• It helps in rapid healing of wounds by forming a clot at the site of injury.

(ii) Lymph

• It cleans the cellular environment.
• It returns proteins and tissue fluids to the blood (drainage)
• It provides a pathway for the absorption of fats and fat-soluble vitamins into the bloodstream.
• It defends the body against disease.

38.(a) Draw a labelled diagram of the respiratory system of human beings with diaphragm at the end of expiration.
(b) List four conditions required for efficient gas exchange in an organism.
Answer.(a)

(b) (i) A large surface area over which exchange can take place.
(ii) A concentration gradient without which nothing will diffuse.
(iii) A thin surface across which gases diffuse.
(iv) Warm conditions.

39.(a) Draw a diagram to show the nutrition in Amoeba and label the parts used for this purpose. Mention any other purpose served by this part other than nutrition.
(b) Name the glands associated with digestion of starch in human digestive tract and mention their role.
(c) How is required pH maintained in the stomach and small intestine?
Answer.(a)

Pseudopodia serves the purpose of locomotion apart from nutrition.
(b) The salivary gland is associated with digestion of starch in human digestive tract.
It secretes saliva which contains enzyme salivary amylase. This enzyme converts starch into maltose (sugar).
(c) Gastric glands present on the walls of the stomach release HC1. HC1 creates an acidic medium, which facilitates the action of enzyme pepsin. Bile juice from liver makes the food alkaline in small intestine for the pancreatic enzymes to act.

40. (a) Draw a labelled diagram of stomata. List two functions of stomata.
(b) What are the raw materials used during photosynthesis? Write chemical equation for photosynthesis.
Answer.(a)

(i) Gaseous exchanges between plant and the atmosphere.
(ii) Plant loses water through stomata which helps in movement of minerals from soil to leaves.
(b) Raw materials for photosynthesis: Carbon dioxide, water, chlorophyll and sunlight are the essential raw materials for photosynthesis.
(i) Carbon dioxide is a gas, which is released into the atmosphere during respiration. This gas is utilised by the autotrophic plants which enters the leaf through the stomata present on its surface during the process of photosynthesis.
(ii) Water is another requirement for photosynthesis, which is transported upward through xylem tissues to the leaves, from where it reaches the photosynthetic cells. This water then splits in the presence of sunlight and chlorophyll.
(iii) Chlorophyll is a green pigment in plants, which acts as a catalyst. It is responsible for absorption of the sun’s energy by the plant. The chlorophyll pigments are photoreceptor molecules which play a key role in the photosynthetic process. The different types of chlorophyll molecules are chlorophyll a, b, c, d, e and bacteriochlorophyll; of which chlorophyll a and b are the most common.
(iv) Light affects photosynthesis by its intensity, quality and duration. In green light, the rate of photosynthesis is minimum, while in red and blue lights the rate of photosynthesis is maximum.
Rate of photosynthesis is higher in plants getting average light of 10-12 hrs a day.
The chemical equation for photosynthesis is as follows:

41.(a) Leaves of a healthy potted plant were coated with vaseline to block the stomata. Will this plant remain healthy for long? State three reasons for your answer.
(b) State any two differences between autotrophic nutrition and heterotrophic nutrition.
Answer. (a) No, this plant will not remain healthy for long. The plant will begin to die because
(i) Gaseous exchange will not take place.
(ii) No absorption of C02, hence no photosynthesis.
(iii) Transpiration will not occur; hence no transportation of water.

Very Short Answer Type Questions [1 Mark] -Year 2010

42.Name the green dot like structures in some cells observed by a student when a leaf peel was viewed under a microscope. What is this green colour due to?
Answer. The green dot-like structures in some cells observed by a student when a leaf peel is viewed under a microscope are chloroplasts. The green colour is due to the presence of green pigment, chlorophyll.

43. State any one difference between autotrophic and heterotrophic modes of nutrition.
Answer. In autotrophic nutrition, organisms obtain their food from inorganic substances. In heterotrophic nutrition, organisms derive their food from organic substances.

44. A Give one reason why multicellular organisms require special organs for exchange of gases between their body and their environment. :
Answer. In unicellular organisms the entire body of the organism is in contact with the environment hence exchange of materials can take place but, in multicellular organisms the entire body of the organism is not in contact with the environment and hence simple diffusion is not helpful.

45. Name the process in plants where water is lost as water vapour. :
Answer. Transpiration is the process when plants loose water as vapour.

46. What is‘translocation’in plants?
Answer. Translocation is the movement of soluble materials, products of photosynthesis from leaves to other tissues throughout the plant.

47.State the basic difference between the process of respiration and photosynthesis.
Answer.

48. Name the intermediate and the end products of glucose breakdown in aerobic respiration.
Answer.

49. In the experiment “Light is essential for photosynthesis”, why does the uncovered part of the leaf turn blue-black after putting iodine solution?
Answer. Starch is produced in the uncovered part of the leaf because it is exposed to sunlight allowing it to photosynthesize, which turns blue-black in presence of iodine solution.

50.Name the component of blood that helps in the formation of blood clot in the event of a cut.
Answer. Platelets help in clotting of blood in the event of a cut.

51.Mention how organisms like bread moulds and mushrooms obtain their food.
Answer. Organisms like bread moulds and mushrooms breakdown the food materials outside the body and then absorb the nutrients of the bread.

Short Answer Type Questions[ll] [3 Marks] -Year 2010

52.Describe in brief the function of kidneys, ureters, urinary bladder and urethra.
Answer.The Kidneys filter the blood and concentrate the filtrate to make urine. They also help regulate blood pressure.
Ureters transport the urine to the urinary bladder.
Urinary bladder is like a holding tank for the urine until it’s ready to be excreted. Urethra is the tube that connects the urinary bladder to the outside of the body for excretion.

53. Explain the process of breakdown of glucose in a cell (ii) in the absence of oxygen.
Answer. The process of breakdown of glucose in a cell are as follows:

The first step in the breakdown of glucose both in presence of 02 and in absence of Os is same. In this step, glucose is broken down into pyruvate.
Second step which involves further breakdown of private into simple compounds can take place in two different ways:
(i) In presence of Oz: In the presence of 02, private is converted into C02 and water. Energy released during aerobic respiration is much greater than that released during an anaerobic respiration.
(ii)In absence of Oz: In the absence of 02 in yeast, pyruvate is converted into ethanol and C02 and the process is called fermentation. In absence of 02, in our muscle cells, pyruvate is converted into lactic acid. The build up of lactic acid in muscle cells causes cramps.

Long Answer Type Question [5 Marks] -Year 2010

54.(a) List the three events that occur during the process of photosynthesis.
Explain the role of stomata in this process.
(b) Describe an experiment to show that “sunlight is essential for photosynthesis.”
Answer.
(a) The three events that occur during the process of photosynthesis are:
(i) Absorption of light energy by the green pigment chlorophyll.
(ii) Conversion of light energy into chemical energy and the splitting of water molecule into hydrogen and oxygen.
(iii) Reduction of carbon dioxide into carbohydrate.
Role of Stomata
Stomata are tiny pores present on the surface of leaves. They are also present on the surface of young stems. Stomata are mainly engaged in the exchange of gases (entry of CO₂ and release of O₂ ) associated with photosynthesis. Plant closes the stomata when it does not need CO₂ for photosynthesis.
(b) Sunlight is essential for photosynthesis

Procedure:
(i) Place a healthy green potted plant in a dark room for 1-2 days. This is done to ensure
that the plant consumes all its reserve food and the leaves do not contain any starch.
(ii) Then, cover a portion of a leaf of this plant on both sides with two uniform pieces of black paper, fixed in position with two paper clips.
(iii) Now, expose this plant to bright light. After a few hours, remove the leaf and decolorize it with alcohol and test the presence of food (starch) with iodine solution.
Observation: It can be observed that the portion of the leaf covered with black paper does not (food),
Conclusion: This is because the food prepared by plants through the process of photosynthesis is stored as starch. Starch reacts with the iodine solution to give blue-black colour. Only those portions of the leaf that were exposed to sunlight could photosynthesise. Hence, gives blue- black colour when tested with iodine. The portion of the leaf covered with black paper did not receive sunlight. Hence, starch was not produced. Thus, it can be
concluded that the sunlight is essential for photosynthesis.

Very Short Answer Type Questions [1 Mark] -Year 2009

55. What will happen to a plant if its xylem is removed?
Answer. Xylem in plant transports water and dissolved mineral nutrients from the roots to all parts of the vascular plant. So, if xylem is removed from the plant, the water and mineral supply to the plant will stop and therefore, the plant will die.

56. Where does digestion of fat take place in our body?
Answer. Digestion of fat takes place in the small intestine of our body.

57. What is the mode of nutrition in human beings?
Answer. Holozoic nutrition.

Long Answer Type Questions [5 Marks] -Year 2009

58.(a) What is meant by breathing? What happens to the rate of breathing during
vigorous exercise and why?
(b) Define translocation with respect to transport in plants. Why is it essential for plants? Where in plants are the following synthesised?
(i) Sugar (iii) Hormone
Answer.(a) The process of taking in of oxygen from air in to the lungs and expulsion of carbon dioxide out of the lungs is called breathing. The rate of breathing during vigorous exercise increases by about 20 to 25 times per minute. It is because, during vigorous exercise the demand for oxygen increases. Breathing occurs involuntarily but its rate is controlled by the respiratory center of the brain.
(b) Translocation is the transport of food from the leaves to other parts of the plant and occurs in the part of the vascular tissue known as phloem.
It is essential for plants because every part of the plant needs food for obtaining energy for building its parts and maintaining its life.
(i) Sugar is synthesised in the leaves of the plant.
(ii) Hormones are synthesised at the tips of roots and stems of a plant.

59.(a) Draw the structure of a nephron and label the following on it:
Glomerulus, Bowman’s capsule, Renal artery, Collecting duct.
(b) What happens to glucose that enters the nephron along with filtrate?
Answer.(a)

(b) During excretion in human beings, glucose which enters the nephron along with filtrate gets reabsorbed by blood capillaries surrounding the nephron.

NCERT Solutions for Class 10

The post Life Processes Chapter Wise Important Questions Class 10 Science appeared first on Learn CBSE.

 NCERT Solutions for Class 11 Political Science: Political Science Class 11 Solutions are solved by experts of LearnCBSE.in in order to help students to obtain excellent marks in their Exams. All the questions and answers that are present in the CBSE NCERT Books has been included in this page.

CBSE Political Science Class 11 NCERT Solutions, Extra Questions and Important Questions pdf download of Textbook Indian Constitution at Work and Political Theory are designed by subject experts to prepare students to score well. So students having great knowledge over NCERT Solutions Class 11 Political Science can easily make a grade in their board exams. Read on to find out more about NCERT Solutions for Class 11 Textbook Indian Constitution at Work and Political Theory PDF.

Political Science Class 11 NCERT Solutions

We have provided all the Class 11 Political Science NCERT Solutions with a detailed explanation i.e., we have solved all the questions with step by step solutions in understandable language.

Political Science Class 11 NCERT Solutions: Indian Constitution at Work

NCERT Solutions for Class 11 Political Science Indian Constitution at Work

• Chapter 1 Constitution: Why and How?
• Chapter 2 Rights and Duties in the Indian Constitution
• Chapter 3 Election and Representation
• Chapter 4 Executive
• Chapter 5 Legislature
• Chapter 6 Judiciary
• Chapter 7 Federalism
• Chapter 8 Local Governments
• Chapter 9 Constitution as a Living Document
• Chapter 10 The Philosophy of the Constitution

Political Science Class 11 NCERT Solutions: Political Theory

NCERT Solutions for Class 11 Political Science Political Theory

• Chapter 1 Political Theory: An Introduction
• Chapter 2 Freedom
• Chapter 3 Equality
• Chapter 4 Social Justice
• Chapter 5 Rights
• Chapter 6 Citizenship
• Chapter 7 Nationalism
• Chapter 8 Secularism
• Chapter 9 Peace

Advantages of Solving Class 11 Political Science from LearnCBSE.in

•  All the Class 11 NCERT Textbook Solutions provided in this page are clear and concise in nature.
• NCERT Solutions for Class 11 Books are solved in easily understandable language to help students to grasp everything on the go.
• Accessible to everyone at any time anywhere without any difficulty.
• All the questions are solved strictly based on the NCERT (CBSE) Syllabus and Books. So mastering these solutions will definitely help students to score good marks in the examination.
• NCERT Solutions for Class XI  given in this page are of free of cost.

We hope the NCERT Solutions for Class 11 provided in this page helps in your board exam preparation.

The post NCERT Solutions for Class 11 Political Science appeared first on Learn CBSE.

NCERT Solutions for Class 9 Science are given for the students so that they can get to know the in text question and chapter end question answers in case they are not able to find it. It is important for all the students who are in Class 9th currently. Here we are providing the solutions to all the chapters of Class 9 Science Textbook for the students. They can refer to samajik vigyan class 9 Physics, Chemistry and Biology Solutions while they are solving the questions from the Science Book.

The students who are in class 9th and have the NCERT Science Books can check this page to know the answers for all the chapters of Science book. They need to ensure that they are checking the solutions for the chapter which they intend to check.

FAQs on NCERT Solutions for Class 9 Science

1. What’s the best way to study Class 9 Science CBSE?

The best way to study Class 9 Science is through NCERT. Don’t just mug up the concepts instead go through all the concepts thoroughly. Make notes and practice as much as you can.

2. Where can I get the NCERT solutions for Class 9 Science?

You need not panic as we will give you all the NCERT Solutions for Class 9 Science all at one place on our site. View or download them and practice as and when you want.

3. Which is the best book of science for Class 9 CBSE?

We will not specifically say one book is the best for preparing Class 9 Science. Instead, we will ay try covering all the concepts in NCERT Textbooks prescribed for Class 9 Science that can help in your Exams.

4. How do I score good marks in science class 9?

The only key for scoring more marks in the CBSE Class 9 Science Exam is through sheer practice. It’s the best way to score well in the exams as well as get command on the subject.

5. Where can I get CBSE Class 9 Sample Papers?

Aspirants can get the CBSE Class 9 Sample Papers that can ace up preparation for free of cost from our page. Check out the links available on our site and use them as a reference to score better grades in the exam.

6. What are the study tips for Class 9 Science Exam?

Some of the best tips that help you to score more in your Class 9 Science Exam are as follows

• Make a proper plan and schedule to study
• Focus on the basic concepts and try concentrating on weaker areas.
• Dedicate time to each of the concepts.
• Practice as much as you can to score well.

Important Questions For Class 9 Science

Importance of the Class 9 Physics, Chemistry, Biology NCERT Solutions

The NCERT Class 9 Science Textbook Solutions are important for the students to get good marks in their exams. Through these, they can get solutions to those questions on which they get stuck. These are the valid solutions and students can check these whenever they face any confusion in the questions.

The post NCERT Solutions for Class 9 Science appeared first on Learn CBSE.

Download MCQ Questions of all subjects for Classes 6 to 12 from here. CBSE Board has brought a major modification in the Exam Pattern and plans to introduce more MCQs in the Board Exams. Multiple Choice Questions will have considerable weightage in the Board Exams. Students might feel it unchallenging and feel relaxed as they don’t have to attempt the lengthy descriptive answers. Cracking the MCQs still needs a lot of practice and hard work.  Answering the Objective Type Questions needs a lot of Critical Thinking.

We will help students to prepare effectively with the Classwise MCQ Quiz Questions for all subjects. CBSE board combines questions from different topics of the subjects covered in NCERT Textbooks. Thus, in order to score good marks in the exam, be thorough with the textbooks, and practice the list of Objective Questions Compiled in PDF links.

• MCQ Questions with Answers PDF Download
  • MCQ Questions for Class 6 Maths with Answers
  • MCQ Questions for Class 6 Science with Answers
  • MCQ Questions for Class 6 Social Science
  • MCQ Questions for Class 7 Maths with Answers
  • MCQ Questions for Class 7 Science with Answers
  • MCQ Questions for Class 7 Social Science
  • MCQ Questions for Class 8 Maths with Answers
  • MCQ Questions for Class 8 Science with Answers
  • MCQ Questions for Class 8 Social Science
  • MCQ Questions for Class 9 Maths
  • MCQ Questions for Class 9 Science
  • MCQ Questions for Class 10 Maths
  • MCQ Questions for Class 10 Science
  • MCQ Questions for Class 10 Social Science
  • Maths MCQ Questions for Class 12 Chapterwise with Answers
  • Physics MCQ Questions for Class 12 Chapterwise with Answers
  • Chemistry MCQ Questions for Class 12 Chapterwise with Answers
  • Biology MCQ Questions for Class 12 Chapterwise with Answers
  • Accountancy MCQ Questions for Class 12 Chapterwise with Answers
  • Business Studies MCQ Questions for Class 12 Chapterwise with Answers
• MCQ Quiz Questions for CBSE Class 6 to Class 12
• Tips to keep in mind while answering MCQ Questions
• What is the best strategy while answering Multiple Choice Questions of Board Exams?
• Where can I get CBSE Classwise Multiple Choice Questions and Answers?
• How should I Prepare for the MCQs of CBSE Board Exams?
• How do I study for Objective Questions or MCQs?
• Where can I find all of the MCQ questions for Class 6 to Class 12?

Download CBSE Board Multiple Choice Questions Classwise

Practice the Multiple Choice Questions of all the Subjects to test your understanding of important topics in Chapters. Download Latest Multiple Choice Questions with Answers for Classes 6, 7, 8, 9, 10, 11, 12 in PDF for free. The Objective type questions over here will cover the important topics of all the subjects.

Click on the direct links to download CBSE Objective Type Questions Subjectwise for various Classes. You can use them during your preparation as a reference and score max. marks in the exam. The Multiple Choice Questions with Answers are as per the Latest Guidelines issued by the CBSE. Objective Questions prevailing here are as per the Latest NCERT Syllabus. You can rely on them completely as they are present to you after enormous research.

MCQ Questions with Answers PDF Download

We in this article will cover MCQ Questions for Class 12 with Answers PDF Download for subjects Maths, Physics, Chemistry, Biology, Accountancy, and Business Studies. Multiple Choice Questions prevailing here are based on the CBSE Syllabus and NCERT Textbooks. Download Objective Type Questions for all subjects for Classes 6 to 12 in PDF format.

When it comes to CBSE Class 10 we have Multiple Choice Questions for Maths, Science, and Social. Get to know the CBSE Objective Type Questions for Classes 6, 7, 8, 9, 10, 11, 12 in PDF Format of all subjects by referring to the further modules. You can download the necessary study material & notes for free of cost by accessing the direct links on our page.

MCQ Quiz Questions for CBSE Class 6 to Class 12

If you are preparing for CBSE Board Exams try having a look at the Objective Type Questions existing. You can rely on them as they are collected from the Past CBSE Sample Question Papers. Give them the utmost importance and cover all the topics in a smart way. Access the direct links to download MCQ Questions for CBSE from here and download them for free. If you wish you can visit our site and view them online too during your preparation.

Tips to keep in mind while answering MCQ Questions

The following are the Tips and Tricks while attempting the Objective Type Questions for Classes 6 to 12. These handy and useful tips can definitely help you to stand out from the crowd. Use these strategies to solve MCQ’s efficiently and fast, answer like a pro.

• Accelerate your mind before the start.
• Read the questions cautiously and then answer.
• Do not try to solve everything in one go.
• Take half to one-minute breaks if possible.
• Learn as many short-cuts as you can and apply them in the examination.
• Use the method of elimination.

Frequently Asked Questions

1. What is the best strategy while answering Multiple Choice Questions of Board Exams?

The best strategy to keep in mind while attempting MCQs in board exams is to try reading the question several times. Use the method of elimination and answer the question correctly.

2. Where can I get CBSE Classwise Multiple Choice Questions and Answers?

There are plenty of resources out there on the web that provides you MCQ Questions and Answers. However, you can find them on our site and direct links to access them are given for your convenience.

3. How should I Prepare for the MCQs of CBSE Board Exams?

Just reading will never reach you to your goal. Hence, ensure you read the Multiple Choice Quiz Questions well and practice them ahead by writing down the answer. Sample papers can guide in understanding how to speed up your writing. Moreover, answering MCQs and practicing continuously will help in gaining a grip on the answers and solving them better.

4. How do I study for Objective Questions or MCQs?

Firstly, try solving as many questions as you can. This is only possible if you are thorough with the concepts. Secondly, if you are unable to solve any quiz questions try eliminating the options and this will also work great while studying MCQs.

5. Where can I find all of the MCQ Questions for Class 6 to Class 12?

You can get the Objective Questions for Classes 6 to 12 from our page and use them as a reference to score well.

Conclusion

Hope, you got enough idea on the MCQ Questions for Classes 6, 7, 8, 9, 10, 11, 12. Aid your preparation with the Objective Questions listed and score more marks. For any other queries feel free to reach us and we will assist you at the earliest possible. Bookmark our site for more information on Objective Type or Quiz Questions.

The post MCQ Questions of all subjects for Classes 6, 7, 8, 9, 10, 11 and 12 (2020-21 Syllabus) appeared first on Learn CBSE.

Download Social Science SST Notes for CBSE Class 6th to 10th: Students who are planning for their upcoming CBSE Class 10 to 6 board exams need to work hard with dedication to secure a good score in the exam. Therefore, to help class 6th to 10th students study adequately, we have presented CBSE Class 10, 9, 8, 7, 6 Social Science Revision Notes of all sub-subjects History, Geography, Political Science, and Economics.

These Chapterwise SST Notes helps students to understand all the concepts easily and revise them quickly during the time of the exam to score better marks. Students can quickly revise the complete NCERT Class 10 to 6 Social Science Syllabus by referring to these SST Notes. So, access CBSE Class 6th to 10th Social Science Notes from this page and download them for free in pdf format.

Quick Revision Notes for Social Science (SST) CBSE Class 6, 7, 8, 9,10

Social Studies subject teachers are prepared these SST Notes for CBSE class 6 to 10 students as per the latest CBSE Social Science Syllabus. So, students can read these revision notes for social studies on a regular basis to grasp the concepts thoroughly. As it aids to score well in the final CBSE sst exams. Use these Social Science handwritten & revision notes and cover the entire syllabus in a smart way.

These SST Key Notes contain important points of all chapter’s along with brief explanations for each concept. Any level of students can easily understand the concepts of Social science by using these notes. As it was written in an easy to understand language and will save students valuable time and helps to attain more knowledge about the concepts and its important points.

To help students, we have provided Chapter-wise CBSE Class 6th to 10th Social Science Notes in Pdf download links for all sub-subjects like History, Geography, Political Science, and Economics in the below modules. Hence, check out these Class 10 to 6 SST revision and handwritten notes and perform well in both CBSE Board exams and Competitive exams.

• Class 10 Social Science Notes
• Class 9 Social Science Notes
• Class 8 Social Notes
• Class 7 Social Notes
• Class 6 Social Notes

Advantages of studying Social Science Revision Notes for CBSE 6th to 10th Board Exams

The following are some of the benefits that students will profit from using NCERT Class 6 to 10 Social Science Revision Notes while studying for the CBSE board exams.

• SST Revision Notes & short keynotes will assist students to perform a quick revise of the entire subject before the exam.
• Well, Social Studies notes contain all important points of topics and concepts covered in the chapters of social science CBSE class 6th to 10th.
• We have prepared this SST handwritten and Revision Notes in a transparent manner so that students of class 6 to 10 can grasp the topics easily.
• Thus, these quick revision notes for SST are considered as a primary source of information to prepare effectively for the forthcoming board exam.
• Make use of these CBSE class 10, 9, 8, 7, 6 Chapterwise social studies notes and understand all the topics which are needed to be covered for the final examinations.

FAQs on Social Science Handwritten Notes for Class 6th to 10th CBSE

1. How do I make notes for social science in the CBSE Class 6, 7, 8, 9, 10?

It’s very important to make SST Notes for a quick revision of all chapters at the time of CBSE Class 6th to 10th final exams. So, write down key points in a coherent manner instead of the whole sentences while making notes. When you use this handy revision notes for SST board exam, it helps you understand and retain the subject matter far longer.

2. Which is the best site for class 6 to 10 SST Notes?
As of now, I feel the best site for downloading or accessing the CBSE class 6th to 10th Social Science Revision Notes Pdf is LearnCBSE.in. It covers CBSE Class 10 to Class 6 Chapter-wise SST Notes as per the latest syllabus & NCERT SST Books. So, refer to this best site and collect whatever you need for SST board exam preparation.

3. Where can I get social science notes for class 10 to 6 board exams?
You can get the SST Notes & Study Material for free of cost. Access the Social Science Revision Notes Chapterwise for Class 6 to 10 via available quick links and make use of them for last-minute exam preparation.

4. From where should I download social science handwritten & revision notes for grade 10th to 6th of the CBSE?
You can download SST Revision Notes and Handwritten Notes pdf for grade 6 to 10 CBSE form our site i.e., LearnCBSE.in and revise all the chapters’ before the exam and score more marks.

The post Social Science Notes for Quick Revision Class 10, 9, 8, 7, 6 PDF Download appeared first on Learn CBSE.

NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State:

NCERT TEXTBOOK QUESTIONS SOLVED

1.1. Why are solids rigid?
Ans: The constituent particles in solids have fixed positions and can oscillate about their mean positions. Hence, they are rigid.

1.2. Why do solids have definite volume ?
Ans: Solids keep their volume because of rigidity in their structure. The interparticle forces are very strong. Moreover, the interparticle spaces are very few and small as well. As a result, their volumes cannot change by applying pressure.

1.3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fibreglass, copper
Ans: Crystalline solids: Benzoic acid, potassium nitrate, copper Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fibreglass

1.4. Why is glass considered as super cooled liquid ? (C.B.S.E. Delhi 2013)
Ans: Glass is considered to be super cooled liquid because it shows some of the characteristics of liquids, though it is an amorphous solid. For example, it is slightly thicker at the bottom. This can be possible only if it has flown like liquid, though very slowly.

1.5. Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?
Ans: As the solid has same value of refractive index along all directions, it is isotropic in nature and hence amorphous. Being amorphous solid, it will not show a clean cleavage and when cut, it will break into pieces with irregular surfaces.

1.6. Classify the following solids in different categories based on the nature of the inter molecular forces : sodium sulphate, copper, benzene, urea, ammonia, water, zinc sulphide, diamond, rubedium, argon, silicon carbide.
Ans: Ionic, metallic, molecular, molecular, molecular (hydrogen bonded), molecular (hydrogen bonded), ionic, covalent, metallic, molecular, covalent (net work).

1.7. Solid A is a very hard electrical insulator in. solid as well as in molten state and melts at extremely high temperature. What type of solid is it?
Ans: It is a covalent or network solid.

1.8. Why are ionic solids conducting in the molten state and not in the solid state ?
Ans: In the ionic solids, the electrical conductivity is due to the movement of the ions. Since the ionic mobility is negligible in the solid state, these are non-conducting in this state. Upon melting, the ions present acquire some mobility. Therefore, the ionic solids become conducting

1.9. What type of solids are electrical conductors, malleable and ductile?
Ans: Metallic solids

 1.10. Give the significance of a lattice point.
Ans: The lattice point denotes the position of a particular constituent in the crystal lattice. It may be atom, ion or a molecule. The arrangement of the lattice points in space is responsible for the shape of a particular crystalline solid.

1.11. Name the parameters that characterise a unit cell.
Ans: A unit cell is characterised by following parameters:
(i)the dimensions of unit cell along three edges: a, b and c.
(ii)the angles between the edges: α (between b and c); β (between a and c) and γ (between a and b)

1.12. Distinguish between :
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Ans:
(i) In a hexagonal unit cell :
a = b # c; α = β = 90° and γ = 120°
In a monoclinic unit cell :
a # b # c and α = γ = 90° and β # 90°
(ii) In a face centred unit cell, constituent particles are located at all the corners as well as at the centres of all the faces.
In end-centred unit cell, constituent particles are located at all the corners as well as at the centres of two opposite faces. (C.B.S.E Foreign 2015)

1.13. Explain how much portion of an atom located at
(i)corner and (ii)body centre of a cubic unit cell is part of its neighbouring unit cell.
Ans: (i) An atom at the comer is shared by eight adjacent unit cells. Hence, portion of the atom at the comer that belongs to one unit cell=1/8.
(ii)An atom at the body centre is not shared by any other unit cell. Hence, it belongs fully io unit cell.

1.14. What is the two dimensional co-ordination number of a molecule in a square close packed layer ?
Ans: In the two dimensional square close packed layer, a particular molecule is in contact with four molecules. Hence, the co-ordination number of the molecule is four.

1.15. A compound forms hexagonal close-packed. structure. What is the total number of voids in 0. 5 mol of it? How many of these are tetrahedral voids?
Ans:
No. of atoms in close packings 0.5 mol =0.5 x 6.022 x 10²³ =3.011 x 10²³
No. of octahedral voids = No. of atoms in packing =3.011 x 10²³
No. of tetrahedral voids = 2 x No. of atoms in packing
= 2 x 3.011 x 10²³ = 6.022 x 10²³
Total no. of voids = 3.011 x 10²³ + 6.022 x 10²³
= 9.033 x 10²³

1.16. A compound is formed by two elements M and N. The element N forms ccp and atoms of the element M occupy 1/3 of the tetrahedral voids. What is the formula of the compound ? (C.B.S.E. Foreign 2015)
Ans: Let us suppose that,
the no. of atoms of N present in ccp = x
Since 1/3rd of the tetrahedral voids are occupied by the atoms of M, therefore,
the no. of tetrahedral voids occupied = 2x/3
The ratio of atoms of N and M in the compound = x : 2x/3 or 3 : 2
∴ The formula of the compound = N₃M₂ or M₂N₃

1.17. Wh ich of the following lattices has the highest packing efficiency (i) simple cubic (ii) body- centred cubic and (iii) hexagonal close-packed lattice?
Ans: Packing efficiency of:
Simple cubic = 52.4% bcc = 68% hcp = 74%
hcp lattice has the highest packing efficiency.

1.18. An element with molar mass 2:7 x 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2:7 x 103 kg m^-3, what is the nature of the cubic unit cell ? (C.B.S.E. Delhi 2015)
Ans: 

Since there are four atoms per unit cell, the cubic unit cell must be face centred (fcc) or cubic close packed (ccp).

1.19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Ans: When a solid is heated, vacancy defect is produced in the crystal. On heating, some atoms or ions leave the lattice site completely, i.e., lattice sites become vacant. As a result-of this defect, density of the substances decreases.

1.20. What types of stoichiometric defects are shown by (C.B.S.E. Delhi 2013)
(i) ZnS
(ii) AgBr?
Ans:
(i) ZnS crystals may show Frenkel defects since the cationic size is smaller as compared to anionic size.
(ii) AgBr crystals may show both Frenkel and Schottky defects.

1.21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Ans: Let us take an example NaCl doped with SrCl, impurity when SrCl2 is added to NaCl solid as an impurity, two Na+ ions will be replaced and one of their sites will be occupied by Sr21- while the other will remain vacant. Thus, we can say that when a cation of higher valence is added as an impurity to an ionic solid, two or more cations of lower valency are replaced by a cation of higher valency to maintain electrical neutrality. Hence, some cationic vacancies are created.

1.22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Ans: Let us take an example of NaCl. When NaCl crystal is heated in presence of Na vapour, some Cl^–ions leave their lattice sites to combine with Na to form NaCl. The e^-1 s lost by Na to form Na⁺ (Na⁺ + Cl^–—> NaCl) then diffuse into the crystal to occupy the anion vacancies. These sites are called F-centres. These e-s absorb energy from visible light, get excited to higher energy level and when they fall back to ground state, they impart yellow colour to NaCl crystal.

1.23. A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
Ans: Impurity from group 15 should be added to get n-type semiconductor.

1.24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Ans: Ferromagnetic substances make better permanent magnets. This is because when placed in magnetic field, their domains get oriented in the directions of magnetic field and a strong magnetic field is produced. This ordering of domains persists even when external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

NCERT EXERCISES

1.1. Define the term ‘amorphous’. Give a few examples of amorphous solids.
Sol. Amorphous solids are those substances, in which there is no regular arrangement of its constituent particles, (i.e., ions, atoms or molecules). The arrangement of the constituting particles has only short range order, i.e., a regular and periodically repeating pattern is observed over short distances only, e.g., glass, rubber and plastics.

1.2. What makes glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass?
Sol. Glass is a super cooled liquid and an amorphous substance. Quartz is the crystalline form of silica (SiO₂) in which tetrahedral units SiO₄ are linked with each other in such a way that the oxygen atom of one tetrahedron is shared with another Si atom. Quartz can be converted into glass by melting it and cooling the melt very rapidly. In glass, Si0₄ tetrahedra are joined in a random manner.

1.3 Classify each of the following solids as ionic, metallic, modular, network (covalent) or amorphous:
(i) Tetra phosphorus decoxide (P₄O₁₀) (ii) Ammonium phosphate, (NH₄)₃P0₄ (iii) SiC (iv) I₂ (v) P₄ (vii) Graphite (viii), Brass (ix) Rb (x) LiBr (xi) Si
Sol.

1.4 (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atom
(a) in a cubic close-packed structure?
(b) in a body centred cubic structure?
Sol. (i) The number of nearest neighbours of a particle are called its coordination number.
(ii) (a) 12 (b) 8

1.5. How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer. (C.B.S.E. Outside Delhi 2011)
Sol.

1.6 ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?
Sol. Higher the melting point, greater are the forces holding the constituent particles together and thus greater is the stability of a crystal. Melting points of given substances are following. Water = 273 K, Ethyl alcohol = 155.7 K, Diethylether = 156.8 K, Methane = 90.5 K.
The intermoleoilar forces present in case of water and ethyl alcohol are mainly due to the hydrogen bonding which is responsible for their high melting points. Hydrogen bonding is stronger in case of water than ethyl alcohol and hence water has higher melting point then ethyl alcohol. Dipole-dipole interactions are present in case of diethylether. The only forces present in case of methane is the weak van der Waal’s forces (or London dispersion forces).

1.7. How will you distinguish between the following pairs of terms :
(a) Hexagonal close packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
Sol.
(a) In hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer
(ABABAB……. type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the spheres of the first layer (ABCABC…..type).
(b) Crystal lattice: It deplicts the actual shape as well as size of the constituent particles in the crystal. It is therefore, called space lattice or crystal lattice.
Unit cell: Each bricks represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.
(c) Tetrahedral void: A tetrahedral void is formed when triangular void made by three spheres of a particular layer and touching each other.

Octahedral void: An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over anothet set of spheres.

1.8 How many lattice points are there is one unit cell of each of the following lattices?
(i) Face centred cubic (if) Face centred tetragonal (iii) Body centred cubic
Sol.

1.9 Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Sol. (i) Metallic and ionic crystals
Similarities:
(a) There is electrostatic force of attraction in both metallic and ionic crystals.
(b) Both have high melting points.
(c) Bonds are non-directional in both the cases.
Differences:
(a) Ionic crystals are bad conductors of electricity in solids state as ions are not free to move. They can conduct electricity only in die molten state or in aqueous solution. Metallic crystals are good conductors of electricity in solid state as electrons are free to move.
(b) Ionic bond is strong due to strong electrostatic forces of attraction.
Metallic bond may be strong or weak depending upon the number of valence electrons and the size of the kernels.
(ii) Ionic solids are hard and brittle.Ionic solids are hard due to the presence of strong electrostatic forces of attraction. The brittleness in ionic crystals is due to the non- directional bonds in them.

1.10 Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic, (ii) body centred cubic, and (iii) face centred cubic (with the assumptions that atoms are touching each other).
Sol. Packing efficiency: It is the percentage of total space filled by the particles.

1.11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10^-8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
Sol.

1.12. A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q?
Sol. Contribution by atoms Q present at the eight corners of the cube = \(\frac { 1 }{ 8 } \)= x 8 = 1
Contribution by atom P present at the body centre = 1
Thus, P and Q are present in the ratio 1:1.
∴ Formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

1.13 Niobium crystallises in a body centred cubic structure. If density is 8.55 g cm^-3, calculate atomic radius of niobium, using its atomic mass 93u.
Sol.

1.14 If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between rand R.
Sol. A sphere is fitted into the octahedral void as shown in the diagram.

1.15 Copper crystallises into a fee lattice with edge length 3.61 x 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm^-3.
Sol.

This calculated value of density is closely in agreement with its measured value of 8.92 g cm³.

Question 16.
Analysis shows that nickel oxide has the formula Ni_0.98 O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
Solution:
98 Ni-atoms are associated with 100 O – atoms. Out of 98 Ni-atoms, suppose Ni present as Ni²⁺ = x
Then Ni present as Ni³⁺ = 98 – x
Total charge on x Ni²⁺ and (98 – x) Ni³⁺ should
be equal to charge on 100 O^2- ions.
Hence, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94
∴ Fraction of Ni present as Ni²⁺ = \(\frac { 94 }{ 98 } \) × 100 = 96%
Fraction of Ni present as Ni³⁺ = \(\frac { 4 }{ 98 } \) × 100 = 4%

Question 17.
What are semi-conductors ? Describe the two main types of semiconductors and contrast their conduction mechanisms.
Solution:
Semi-conductors are the substances whose conductivity lies in between those of conductors and insulators. The two
main types of semi-conductors are n-type and p-type.
(i) n-type semiconductor: When a silicon or germanium crystal is doped with group 15 element like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalised and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.

(ii) p-type semiconductor: When a silicon or germanium is doped with group 13 element like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of missing fourth electron. Here, this hole moves throughout the crystal like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-type semiconductor, p stands for positive hole, since it is the positive hole that is responsible for conduction.

Question 18.
Non-stoichiometric cuprous oxide, Cu₂O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
Solution:
The ratio less than 2 : 1 in Cu₂0 shows cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. For maintaining electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal dose- packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Solution:
Suppose the number of oxide ions (O^2-) in the packing = 90
∴ Number of octahedral voids = 90
As 2/3^rd of the octahedral voids are occupied by ferric ions, therefore, number of ferric ions 2 present = \(\frac { 2 }{ 3 } \) × 90 = 60
∴ Ratio of Fe³⁺ : O^2- = 60 : 90 = 2 : 3
Hence, the formula of ferric oxide is Fe₂O₃.

Question 20.
Classify each of the following as being either a p-type or n-type semiconductor :

1. Ge doped with In
2. B doped with Si.

Solution:

1. Ge is group 14 element and In is group 13 element. Hence, an electron deficient hole is created and therefore, it is a p – type semiconductor.
2. B is group 13 element and Si is group 14 element, there will be a free electron, So, it is an n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?
Solution:
For a face centred cubic unit cell (fcc)
Edge length (a) = \(2\sqrt { 2 } r\) = 2 x 1.4142 x 0.144 mm = 0.407 nm

Question 22.
In terms of band theory, what is the difference

1. between a conductor and an insulator
2. between a conductor and a semiconductor?

Solution:
In most of the solids and in many insulating solids conduction takes place due to migration of electrons under the influence of electric field. However, in ionic solids, it is the ions that are responsible for the conducting behaviour due to their movement.

(i) In metals, conductivity strongly depends upon the number of valence electrons available in an atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other, as to form a band. If this band is partially filled or it overlaps with the higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal behaves as a conductor.

If the gap between valence band and next higher unoccupied conduction band is large, electrons cannot jump into it and such a substance behaves as insulator.

(ii) If the gap between the valence band and conduction band is small, some electrons may jump from valence band to the conduction band. Such a substance shows some conductivity and it behaves as a semiconductor. Electrical conductivity of semiconductors increases with increase in temperature, since more electrons can jump to the conduction band. Silicon and germanium show this type of behaviour and are called intrinsic semiconductors. Conductors have no forbidden band.

Question 23.
Explain the following terms with suitable examples :

1. Schottky defect
2. Frenkel defect
3. Interstitial defect
4. F-centres.

Solution:
(i) Schottky defect : In Schottky defect a pair of vacancies or holes exist in the crystal lattice due to the absence of equal number of cations and anions from their lattice points. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr, etc. Due to this defect density of crystal decreases and it begins to conduct electricity to a smaller extent.

(ii) Frenkel defect : This defect arises when some of the ions in the lattice occupy interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation, e.g., AgBr, ZnS, etc. Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall chemical composition of the crystal.

(iii) Interstitial defect : When some constituent particles (atoms or molecules) occupy an interstitial site of the crystal, it is said to have interstitial defect. Due to this defect the density of the substance increases.

(iv) F-Centres : These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

Question 24.
Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.

1. What is the length of the side of the unit cell?
2. How many unit cells are there in 1.00 cm³ of aluminium?

Solution:

Question 25.
If NaCI is doped with 10^-3 mol % SrCl₂, what is the concentration of cation vacancies ?
Solution:
Let moles of NaCI = 100
∴ Moles of SrCl₂ doped = 10^-3
Each Sr²⁺ will replace two Na+ ions. To maintain electrical neutrality it occupies one position and thus creates one cation vacancy.
∴ Moles of cation vacancy in 100 moles NaCI = 10^-3
Moles of cation vacancy in one mole
NaCI = 10^-3 × 10^-2 = 10^-5
∴ Number of cation vacancies
= 10^-5 × 6.022 × 10²³ = 6.022 × 10¹⁸ mol^-1

Question 26.
Explain the following with suitable example:

1. Ferromagnetism
2. Paramagnetism
3. Ferrimagnetism
4. Antiferromagnetism
5. 12-16 and 13-15 group compounds.

Solution:
(i) Ferromagnetic substances : Substances which are attracted very strongly by a magnetic field are called ferromagnetic substances, e.g., Fe, Ni, Co and CrO₂ show ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. This type of magnetic moments are due to unpaired electrons in the same direction.

The ferromagnetic material, CrO₂, is used to make magnetic tapes used for audio recording.

(ii) Paramagnetic substances : Substances which are weakly attracted by the external magnetic field are called paramagnetic substances. The property thus exhibited is called paramagnetism. They are magnetised in the same direction as that of the applied field. This property is shown by those substances whose atoms, ions or molecules contain unpaired electrons, e.g., O₂, Cu²⁺, Fe³⁺, etc. These substances, however, lose their magnetism in the absence of the magnetic field.

(iii) Ferrimagnetic substances : Substances which are expected to possess large magnetism on the basis of the unpaired electrons but actually have small net magnetic moment are called ferrimagnetic substances, e.g., Fe₃O₄, ferrites of the formula M²⁺Fe₂O₄ where M = Mg, Cu, Zn, etc. Ferrimagnetism arises due to the unequal number of magnetic moments in opposite direction resulting in some net magnetic moment.

(iv) Antiferromagnetic substances : Substances which are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but actually they possess zero net magnetic moment are called antiferromagnetic substances, e.g., MnO. Antiferromagnetism is due to the presence of equal number of magnetic moments in the opposite directions

(v) 13-15 group compounds : When the solid state materials are produced by combination of elements of groups 13 and 15, the compounds thus obtained are called 13-15 compounds. For example, InSb, AlP, GaAs, etc.

12-16 group compounds : Combination of elements of groups 12 and 16 yield some solid compounds which are referred to as 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe, etc. In these compounds, the bonds have ionic character.

More Resources for CBSE Class 12:

NCERT SolutionsMathsPhysicsChemistryBiologyScience

The post NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State appeared first on Learn CBSE.

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions:

NCERT TEXTBOOK QUESTIONS SOLVED

2.1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans: Mass of solution = Mass of C₆H₆ + Mass of CCl₄
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl₄ = 122/144 x 100 = 84.72 %

2.2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans: 30% by mass of C₆H₆ in CCl₄ => 30 g C₆H₆ in 100 g solution
.’. no. of moles of C₆H₆,(ⁿC₆h₆) = 30/78 = 0.385

2.3. Calculate the molarity of each of the following solutions
(a) 30 g of Co(NO₃)26H₂O in 4·3 L of solution
(b) 30 mL of 0-5 M H₂SO₄ diluted to 500 mL.
Ans:

2.4. Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH₂CONH₂) = 60 g mol^-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g

2.5. Calculate
(a) molality
(b) molarity and
(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI solution is 1·202 g mL^-1.
Ans:
Step I. Calculation of molality of solution
Weight of KI in 100 g of the solution = 20 g
Weight of water in the solution = 100 – 20 = 80 g = 0-08 kg
Molar mass of KI = 39 + 127 = 166 g mol^-1.

Step II. Calculation of molarity of solution

Step III. Calculation of mole fraction of Kl

2.6. H₂ S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Ans: Solubility of H₂S gas = 0.195 m
= 0.195 mole in 1 kg of solvent
1 kg of solvent = 1000g

2.7. Henry’s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.
Ans.:

2.8 The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition in the vapour phase.
Ans:
Vapour pressure of pure liquid A (\({ P }_{ A }^{ \circ }\)) = 450 mm
Vapour pressure of pure liquid B (\({ P }_{ B }^{ \circ }\)) = 700 mm
Total vapour pressure of the solution (P) = 600 mm

2.9. Vapour pressure of pure water at 298 K is 23.8 m m Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Ans:

2.10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Ans:

2.11 Calculate the mass of ascorbic acid (vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1·5°C. (K_f for CH₃COOH) = 3·9 K kg mol^-1)
Ans:

2.12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Ans:

NCERT EXERCISES

2.1. Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.
Sol: A solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solutions: There are nine types of solutions.
Types of Solution Examples
Gaseous solutions
(a) Gas in gas Air, mixture of 0₂ and N₂, etc.
(b) Liquid in gas Water vapour
(c) Solid in gas Camphor vapours in N2 gas, smoke etc.
Liquid solutions
(a) Gas in liquid C02 dissolved in water (aerated water), and 02 dissolved in water, etc.
(b) Liquid in liquid Ethanol dissolved in water, etc.
(c) Solid in liquid Sugar dissolved in water, saline water, etc.
Solid solutions
(a) Gas in solid Solution of hydrogen in palladium
(b) Liquid in solid Amalgams, e.g., Na-Hg
(c) Solid in solid Gold ornaments (Cu/Ag with Au)

2.2. Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What type of solid solution is this likely to be ?
Sol: The solution likely to be formed is interstitial solid solution.

2.3 Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Sol: (i) Mole fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute


(ii) Molality: It is defined as die number of moles of a solute present in 1000g (1kg) of a solvent.

NOTE: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not change with change in temperature since the mass of solvent does not vary with temperature,
(iii) Molarity: It is defined as the number of moles of solute present in one litre of solution.

NOTE: Molarity is the most common way of expressing concentration of a solution in laboratory. However, it has one disadvantage. It changes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.
(iv) Mass percentage: It is the amount of solute in grams present in 100g of solution.

2.4. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1·504 g mL^-1 ?
Sol: Mass of HNO₃ in solution = 68 g
Molar mass of HNO₃ = 63 g mol^-1
Mass of solution = 100 g
Density of solution = 1·504 g mL^-1

2.5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1 .2 g m L^-1, then what shall be the molarity of the solution?
Sol: 10 percent w/w solution of glucose in water means 10g glucose and 90g of water.
Molar mass of glucose = 180g mol^-1 and molar mass of water = 18g mol^-1

2.6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂C0₃ and NaHCO₃ containing equimolar amounts of both?
Sol: Calculation of no. of moles of components in the mixture.

2.7. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.
Sol:

2.8. An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C₂ H₆O₂ ) and200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^-1, then what shall be the molarity of the solution?
Sol:

2.9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).
(i) express this in percent by mass.
(ii) determine the molality of chloroform in the water sample.
Sol: 15 ppm means 15 parts in million (10⁶) by mass in the solution.

2.10. What role does the molecular interaction play in solution of alcohol in water?
Sol: In case of alcohol as well as water, the molecules are interlinked by intermolecular hydrogen bonding. However, the hydrogen bonding is also present in the molecules of alcohol and water in the solution but it is comparatively less than both alcohol and water. As a result, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult’s Law. This will lead to increase in vapour pressure of the solution and also decrease in its boiling point.

2.11. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Sol: When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.

2.12. State Henry’s law and mention some of its important applications.
Sol:
Henry’s law: The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.
or
The partial pressure of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution. p = KHX
where KH is Henry’s law constant.
Applications of Henry’s law :
(i) In order to increase the solubility of CO₂ gas in soft drinks and soda water, the bottles are normally sealed under high pressure. Increase in pressure increases the solubility of a gas in a solvent according to Henry’s Law. If the bottle is opened by removing the stopper or seal, the pressure on the surface of the gas will suddenly decrease. This will cause a decrease in the solubility of the gas in the liquid i.e. water. As a result, it will rush out of the bottle producing a hissing noise or with a fiz.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium inorder to reduce or minimise the painfril effects during decompression.
(iii) As the partial pressure of oxygen in air is high, in lungs it combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low. Therefore, oxyhaemoglobin releases oxygen in order to carry out cellular activities.

2.13. The partial pressure of ethane over a solution containing 6.56 × 10^-3 g of ethane is 1 bar. If the solution contains 5.00 × 10^-2 g of ethane, then what shall be the partial pressure of the gas?
Sol:

2.14. According to Raoult’s law, what is meant by positive and negative deviaitions and how is the sign of ∆_solH related to positive and negative deviations from Raoult’s law?
Sol: Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆_solH is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall ∆_sol H is positive. Similarly ∆_solV is positive i.e. the volume of solution is some what more than sum of volumes of solvent and solute.
So there is expansion in volume on solution formation.
Similarly in case of solutions exhibiting negative deviations, A – B interactions are stronger than A-A&B-B. So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. ∆_sol H is negative.

2.15. An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1·004 bar at the boiling point of the solvent. What is the molecular mass of the solute ?
Sol:
According to Raoult’s Law,

2.16  Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?
Sol.

2.17.  The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it
Sol: 1 molal solution of solute means 1 mole of solute in 1000g of the solvent.

2.18. Calculate the mass of a non-volatile solute (molecular mass 40 g mol^-1) that should be dissolved in 114 g of octane to reduce its pressure to 80%. (C.B.S.E. Outside Delhi 2008)
Sol: According to Raoult’s Law,

2.19. A solution containing 30g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new of vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) molar mass of the solute.
(ii) vapour pressure of water at 298 K.
Sol: Let the molar mass of solute = Mg mol^-1

2.20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Sol: Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)

2.21. Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Sol:

2.22. At 300 K, 36 g glucose present per litre in its solution has osmotic pressure of 4·98 bar. If the osmotic pressure of the solution is 1·52 bar at the same temperature, what would be its concentration?
Sol:

2.23. Suggest the most important type of intermolecular attractive interaction in the following pairs:
(i) n-hexane and n-octane
(ii) I₂ and CCl₄.
(iii) NaCl0₄ and water
(iv) methanol and acetone
(v) acetonitrile (CH₃CN) and acetone (C₃H₆0)
Sol: (i) Both w-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(ii) Both I₂ and CCl₄ are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(iii) NaCl0₄ is an ionic compound and gives Na⁺ and Cl0₄^– ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.
(iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
(v) Both CH₃CN and C₃H₆O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.

2.24. Based on solute solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Sol: n-octane (C₈H₁₈) is a non-polar liquid and solubility is governed by the principle that like dissolve like. Keeping this in view, the increasing order of solubility of different solutes is:
KCl < CH₃OH < CH₃C=N < C₆H₁₂ (cyclohexane).

2.25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol
Sol: (i) Phenol (having polar – OH group) – Partially soluble.
(ii) Toluene (non-polar) – Insoluble.
(iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble.
(iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble.
(v) Chloroform (non-polar)- Insoluble.
(vi) Pentanol (having polar -OH) – Partially soluble.

2.26. If the density of lake water is 1·25 g mL-1, and it contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake. (C.B.S.E. Outside Delhi 2008)
Sol:

2.27. If the solubility product of CuS is 6 x 10^-16, calculate the maximum molarity of CuS in aqueous solution.
Sol:

2.28. Calculate the mass percentage of aspirin (C₉H₈O₄ in acetonitrile (CH₃CN) when 6.5g of CHO is dissolved in 450 g of CH3CN.
Solution:

2.29. Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10^-3 m aqueous solution required for the above dose.

Solution:

2.30. Calculate the amount of benzoic acid (C₅H₅COOH) required for preparing 250 mL of 0· 15 M solution in methanol.
Solution:

2.31. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

Solution:

Fluorine being more electronegative than chlorine has the highest electron withdrawing inductive effect. Thus, triflouroacetic acid is the strongest trichloroacetic acid is second most and acetic acid is the weakest acid due to absence of any electron withdrawing group. Thus, F₃CCOOH ionizes to the largest extent while CH₃COOH ionizes to minimum extent in water. Greater the extent of ionization greater is the depression in freezing point. Hence, the order of depression in freezing point will be CH₃COOH < Cl₃CCOOH < F₃CCOOH.

2.32. Calculate the depression in the freezing point of water when 10g of CH₃CH₂CHClCOOH is added to 250g of water. Ka = 1.4 x 1o^-3 Kg = 1.86 K kg mol^-1.

Solution:

2.33. 19.5g of CH₂FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’s Hoff factor and dissociation constant of fluoroacetic acid.

Solution:

2.34. Vapour pressure of water at 293 K is 17·535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Solution:
According to Raoult’s Law,

2.35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution:

2.36. 100g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000g of liquid B (molar mass 180g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Solution:

2.37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot P_total, P_chlroform and P_acetone as a function of χ_acetone. The experimental data observed for different compositions of mixtures is:

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Solution:

2.38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.

Solution:

2.39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen are 3.30 x 10⁷ mm and 6.51 x 10⁷ mm respectively, calculate the composition of these gases in water.

Solution:

2.40. Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Solution:

2.41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming that it is completely dissociated. (C.B.S.E. 2013)
Solution:
Step I. Calculation of Van’t Hoff factor (i)
K₂SO₄ dissociates in water as :

More Resources for CBSE Class 12:

NCERT SolutionsMathsPhysicsChemistryBiologyScience

The post NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions appeared first on Learn CBSE.

NCERT Solutions For Class 12 Chemistry Chapter 3 Electrochemistry

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry:

NCERT TEXTBOOK QUESTIONS SOLVED

3.1. How would you determine the standard electrode potential of the system Mg²⁺¹ Mg?
Ans: A cell will be set up consisting of Mg/MgSO₄ (1 M) as one electrode and standard hydrogen electrode Pt, H, (1 atm)H⁺/(l M) as second electrode, measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that e^-1 s flow from mg electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as follows :

3.2. Can you store copper sulphate solutions in a zinc pot?
Ans: Zn being more reactive than Cu, displaces Cu from CuSO₄ solution as follows:Zn (s) + CuSO₄ (aq) –> ZnSO₄(ag)+Cu (s)

3.3. Consult the table on standard electrode potentials and suggest three substances that can oxidise Fe2+ ions under suitable conditions.
Ans. The oxidation of Fe2+ ions to Fe3+ ions proceeds as follows :
Fe²⁺ → ³⁺ + e^– ; \({ E }_{ OX }^{ \circ }\) = – 0·77 V
Only those substances can oxidise Fe²⁺ ions to Fe³⁺ ions which can accept electrons released during oxidation or are placed above iron in electrochemical series. These are : Cl₂(g), Br₂(g) and Cr₂\({ O }_{ 7 }^{ 2- }\) ions (in the acidic medium).

3.4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Ans. For hydrogen electrode,H⁺ + e^– —>1/2 H₂,

3.5. Calculate the emf of the cell in which the following reaction takes place:
Ni(s)+2Ag⁺ (0.002 M) -> Ni²⁺ (0.160 M)+2Ag(s) Given that E^(-)_(cell) = 1.05 V .
Ans:

3.6. The cell in which the following reaction occurs: 2Fe³⁺ (aq) + 2I^– (aq) —> 2Fe²⁺ (aq) +I² (s) has E^°_cell=0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Ans:

3.8. Why does the conductivity of a solution decrease with dilution?
Ans: The conductivity of a solution is linked with the number of ions present per unit volume. With dilution, these decrease and the corresponding conductivity or specific conductance of the solution decreases.

3.8.  Suggest a way to determine the value of water.
Ans:

3.9. The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation and dissociation constant Given λ^°(H⁺)=349.6 S cm² mol^-1 andλ^°(HCOO-) = 54.6 S cm² mol^-1
Ans:

3.10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Ans:

3.11. Suggest a list of metals which can be extracted electrolytically.
Ans: The highly reactive metals having large -ve E° values, which can themselves act as powerful reducing agents can be extracted electrolytically. The process is known as electrolytic reduction. For details, consult Unit-6. For example, sodium, potassium, calcium, magnesium etc.

3.12. Consider the reaction: Cr₂O₇^2--+ 14H⁺ + 6e- -> 2Cr₃+ + 7H₂O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇^2- ?
Ans:

3.13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Ans: A lead storage battery consists of anode of lead, cathode of a grid of lead packed with lead dioxide (PbO₂) and 38% H₂SO₄ solution as electrolyte. When the battery is in use, the reaction taking place are:

On charging the battery, the reverse reaction takes place, i.e., PbSO₄ deposited on electrodes is converted back to Pb and PbO₂ and H₂SO₄ is regenerated.

3.14. Suggest two materials other than hydrogen that can be used as fuels in the fuel cells.
Ans: Methane (CH₄) and methanol (CH₃OH) can also be used as fuels in place of hydrogen in the fuel cells.

3.15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Ans: The water present on the surface of iron dissolves acidic oxides of air like CO₂ , SO₂ , etc. to form acids which dissociate to give H⁺ ions :

Thus, an electrochemical cell is set up on the surface.
Ferrous ions are further oxidised by atmospheric oxygen to ferric ions which combine with water to form hydrated ferric oxide, Fe₂O₃. xH₂O, which is rust.

NCERT EXERCISES

3.1. Arrange the following metals in the order in which they displace each other from their salts.
Al, Cu, Fe, Mg and Zn
Sol: Mg, Al, Zn, Fe, Cu.

3.2. Given the standard electrode potentials, K⁺/K=-2. 93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg =0.79V, Mg²⁺/Mg=-2.37V, Cr³⁺/Cr=0.74V.
Arrange these metals in their increasing order of reducing power.
Sol: Higher the oxidation potential more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be Ag<Hg<Cr<Mg<K.

3.3. Depict the galvanic cell in which the reaction
Zn(s) + 2Ag⁺(aq) —-> 7M²⁺(aq) + 2Ag (s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Sol. The set-up will be similar to as shown below,

(i) Anode, i. e, zinc electrode will be negatively charged.
(ii) The current will flow from silver to copper in the external circuit.
(iii) At anode: Zn(s) ——–> Zn²⁺(aq) + 2e^–
At cathode: 2Ag⁺(aq) + 2e^– ——–> 2Ag(s)

3.4. Calculate the standard cell potentials of the galvanic cells in which the following reactions take place.

Also calculate ∆G° and equilibrium constant for the reaction. (C.B.S.E. Outside Delhi 2008)
Sol:

3.5. Write the Nernst equation and emf of the following cells at 298 K:


Sol:

3.6. In the button cells widely used in watches and other devices the following reaction takes place:

Sol:

3.7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Sol: The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by K (kappa). Thus, if K is the specific conductance and G is the conductance of the solution, then
Now, if I = 1 cm and A = lsq.cm, then K = G.

Hence, conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte.

Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by ∆_m.
Variation of conductivity and molar conductivity with concentration: Conductivity always decreases with decrease in concentration, for both weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity increases with decrease in concentration. This is because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume.

3.8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.
Sol:

3.9. The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω What is the cell constant if conductivity of 0.001 M KCI solution at 298 K is 0.146 x 10^-3 S cm^-1?
Sol:

3.10. The conductivity of NaCl at 298 K has been determined at different concentrations and the results are given below:

Sol:

3.11. Conductivity of 0.00241 M acetic acid is 7.896 x 10^-5 S cm^-1. Calculate its molar conductivity. If Λ_m0, for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?
Sol:

3.12. How much charge is required for the following reductions:
(i) 1 mol of Al³⁺ to Al?
(ii) 1 mol of Cu²⁺ to Cu ?
(iii) 1 mol of Mn04- to Mn²⁺?
Sol: (i) The electrode reaction is Al³⁺ + 3e ——> Al
∴ Quantity of charge required for reduction of 1 mol of Al³⁺=3F=3 x 96500C=289500C.
(ii) The electrode reaction is Cu²⁺ + 2e^– ——–> Cu
∴ Quantity of charge required for reduction of 1 mol of Cu²⁺=2F=2 x 96500=193000 C.
(iii) The electrode reaction is Mn04- ———-> Mn²⁺.
i.e., Mn⁷⁺ + 5e^–——-> Mn²⁺.
∴ Quantity of charge required = 5F
=5 x 96500 C=4825000.

3.13. How much electricity in terms of Faraday is required to produce :
(i) 20·0 g of Ca from molten CaCl₂
(ii) 40·0 g of Al from molten Al₂O₃ ?
Sol:

3.14. How much electricity is required in coulomb for the oxidation of (i) 1 mol of H₂O to 0₂ (ii) 1 mol of FeO to Fe₂0₃
Sol:

3.15. A solution of Ni(N0₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Sol:

3.16. Three electrolytic cells A, B, C containing solutions of ZnS0₄, AgNO₃ and CuS0₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 45 g of silver deposited at the cathode of call B. How long did the current flow? What mass of copper and zinc were deposited?
Sol:

3.17. Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible.
(a) Fe³⁺(aq) and I^–(aq)
(b) Ag⁺(aq) and Cu(s)
(c) Fe³⁺(aq) and Br^–(aq)
(d) Ag(s) and Fe³⁺(aq)
(e) Br₂(aq) and Fe²⁺(aq).
Sol:
A particular reaction can be feasible if e.m.f. of the cell based on the E° values is positive. Keeping this in mind, let us predict the feasibility of the reactions.

3.18. Predict the products of electrolysis in each of the following.
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes.
(iii) A dilute solution of H₂S0₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes.
Sol:

More Resources for CBSE Class 12:

NCERT SolutionsMathsPhysicsChemistryBiologyScience

The post NCERT Solutions For Class 12 Chemistry Chapter 3 Electrochemistry appeared first on Learn CBSE.

NCERT Solutions For Class 12 Chemistry Chapter 4 Chemical Kinetics

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics:

NCERT TEXTBOOK QUESTIONS SOLVED

4.1.For the reaction R—>P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Ans.

4.2.In a reaction, 2A —-> Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 molL^-1 in 10 minutes. Calculate the rate during this interval?
Ans.

4.3. For a reaction, A + B → Products, the rate law is given by : r = k [A]^1/2[B]². What is the order of reaction?
Ans. Rate law(r) = k [A]^1/2[B]²
order of reaction = \(\frac { 1 }{ 2 } +2=2\frac { 1 }{ 2 } \quad or\quad 2.5\)

4.4.The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ?
Ans.The reaction is : X—>Y
According to rate law,
rate = k[X]²
If [X] is increased to 3 times, then the new rate is
rate’ = k[3X]²
rate’ = 9 k [X]² = 9 rate
Thus, rate of reaction becomes 9 times and hence rate of formation of Y increases 9-times.

4.5. A first order reaction has a rate constant 1.15 x 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3 g?
Ans.

4.6.Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Ans. For 1st order reaction,

4.7. What will be the effect of temperature on rate constant?
Ans. In general, the rate constant for a reaction nearly becomes double with about 10° rise in temperature because of the fact that the effective collisions become almost double. The exact dependence of the reaction rate on temperature is given by Arrhenius equation; \(k=A{ e }^{ -{ E }_{ a }/Rt }\).
Where A is the Arrhenius factor or the frequency factor. It is also called pre exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol^-1).

4.8.The rate of the chemical reaction doubles for and increase of 10 K in absolute temperature from 298 K. Calculate E_a.
Ans.

4.9.The activation energy for the reaction, 2 HI(g) —-> H₂+I₂ (g) is 209.5 k J mol^-1 at 581 K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Ans.Fraction of molecules having energy equal to or greater than activation energy is given by:

NCERT EXERCISES

4.1. From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants:

Sol:

4.2. For the reaction ; 2A + B → A₂B, the reaction rate = k [A][B]² with k = 2·0 x 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0·1 mol L^-1; [B] = 0·2 mol L^-1. Also calculate the reaction rate when [A] is reduced to 0·06 mol L^-1.
Sol:

4.3. The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if Ar=2.5 x 10^-4 mol-1 Ls^-1.
Sol:

4.4. The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and die reaction, rate is given by Rate=k [CH₃OCH₃]^3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also, be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate= k (PCH₃OCH₃)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Sol: As the concentration in the rate law equation is given in terms of pressure,

4.5. Mention the factors which affect the rate of a chemical reaction.
Sol: The rates of chemical reactions are influenced by a number of factors. These are :
(i) Concentration of reactants. The rate of a chemical reaction is proportional to the concentration of the reacting species taking part in the reaction. It is maximum to start with and slowly decreases since the concentration of the reacting species decreases accordingly. In case of reversible chemical reactions, the rate of chemical reaction can be studied separately for both the forward and backward reactions. In case of gaseous reactions, the increase in pressure increases the reaction rate.

(ii) Temperature. In general, the increase in temperature increases the reaction rate (there are a few exceptions as well). Actually, the energy of the reactant species increases with the increase in temperature and so will be number of collisions. It has been observed that in most of the cases, about 10° increase in temperature makes reaction rate double. Please note that the effect of temperature is quite independent of the concentration of the reactant species.

(iii) Presence of catalyst. In many chemical reactions, the reaction rate can be enhanced by certain foreign substances called catalysts. These are actually not consumed in the reactions and also donot undergo any change in chemical characteristics. However, their physical states such as colour, particle size etc., might change. Certain catalysts may have adverse effect as well as the reaction rate. They result in decreasing the reaction rate instead of increasing it. These are called negative catalysts or inhibitors.

(iv) Nature of reactants. The nature of the reacting species may also the influence the reaction rate. For example, combustion of nitric oxide (NO) is faster as compared to that of carbon monoxide (CO)

(v) Surface area. Increase in surface area provides more opportunity for the reactants to come in contact or collide resulting in increased reaction rate. For example, in laboratory. We quite often prefer granulated zinc lump of the metal while preparing hydrogen gas on reacting with dilute hydrochloric acid or dilute sulphuric acid. Actually, granulated zinc has greater surface area available for the attack by the acid than lump of zinc. Therefore, it reacts at a faster rate.

(vi) Exposure to radiations. Many chemical reactions known as photochemical reactions are carried in the presence of sun light. For example,

In these reactions, the photons of light are the source of energy which helps in breaking the bonds in the reacting molecules so that may react and form molecules of products.

4.6. A reaction is second order with respect to a reactant How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Sol:

4.7. What is the effect of temperature on the rate constant of reaction? How can this temperature effect on the rate constant be represented quantitatively?
Sol: The rate constant (k) for a reaction increases with rise in temperature and becomes nearly double with about every 10° rise in temperature. The effect is expressed with Arrhenius equation.
\(k=A{ e }^{ -{ E }_{ a }/Rt }\)

4.8. In a pseudo first order hydrolysis of ester in water, the following results were obtained:
t/s 0 30 60 90
[Ester] mol L^-1 0-55 0-31 0 17 0 085
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Sol:

4.9. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B is doubled?
Sol:

4.10. In a reaction between A and B, the initial rate of reaction (r₀ ) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?
Sol: