Get Free NCERT Solutions for Class 10 Maths Chapter 13 Ex 13.4 PDF. Surface Areas and Volumes Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 13.4 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 13 Maths Class 10 Surface Areas and Volumes Exercise 13.4 provided in NCERT TextBook.

Surface Areas and Volumes Class 10 Formulas PDF

Topics and Sub Topics in Class 10 Maths Chapter 13 Surface Areas and Volumes:

• Surface Areas and Volumes Class 10 Ex 13.1
• प्रश्नावली 13.1 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.2
• प्रश्नावली 13.2 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.3
• प्रश्नावली 13.3 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.4
• प्रश्नावली 13.4 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.5
• प्रश्नावली 13.5 का हल हिंदी में
• Surface Areas and Volumes Class 10 Extra Questions

You can also download the free PDF of  Ex 13.4 Class 10 Surface Areas and Volumes NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Ex 13.4 Class 10 Maths Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:

Ex 13.4 Class 10 Maths Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:

Download NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas and Volumes PDF

Ex 13.4 Class 10 Maths Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:

Ex 13.4 Class 10 Maths Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.
Solution:

Ex 13.4 Class 10 Maths Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Solution:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Hindi Medium) Ex 13.4

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4, drop a comment below and we will get back to you at the earliest.

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Get Free NCERT Solutions for Class 10 Maths Chapter 13 Ex 13.5 PDF. Surface Areas and Volumes Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 13.5 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 13 Maths Class 10 Surface Areas and Volumes Exercise 13.5 provided in NCERT TextBook.

Surface Areas and Volumes Class 10 Formulas PDF

Topics and Sub Topics in Class 10 Maths Chapter 13 Surface Areas and Volumes:

• Surface Areas and Volumes Class 10 Ex 13.1
• प्रश्नावली 13.1 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.2
• प्रश्नावली 13.2 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.3
• प्रश्नावली 13.3 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.4
• प्रश्नावली 13.4 का हल हिंदी में
• Surface Areas and Volumes Class 10 Ex 13.5
• प्रश्नावली 13.5 का हल हिंदी में
• Surface Areas and Volumes Class 10 Extra Questions

You can also download the free PDF of  Ex 13.5 Class 10 Surface Areas and Volumes NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Ex 13.5 Class 10 Maths Question 1.
A copper wire, 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.
Solution:

Ex 13.5 Class 10 Maths Question 2.
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.
Solution:

Download NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas and Volumes PDF

Ex 13.5 Class 10 Maths Question 3.
A cistern measuring 150 cm x 120 cm x 110 cm has 129600 cm³ water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?
Solution:

Ex 13.5 Class 10 Maths Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km, show that the total rainfall was approximately equivalent to addition to the normal water of three rivers each 1072 km long. 75 m wide and 3 m deep.
Solution:

Ex 13.5 Class 10 Maths Question 5.
An oil funnel made of tin sheer consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Solution:

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Hindi Medium) Ex 13.5

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5, drop a comment below and we will get back to you at the earliest.

The post NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 appeared first on Learn CBSE.

Knowing Our Numbers Class 6 Extra Questions Maths Chapter 1

Extra Questions for Class 6 Maths Chapter 1 Knowing Our Numbers

Knowing Our Numbers Class 6 Extra Questions Very Short Answer Type

Question 1.
Write the smallest three digit number whose value does not change on reversing its digits.
Solution:
The required number is 101.

Question 2.
Write the greatest three digit number which does not change on reversing its digits.
Solution:
The required number is 999.

Question 3.
What must be added to 203 to get a number whose digits are reversed of the given number?
Solution:
The number obtained by reversing the digits of 203 = 302.
∴ Difference = 302 – 203 = 99
Hence, the required number is 99.

Question 4.
Write the following in Roman numerals:
(a) 72
(b) 38
Solution:
(a) 72 = LXXII
(b) 38 – XXXVIII

Question 5.
Write 438 in its expanded form.
Solution:
438 = 4 x 100 + 3 x 10 + 8.

Question 6.
Write the greatest five-digit number using the digits 4, 2 and 0.
Solution:
The greatest five-digit number using the digits 4, 2 and 0 is 44420.

Question 7.
The capacity of a water tank is 300 litres. Express its capacity in millilitres.
Solution:
We know that
1 litre = 1000 mL
∴ 300 litres = 300 x 1000 mL = 3,00,000 mL
Hence, the capacity of water tank = 3 lakh millilitres.

Question 8.
What is the successor of greatest 6-digit number?
Solution:
Greatest 6-digit number = 999999
Successor of it = 999999 + 1 = 1000000
i. e., smallest 7-digit number.
Hence, the required successor = 10,00,000.

Question 9.
What is the place value of 7 in 1743?
Solution:
Let us write 1743 in its expanded form
1743 = 1000 + 700 + 40 +3
Place value of 7 = 700
Hence, the place value of 7 = 700.

Knowing Our Numbers Class 6 Extra Questions Short Answer Type

Question 10.
Of 7,12,540 and 71,25,400 which number is greater and by how much?
Solution:
Since 71,25,400 is a seven-digit number and 7,12,540 is a six-digit number.
So 71,25,400 is greater than 7,12,540.

Hence 71,25,400 is greater than 7,12,540 by 64,12,860.

Question 11.
Write the smallest and the greatest 5-digit numbers using the digits 0,2,4,6,8 (Repetition of digits is not allowed).
Solution:
Given digits are 0, 2, 4, 6, 8
5 – digit greatest number = 86420;
5 – digit smallest number = 20468.

Question 12.
Write the following numbers in ascending order. How many of them are even numbers?
63,854, 63,584, 65,348, 68,543, 64,835
Solution:
The given numbers are 63,854, 63,584, 65,348, 68,543 and 64,835.
Ascending order is 63,584 ; 63,854 ; 64,835 ; 65,348 ; 68,543
Even numbers are 63,584, 63,854 and 65,348.

Question 13.
Round the given numbers to the nearest tens.
(a) 48
(b) 59
(c) 64
(d) 215
Solution:
Given number Rounded off to tens
(a) 48 → 50
(6) 59 → 60
(c) 64 → 60
(d) 215 → 220

Question 14.
Estimate the following products:
(а) 86 x 316
(b) 898 x 786
Solution:
(a) 86 x 316
∵ 86 → 90 [Rounding off to tens] and 316 → 320 [Rounding off to tens]
So, the estimated product is 90 x 320 = 28800

(b) 898 x 786
∵ 898 → 900 [Rounding off to hundreds] and 786 → 800 [Rounding off to hundreds]
So, the estimated product is 900 x 800 = 720000.

Question 15.
Divide 2,63,175 by 275.
Solution:
We have

Hence, quotient = 957 and remainder = 0.

Question 16.
A student multiplied 3759 by 231 instead of multiplying by 213. How much was his product greater than the correct product?
Solution:
First Method:
(3759 x 231)-(3759 x 213) = 868329 – 800667 = 67662
Second Method: 3759 x (231 – 213) = 3759 x 18 = 67662
Hence, the product difference is 67662.

Question 17.
Estimate: 25,148 + 7394 + 9343 + 752
Solution:
Estimated values are
25,148     →    25100
7394        →     7400
9343       →      9300
752          →      800
So, the estimated sum is 25100 + 7400 + 9300 + 800 = 42600
Hence, the estimated sum is 42600.

Question 18.
Write all the even numbers between 90 and 100 in Roman Numerals.
Solution:
Even numbers between 90 and 100, we have 92, 94, 96, 98.
∴ 92 = XCII,
94 = XCIV,
96 = XCVI,
98 = XCVIII

Knowing Our Numbers Class 6 Extra Questions Long Answer Type

Question 19.
Write the missing digits in the following sums:

Solution:

Question 20.
Write Hindu-Arabic numerals for:
(a) LXXXVI
(b) LXXV
(c) XCIX
(d) XCI
Solution:
(a) LXXXVI = 50 + 30 + 6 = 86
(b) LXXV = 50 + 20 + 5 = 75
(c) XCIX = (100 – 10) + 9 = 99
(d) XCI = (100 – 10) + 1 = 91

Question 21.
The distance between the school and Reena’s house is 1 km 480 m. Everyday she walks both ways. What distance does she cover in 6 days of a week?
Solution:
Distance covered when she walks one way = 1 km 480 m = 1480 m
Therefore, the distance covered when she walk both ways in a day = 1480 x 2 m = 2960 m
Total distance covered by Reena in 6 days = 2960 x 6 m = 17760 m or 17 km 760 m.

Question 22.
Simplify: 36 ÷ [5 + {4 x 5 ÷ 2}]
Solution:
Given:
36 ÷ [5 + {4 x 5 ÷ 2}]
Using B, O, D, M, A, S

Question 23.
To stitch a pant 1 m 15 cm cloth is needed. Out of 36 m cloth, how many pants can be stitched and how much cloth will remain?
Solution:

Cloth required to stitch 1 pant = 1 m 15 cm
= 100 cm + 15 cm [∵ 1 m = 100 cm]
= 115 cm
Total cloth = 36 m = 36 x 100 cm = 3600 cm
Therefore number of pants stitched = \(\frac { 3600 }{ 115 }\)
Hence, 31 pants can be stitched and cloth left over is 35 cm.

Question 24.
Write each of the following numbers in figures:
(a) Eighty-one million four hundred twelve thousand six hundred fifty.
(b) Twenty million three hundred eighty thousand one hundred.
(c) Ninety million nine.
(d) Forty-nine million seven hundred eighty two thousand fifty eight.
(e) Six millions three hundred fifty-two thousand nine hundred forty-six.
(f) Seven crore twenty-three lakh eighty-six thousand, five hundred ninety-four.
(g) Fifty crore forty lakh sixty thousand nine.
(h) Nineteen crore, ninety lakh, fourteen thousand, six hundred eighty.
Solution:
In words                                                                                                                         In figure
(а) Eighty-one millions four hundred twelve thousand, six hundred fifty.     81,412,650
(b) Twenty million three hundred eighty thousand one hundred                     20,380,100
(c) Ninety million nine                                                                                               90,000,009
(d) Forty-nine million seven hundred eighty-two thousand fifty-eight           49,782,058
(e) Six-millions three hundred fifty-two thousand nine hundred forty-six    6,352,946
(f) Seven crore, twenty-three lakh eighty-six thousand five hundred ninety-four 7,23,86,594
(g) Fifty crore forty lakh sixty thousand nine                                                        50,40,60,009
(h) Nineteen crore ninety lakh fourteen thousand six hundred eighty.           19,90,14,680

Question 25.
Write True/False for the following statements:
(a) Roman symbol X cannot be repeated more than three times
(b) VXXX = 25 …….. .
(c) Estimate value of274 rounding off to nearest hundreds = 200 …….
(d) I and X can repeat at the most three times …….
(e) V, L and D are neither, repeated nor written to the left of greater value symbol ……..
(f) There are six basic symbols in Roman Numeration system ……..
Solution:
(a) True
(b) False
(c) False
(d) True
(e) True
(f) False.

Knowing Our Numbers Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 26.
There are two factories located at place P and the other at place Q. From these factories, a certain commodity is to be delivered to each of the depots situated at A, B and C. Weekly production of commodity by P and Q are 120 kg and 150 kg respectively. Weekly requirement of commodity by A, B and C are 80 kg, 90 kg and 100 kg respectively. P delivers 60 kg to A, 40 kg to B and 20 kg to C. How much amount of the commodity should Q deliver to A, B and C to meet their requirement? If the rate of the commodity is ? 20 per kg, find the total amount to be paid to P and Q.
Solution:

Amount of commodity delivered by P to A = 60 kg
Amount of commodity delivered by Q to A = 80 – 60 = 20 kg
Amount of commodity delivered by P to B = 40 kg
Amount of commodity delivered by Q to B – 90 – 40 = 50 kg
Amount of commodity delivered by P to C = 20
Amount of commodity delivered by Q to C = 100 – 20 = 80 kg.
Now Amount of money to be paid to P by A, B and C = ₹( 60 x 20 + 40 x 20 + 20 x 20)
= ₹ (1200 + 800 + 400)
= ₹ 2400
and amount of money to be paid to Q by A, B and C
= ₹ (20 x 20 + 50 x 20 + 80 x 20)
= ₹ (400 + 1000 + 1600) = ₹ 3000
Hence, the total amount
= ₹ 2400 + ₹ 3000 = ₹ 5400.

Extra Questions for Class 6 Maths

NCERT Solutions for Class 6 Maths

The post Knowing Our Numbers Class 6 Extra Questions Maths Chapter 1 appeared first on Learn CBSE.

In Maths, a Percentage is a Number or Ratio Expressed as a Fraction Over 100. In other words, percent means parts per hundred and is given by the symbol %. If we need to calculate the percent of a number you just need to divide the number by whole and multiply with 100. Percentages can be represented in any of the forms like decimal, fraction, etc.

You can get entire information regarding Percentage like Definition, Formula to Calculate Percentage, Conversions from Percentage to another form, and vice versa in the coming modules. Learn the Percentage Difference, Increase or Decrease Concepts too that you might need during your academics or in your day to day calculations.

List of Percentage Concepts

Access the concepts that you want to learn regarding the Percentage through the quick links available. Simply tap on the concept you wish to prepare and get the concerned information explained step by step. Clarify all your queries and be perfect in the corresponding topics.

• Fraction to Percentage
• Percentage to Fraction
• Percentage to Ratio
• Ratio to Percentage
• Percentage to Decimal
• Decimal to Percentage
• Percentage of the given Quantity
• How much Percentage One Quantity is of Another?
• Percentage of a Number
• Percentage Increase
• Percentage Decrease
• Basic Problems on Percentage
• Solved Examples on Percentage
• Problems on Percentage
• Real-Life Problems on Percentage
• Word Problems on Percentage
• Application of Percentage

Percentage Formula

To make your calculations quite simple we have provided the Percentage Formula here. Make use of it during your calculations and arrive at the solution easily.

Formula to Calculate Percentage is given by = (Value/Total Value) *100

How to Calculate Percentage of a Number?

To find the Percent of a Number check the following procedure

Consider the number to be X

P% of number = X

Removing the % sign we have the formula as under

P/100*Number = X

Percentage Change

% Change = ((New Value – Original Value)*100)/Original Value

There are two different types when it comes to Percent Change and they are given as under

• Percentage Increase
• Percentage Decrease

Percentage Increase

If the new value is greater than the original value that shows the percentage change in the value is increased from the original number. Percentage Increase is nothing but the subtraction of the original number from the new number divided by the original number.

% increase = (Increase in Value/Original Value) x 100

% increase = [(New Number – Original Number)/Original Number] x 100

Percentage Decrease

When the new value is less than the original value, that indicates the percentage change in the value shows the percent decrease in the original number. Percentage Decrease is nothing but the subtraction of new number from the original number.

% Decrease = (Decrease in Value/Original Value) x 100

% Decrease = [( Original Number – New Number)/Original Number] x 100

Percentage Difference

If you need to find the Percentage Difference if two values are known then the formula to calculate Percentage Difference is given by

Percentage Difference = {|N1 – N2|/(N1+N2/2)}*100

Conversion of Fraction to Percentage

To convert fraction to percentage follow the below-listed guidelines.

• Divide the numerator with the denominator.
• Multiply the result with 100.
• Simply place the % symbol after the result and that is the required percentage value.

Conversion of Decimal to Percentage

Follow the easy steps provided to change between Decimal to Percentage. They are as such

• Obtain the decimal number.
• Simply multiply the decimal value with 100 to get the percentage value.

Solved Examples on Percentage

1. What is 50% of 30?

Solution:

Given 50% of 30

= (50/100)*30

= 1500/100

= 15

Therefore, 50% of 30 is 15.

2. Find 20% of 40?

Solution:

Given 20% of 40

= (20/100)*40

= (20*40)/100

= 800/100

= 8

3. What is 15% of 60 equal to?

Solution:

= (15/100)*60

= (15*60)/100

= 900/100

= 9

4. There are 120 people present in an examination hall. The number of men is 50 and the number of women is 70 in the examination hall. Calculate the percentage of women present in the examination hall?

Solution:

Number of Women = 70

the percentage of women present in the examination hall = 70 is what percent of 120

Percentage of Women = (70/120)*100

= (7000)/120

= 700/12

= 58.3333 %

The Percentage of Women in the Examination Hall is 84%.

5. What is the percentage change in the rent of the house if in the month of January it was Rs. 20,000 and in the month of March, it is Rs. 15,000?

Solution:

We can clearly say that there is a decrease in the rent

Decreased Value  = 20,000 – 15, 000

= 5, 000

Percent Change = (Decreased Value/Original Value)*100

= (5000/20,000)*100

= (1/4)*100

= 25%

Hence, there is a 25% decrease in the rent.

FAQs on Percentage

1. What is meant by Percentage?

A percentage is a Number or Ratio Expressed as a Fraction Over 100.

2. What is the Formula for Percentage?

The formula for Percentage is (Value/Total Value) *100

3. What is the Symbol of Percentage?

The percentage is denoted by the symbol %.

4. What is 10% of 45?

10% of 45 is given by 10/100*45 i.e. 4.5

The post Percentage (How to Calculate, Formula and Tricks) appeared first on Learn CBSE.

CBSE Class 10 Maths Notes Chapter 10 Circles Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 10 Circles. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 10 Circles

Circle: A circle is a collection of all points in a plane which are at a constant distance from a fixed point.

Centre: The fixed point is called the centre.

Radius: The constant distance from the centre is called the radius.

Chord: A line segment joining any two points on a circle is called a chord.

Diameter: A chord passing through the centre of the circle is called diameter. It is the longest chord.

Tangent: When a line meets the circle at one point or two coincidings The line is known as points, a tangent.
The tangent to a circle is perpendicular to the radius through the point of contact.
⇒ OP ⊥ AB

The lengths of the two tangents from an external point to a circle are equal.
⇒ AP = PB

Length of Tangent Segment
PB and PA are normally called the lengths of tangents from outside point P.

Properties of Tangent to Circle

Theorem 1: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: XY is a tangent at point P to the circle with centre O.
To prove: OP ⊥ XY
Construction: Take a point Q on XY other than P and join OQ
Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle
OQ > OP

This happens with every point on line XY except point P. OP is the shortest of all the distances of point O the points of XY
OP ⊥ XY …[Shortest side is the perpendicular]

Theorem 2: A line drawn through the endpoint of a radius and perpendicular to it, is tangent to the circle.
Given: A circle C(O, r) and a line APB is perpendicular to OP, where OP is the radius.
To prove: AB is tangent at P.
Construction: Take a point Q on line AB, different from P, and join OQ.
Proof: Since OP ⊥ AB
OP < OQ ⇒ OQ > OP

Point Q lies outside the circle.
Therefore, every point on AB, other than P, lies outside the circle.
This shows that AB meets the circle at point P.
Hence, AP is tangent to the circle at P.

Theorem 3: Prove that the lengths of tangents drawn from an external point to a circle are equal
Given: PT and PS are tangents from an external point P to the circle with centre O.
To prove: PT = PS
Construction: Join O to P, T and S.

Proof: In ∆OTP and ∆OSP.
OT = OS …[radii of the same circle]
OP = OP …[common]
∠OTP = ∠OSP …[each 90°]
∆OTP = ∆OSP …[R.H.S.]
PT = PS …[c.p.c.t.]

Note: If two tangents are drawn to a circle from an external point, then:

• They subtend equal angles at the centre i.e., ∠1 = ∠2.
• They are equally inclined to the segment joining the centre to that point i.e., ∠3 = ∠4.
  ∠OAP = ∠OAQ
  

Class 10 Maths Notes

NCERT Solutions

The post Circles Class 10 Notes Maths Chapter 10 appeared first on Learn CBSE.

CBSE Class 10 Maths Notes Chapter 9 Some Applications of Trigonometry Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 9 Some Applications of Trigonometry. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 9 Some Applications of Trigonometry

Line of Sight
When an observer looks from a point E (eye) at object O then the straight line EO between eye E and object O is called the line of sight.

Horizontal
When an observer looks from a point E (eye) to another point Q which is horizontal to E, then the straight line, EQ between E and Q is called the horizontal line.

Angle of Elevation
When the eye is below the object, then the observer has to look up from point E to object O. The measure of this rotation (angle θ) from the horizontal line is called the angle of elevation.

Angle of Depression
When the eye is above the object, then the observer has to look down from point E to the object. The horizontal line is now parallel to the ground. The measure of this rotation (angle θ) from the horizontal line is called the angle of depression.

How to convert the above figure into the right triangle.
Case I: Angle of Elevation is known
Draw OX perpendicular to EQ.
Now ∠OXE = 90°
ΔOXE is a rt. Δ, where
OE = hypotenuse
OX = opposite side (Perpendicular)
EX = adjacent side (Base)

Case II: Angle of Depression is known
(i) Draw OQ’parallel to EQ
(ii) Draw perpendicular EX on OQ’.
(iii) Now ∠QEO = ∠EOX = Interior alternate angles
ΔEXO is an rt. Δ. where
EO = hypotenuse
OX = adjacent side (base)
EX = opposite side (Perpendicular)

• Choose a trigonometric ratio in such a way that it considers the known side and the side that you wish to calculate.
• The eye is always considered at ground level unless the problem specifically gives the height of the observer.

The object is always considered a point.
Some People Have
Sin θ = \(\frac { Perpendicular }{ Hypotenuse }\)
Curly Black Hair
Cos θ = \(\frac { Base }{ Hypotenuse }\)
Turning Permanent Black.
Tan θ = \(\frac { Perpendicular }{ Base }\)

Class 10 Maths Notes

NCERT Solutions

The post Some Applications of Trigonometry Class 10 Notes Maths Chapter 9 appeared first on Learn CBSE.

HCF and LCM Word Problems : In this section, we will learn how to solve word problems involving highest common factor and lowest common multiple.

HCF and LCM Word Problems

Problem 1 :

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? (excluding the one at start)

Solution :

For example, let the two bells toll after every 3 secs and 4 secs respectively.

Then the first bell tolls after every 3, 6, 9, 12 seconds…
Like this, the second bell tolls after every 4, 8, 12 seconds…

So, if the two bell toll together now, again they will toll together after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds

The same thing happened in our problem. To find the time, when they will all toll together, we have to find the L.C.M of (2, 4, 8, 6, 10, 12).

L.C.M of (2, 4, 8, 6, 10, 12) is 120 seconds  =  2 minutes.

So, after every two minutes, all the bell will toll together.

For example, in 10 minutes, they toll together :

10/2  =  5 times

That is, after 2,4,6,8,10 minutes. It does not include the one at the start

Similarly, in 30 minutes, they toll together :

30/2  =  15 times

(excluding one at the start).

Problem 2 :

The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hrs, when will they again change simultaneously ?

Solution :

For example, let the two signals change after every 3 secs and 4 secs respectively.

Then the first signal changes after 3, 6, 9, 12 seconds…

Like this, the second signal changes after 4, 8, 12 seconds…

So, if the two signals change simultaneously now, again they will change simultaneously after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds

The same thing happened in our problem. To find the time, when they will all change simultaneously, we have to find the L.C.M of (48, 72, 108).

L.C.M of (48,72,108) is 432 seconds  =  7 min 12 sec

So, after every 7 min 12 sec, all the signals will change simultaneously.

At 8:20:00 hrs, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hrs.

Hence, three signals will change simultaneously at 8:27:12 seconds.

Problem 3 :

A merchant has 120 ltrs of and 180 ltrs of two kinds of oil. He wants the sell oil by filling the two kinds of oil in tins of equal volumes. What is the greatest of such a tin.

Solution :

The given two quantities 120 and 180 can be divided by 10, 20,… exactly. That is, both the kinds of oils can be sold in tins of equal volume of 10, 20,… ltrs.

But, the target of the question is, the volume of oil filled in tins must be greatest.

So, we have to find the largest number which exactly divides 120 and 180.That is nothing but the H.C.F of (120, 180)

H.C.F of (120, 180)  =  60

The 1st kind 120 ltrs is sold in 2 tins of of volume 60 ltrs in each tin.

The 2nd kind 180 ltrs is sold in 3 tins of volume 60 ltrs in each tin.

Hence, the greatest volume of each tin is 60 ltrs.

Problem 4 :

Find the least number of soldiers in a regiment such that they stand in rows of 15, 20, 25 and form a perfect square.

Solution :

To answer this question, we have to find the least number which is exactly divisible by the given numbers 15,20 and 25.That is nothing but the L.C.M of (15, 20, 25)

L.C.M of (15, 20, 25)  =  300

So, we need 300 soldiers such that they stand in rows of 15, 20 , 25.

But, it has to form a perfect square (as per the question)

To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square.

To make 300 as perfect square, we have to multiply 300 by 3.
Then, it is 900 which is a perfect square.

Hence, the least number of soldiers required is 900.

Problem 5 :

Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely.

Solution :

We require the least number of square tiles. So, each tile must be of maximum dimension.

To get the maximum dimension of the tile, we have to find the largest number which exactly divides 16.58 and 8.32. That is nothing but the H.C.F of (16.58, 8.32).

To convert meters into centimeters, we have to multiply by 100.

16.58 ⋅ 100  =  1658 cm

8.32 ⋅ 100  =  832 cm

H.C.F of (1658, 832)  =  2

Hence the side of the square tile is 2 cm

Required no. of tiles :

=  (Area of the floor) / (Area of a square tile)

=  (1658 ⋅ 832) / (2 ⋅ 2)

=  344,864

Hence, the least number of square tiles required is 344,864.

Problem 6 :

A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing.

Solution :

For the least possible number of casks of equal size, the size of each cask must be of the greatest volume.

To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465. That is nothing but the H.C.F of (403, 434, 465)

The H.C.F of (403, 434, 465)  =  31 liters

Each cask must be of the volume 31 liters.

Req. No. of casks is

=  (403/31) + (434/31) + (465/31)

=  13 + 14 + 15

=  42

Hence, the least possible number of casks of equal size required is 42.

Problem 7 :

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed ?

Solution :

Because the H.C.F is 49, the two numbers can be assumed as 49x and 49y

Their sum is 588. So, we have

49x + 49y  =  588

Divide each side 49.

x + y  =  12

We have to find the values of “x” and “y” such that their sum is 12.

The possibles pairs of values of (x, y) are

(1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6)

Here, we have to check an important thing. That is, in the above pairs of values of (x, y), which are all co-primes ?

[Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1]

Therefore in the above pairs, (1, 11) and (5, 7) are the co-primes.

Hence, the number of pairs is 2.

Problem 8 :

The product of two numbers is 2028 and their H.C.F. is 13. Find the number of such pairs

Solution :

Since the H.C.F is 13, the two numbers could be 13x and 13y

Their product is 2028.

So, we have

(13x) ⋅ (13y)  =  2028

169xy  =  2028

Divide each side by 169.

xy  =  12

We have to find the values of “x” and “y” such that their product is 12.

The possibles pairs of values of (x, y) are

(1, 12), (2, 6), (3, 4)

Here, we have to check an important thing. That is, in the above pairs of values of (x, y), which are all co-primes?

[Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1]

Therefore in the above pairs, (1, 12) and (3, 4) are the co-primes.

Hence, the number of pairs is 2

Problem 9 :

Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls. If he wants to make all the plates identical without any food left over, what is the greatest number of plates Lenin can prepare ?

Solution :

To make all the plates identical and find the greatest number of plates, we have to find the greatest number which can divide 12 and 16 exactly.

That is nothing but H.C.F of 12 and 16.

H.C.F of (12, 16)  =  4

That is, 12 pieces of chicken would be served in 4 plates at the rate of 3 pieces per plate.

And 16 rolls would be served in 4 plates at the rate of 4 rolls per plate.

In this way, each of the 4 plates would have 3 pieces of chicken and 4 rolls. And all the 4 plates would be identical.

Hence, the greatest number of plates Lenin can prepare is 4

Problem 10 :

The drama club meets in the school auditorium every 2 days, and the choir meets there every 5 days. If the groups are both meeting in the auditorium today, then how many days from now will they next have to share the auditorium ?

Solution :

If the drama club meets today, again they will meet after 2, 4, 6, 8, 10, 12…. days.

Like this, if the choir meets today, again they will meet after 5, 10, 15, 20 …. days.

From the explanation above, If both drama club and choir meet in the auditorium today, again, they will meet after 10 days.

And also, 10 is the L.C.M of (2, 5).

Hence, both the groups will share the auditorium after ten days.

Problem 11 :

John is printing orange and green forms. He notices that 3 orange forms fit on a page, and 5 green forms fit on a page. If John wants to print the exact same number of orange and green forms, what is the minimum number of each form that he could print ?

Solution :

The condition of the question is, the number of orange forms taken must be equal to the number of green forms taken.

Let us assume that he takes 10 orange and 10 green forms.

10 green forms can be fit exactly on 2 pages at 5 forms/page. But,10 orange forms can’t be fit exactly on any number of pages.

Because, 3 orange forms can be fit exactly on a page. In 10 orange forms, 9 forms can be fit exactly on 3 pages and 1 form will be remaining.

To get the number of forms in orange and green which can be fit exactly on some number of pages, we have to find L.C.M of (3,5). That is 15.

15 orange forms can be fit exactly on 5 pages at 3 forms/page.

15 green forms can be fit exactly on 3 pages at 5 forms/page.

Hence,the smallest number of each form could be printed is 15.

Problem 12 :

Lily has collected 8 U.S. stamps and 12 international stamps. She wants to display them in identical groups of U.S. and international stamps, with no stamps left over. What is the greatest number of groups Lily can display them in ?

Solution :

To make all the groups identical and find the greatest number of groups, we have to find the greatest number which can divide 8 and 12 exactly.

That is nothing but H.C.F of 8 and 12.

H.C.F of (8, 12) = 4

That is, 8 U.S stamps can be displayed in 4 groups at 2 stamps/group.

And 12 international stamps can be displayed in 4 groups at 3 stamps/group.

In this way, each of the 4 groups would have 2 U.S stamps and 3 international stamps. And all the 4 groups would be identical.

Hence, the greatest number of groups can be made is 4

Problem 13 :

Abraham has two pieces of wire, one 6 feet long and the other 12 feet long. If he wants to cut them up to produce many pieces of wire that are all of the same length, with no wire left over, what is the greatest length, in feet, that he can make them ?

Solution :

When the two wires are cut in to small pieces, each piece must of same length and also it has to be the possible greatest length.

6 feet wire can be cut in to pieces of (2, 2, 2) or (3, 3)

12 feet wire can be cut in to pieces of (2, 2, 2, 2, 2, 2 ) or (3, 3, 3, 3)

The length of each small piece must be of possible greatest length.

To find the possible greatest length, we have to find the greatest number which can divide both 6 and 12. That is H.C.F of (6, 12).

H.C.F of (6, 12) = 6.

Hence, the greatest length of each small piece will be 6 ft.

(That is, 6 feet wire is not cut in to small pieces and it is kept as it is. Only the 12 feet wire is cut in to 2 pieces at the length of 6 feet/piece)

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CBSE Class 10 Maths Notes Chapter 11 Constructions Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 11 Constructions. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 11 Constructions

Determining a Point Dividing a given Line Segment, Internally in the given Ratio M : N

Let AB be the given line segment of length x cm. We are required to determine a point P by dividing it internally in the ratio m : n.

Steps of Construction:

• Draw a line segment AB = x cm.
• Make an acute ∠BAX at the end A of AB.
• Use a compass of any radius and mark off arcs. Take (m + n) points A₁, A₂, … A_m, A_m+1, …, A_m+n along AX such that AA₁ = A₁A₂ = … = A_m+n-1 , A_m+n
• Join A_m+nB.
• Passing through A_m, draw a line A_mP || A_m+nB to intersect AB at P. The point P so obtained is the A required point which divides AB internally in the ratio m : n.
  

Construction of a Tangent at a Point on a Circle to the Circle when its Centre is Known

Steps of Construction:

• Draw a circle with centre O of the given radius.
  
• Take a given point P on the circle.
• Join OP.
  
• Construct ∠OPT = 90°.
  
• Produce TP to T’ to get TPT’ as the required tangent.
  

Construction of a Tangent at a Point on a Circle to the Circle when its Centre is not Known

If the centre of the circle is not known, then we first find the centre of the circle by drawing two non-parallel chords of the circle. The point of intersection of perpendicular bisectors of these chords gives the centre of the circle. Then we can proceed as above.

Construction of a Tangents from an External Point to a Circle when its Centre is Known

Steps of Construction:

• Draw a circle with centre O.
• Join the centre O to the given external point P.
• Draw a right bisector of OP to intersect OP at Q.
• Taking Q as the centre and OQ = PQ as radius, draw a circle to intersect the given circle at T and T’.
• Join PT and PT’ to get the required tangents as PT and PT’.
  

Construction of a Tangents from an External Point to a Circle when its Centre is not Known

If the centre of the circle is not known, then we first find the centre of the circle by drawing two non-parallel chords of a circle. The point of intersection of perpendicular bisectors of the chords gives the centre of the circle. Then we can proceed as above.

Construction of a Triangle Similar to a given Triangle as per given Scale Factor \(\frac { m }{ n }\) , m < n.

Let ΔABC be the given triangle. To construct a ΔA’B’C’ such that each of its sides is \(\frac { m }{ n }\) (m < n) of the corresponding sides of ΔABC.

Steps of Construction:

• Construct a triangle ABC by using the given data.
• Make an acute angle ∠BAX, below the base AB.
• Along AX, mark n points A₁, A₂ …, A_n, such that AA₁ = A₁A₂ = … = A_m-1 A_m = … A_n-1 A_n.
• Join A_nB.
• From A_m, draw A_mB’ parallel to AnB, meeting AB at B’.
• From B’, draw B’C’ parallel to BC, meeting AC at C’.
  Triangle AB’C’ is the required triangle, each of whose sides is \(\frac { m }{ n }\) (m < n) of the corresponding sides of ΔABC.
  

Construction of a Triangle Similar to a given Triangle as per given Scale Factor \(\frac { m }{ n }\) , m > n.

Let ΔABC be the given triangle and we want to construct a ΔAB’C’, such that each of its sides is \(\frac { m }{ n }\) (m > n) of the corresponding side of ΔABC.

Steps of Construction:

• Construct a ΔABC by using the given data.
• Make an acute angle ∠BAX, below the base AB. Extend AB to AY and AC to AZ.
• Along AX, mark m points A₁, A₂ …, A_n, ..A_m, such that AA₁ = A₁A₂ = A₂A₃ = … = A_n-1 A_n = … = A_m-1 A_m
• Join A_nB.
• From A_m, draw A_mB’ parallel to A_nB, meeting AY produced at B’.
• From B’, draw B’C’ parallel to BC, meeting AZ produced at C’.
• Triangle AB’C’ is the required triangle, each of whose sides is (\(\frac { m }{ n }\)) (m > n) of the corresponding sides of ΔABC.
  

We hope the given CBSE Class 10 Maths Notes Chapter 11 Constructions Pdf free download will help you. If you have any query regarding NCERT Class 10 Maths Notes Chapter 11 Constructions, drop a comment below and we will get back to you at the earliest.

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NCERT Exemplar Class 9 Science Chapter 8 Motion are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 8 Motion.

NCERT Exemplar Class 9 Science Solutions Chapter 8 Motion

Multiple Choice  Questions

Question 1.
A particle is moving in a circular path of radius r. The displacement after half a circle would be
(a) Zero
(b) πr
(c) 2r
(d) 2πr
Solution:
(c) Displacement is the shortest distance measured from the initial position to final position.

Therefore, displacement after jralf a circle is the diameter of the circular path, i.e., 2r

Question 2.
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) u/g
(b) u²/2g
(c) u²/g
(d) u/2g
Solution:
(b) At highest point, v = 0
Using v²– u² = 2as, we get
0- u²=2(-g)h( ∵ a=-g and s=h)
or h=u²/2g

Question 3.
Thenumerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1
Solution:
(d) As displacement is the shortest distance between two positions so it is generally less than distance. But displacement can be equal to distance when path taken is a straight line. So, the numerical ratio of displacement to distance for a moving object is equal or less than 1.

Question 4.
If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration
Solution:
(b) As s α t², ∴ v α t and a = constant i.e., the object moves with uniform acceleration.

Question 5.
From the given v – t graph it can be inferred that the object is
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration

Solution:
(a) As the given v-t graph shows that velocity is same for all time values, so, the object is in uniform motion.

Question 6.
Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m s~\ It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity
Solution:
(c) The merry-go-round moves in a circular path with constant speed, but its velocity changes due to the change in the direction of motion at every point. It implies that the boy on merry-go-round is in accelerated motion.

Question 7.
Area under a v – t graph represents a physical quantity which has the unit
(a) m²
(b) m
(c) m³
(d) m s^-1
Solution:
(b) Area under v-t graph represents distance and unit of distance is metre (m).

Question 8.
Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in figure. Choose the correct statement
(a) Car A is faster than car D.
(b) Car 6 is the slowest.
(c) Car O is faster than car C.
(d) Car C is the slowest.
Solution:

Question 9.
Which of the following figures represents uniform motion of a moving object correctly?

Solution:
(a) In uniform motion, object covers equal distances in equal intervals of time. Therefore, the distance-time graph must be a straight line inclined to time axis. Hence, graph (a) is correct.

Question 10.
Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed
Solution:
(c) Slope of a velocity-time graph gives the rate of change of velocity or the acceleration.

Question 11.
In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the sun
Solution:
(a) If the car is moving on straight road, the distance moved and the magnitude of displacement are equal.

Short Answer Type Questions

Question 12.
The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify your answer.
Solution:
In a given interval of time, displacement of a moving object is zero when its final position is the same as initial position, whereas distance travelled by the object is not zero. For example, an athlete moving on a circular track. If he starts from A and completes one round and reaches back to point A, his displacement is zero whereas distance travelled by him is not zero but 2πr.

Question 13.
How will the equations of motion for an object moving with a uniform velocity change?
Solution:
For an object moving with a uniform velocity, acceleration a = 0.

Question 14.
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in figure. Plot a velocity-time graph for the same

Solution:
From displacement – time graph,

Symmetry of graph shows that the girl comes back to her initial position with same velocity (2 m s^-1) but in opposite direction. So, the velocity-time graph will look like

Question 15.
A car starts from rest and moves along the x-axis with constant acceleration 5 m s~2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?
Solution:

Question 16.
A motorcyclist drives from A to B with a uniform speed of 30 km h’1 and returns back with a speed of 20 km h_1. Find its average speed.
Solution:

Question 17.
The velocity-time graph shows the motion of a cyclist. Find
(i) its acceleration
(ii) its velocityand
(iii) the distance covered by the cyclist in 15 seconds

Solution:
Velocity-time graph is a straight line parallel to time axis, so, velocity of the cyclist is constant.
(i) Acceleration = 0
(ii) At, f = 15 s velocity = 20 m s’1 (from the given graph)
(iii) Distance covered by the cyclist in 15 s
= Area under v-t graph during that time interval
= 20 m s’1 x 15 s = 300 m.

Question 18.
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Solution:
When a stone is thrown vertically upwards, its velocity at highest point is zero. As acceleration due to gravity (g) acts vertically downwards, so the upward motion of stone is uniformly decelerated and the downward motion is uniformly accelerated. This makes the velocity of stone while reaching at ground equal to initial velocity of stone. So the velocity-time graph will look like.

Here, PQ corresponds to upward motion and QR corresponds to downward motion of stone.

Maximum Height Calculator – Projectile Motion is a free online tool that computes the maximum height of the projectile. It takes velocity, initial height and angle of launch as input and produces the projectile maximum height in a fraction of seconds.

Long Answer Type Questions

Question 19.
An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations?How does the difference in heights vary with time?
Solution:

Initially, difference in heights of two objects
= 150 m – 100 m = 50 m
Distance travelled by first object in 2 s

Difference in heights does oot vary with time as long as both the objects are in motion. However, when second object reaches ground and first one is still in motion, then it decreases.

Question 20.
An object starting from rest travels 20m in first 2 s and 160m in next 4 s. What will be the velocity after 7 s from the start.
Solution:
For first 2 s motion of object,
u = 0, f = 2 s, s = 20 m.

Question 21.
Using following data, draw time – displacement graph for a moving object:

Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.
Solution:
From the given data, displacement-time graph is shown as

Question 22.
An electron moving with a velocity of 5 x 104 ms-¹ enters into a uniform electric field and acquires a uniform acceleration of 104 m s~2 in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time?
Solution:
Here initial velocity of electron ,

Question 23.
Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Solution:
Let the object be moving with initial velocity u m s-¹ and uniform acceleration a ms-²

Question 24.
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u2: uf (Assume upward acceleration is -g and downward acceleration to be +g
Solution:
At the highest point, v=0
For the sone thrown with velocity u₁

We hope the NCERT Exemplar Class 9 Science Chapter 8 Motion will help you. If you have any query regarding NCERT Exemplar Class 9 Science Solutions Chapter 8 Motion, drop a comment below and we will get back to you at the earliest.

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CBSE Class 10 Maths Notes Chapter 12 Areas related to Circles Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 12 Areas related to Circles. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 12 Areas related to Circles

Circumference of a circle = 2πr
Area of a circle = πr² …[where r is the radius of a circle]
Area of a semi-circle = \(\frac { { \pi r }^{ 2 } }{ 2 }\)
Area of a circular path or ring:

Let ‘R’ and ‘r’ he radii of two circles
Then area of shaded part = πR² – πr² = π(R² – r²) = π(R + r)(R – r)

Minor arc and Major Arc: An arc length is called a major arc if the arc length enclosed by the two radii is greater than a semi-circle.

If the arc subtends angle ‘θ’ at the centre, then the
Length of minor arc = \(\frac { \theta }{ 360 } \times 2\pi r=\frac { \theta }{ 180 } \times \pi r\)
Length of major arc = \(\left( \frac { 360-\theta }{ 360 } \right) \times 2\pi r\)

Sector of a Circle and its Area
A region of a circle is enclosed by any two radii and the arc intercepted between two radii is called the sector of a circle.
(i) A sector is called a minor sector if the minor arc of the circle is part of its boundary.
\(\hat { OAB }\) is minor sector.

Area of minor sector = \(\frac { \theta }{ 360 } \left( { \pi r }^{ 2 } \right)\)
Perimeter of minor sector = \(2r+\frac { \theta }{ 360 } \left( { 2\pi r } \right) \)

(ii) A sector is called a major sector if the major arc of the circle is part of its boundary.
\(\hat { OACB }\) is major sector
Area of major sector = \(\left( \frac { 360-\theta }{ 360 } \right) \left( { \pi r }^{ 2 } \right)\)
Perimeter of major sector = \(2r+\left( \frac { 360-\theta }{ 360 } \right) \left( { 2\pi r } \right)\)

Minor Segment: The region enclosed by an arc and a chord is called a segment of the circle. The region enclosed by the chord PQ & minor arc PRQ is called the minor segment.

Area of Minor segment = Area of the corresponding sector – Area of the corresponding triangle

Major Segment: The region enclosed by the chord PQ & major arc PSQ is called the major segment.
Area of major segment = Area of a circle – Area of the minor segment
Area of major sector + Area of triangle

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CBSE Class 10 Maths Notes Chapter 13 Surface Areas and Volumes Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 13 Surface Areas and Volumes. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 13 Surface Areas and Volumes

SURFACE AREA AND VOLUME OF COMBINATIONS

Cone on a Cylinder.

r = radius of cone & cylinder;
h₁ = height of cone
h₂ = height of cylinder
Total Surface area = Curved surface area of cone + Curved surface area of cylinder + area of circular base
= πrl + 2πrh₂ +πr²;
Slant height, l = \(\sqrt { { r }^{ 2 }+{ { { h }_{ 1 }^{ 2 } } } }\)
Total Volume = Volume of cone + Volume of cylinder
= \(\frac { 1 }{ 3 } { \pi r }^{ 2 }{ h }_{ 1 }+{ \pi r }^{ 2 }{ h }_{ 2 }\)

Cone on a Hemisphere:

h = height of cone;
l = slant height of cone = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } }\)
r = radius of cone and hemisphere
Total Surface area = Curved surface area of cone + Curved surface area of hemisphere = πrl + 2πr²
Volume = Volume of cone + Volume of hemisphere = \(\frac { 1 }{ 3 } { \pi r }^{ 2 }h+\frac { 2 }{ 3 } { \pi r }^{ 3 }\)

Conical Cavity in a Cylinder

r = radius of cone and cylinder;
h = height of cylinder and conical cavity;
l = Slant height
Total Surface area = Curved surface area of cylinder + Area of the bottom face of cylinder + Curved surface area of cone = 2πrh + πr² + πrl
Volume = Volume of cylinder – Volume of cone = \({ \pi r }^{ 2 }h-\frac { 1 }{ 3 } { \pi r }^{ 2 }h=\frac { 2 }{ 3 } { \pi r }^{ 2 }h\)

Cones on Either Side of the Cylinder.

r = radius of cylinder and cone;
h₁ = height of the cylinder
h₂ = height of cones
Slant height of cone, l = \(\sqrt { { h }_{ 2 }^{ 2 }+{ r }^{ 2 } }\)
Surface area = Curved surface area of 2 cones + Curved surface area of cylinder = 2πrl + 2πrh₁
Volume = 2(Volume of cone) + Volume of cylinder = \(\frac { 2 }{ 3 } { \pi r }^{ 2 }{ h }_{ 2 }+{ \pi r }^{ 2 }{ h }_{ 1 }\)

Cylinder with Hemispherical Ends.

r = radius of cylinder and hemispherical ends;
h = height of cylinder
Total surface area= Curved surface area of cylinder + Curved surface area of 2 hemispheres = 2πrh + 4πr²
Volume = Volume of cylinder + Volume of 2 hemispheres = \({ \pi r }^{ 2 }h+\frac { 4 }{ 3 } { \pi r }^{ 3 }\)

Hemisphere on Cube or Hemispherical Cavity on Cube

a = side of cube;
r = radius of hemisphere.
Surface area = Surface area of cube – Area of hemisphere face + Curved surface area of hemisphere
= 6a² – πr² + 2πr² = 6a² + πr²
Volume = Volume of cube + Volume of hemisphere = \({ a }^{ 3 }+\frac { 4 }{ 3 } { \pi r }^{ 3 }\)

Hemispherical Cavity in a Cylinder

r = radius of hemisphere;
h = height of cylinder
Total surface area = Curved surface area of cylinder + Surface area of base + Curved surface area of hemisphere
= 2πrh + πr² + 2πr² = 2πrh + 3πr²
Volume = Volume of cylinder – Volume of hemisphere = \({ \pi r }^{ 2 }h-\frac { 2 }{ 3 } { \pi r }^{ 3 }\)

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NCERT Solutions For Class 10 Science Chapter 5 Periodic Classification Of Elements: Students who are in search for periodic classification of elements class 10 question and answers can refer to this article. Working on NCERT Solutions for class 10 science chapter 5 notes will help candidates to build a strong foundation over the subject Science. NCERT Solutions For Class 10 Science Chapter 5 Periodic Classification Of Elements will further help in clearing the UG competitive exams.

Going through this periodic classification of elements class 10 NCERT activity solutions will help candidates to understand the theory behind every concept which in turn helps candidates to solve the questions at the end of every topics and subtopic. Read on to find everything about the periodic classification of elements class 10 question and answers.

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

Before getting into the details of NCERT Solutions For Class 10 Science Chapter 5 Periodic Classification Of Elements, let’s have an overview of topics & subtopics under NCERT Solutions for class 10 science chapter 5 notes:

1. Periodic Classification Of Elements
2. Making Order Out Of Chaos – Early Attempts At The Classification Of Elements
3. Making Order Out Of Chaos – Mendeléev’S Periodic Table
4. Making Order Out Of Chaos – The Modern Periodic Table

Free download NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification Of Elements 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.

• तत्वों के आवर्त वर्गीकरण कक्षा 10 विज्ञान हिंदी में
• Class 10 Periodic Classification of Elements Important Questions
• Periodic Classification of Elements Class 10 Notes
• Periodic Classification of Elements NCERT Exemplar Solutions
• Periodic Classification of Elements Class 10 Extra Questions
• Class 10 Periodic Classification of Elements Mind Map

NCERT Solutions for Class 10 Science Chapter 5 Intext Questions

Page Number: 81

Question 1
Did Dobereiner’s triads also exist in the columns of Newlands’ Octaves ? Compare and find out.
Answer:
Yes, Dobereiner’s triads also existed in the columns of Newland’s Octaves.
For example, Li, Na, K.
If we consider lithium (Li) as the first element, then sodium (Na) is eighth element. If we consider sodium as the first element, then potassium is the eighth element.

Question 2
What were the limitations of Dobereiner’s classification ?
Answer:
It failed to arrange all the then known elements in the form of triads of elements having similar chemical properties. Dobereiner could identify only three triads from the elements known that time.

Question 3
What were the limitations of Newlands’ law of octaves ?
Answer:
(i) Newlands law of octaves was applicable to the classification of elements upto calcium only. After calcium every eighth element did not possess the properties similar to that of the first element.
(ii) Newlands assumed that only 56 elements existed in nature and no more elements would be discovered in the future. But later on, several new elements were discovered whose properties did not fit into Newlands’ law of Octaves.
(iii) In order to fit elements into his table, Newlands put even two elements together in one slot and that too in the column of unlike elements having very different properties.
For example, the two elements cobalt (Co) and nickel (Ni) were put together in just one slot and that too in the column of elements like fluorine, chlorine and bromine which have very different properties from these elements.
(iv) Iron (Fe) element which resemble elements like cobalt and nickel in properties, was placed far away from these elements.

Page Number: 85

Question 1
Use Mendeleev’s Periodic Table to predict the formulae for the oxides of the following elements : K, C, Al, Si, Ba
Answer:
K₂O, CO₂, Al₂O₃, SiO₂, BaO.

Question 2
Besides gallium, which other elements have since been discovered that were left by Mendeleev in his periodic table ? (any two)
Answer:
Scandium and Germanium.

Question 3
What were the criteria used by Mendeleev in creating his Periodic Table ?
Answer:
Mendeleev used the relationship between the atomic masses of the elements and their physical and chemical properties. He used similarity in physical properties, similarity in the formation of hydrides and oxides of element.

Question 4
Why do you think the noble gases are placed in a separate group ?
Answer:
Noble gases are chemically inert and are present in atmosphere in extremely low concentrations. Therefore, owing to their similar inert behaviour and similar electronic configuration, they are justified to be placed in a separate group.

Page Number: 90

Question 1
How could the modern periodic table remove various anomalies of Mendeleev’s periodic Table ?
Answer:
(i) The modern periodic table is based on atomic number, while Mendeleev’s periodic table was based on atomic mass.
(ii) The isotopes of an element have same number of protons (or atomic number). So they are alloted the same position in modern periodic table.
(iii) Cobalt and nickel are placed at 9th and 10th position respectively.
(iv) Hydrogen has been alloted special position, i.e., it is placed at the top of alkali metals in the first group.

Question 2
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice ?
Answer:
Beryllium (Be) and Calcium (Ca).
Both Be (atomic number 4) and Ca (atomic number 20) have similar electronic configuration, i.e. two electrons in outermost shells.
Be               2,2
Ca             2, 8, 8, 2
Both Be and Ca react with oxygen to give basic oxides, BeO and MgO.

Question 3
Name :
(a) three elements that have a single electron in their outermost shells.
(b) two elements that have two electrons in their outermost shells.
(c) three elements with filled outer most shells.
Answer:
(a) Lithium : Atomic number – 3(2, 1); Sodium : Atomic number – 11(2, 8, 1); Potassium : Atomic number – 19(2, 8, 8, 1).
(b) Beryllium : Atomic number – 4(2, 2); Calcium : Atomic number – 20(2, 8, 8, 2)
(c) Helium : Atomic number – 2(2); Neon : Atomic number – 10(2, 8); Argon : Atomic number – 18(2, 8, 8).

Question 4
(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements ?
(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common ?
Answer:
(a) Lithium, sodium and potassium all belong to the same group. The atoms of lithium, sodium and potassium all have only one electron in their outermost shells and all of these are metals. All of these react with water to form alkalies.
(b) The atoms of helium and neon have their outermost shells completely filled. Helium has its first shell completely filled, while neon has its first and second shells (K and L) completely filled.

Question 5
In the modern periodic table, which are the metals among the first ten elements ?
Answer:
The first ten elements in modern periodic table are hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon. Out of these, lithium, beryllium and boron are metals, because they have 1, 2 and 3 electrons respectively in their outermost shells.

Question 6
By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristics ?
Ga, Ge, As, Se, Be
Answer:
Beryllium (Be). In the periodic table, the elements placed on the left show maximum metallic characteristics. Since beryllium occupies the most left position in comparison to other elements, hence it shows maximum metallic characteristics.

NCERT Solutions for Class 10 Science Chapter 5 Textbook Chapter End Questions

Question 1
Which of the following statements is not a correct statement about the trends wlien going from left to right across the periods of Periodic Table.
(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c) The atoms lose their .electrons more easily.

Question 2
Element X forms a chloride with the formula XCl₂, which is solid with a high melting point. X would most likely to be in the same group of the periodic table as
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) Mg

Question 3
Which element has
(a) two shells, both of which are completely filled with electrons ?
(b) the electronic configuration 2, 8, 2 ?
(c) a total of three shells, with four electrons in its valence shell ?
(d) a total of two shells with three electrons in its valence shell. v
(e) twice as many electrons in its second shell as in its first shell ?
Answer:
(a) Neon (2, 8)
(b) Magnesium
(c) Silicon (2, 8, 4)
(d) Boron (2, 3)
(e) Carbon (2, 4)

Question 4
(a) What property do all elements in the same column of the Periodic Table as boron have in common ?
(b) What property do all elements in the same column of the Periodic Table . as fluorine have in common ?
Answer:
(a) Elements in the same column or group as boron have valency of three and have three valence electrons.
(b) Elements in the same column or group as fluorine form acidic oxides and have seven electrons in their outermost shells and have valency of one.

Question 5
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element ?
(b) To which of the following elements would it be chemically similar ? (Atomic numbers are given in parentheses.)
N (7), F (9), P (15), Ar (18)
Answer:
(a) The atomic number of the given element is 2 + 8 + 7(= 17).
(b) It would be chemically similar to fluorine [F(9)] because its electronic configuration is 2, 7.

Question 6
The positions of three elements A, B and C in the periodic table are shown below :
(a) State whether A is a metal or non-metal.

(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B ?
(d) Which type of ion, cation or anion, will be formed by element A ?
Answer:
(a) Since the valency of group 17 elements is 1 and all these elements accept electrons, thus A is a non-metal.
(b) C is less reactive than A because as we move down in a group, the reactivity of non-metals increases.
(c) C is smaller in size than B because B and C both are related to the same period and the size decreases as one moves from left to right in a period.
(d) A will form anion because it is a non-metal.

Question 7
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write the electronic configuration of these two elements. Which of these will be more electronegative ? Why ?
Answer:
Electronic configuration of nitrogen -2,5
Electronic configuration of phosphorus = 2, 8, 5
Nitrogen will be more electronegative because outermost shell is nearer to nucleus and therefore nucleus will attract electrons more strongly. In a group of the periodic table, electron attracting tendency decreases as we move from top to bottom.

Question 8
How does the electronic configuration of an atom relate to its position in the Modern Periodic Table ?
Answer:
Modern periodic table is based on the atomic number and atomic number is directly related to the electronic configuration. One can find the group number and period number of an element on the basis of electronic configuration. For example, if an element has 1 or 2 electrons in its outermost shell, then it would belong to group 1 or group 2. And if it has 3 or more electrons in its outermost shell, then it would belong to group 10 4- the number of electrons in the outermost shell.

All the alkali metals have one electron in their outermost shell, so they are placed in group 1. Thus, all the group 2 elements have 2 electrons in their outermost shell. In group 15 elements, there are 5 electrons in their outermost shell. Similarly, the number of shells in an element indicates its period number. For example, the atomic number of magnesium is 12 and its electronic configuration is 2, 8, 2. Thus it is an element of 3rd period.

Question 9
In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic number 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium ?
Answer:
The electronic configuration of elements with :
Atomic number 12 = 2, 8, 2
Atomic number 19 = 2, 8, 8, 1
Atomic number 20 = 2, 8, 8, 2
Atomic number 21 = 2, 8, 9, 2
Atomic number 38 = 2, 8, 18, 8, 2
Elements with atomic number 12 i.e., magnesium (Mg) and 38 i.e., strontium (Sr) will have similar physical and chemical properties as element with atomic numbers 20 i.e., calcium (Ca).

Question 10
Compare and contrast the arrangement of elements in Mendeleev’s Periodic Table and the Modern Periodic Table.
Answer:

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

Periodic classification of elements: Needs for classification, Modern Periodic table, gradation in properties, valency, atomic number, metallic and non-metallic properties.

Formulae Handbook for Class 10 Maths and Science

Question 1
Did Dobereiner’s triads also exist in the columns of Newlands Octaves? Compare and find out.
Solution:
Yes, Dobereiner’s triads also exist in the columns of Newlands Octaves. For example, the second column of Newlands classification has the elements lithium (Li), sodium (Na) and Potassium (K), which constitute a Dobereiner’s triad.

Question 2
What were the limitations of Dobereiner’s classification?
Solution:
All the known elements could not be arranged in the form of triads. For very low mass or for very high mass elements, the Dobereiner’s triads are not applicable. Take the example of F, Cl and Br. Atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br. As the techniques for measuring atomic masses accurately, improved, the Dobereiner’s triad was unable to remain strictly valid.

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Question 3
What were the limitations of Newland’s Law of Octaves?
Solution:
It was not valid for elements that had atomic masses higher than Ca.When more elements were discovered, such as elements from the noble gases such as He, Ne, Ar, they could not be accommodated in his table.

Download NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements PDF

Question 4
Use Mendeleev’s periodic table to predict the formulae for the oxides of the following elements: K, C, Al, Si, Ba.
Solution:
K₂O – Potassium Oxide
CaO – Calcium Oxide
Al₂O₃ – Aluminium Oxide
SiO₂ – Silicon dioxide
BaO – Barium Oxide.

Question 5
Besides gallium, which other elements have been left by Mendeleev in his periodic table, since the time they were discovered? (Any two)
Solution:
Scandium and Germanium.

Question 6
What were the criteria used by Mendeleev in creating his periodic table?
Solution:
He observed the relationship between the atomic masses of the elements and their physical properties. Among chemical properties, he concentrated on the compounds formed by elements with oxygen and hydrogen.

Question 7
Why do you think, the noble gases are placed in a separate group?
Solution:
Due to its inert and low concentration in our atmosphere, they could be placed in a new group without disturbing the existing order.

Question 8
How could the Modern periodic table remove various anomalies of Mendeleev’s periodic table?
Solution:
When the elements are arranged according to their atomic numbers on the basis of modern periodic law, all the anomalies (defects) of Mendeleev’s classification disappear. For example, Position of isotopes: All the isotopes of an element have the same number of protons, so their atomic number is also the same. Since all the isotopes of an element have the same atomic number, they can be put at one place in the same group of the periodic table.

Question 9
Name two elements, which you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Solution:
Calcium and Beryllium are the elements that will show chemical reactions similar to magnesium. This is because beryllium and calcium belong to the same group of periodic table as magnesium. All of them have similar electronic configurations with 2 valence electrons each.

Question 10
Name:
a. Three elements that have a single electron in their outermost shell.<
b. Two elements that have two electrons in their outermost shell.<
c. Three elements with filled outermost shell.
Solution:
a. Three elements that have a single electron in their outermost shell are:
1. Lithium
2. Sodium
3. Potassium
b. Two elements that have two electrons in their outermost shell are:
1. Magnesium
2. Calcium
c. Three elements with filled outermost shell are:
1. Argon
2. Helium
3. Neon.

Question 11
a. Lithium, sodium, potassium are metals that react with water to liberate hydrogen. Is there any similarity in the atoms of these elements?
b. Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common
Solution:
a. These elements are alkali metals and they have 1 valence electron in their outermost shell and are therefore very unstable and reactive.
b. These elements each have full outermost subshell, which results in high stability. They only react with other elements in extreme circumstances, the trait for which they are named.

Question 12
In the Modern periodic table, which are the metals among the first ten elements?
Solution:
The metals are Lithium and Beryllium.

Question 13
By considering their position in the periodic table, which one of the following elements would you expect to have maximum metallic characteristic?
Ga, Ge, As, Se, Be
Solution:
Beryllium

Question 14
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of the periodic table?<
(i)The elements become less metallic in nature
(ii) The number of valence electrons increases
(iii) The atoms lose their electrons more easily
(iv) The oxides become more acidic
Solution:
(iii) The atoms lose their electrons more easily – Incorrect statement.

Question 15
Element X forms a chloride with the formula XCl₂, which is a solid with a high melting point. X would most likely be in the same group of the periodic table as
a. Na b. Mg c. Al d. Si
Solution:
b. Mg

Question 16
Which element has?
a.Two shells, both of which are completely filled with electrons?<
b. The electronic configuration of 2,8,2?<
c. A total of three shells, with four electrons in its valence shell?<
d. A total of two shells, with three electrons in its valence shell?<
e. Twice as many electrons in its second shell, as in its first shell?
Solution:
a. Neon (2,8)
b. Magnesium (2,8,2)
c. Silicon (2,8,4)
d. Boron (2,3)
e. Carbon (2,4)

Question 17
What property do all elements in the same column of the periodic table as fluorine have in common?
Solution:
These elements all have 7 electrons in their outermost shells and these often exist as salts, combined with elements from the Alkali metal group.

Question 18
An atom has electronic configuration 2,8,7.
i. What is the atomic number of this element?
ii. To which of the following elements would it be chemically similar<
N (7) F (9) P (15) Ar (18)
Solution:
(i) Chlorine – 17
(ii) F (9)

Question 19
Which type of ion, cation or anion, will be formed by element A?
Solution:
C is less reactive than A
“C” will be smaller in size than “B” as the atomic size decreases as we go across a period.
Anion will be formed by element A

Question 20
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Solution:
Electronic configuration – Nitrogen – 2s² 2p³ and Phosphorus – 1s² 2s² 2p⁶ 3s² 3p³. Nitrogen will be more electronegative; this is because its atom has small size due to which the attraction of its nucleus for the incoming electron is more.

Question 21
How does the electronic configuration of an atom relate to its position in the Modern periodic table?
Solution:
The electronic configuration of an atom increases in the outermost valence shell which relates to its position in the Modern periodic table.

Question 22
In the Modern periodic table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium?
Solution:
The atomic number of calcium is 20, so its electronic configuration is 2, 8, 8, 2. Thus, calcium has 2 valence electrons (in its outermost shell). Now, the element which has 2 valence electrons, will have physical and chemical properties resembling to that of calcium. The electronic configuration of element having atomic number 12 is 2, 8, 2. It has 2 valence electrons just like calcium. So, the element having atomic number 12 will have physical and chemical properties resembling that of calcium.

Question 23
Where do you think should hydrogen be placed in the Modern periodic table?
Solution:
Hydrogen element has been placed at the top of group 1, above the alkali metals because the electronic configuration of hydrogen is similar to those of alkali metals.

Multiple Choice Questions (MCQs) [1 Mark each]

Question 1.
Which of the following is the outermost shell for elements of 2nd period? [NCERT Exemplar]
(a) K-shell
(b) L-shell
(c) M-shell
(d) N-shell
Answer:
(b) The elements of 2nd period involve the filling of 2nd shell, i.e. Z-shell. Because in period 2, there are two shell, K and L

Question 2.
An element which is an essential constituent of all organic compounds belongs to [NCERT Exemplar]
(a) group 1
(b) group 14
(c) group 15
(d) group 16
Answer:
(b) Constituent of all organic compounds is carbon. It belongs to group 14.

Question 3.
Which one of the following elements exhibit maximum number of valence electrons? [NCERT Exemplar]
(a) Na
(b) Al
(c) Si
(d) P
Answer:
(d) Na (group 1) has one, A1 (group 13) has three (13 -10), Si (group 14) has four (14-10) and P (group 15) has five (15 – 10) valence electrons. Therefore, P has maximum number of valence electrons, i.e.3 (maximum among the given elements).

Question 4.
Which of the given elements A, B, C, D and E with atomic number 2, 3, 7, 10 and 30 respectively belong to the same period? [NCERT Exemplar]
(a) A, B, C
(b) B, C, D
(c) A, D, E
(d) B, D, E
Answer:
(b) 2nd period contains elements with atomic number 3(Li), 7(N), 10(Ne). Since, 2nd period has elements having atomic number 3 to 10.

Question 5.
The elements A, B, C, D and E have atomic number 9, 11, 17, 12 and 13 respectively. Which pair of elements belong to the same group? [NCERT Exemplar]
(a) A and B
(b) B and D
(c) A and C
(d) D and E
Answer:
(c)Electronic configuration of A (atomic number = 9) is 2, 7.
Electronic configuration of B (atomic number = 11) is 2, 8,1.
Electronic configuration of C (atomic number = 17) is 2, 8, 7.
Electronic configuration of D (atomic number =12) is 2, 8, 2.
Electronic configuration of E (atomic number =13) is 2, 8, 3.
Elements which differ in atomic number by 8, i.e. 9 (A, fluorine) and 17 (C, chlorine) lie in the same group, i.e. group 17 (halogen).

Question 6.
In Mendeleev’s periodic table, gaps were left for the elements to be discovered later. Which one of the following elements found a place in the periodic table later? [NCERT Exemplar]
(a) Germanium
(b) Chlorine
(c) Oxygen
(d) Silicon
Answer:
(a) Mendeleev’s left some gaps in the periodic table for those elements which were not known at that time. Germanium element found a place in the periodic table later and Mendeleev’s predictions were found to be remarkably correct

Question 7.
Which of the following are the characteristics of isotopes of an element? [NCERT Exemplar]
(i) Isotopes of an element have same atomic masses.
(ii) Isotopes of an element have same atomic number.
(iii) Isotopes of an element show same physical properties.
(iv) Isotopes of an element show same chemical properties.
(a) (i), (iii) and (iv)
(b) (ii), (iii) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(d) Isotopes of an element have same atomic number and show same chemical properties.

Question 8.
Which of the following elements would lose an electron easily? [NCERT Exemplar]
(a) Mg
(b) Na
(c) K
(d) Ca
Answer:
(c) Electronic configuration of Mg (atomic number = 12) 2, 8, 2
Electronic configuration of Na (atomic number = 11)2, 8, 1
Electronic configuration of K (atomic number=19) 2, 8, 8, 1
Electronic configuration of Ca (atomic number =20) 2, 8, 8, 2
From the above electronic configurations, it is clear that K and Na will lose electron easily to achieve stable configuration. But out of K and Na, K will lose electron more easily because the force of, attraction on valence electron of K is least among the given elements.

Question 9.
Which among the following elements has the largest atomic radii? [NCERT Exemplar]
(a) Na
(b) Mg
(c) K
(d) Ca
Answer:
(c) Atomic radius increases on moving down in a group. Na and K are in the same group and K is below Na, so K will have higher atomic radius, i.e. K > Na. In a period on moving left to right, atomic radius decreases. Since, K and Ca are in the same period and K is in 1st group and Ca is in 2nd group, so atomic radius of K will be more than Ca, i.e. K > Ca.
Also Na and Mg are in the same period, but Na belongs to 1st group and Mg belongs to 2nd group, so atomic radius of Na is more than Mg, i.e. Na > Mg. Thus, if we take all these together we get K > Na >Mg and K > Ca > Mg. Hence, we can say that the atomic radius of K is largest.

Question 10.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic table? [NCERT]
(a) The elements become less metallic in nature
(b) The number of valence electrons increases
(c) The atoms lose their electrons more easily
(d) The oxides become more acidic
Answer:
(c) On moving from left to right, the atomic number increases and hence, the nude charge increases. With the increase of nudear charge, the force binding the electron increases so the atom lose the electrons with more difficulty, not easily.

Question 11.
Which one of the following depicts the correct representation of atomic radius (r) of an atom? [NCERT Exemplar]

(a) (i) and (i)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) Atomic radius is the distance between nucleus and outermpst shell [consisting electron(s)]. Hence (ii) and (iii) are the correct representations.

Question 12.
Which of the following statement(s) about the modern periodic table are incorrect?
(i) The elements in the modern periodic table are arranged on the basis of their decreasing atomic numbers.
(ii) The elements in the modern periodic table are arranged on the basis of their increasing atomic masses.
(iii) Isotopes are placed in adjoining group(s) in the periodic table.
(iv) The elements in the modern periodic table are arranged on the basis of their increasing atomic number. [NCERT Exemplar]
(a) Only (i)
(b) (i), (ii) and (iii)
(c) (i), (ii) and (iv)
(d) Only (iv)
Answer:
(b) Only statement (iv) is correct. All the elements in the modern periodic table are arranged on the basis of their increasing atomic number. All the isotopes can be placed at one place in the same group of the periodic table.

Question 13.
The element with atomic number 14 is hard and forms acidic oxide and a covalent halide. To which of the following categories does the element belong? [NCERT Exemplar]
(a) Metal
(b) Metalloid
(c) Non-metal
(d) Left-hand side element
Answer:
(c) Its outermost shell has 4 electrons. So, it is a non-metal. Non-metal forms acidic oxide and by sharing of electrons with halogen, it forms covalent halide.

Question 14.
Arrange the following elements in the order of their decreasing metallic character Na, Si, Cl, Mg, Al. [NCERT Exemplar]
(a) Cl > Si > Al > Mg > Na
(b) Na > Mg > Al > Si > Cl
(c) Na > Al > Mg > Cl > Si
(d) Al > Na > Si > Ca > Mg
Answer:
(b) Metals lie on the extreme left side of the periodic table. Metallic character decreases from left to right in a period. Na, Si, Cl, Mg, Al belong to same period in the order Na, Mg, Al, Si, Cl. On moving in a i period from left to right, the metallic character decreases. Thus, the order of decreasing metallic character is: Na > Mg > Al > Si > Cl

Question 15.
Which of the following set of elements is written in order of their increasing metallic character? [NCERT Exemplar]
(a) Be, Mg, Ca
(b) Na, Li, K
(c) Mg, Al, Si
(d) C, O, N
Answer:
(a) Metallic character increases as we go down in a group.

Question 16.
Which one of the following does not increase while moving down the group of the periodic table? [NCERT Exemplar]
(a) Atomic radius
(b) Metallic character
(c) Valence electrons
(d) Number of shells in an element
Answer:
(c) Elements in each group has some number of valence electrons hence have same valency and thus exhibit similar chemical properties.

Question 17.

The elements X, Y and Z are shown in a portion of periodic table. What would be the representations of ionic forms of X and Z respectively?
(a) X^– and Z⁺
(b) X⁺ and Z^–
(c) X^2- and Z²⁺
(d) X²⁺and Z^2-
Answer:
(a) X having 1 electron less than that of Y, which is a noble gas (having complete octet), will tend to gain 1 electron while Z will tend to lose 1 electron to achieve complete octet configuration. Hence, they will have ionic formula as X^– and Z⁺.

Question 18.
which of the following elements will form an acidic oxide? [NCERT Exemplar]
(a) An element with atomic number = 7
(b) An element with atomic number = 3
(c) An element with atomic number = 12
(d) An element with atomic number = 19
Answer:
(a) Non-metals form acidic oxides in general. Non-metals have 4 to 8 electrons in the outermost shell. The electronic configuration of given elements are (a) i.e. 7 = 2, 5 (b) i.e. 3 =2, 1 (c) i.e. 12 =2, 8, 2 (d)i.e. 19=2, 8, 8,1.
So, element with atomic number = 7 (electronic configuration = 2, 5) with is non-metal (N) and it will form an acidic oxide. Rest three elements with atomic- numbers, 3 (Li), 12 (Mg) and 19 (K) are metals and hence, form basic oxides.

Question 19.
what type of oxide would Eka-aluminium form? [NCERT Exemplar]
(a) EO₃
(b) E₃O₂
(C) E₂O₃
(d) EO
Answer:
(c) Gallium has a valency of 3. Hence, it forms an oxide having molecular formula E₂O₃. In other options, valency of E is not 3.

Question 20.
The diagram given below shows the position of elements in a portion of the periodic table. Ionic compound is formed between ………. and ………

(a) A and B
(b) B and E
(c) C and D
(d) D and E
Answer:
(D) being in group IIA is most electropositive among given elements while E being in group VIIA is most electronegative among the given elements. Hence, both of these will form ionic compound, more readily than the other given elements.

Question 21.
Element X forms a chloride with formula, XCl₂, which is a solid with a high melting point. X would most likely to be in the same group of the periodic table as [NCERT]
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) The formula of chloride is XCl₂, that means the valency of the element X is 2. The element having valency 2 will be present in group 2. Out of the given choices magnesium (Mg) belongs to group 2.

Question 22.
The adjacent diagram represents the arrangement of the atoms of an element (having valency = 4) forming giant covalent network.

Identify the heavy metal which belongs to same group as the element discussed above.
(a) As
(b) Bi
(c) Pb
(d) Hg
Answer:
(c) The element discussed in diagram is carbon. Carbon (C) and lead (Pb) both belong to group IVA.

Question 23.

Arrange the elements, represented by alphabets p, q, r, s and t in the above shown outline of periodic table, in increasing order of their valency.
(a) t < q < r < s < p
(b) s < t < q < r < p
(c) s < q < t < r < p
(d) q < s < t < p < r
Answer:
(c) p(group IVA), q(group VIIA), r(group IILA) s(group VIII) and t(group IIA) has valency 4, 1, 3, 0 and 2 respectively hence the correct order of increasing valency is: s(0) < q(1) < t(2) < r(3) < p(4)

Question 24.

The diagram given above represents outline of the periodic table. The alphabets p, q, r, s and t represent elements. Which one of the following pairs of alphabets represents elements which consists of same number of shells in their atom?
(a) p and q
(b) r and t
(c) p and s
(d) g and s
Answer:
(d) Elements q and s belong to same period of the periodic table and hence, will have same number of shell.

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements (Hindi Medium)

Class 10 Science Periodic Classification of Elements Mind Map

Dobereiners’s Triads

• Dobereiner found that when elements are arranged into groups of three in the order of their increasing atomic mass, the atomic mass of the element; which comes in the middle, is the arithmetic mean of rest of the two.
• He arranged three elements in one group which is known as Dobereiner’s Triads. For e.g.:
  (Li) 7.0 (Na) 23.0 (K) 39.0 (Ca) 40.0 (Sr) 87.5 (Ba) 137.0
• Here atomic mass of sodium is equal to arithmetic mean of atomic masses of lihtium and potassium. Similarly, atomic mass of strontium is equal to arithmetic mean of atomic masses of calcium and barium

Limitations of Doberiner’s Triads

• He could identify only a few such triads and so the law could not gain importance.
• In the triad of Fe, Co, Ni, all the three elements have a nearly equal atomic mass and thus it does not follow the above law.

Mendeleev’s Periodic Classification

• Mendeleev periodic law, states that ‘the properties of elements are the periodic function of their atomic masses’.
• Mendeleev’s periodic table contains vertical columns called ‘groups’ and horizontal rows called ‘periods’.

Characteristics of the Mendeleev’s Periodic Table

• The elements are arranged in vertical rows called groups and horizontal rows called periods.
• There are eight groups indicated by Roman Numerals I, II, III, IV, V, VI, VII, VIII. The elements belonging to first seven groups have been divided into sub-groups designated as A and B on the basis of similarities. Group VIII consists of nine elements arranged in three triads.
• There are six periods (numbered 1,2,3,4,5 and 6).

Limitation of Mendeleev’s Periodic Table

• Some elements in Mendeleev’s Table have not been arranged in the increasing order of their atomic masses. For example, Co and Ni.
• Hydrogen forms similar compounds as Group 1 elements. However, it also forms similar diatomic molecules as Group 7 elements (H2, F2, C12, Br2, 12). Hence, it could not be assigned a fixed position in the table.
• Isotopes posed a challenge to Mendeleev’s table. For example, Cl has two major isotopes – Cl-35 and Cl-37.

Merits of Mendeleev’s Periodic Table

• Mendeleev left some blank spaces in his periodic table in order to place the elements having similar properties in the same group.
• Mendeleev predicted the discovery of some elements and named them as eka-boron, eka- aluminium and eka-silicon.
• One of the strengths of Mendeleev, s periodic table was that, when inert gases were discovered they could be placed in a new group without disturbing the existing order.

Newlands’ Law of Octaves

• According to this law “if elements are arranged by the increasing order of their atomic masses, property of every eighth element repeats.”
• The arrangement of elements in Newlands’ Octave resembles the musical notes.

Limitation of Newlands’ Octaves

• Law of Octaves could be valid up to calcium only; as after calcium, elements do not obey the rules of Octaves.
• It was assumed by Newlands that only 56 elements existed in nature and no more elements would be discovered in the future
• More than one element had to be placed in some of the groups; in order to place the elements having similar properties in one group. But in order to do so, he also put some dissimilar elements in same group.
• Iron; which has similar property as cobalt and nickel, was placed far from them.
• Cobalt and nickel were placed in the group with chlorine and fluorine in spite of having different properties.

Modern Periodic Table
In 1913, Henry Moseley showed that atomic number of an element is a more fundamental property than its atomic mass. According to this law “’properties of elements are a periodic function of their atomic number”.

Trends in Modern Periodic Table

Position of Elements in the Modern Periodic Table
Position of Elements

Elements are placed in groups according to the number of valence electrons and placed in periods according to the number of shells present in them.

Helium has valence electrons equal to 2, but it is placed in group number 18 because it is a noble gas and has completely filled outermost shell.

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CBSE Class 10 Maths Notes Chapter 14 Statistics Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 14 Statistics. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 14 Statistics

MEAN (AVERAGE): Mean [Ungrouped Data] – Mean of n observations, x₁, x₂, x₃ … x_n, is

MEAN [Grouped Data]: The mean for grouped data can be found by the following three methods:
(i) Direct Mean Method:

Class Mark = \(\frac { Upper\quad Class\quad Limit+Lower\quad Class\quad Limit }{ 2 }\)

Note: Frequency of a class is centred at its mid-point called class mark.

(ii) Assumed Mean Method: In this, an arbitrary mean ‘a’ is chosen which is called, ‘assumed mean’, somewhere in the middle of all the values of x.

…[where d_i = (x_i – a)]

(iii) Step Deviation Method:

….. [where \({ u }_{ i }=\frac { { d }_{ i } }{ h }\) , where h is a common divisor of d_i]

MEDIAN: Median is a measure of central tendency that gives the value of the middle-most observation in the data.

…where[l = Lower limit of median class; n = Number of observations; f = Frequency of median class; c.f. = Cumulative frequency of preceding class; h = Class size]

(iii) Representing a cumulative frequency distribution graphically as a cumulative frequency curve, or an ogive of the less than type and of the more than type. The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for this data.

Mode:
(i) Ungrouped Data: The value of the observation having maximum frequency is the mode.
(ii) Grouped Data:

…where[l = Lower limit of modal class; f₁ = Frequency of modal class; f₀ = Frequency of the class preceding the modal class; f₂ = Frequency of the class succeeding the modal class; h = Size of class interval. c.f. = Cumulative frequency of preceding class; h = Class size]

Mode = 3 Median – 2 Mean
Median = \(\frac { Mode+2Mean }{ 3 }\)
Mean = \(\frac { 3Median-Mode }{ 2 }\)

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CBSE Class 10 Maths Notes Chapter 15 Probability Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 15 Probability. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 15 Probability

Probability: It is the numerical measurement of the degree of certainty.

• The theoretical probability associated with an event E is defined as “If there are ‘n’ elementary events associated with a random experiment and m of these are favourable to the event E then the probability of occurrence of an event is defined by P(E) as the ratio \(\frac { m }{ n }\) “.
  
• If P(E) = 1, then it is called a ‘Certain Event’.
• If P(E) = 0, then it is called an ‘Impossible Event’.
• The probability of an event E is a number P(E) such that: 0 ≤ P(E) ≤ 1
• An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.
• For any event E, P(E) + P(\(\bar { E }\)) = 1, where \(\bar { E }\) stands for ‘not E’. E and \(\bar { E }\) is called complementary events.
• Favourable outcomes are those outcomes in the sample space that are favourable to the occurrence of an event.

Sample Space
A collection of all possible outcomes of an experiment is known as sample space. It is denoted by ‘S’ and represented in curly brackets.
Examples of Sample Spaces:
A coin is tossed = Event
E₁ = Getting a head (H) on the upper face
E₂ = Getting a tail (T) on the upper face
S = {H, T}
Total number of outcomes = 2

Two coins are tossed = Event = E
E₁ = Getting a head on coin 1 and a tail on coin 2 = (H, T)
E₂ = Getting a head on both coin 1 and coin 2 = (H, H)
E₃ = Getting a tail on coin 1 and a head on coin 2 = (T, H)
E₄ = Getting a tail on both, coin 1 and coin 2 = (T, T)
S = {(H, T), (H, H), (T, H), (T, T)}.
Total number of outcomes = 4

NOTE: In probability the order in which events occur is important
E₁ & E₃ are treated as different outcomes.

Important Tips

• Coin: A coin has two faces termed a Head and Tail.
• Dice: A dice is a small cube that has between one to six spots or numbers on its sides, which is used in games.
• Cards: A pack of playing cards consists of four suits called Hearts, Spades, Diamonds, and Clubs. Each suite consists of 13 cards.

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Babulal Nagarmal Satnalika Foundation Scholarship 2022-2023: Babulal Nagarmal Satnalika Foundation Scholarship offered by the Babulal Nagarmal Satnalika Foundation (BNSF), West Bengal. Babulal scholarship is offered to the economically backward students up to class 12th to complete their education. This scholarship provides financial support to underprivileged students to continue their studies. The BNS Foundation organized this scholarship to provide financial support to talented students. This helps students who cannot continue their education due to financial problems.

The application form for this scholarship is available online on the official website of the BNS Foundation. There is no deadline for the application submission as it is open throughout the year. So, all the eligible students can fill and apply for this scholarship at any time throughout the year. Babulal Scholarship online apply 2023 last date is 31st December 2019. Read this article below to know the complete information about the Babulal Nagarmal Satnalika Foundation Scholarship.

Babulal Nagarmal Satnalika Foundation Scholarship Details

The Babulal Nagarmal Satnalika Foundation Scholarship can be applied in both the online and offline mode. This scholarship is available for the underprivileged students across India. So that they can continue their education and fulfill their dreams. This scholarship aims to encourage meritorious and needy students. So that they can continue their education without financial hassle. Babulal Nagarmal Satnalika Foundation Scholarship 2019 last date is 31st December 2019.

BNS Foundation is a non-profitable and non-government organization (NGO). It aimed to improve the learning and education within the underprivileged children of our society. The foundation works for the welfare and progress of the children. It supports them with financial assistance on a monthly basis. So that they will not deprive education in today’s world.

The BNS Foundation founded in 2011. It was named after the Late Shri Babulal Nagarmal Satnalika. He was from a small town of Raniganj, West Bengal and became a successful businessman. The application form is available throughout the year on the BNSF Portal. So, the eligible meritorious and needy students can apply for this scholarship.

Babulal Nagarmal Satnalika Foundation Scholarship Overview

Babulal Nagarmal Satnalika Foundation Scholarship 2020 Eligibility Criteria

Students must fulfill the following eligibility criteria to apply for the BNSF Scholarship.

• The student must be a resident of India.
• Both boys and girls students can apply for this scholarship.
• The students up to class 12th from any recognized school/college in India are eligible.
• The student must have scored more than 80% of marks in the previous examination.
• The student’s family income should not be more than 60,000 per annum.

Babulal Nagarmal Satnalika Foundation Scholarship Benefits

The selected students will receive the tuition fees for 10 months in a year.

Babulal Nagarmal Satnalika Foundation Scholarship Selection Criteria

Students who fulfill the eligibility criteria will get selected for this scholarship.

Babulal Nagarmal Satnalika Foundation Scholarship Application Process

Students can apply for this scholarship in both online and offline mode. Refer to the following points to apply online and offline for the BNSF Scholarship.

How to Apply Online?

• Go to the official website of the BNS Foundation.
• Click on the ‘Apply Online’ link that appeared on the homepage.
• The application form will appear on your screen to fill in the details.
• Enter the personal, educational, and bank details in the application form.
• Fill the correct and valid information in the application. Otherwise, you may not get contacted due to the provided wrong information.
• Upload the scanned copies of the necessary documents in .jpeg format.
• Submit the BNSF Scholarship application form along with the necessary documents.

How to Apply Offline?

• Go to the official website of the BNS Foundation.
• Download the application form that appeared on the homepage of the portal.
• Enter the personal, educational, and bank details as per the document.
• Fill the valid and correct information by using only the black or blue ballpoint pen.
• Make sure to fill the application form only in capital letters.
• Send the filled application form along with the documents by post/courier to the below address:
  Babulal Nagarmal Satnalika Foundation
  C/o. Mr. Devi Prasad Satnalika
  No. 92, M.G. Road, Raniganj – 713347,
  Dist: Burdwan, West Bengal

Documents Required for Babulal Nagarmal Satnalika Foundation Scholarship

The following documents have to be submitted to apply for the BNSF Scholarship.

• Marksheets and certificates of the previous qualifying examination
• Address proof (Birth certificate or ration card or electricity bill)
• Parents income certificate
• Recent passport size photograph of the student

Babulal Nagarmal Satnalika Foundation Scholarship Result

The BNSF Scholarship result will be released by BNSF every year in February month. The selected students will be informed about the result through the registered email Id and mobile number. However, BNSF has the power to withdraw the given scholarship at any time. In case of finding any misbehavior from the student.

Babulal Nagarmal Satnalika Foundation Contact Number

For any queries or complaints, students can contact the authority at the following details

• Phone No: +91-7679635152
• Email: niraj.satnalika@bnsatnalikafoundation.org

FAQ’s on Babulal Nagarmal Satnalika Foundation Scholarship

Question 1.
What is the Babulal Nagarmal Satnalika Foundation Scholarship?

Answer:
BNSF Scholarship offered by the Babulal Nagarmal Satnalika Foundation (BNSF), West Bengal. This scholarship offered to the economically backward students up to class 12th. So that they can complete their education without financial problems.

Question 2.
Is the Babulal Nagarmal Satnalika Foundation Scholarship available to all the state students?

Answer:
Yes, the BNSF Scholarship is available to the underprivileged students across India.

Question 3.
What is the educational requirement to apply for the BNSF Scholarship?

Answer:
The students up to class 12th from any recognized school/college in India can apply for this scholarship.

Question 4.
What is the deadline to apply for the Babulal Nagarmal Satnalika Foundation Scholarship?

Answer:
There is no deadline to apply for this scholarship as it is open throughout the year.

Question 5.
Is the Babulal Nagarmal Satnalika Foundation Scholarship applied in both online and offline mode?

Answer:
Yes, the BNSF Scholarship can be applied in both the online and offline mode.

We hope this article will help you to get more information about the Babulal Nagarmal Satnalika Foundation Scholarship. If you have queries related to BNSF Scholarship, then leave it in the comment box.

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

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Scholarships for Students 2023-2024: Details of scholarships for students 2023-2024 are available here. Students who are eligible and willing to continue their studies could check with the scholarship registrations here. Scholarships help students to get further education without any financial restrictions. There are many scholarships available in India. Some are provided by individual states, some scholarships are based on different categories.

Few scholarships are available only for girl students. Scholarships for students after 10 and 12 classes are also available. Here in this article, we will be discussing all types of scholarships provided for students along with scholarship under Karnataka state, popular and best scholarships in India.

There are scholarships available for classes 1 to 10 students, such as pre-matric and post-matric scholarships for SC, ST, OBC and General category students. Let us discuss all types of scholarships in this article.

• State Wise Scholarships
• Category Wise Scholarships
• Scholarships After Class 12
• Scholarships After Class 10
• Scholarship for Girls
• Scholarships Under Karnataka State
• Scholarships Under West Bengal State
• Scholarships Under Kerala State
• Scholarships for Pursuing Students
• Scholarships Under Minority and Backward Community
• Education Loans in India
• Fellowships in India
• Other Scholarships in India
• Best Scholarships in India

State Wise Scholarships 2023

A state-wise scholarship is a widespread category of scholarships that contains the list of scholarships proposed by different state governments and other private organizations for the particular states. candidate can get candidate state-specific scholarship below, with respect to the class, caste or gender to which the candidate belongs. Given hereunder is a collection of all related information about scholarships of various Indian states and UT’s. Depending on the belonging state, the candidate can see for scholarships to finance candidate further education.

Go Through the list of Scholarships are as follows:

• AP ePASS
• Bihar Scholarship
• ePASS Karnataka
• Goa Education Trust Scholarship
• Kerala Scholarship – DCE Scholarship
• Himachal Pradesh Scholarship – HP ePASS Scholarship Portal
• Jammu and Kashmir Scholarship –  JK Scholarshipalso Known as PMSSS
• Maharashtra Scholarship – MahaDBT
• Madhya Pradesh Scholarship – MP Scholarship
• Odisha Scholarship
• Punjab Scholarship
• Rajasthan Scholarship
• Sanskrit Scholarship – Delhi Scholarship
• Telangana ePASS
• UP Scholarship 2023 – Uttara Pradesh Scholarship
• Uttarakhand Scholarship
• West Bengal Scholarship – WB Scholarship for OBC, SC, ST Students
• WBMDFC Scholarship

Category Wise Scholarships 2023

The category wise or SC, ST, OBC Scholarships is governed by the Government of India and the state governments to support students continue their studies at other levels. The centrally funded scholarships are organised by the Government of India and distributed by different states while the state-funded scholarships are organised and disbursed by particular state governments only. An SC, ST, OBC Scholarship offers proper financial support to SC/ST/OBC candidates at both pre-matric and post-matric levels of education.

In this section, we will compile all the important scholarships for SC/ST/OBC categories. Also, we have categorised scholarships for Class 10 and 12 students. This includes a flash of eligibility requirements, types of awards, application procedure and selection method and other conditions. This article will provide the candidate beforehand about each category scholarship.

• e-Kalyan Scholarship
• Google Scholarship
• Haryana Scholarship
• Scholarship for 10th Passed Students
• Scholarship for 12th Passed Students
• ST Scholarship
• Scholarship Form
• Merit Scholarship
• OBC Scholarship
• Commonwealth Scholarship
• Udacity Scholarship
• SC Scholarship
• Scholarship Registration
• Samunnathi Scholarship
• Vidyadhan Scholarship
• IOCL Scholarship
• Medhavi Scholarship – Digital India Programme
• Aditya Birla Scholarship – IITs/IIMs/BITs/XLRI
• GATE Scholarship – Ph.D./M.Tech
• National Fellowship for OBC
• UGC Scholarship – Higher Education

Scholarships After Class 12

Students who have passed their 12th class and want to develop their future in the path chosen career. For them here are the list of scholarships, who are unable to support their education financially due to financial crisis at their family or low income. After class 12th it is a turning point for each student since they must have set some goals for their upcoming life. Here are the scholarships for students after 12th that can financially benefit them in pursuing their dream career.

Towards supporting meritorious and destitute students in persevering their additional studies, the Government of India and the management of different states administer different scholarships after 12th class. While the Centre and state government administer their own scholarship projects separately. They collectively run many scholarship plans where the scheme is supported by the central government and distributed by the state government. Here is a collection of all the government-funded scholarships that you must not avoid.

• Anant Scholarship
• NSF Scholarship
• CCRT Scholarship
• 23rd NEST Scholarship 2023
• Next Genius Scholarship
• Joseph Mundassery Scholarship
• INSPIRE Scholarship
• AICTE Pragati Scholarship for Girls
• PM Narendra Modi Scholarship for 12th Pass Students
• Ishan Uday Scholarship
• VIT Scholarship
• LPU Scholarship
• Kanya Utthan Yojana

Scholarships After Class 10

Education sets the base for a thriving career. Nevertheless, with the growing expense of education in India, several Indian students find it hard to resume their studies after qualifying the Class 10 Board Exam. It doesn’t matter how hard-working or worthy they are; students are often forced to surrender their dream, due to the lack of financial support needed to proceed with their higher studies.

Both the Government and private organizations have presented some scholarship plans to help these students. The scholarships are presented based on the scores achieved at Class 10 Board exams. There are few scholarships schemes which conduct their entrance exam to select the applicants who justify this scholarship apart from being merit further who are really in need of them.

• Atal Bihari Vajpayee Scholarship
• Post Matric Scholarship
• National Talent Search Examination (NTSE)
• IOCL Scholarship
• Jio Scholarship
• CBSE Single Girl Child Scholarship
• Vidyadhan Scholarship
• PM Scholarship for Class 10th Students
• AKASH ANTHE
• MOMA Scholarship
• Ujjwal Bhavishya Scholarship

Scholarship for Girls 2023

Scholarships for girls in India are intended to promote better education and career possibilities for girls who are still to match their boy equivalents on various socio-economic parameters. However, if given a chance, women don’t limp behind men either in the professional field or academics. Here appears the purpose of women-specific scholarships that prompts them to counter their monetary constraints and seek academic and career breaks. Of late, many government and private organizations in various fields like science and research, management, aviation, hospitality and defense have been granting women scholarships and awards towards encouraging half of the community to take educational opportunities.

• CH Muhammed Koya Scholarship
• Indira Gandhi Scholarship
• Kalpana Chawla Scholarship – Meritorious Girl Students
• CBSE Scholarship
• AICTE Pragati Scholarship for Girls
• Maulana Azad Scholarship
• Ladli Yojana
• Savitri Bhai Phule Scholarship – From Class 5th to 10th
• Kanya Utthan Yojana – After Class 12th

Scholarships Under Karnataka State

The government of Karnataka and several other private organizations give a number of scholarship plans specifically for the pupils who are a resident of the state. This section provides a detailed report on top Karnataka scholarships to support the considered students’ defeat financial restrictions towards proceeding their dream educational career. Identified as the educational center of the country, Karnataka is a shelter to around 54,529 primary schools with 252,875 teachers and 8.495 million students.

Moreover, there are 9,499 secondary schools with 92,287 teachers with 1.384 million students. Analyzing this huge student base, the government of Karnataka, its subsidiary offices and many private organizations administer scholarships for Karnataka students.

• Prize Money Scholarship
• Vidyasiri Scholarship
• SSP Scholarship Karnataka
• Shamanur Shivashankarappa Scholarship

Scholarships Under West Bengal State

West Bengal Scholarships are intended for students who are a resident of the state at both school and college level. Administered by the government of West Bengal and its subsidiary boards, the scholarships for West Bengal point at supporting worthy, deserving and impoverished students of the state.

Moreover, the state also possesses various alleged institutions of higher education like; Indian Institute of Management (Kolkata), National Institute of Technology (Durgapur), Indian Institute of Technology (Kharagpur), Indian Institute of Science Education and Research (Kolkata) which makes the learning easier and hassle-free for students of West Bengal.

• Aikyashree Scholarship
• TSP Scholarship
• OASIS Scholarship
• Nabanna Scholarship

Scholarships Under Kerala State

With the maximum number of literacy rates, Kerala stands first in India, pondering the point of how much learning is essential for people of the state. Kerala has its culture and beliefs. Besides all these, Kerala is the various sought-after tourist destination with its peaceful beaches, hill stations, bird, backwaters and wildlife sanctuaries, etc.

Kerala has made many significant accomplishments in the department of social development and the standard of life. The priority for the state is continuously literacy and that is why the state has set up many schools and other educational institutions. Kerala’s city named Kottayam was the primary district to achieve full literacy rate in India. It is where the Directorate of Collegiate Education Kerala is located. The organization here functions out the Government of Kerala Scholarships for helping students to achieve quality education.

• District Merit Scholarship
• E Grantz Scholarship
• Higher Education Scholarship
• State Merit Scholarship Kerala
• Snehapoorvam Scholarship

Scholarships for Pursuing Students

Undergraduate scholarships are meant to promote the education of pupils at the undergraduate level. The students, after passing class 12, look up for undergraduate programs in different streams like Engineering, Medical, Architecture, Arts/Humanities, Science, Commerce, etc. To support the education of these learners at the undergraduate level, many organizations and institutions award undergraduate scholarships. These scholarships give monetary liberty to those students, who face economic limitations in seeking their dream education.

• Bikash Bhavan Scholarship – Class 11, 12, Under Graduate, Post Graduate and Doctoral Courses.
• JSPN Scholarship – Class 11th, 12th, BE, BAMS, Dental, Medical, and Professional Courses
• Swami Vivekananda Scholarship – Class 11 & Class 12.
• Vidyasaarathi Scholarship – ITI, Under Graduate, BE/B.Tech or Diploma Courses.
• Ashirwad Scholarship – UG/PG
• HDFC Scholarship – From Class 6 – PG
• Fair & Lovely Foundation Scholarship – Higher Education
• ONGC Scholarship – Professional Courses
• NEC Scholarship – Higher Studies
• DTE Maharashtra Scholarship – Professional and Technical Courses
• CG Scholarship – From Class 9th – 12th
• Rajiv Gandhi Fellowship – M.Phil & Ph.D. Degree
• Pragati Scholarship – 1st-year degree
• Siemens Scholarship – 1st Year Engineering from Government Colleges
• Jain Scholarship – Pursuing Higher Studies
• SRM Scholarship – Pursuing B.Tech
• Yashad Sumedha Scholarship – Pursuing B.Tech
• Sarla Devi Scholarship – Pursuing Degree in any stream
• Gandhi Fellowship – Pursuing Final Year of Degree
• DARC Fellowship – PG 1st Year
• DU Scholarship – UG and PG Students
• Sathyadeepam Scholarship – From LKG to Class 10 and ICSE and CBSE also.
• Umbrella Scholarship – From Class 9 to PG
• Rhodes Scholarship – UG Outside India
• Swachh Bharat Fellowship – Complete UG and PG

Scholarships Under Minority and Backward Community

The Prime Minister’s New 15 Point Programme for the Welfare of Minorities was announced in June 2006. It provides that a post-matric scholarship scheme for meritorious students from minority communities would be implemented. The objective of the scheme is to award scholarships to meritorious students belonging to economically weaker sections of minority community so as to provide them better opportunities for higher education, increase their rate of attainment in higher education and enhance their employability.

• Devaraj Arasu Scholarship (ದೇವರಾಜ್ ಅರಸು ವಿದ್ಯಾರ್ಥಿವೇತನ)
• Gokdom Scholarship
• Begum Hazrat Mahal Scholarship
• Gujarat Scholarship
• PRERANA Scholarship
• MCM Scholarship
• MOMA Scholarship
• LIC Scholarship
• NMMS Scholarship – Central Government scholarship for backward classes for class 9 students who passed Class 8 with 55%.
• ADW Scholarship – BackWard Classes
• HP Udaan Scholarship Program – Who have completed Class 10 and Class 12.

Education Loans in India

• HDFC Student Loan
• ICICI Bank Education Loan

Fellowships in India

• Swarnajayanti Fellowship
• Junior Research Fellowship
• LAMP Fellowship
• Ramanujan Fellowship
• Young India Fellowship
• Marie Curie Fellowship
• Dalai Lama Fellowship
• Clinton Fellowship
• INSA Fellowship
• Naropa Fellowship
• Shastri Research Student Fellowship
• UGC BSR Fellowship
• IIHS Fellowship
• Eisenhower Fellowship
• Stanford Reliance Dhirubhai Fellowship
• Ramalingaswami Re-entry Fellowship

Other Scholarships in India

Scholarships, in India, are the root for the students to make the best career. Since India is a developing nation, many of the people are under the poverty level. In India, both the government and private institutions help the aspirants in presenting the scholarships to strike their goals. The financial support is given to the applicant in the form of scholarship at a distinct level of education such as Graduate, Postgraduate, doctorate or post-doctorate degree through which economically behind applicant can make there believed career.

The scholarship is not only the origin to render financial assistance to the pupils but also act has information about literacy among different people. Since the charge for higher studies, day by day is growing, rather than being a bright candidate, it’s not likely for some students to go for higher education. Here the scholarships act as a means for them to proceed to higher studies. Indian awards are meant for personal development. Nevertheless, through which overall growth of the country is reached.

• AP Brahmi Pension Scheme
• CCB Scholarship – After Degree for all categories
• Charpak Scholarship – For Bachelors and Masters Students
• Meghalaya Scholarship
• Teach for India Fellowship
• PFMS Scholarship
• IFFCO MD scholarship
• Canara Bank Scholarship
• US Awasthi Scholarship
• Mizoram Scholarship
• PACE Scholarship
• KC Mahindra Scholarship
• Abdul Kalam Scholarship
• Colgate Scholarship
• PMSSS
• Dr. Manmohan Singh Scholarships
• SN BOSE Scholarship
• Tata Scholarship
• Vidya Lakshmi Portal
• Inlaks Scholarship
• Samaj Kalyan Scholarship
• SSO Scholarship
• Azim Premji Fellowship
• CSIR Scholarship
• KIIT Scholarship
• MEA Scholarship
• AFNA Scholarship
• Orange Tulip Scholarship
• Times SPARK Scholarship – Class 9 to 12
• Babulal Nagarmal Satnalika Foundation Scholarship – UP TO Class 12
• IAFBA Scholarship – Class 1 to PG
• Sahapedia Fellowship – For Post Graduate Students
• Gramin Pratibha Khoj Pariksha – For Class 8 Students
• NAROTAM SEKHASRIA Scholarship – For graduation Students
• Deen Dayal SPARSH Yojana – From Class 6 to 9
• Samsung Star Scholarship – B.Tech or Dual Degree (B.Tech + M.Tech)
• Siksha Abhiyan Scholarship – From Class 8th to 12th Passed Students
• Dr. Ambedkar Scholarship – From Class 1 Up to PG
• Sitaram Jindal Foundation Scholarship Only for Class 11
• STS Scholarship – MBBS & BDS Students
• Pre-Matric Scholarship – From Class 1 – 10
• Post-Matric Scholarship – From Class 11 – Ph.D.
• MHRD Scholarship – From Class 1 – Post Graduate
• Narendra Modi Education Scholarship – For Class 10th & 12th
• e-Medhabruthi Scholarship
• Kothari Fellowship – Ph.D. and Degree Holders
• Central Sector Scholarship
• Vodafone Scholarship – From Primary Classes Upto Ph.D.
• LIC HFL Vidyadhan Scholarship – From Class 8 – Post Graduate
• UGAM Scholarship – Pursuing PG
• DST Postdoctoral Fellowship – For PhD
• ICAI Scholarship – For CA Students

Best Scholarships in India

A lot of intelligent students often miss achieving their dream for higher education with the expense of education having grown quite an expensive affair. The requirement for scholarships is compelling and important for our students more than ever since. There are several public and private organizations that support capable students in charge of monetary assistance. Here is the list of best scholarships available for students in India to fulfill their dream education opportunities.

• NTSE – National Talent Search Examination
• KVPY – Kishore Vaigyanik Protsahan Yojana 2019
• NSP – National Scholarship Portal
• NATS – National Accounting Talent Search
• E-District Scholarship

FAQs for Scholarships for Students 2023

1. Where can I find state-wise Scholarships for students in India?

Students who are pursuing their education can get various scholarships offered by the Indian government or private institutions in India. Students can find complete information on the state-wise scholarships 2023 in India from this page.

2. What are the best scholarships in India?

There are numerous scholarships available in India offered by the government or other institutions/companies for students. In plenty of scholarships, we have collected a few best scholarships in India and mentioned below for the sake of students who are financially weak. 

• e-Kalyan Scholarship
• Google Scholarship
• Haryana Scholarship
• Aga Khan Foundation International Scholarship Programme
• Aditya Birla Scholarship
• BEML Scholarships For SC ST Students

If you need to know more about the best scholarships in India for students, then visit our site LearnCBSE.in or click on the links available on this page.

3. How to find Scholarships After Class 10 & 12 in India?

To obtain the scholarships after class 10 and 12, students have to score merit marks in the final examination. For those students who are economically weak and having merit marks can enjoy the various scholarships in India. Both the Government and private organizations have presented some scholarship plans to help these students for their higher education. Students can find scholarship information by tracking the education sites and official portals of the organization.

4. What can I do to apply for college scholarships?

• Students should discover the scholarships at the institutions you want to apply to.
• Then go through the details & eligibility for scholarships that you are applying for.
• Find the application form for a particular scholarship, then fill it.
• Keep track of its application timeline so as to avoid any last-minute rush.

5. What documents are required for a scholarship form?

The required documents may vary from one scholarship to another as per the organization guidelines. So, we have given some of the common documents that students should keep handy when applying for scholarships. They are as under

• Caste/Category certificate (if applicable)
• Income certificate
• Domicile certificate (if applying for a state-specific scholarship)
• Academic transcripts and certificates of previous qualifying examinations
• Photograph of the applicant
• Identity proof
• Aadhaar card
• Age proof (if required)

The post Scholarships for Students 2023 | List of Scholarships for All Types of Students appeared first on Learn CBSE.

AP ePASS 2023: The scholarships in Andhra Pradesh state are intended for learners who are a member of the education system. The main aims of Epass scholarship in Andhra Pradesh are, to help the worthy students who cannot afford learning and the government will pay the expenditure fee designated by the colleges in the form of scholarships.

The Electronic Payment and Application System of Scholarships (Epass is now – Jnanabhumi) is essentially an online scholarship payment system that was conceived by the Andhra Pradesh government to assure the faster disbursal of scholarships to the economically underprivileged students. It is a very good action taken by the administration of Andhra Pradesh to award the scholarship to the students whose families are not able of giving their children for study for obtaining higher knowledge or the candidates who belong to SC/ST/BC class.

It attracts the scholarship schemes for students who belong to SC, ST, BC, EBC, Minority and Disabled categories. Some of the famous scholarship schemes are listed on the Andhra Pradesh ePASS Jnanabhumi website which includes post-matric scholarships, pre-matric scholarships, overseas scholarships, skill up-gradation services, corporate admissions, etc. bc welfare apcfss.in Read the article moreover to get complete information about the ePASS scholarship AP, ePASS Status AP, eligibility, awards, application procedure, Scholarship Status AP and more.

ePASS Status Telangana

AP ePASS Jnanabhumi – List of Scholarship

Let us discuss here, AP Scholarship List 2023-24, who provides these scholarships, kind of financial assistance is given to it when to apply for it, etc. in this section. Check the table below for further details.

• Pre matric Scholarship for SC/ST/BC and Disabled Welfare
• Post Matric Scholarship for SC/ST/BC and Disabled Welfare
• Pre Matric Scholarship for Minority Communities
• Post Matric Scholarship for Minority Communities
• Jagananna Vidya Deevena
• Jagananna Vasathi
• Jagananna Videshi Vidya Deevena
• AP Social Empowerment Fellowship
• Corporate Admission Scholarship
• Veda Vyasa Scheme for Vedic Education (VV-VE)
• Gayathri Scheme for Academic Excellence (GS-AE) Andhra Pradesh
• GRE, TOEFL, IELTS Coaching Registration
• Overseas Scholarship for Kapu Students
• Overseas Scholarship for BC Students
• Overseas Scholarship for EBC Students

AP ePASS Overview – Jnanabhumi

AP ePASS Jnanabhumi Eligibility Criteria

Which Students are Eligible for AP ePASS Jnanabhumi?

• Students who belong to the categories of SC, ST whose annual parental income is Rs. 200000 or below.
• Students who belong to BC, EBC, Disabled Welfare Students whose parental income is RS. 100000 or below.
• Students whose attendance is 75% at the end of each quarter.

Note: The following category students are not eligible for AP Epass Jnanabhumi.

Which Students not eligible for AP ePASS Jnanabhumi?

• Students belonging to the categories other than SC, ST, BC, EBC, and DW(Disabled).
• SC, ST Students whose annual parental income is more than Rs. Two Lakhs and BC, EBC, Disabled Students whose parental income is more than Rupees One lakh.
• All Students pursuing part-time courses, online courses.
• Students admitted under Sponsored seats, Management Quota seats.
• Students drawing the stipend more than the scholarship amount in aggregate per annum.
• Students of BC, EBC and DW students studying the Courses offered by open universities, distant mode, category B seats in MBBS, BDS.
• EBC students studying Intermediate or equivalent courses.

ITI Admission

AP ePASS Jnanabhumi Eligible Courses

Post Matric Courses certified by the concerned University/Board having a term of 1 year and above:

Courses not eligible for AP ePASS Jnanabhumi

Scholarships are not awarded for training courses like

• Aircraft Maintenance Engineer’s Courses
• Private Pilot license Courses.
• Courses at Pre-examination Training Centers of all India and State levels.

AP ePASS Jnanabhumi Eligible Colleges

• All Post Matric Colleges in Andhra Pradesh approved by the Government of Andhra Pradesh/Competent Authority.
• The list of colleges is communicated by Administrative Departments (Departments of Higher Education, Technical Education, School Education, Health Medical and Family Welfare, Employment and Training) to the Commissioner of Social Welfare.

AP ePASS Jnanabhumi Timeline

It is necessary to submit the application form online within one month from the date of admission. The principal of the Colleges or schools should issue the bonafide certificate on the same day of submission of the application form (hard copy) in the college.

Field Officer/ASWO

• Verification would be done physically twice in a year.
• Within one month of re-opening of college.
• Within one month from the last date of closing of Admissions.

An officer from any of the Government Department verifies the Certificates and confirms their candidature in the Starting day of AP Epass.

If any candidate missed the process will have to face many problems. Hence, the Government has introduced the Biometric Process of Verification. The school or college can conduct this at any time. The overall process should be completed within one month of submission of the application.

AP ePASS Jnanabhumi Application Procedure

As you have already given the information on the list of AP Epass scholarship and their eligibility criteria, now it’s time to assemble the information about its application process. Students will be provided by the ePASS application number while registering. Let us now examine how to apply for AP Epas by following these steps given below.

• Visit the official website of the AP Epass, https://epass.apcfss.in/.
• Select the Course for which you want to apply.
• Now, go to the option of pre-matric or post matric as per your demand.
• The students who belong to the scheduled tribe, scheduled caste, BC and Disabled can apply for the pre-matric scholarship program.
• Click on the “Registration button”,  to get admission or registration form.
• Provide your Aadhaar and ration card details and all the necessary details.
• Also give details of your parents/guardian, as per directed.
• Provide details of your school or college as well as the course.
• Mention the State bank account number clearly with IFSC code.
• Mention your caste and income certificate details.
• Upload the required documents, enter the captcha and click on the Submit button.
• Your application is submitted now.
• You can take the print out of the application form for future reference.

Documents Required for AP ePASS Jnanabhumi Scholarship

For AP ePASS Jnanabhumi Fresh Registrations

• Student last year exam certificate
• Caste certificate
• Aadhar card
• Residential certificate
• Ration Card
• Bonafide certificate
• Birth Certificate
• Income Certificate
• Mothers and fathers ID Proof
• Scan Bank Passbook with the bank account and IFCS Code

For AP ePASS Jnanabhumi Renewal Registrations

• Student Previous year application numbers
• Student Scanned Aadhar card
• Previews Semester Mark Sheet
• Fathers Aadhar card
• Mothers Aadhar card
• Ration Card
• Current mobile Number
• Marks List according to qualification
• Bank account number details

AP EAMCET

AP ePASS Jnanabhumi Application Status 2023

Candidates, who have successfully applied for the AP Epass and filled the application form along with the needed documents can check the status of the application for the scholarship provided by Andhra Pradesh Board, time to time, on the university website. AP ePASS Status check online @ official website which is given below.

With this, they will be able to trace the progress 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 application status. Follow the below steps to check the status of Andhra Pradesh State Scholarships for different schemes.

• Visit the official website epass.apcfss.in
• Click on the option “AP Scholarship status tab”
• Next, you will be redirected to the new page.
• Now, fill the required details on this page such as SSC hall ticket number, exam category, the application number, academic year, date of birth.
• Click on the “Get Status” button to know the current status of your application for the scholarship of AP ePASS. Students can also take the print of the application/scholarship status.

Reason for Rejection – AP ePASS Application Rejection

Students application form for scholarships can be rejected for the following condition;

• If the student is not bonafide.
• Incorrect caste or annual Income details.
• Non-submission of caste, Income certificates.
• Non-receiving of renewal proposal.
• Claiming scholarship for same level courses.
• Not recommended by Field Officer.
• Student admitted under management quota .
• Incorrect course & year of study.
• Non-submission of enclosures.
• Incorrect course & year of study.
• Non-submission of hard copy of the application.
• If the student is not physically present.
• Discontinued/Detained students in case of renewals.
• Claiming scholarship for same level courses.
• Previous sanction verification for renewal.
• Non-receiving of hard copy for Fresh.

Important Instructions for AP ePASS

• AP Epass is fairly open for all the candidates belonging to any of the categories such as General, OBC, SC, ST.
• It is important for students to attach a hardcopy of the photocopy of the SBI bank passbook with the application form.
• Candidates must provide their own effective email ID and mobile phone number.
• All the details and documents produced by the applicant must be genuine and authentic. 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 portal have to renew the account by adding the updates. No new registration is required.
• 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.

FAQs on AP ePASS Jnanabhumi 2023

Question 1.
How much reimbursement, a student can get?

Answer:
The reimbursement of tuition will be dependent on the course you pursue.  While most of the courses are eligible for 100% of the tuition fee as fixed by the government. Self-financed courses are eligible only for a maximum of Rs. 20,000 or the actual fee charged by the college, whichever is less.

Question 2.
Can I obtain a scholarship for the previous year?

Answer:
No, the scholarship can be claimed only for the current year. Scholarships for the previous year can not be claimed under any circumstance.

Question 3.
How do I know my verification officer?

Answer:
The District Collector appoints verification officers for one or more colleges depending on the number of students in the college. The verification process is a two-step verification process:

• Verification by the College Principle: All applications to be verified individually by the college and signed by the principal of the college.
• Verification by the verification officer: The verification officer will have to verify all the student in the college an appointed date and time.

The details of the verification officer can be viewed on this website by clicking the verification officer details given on the right side of the web page.

Question 4.
What is the meaning of verification in the scholarship process?

Answer:
The process of verification is essentially meant to check whether the particulars given in the scholarship form are correct as per the documents enclosed.  The verification is done in two steps namely college verification and Independent verification.

• College Verification:  In this verification, the college has been mandated to verify the documents furnished by each student with the entry made in the application form.  Once the verification is completed and all entries are found correct, he would finally sign the same and send it to the department for verification by the verification officer appointed by the District Collector
• Verification by the Verification Officer:  the verification officer appointed by the District Collector will conduct physical and documentary verification and submit his report either accepting or rejecting the scholarship application.
• Scrutiny by the Welfare Officer: before each and every application is taken up for sanction it is the responsibility of the welfare officer to satisfy himself of the verification by the college principal and the verification officer and finally sanction the scholarship.

AP ePASS Jnanabhumi Contact Details

ePASS
Project Monitoring Unit
The Director of Social Welfare
Service Road Opposite to Manipal Hospital,
TG Plaza, Tadepalli,
Vijayawada – 520001.

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

The post AP ePASS 2023 | Jnanabhumi Application Status, Eligibility, Fresh & Renewal Registrations appeared first on Learn CBSE.

Bihar Scholarship 2023-2024: Bihar Scholarship gives a vast range of educational opportunities for the candidates who are residing in the Bihar state permanently. The Bihar Government and different private organizations collectively offer plenty of scholarships for worthy and impoverished scholars of the state. The key purpose of each Bihar Student scholarship is to assist and prompt deserving learners to pursue further studies despite any financial limitation. Students can get e Kalyan Bihar Scholarship list 2023 from here.

बिहार छात्रवृत्ति 2019-20: बिहार छात्रवृत्ति बिहार राज्य में स्थायी रूप से रहने वाले उम्मीदवारों के लिए शैक्षिक अवसरों की एक विशाल श्रृंखला प्रदान करती है। बिहार सरकार और विभिन्न निजी संगठन सामूहिक रूप से राज्य के योग्य और गरीब विद्वानों के लिए बहुत सारी छात्रवृत्ति प्रदान करते हैं। प्रत्येक छात्रवृत्ति का मुख्य उद्देश्य किसी भी वित्तीय सीमा के बावजूद आगे के अध्ययन को आगे बढ़ाने के लिए योग्य शिक्षार्थियों की सहायता और संकेत देना है।

The article provides here all the highlights of Bihar Board scholarships 2023 are presented by the government of Bihar and other firms. Students who have completed class 10 can apply for Bihar scholarship 10th pass. Bihar Scholarship 2023 Last date is  5th March 2023. Get details of the list of scholarship, eligibility criteria, the detailed application process, application timeline, Bihar Scholarship online application form, Bihar Scholarship Date extended and award details.

लेख यहाँ बिहार सरकार और अन्य फर्मों द्वारा प्रस्तुत बिहार छात्रवृत्ति के सभी मुख्य अंश प्रदान करता है। छात्रवृत्ति की सूची, पात्रता मानदंड, विस्तृत आवेदन प्रक्रिया, आवेदन समयरेखा और पुरस्कार विवरण का विवरण प्राप्त करें।

Bihar Scholarship Details 2023

Bihar Scholarship List- बिहार छात्रवृत्ति की सूची

Here you will get the details of a number of Bihar scholarships are available, time duration to apply for the scholarship. The table given here contains the entire Bihar scholarship list 2023, with their provider details and tentative application period. This will present you with an insight into scholarships that are prepared exclusively for the candidates of Bihar. Bihar Board Scholarship 2023 list is as discussed under. Check for the scholarship that suits your requirements and fills the respective application, specified for them. Bihar Board 10th scholarship 2023 list is as follows:

यहां आपको कई बिहार छात्रवृत्ति के विवरण उपलब्ध हैं, छात्रवृत्ति के लिए आवेदन करने की समय अवधि। यहां दी गई तालिका में संपूर्ण बिहार छात्रवृत्ति सूची है, जिसमें उनके प्रदाता विवरण और अस्थायी आवेदन अवधि है। यह आपको बिहार के उम्मीदवारों के लिए विशेष रूप से तैयार की गई छात्रवृत्ति के बारे में जानकारी प्रदान करेगा। छात्रवृत्ति के लिए जांचें जो आपकी आवश्यकताओं के अनुरूप हैं और उनके लिए निर्दिष्ट संबंधित एप्लिकेशन को भरता है।

Bihar Scholarship Eligibility – बिहार छात्रवृत्ति पात्रता

Now, as you are informed of all the list of Bihar Scholarship, 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 Bihar state and no candidate is eligible who reside outside the state.

अब, जैसा कि आपको बिहार छात्रवृत्ति की सभी सूची से अवगत कराया 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 कि, उम्मीदवारों को बिहार राज्य का स्थायी निवासी होना चाहिए और कोई उम्मीदवार योग्य नहीं है जो राज्य के बाहर निवास करता है।

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

अब हम बिहार बोर्ड के द्वारा परिभाषित पात्रता के अनुसार सभी श्रेणी के छात्रों के लिए शैक्षणिक योग्यता और अधिकतम पारिवारिक आय मानदंड की जांच करते हैं।

Post Matric Scholarship (PMS), Bihar for OBC/SC/ST Students

• The students studying at the post-matriculation level (class 11 to postgraduation level) can apply for this scholarship.
• They must have passed the last qualifying examination.
• The annual income of the family should be less than INR 2.50 Lakh (for SC/ST candidates) and INR 1 Lakh (for OBC candidates).
• They must not be in receipt of any other scholarship.

Pre-Matric Scholarship, Bihar, for OBC Students

• The students studying in class 1 to 10 at a government or government-recognized school can apply for this scholarship.

Chief Minister Medhavi Yojana, Bihar, for EBC/BC Students

• The students who have passed their class 10 examinations organized by Bihar School Examination Board with first-class (over 60% marks) can apply for this scholarship.
• The annual income of the family for BC students should be less than INR 1.50 Lakh.

BTSE Bihar Talent Search Examination, Bihar (for all candidates)

• This scholarship examination is open for students of class 7 to 10.

Combined Counselling Board (CCB) Scholarship, Bihar (for all candidates)

• The students studying at college/university level (pursuing diploma level, degree level or postgraduate level courses) can apply for this scholarship.
• They must have passed the last qualifying examination with at least 40% to 50% marks.

How To Apply for Bihar Scholarships?

The application process varies for different Bihar scholarships. We are providing the details here for individual scholarships and how to apply for them. Check the details in the below-given table and proceed with the application procedure as per your requirement.

विभिन्न बिहार छात्रवृत्ति के लिए आवेदन प्रक्रिया भिन्न होती है। हम अलग-अलग छात्रवृत्ति के लिए और उनके लिए आवेदन करने के तरीके के बारे में विवरण प्रदान कर रहे हैं। नीचे दी गई तालिका में विवरण की जाँच करें और अपनी आवश्यकता के अनुसार आवेदन प्रक्रिया के साथ आगे बढ़ें।

Bihar Scholarship – Application Procedure of Post Matric Scholarship

Interested candidates may apply online through the official website of the National Scholarship Portal (Bihar Scholarship Portal) from Feb-2023 to 05-March-2023. To apply for the 10th pass scholarship 2023 Bihar follow the below-given steps:

• Visit the National Scholarship Portal, https://scholarships.gov.in/
• Click on New User/Register.
• Read the guidelines properly.
• Continue to register.
• Once registered, take note of the student application ID and continue to apply.
• The applicants will receive an OTP on their registered mobile number. Confirm the OTP.
• Change the password.
• Log in with your registration number, date of birth and password.
• You will be directed to a webpage comprising important instructions about the Bihar Scholarship Online form Date 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.

बिहार छात्रवृत्ति – पोस्ट मैट्रिक छात्रवृत्ति की आवेदन प्रक्रिया

इच्छुक उम्मीदवार राष्ट्रीय छात्रवृत्ति पोर्टल की आधिकारिक वेबसाइट फरवरी 2023 से 05 मार्च-2023 तक ऑनलाइन आवेदन कर सकते हैं। छात्रवृत्ति के लिए आवेदन करने के लिए नीचे दिए गए चरणों का पालन करें

• राष्ट्रीय छात्रवृत्ति पोर्टल पर जाएं, https://scholarships.gov.in/
• पर क्लिक करें। नए उपयोगकर्ता / रजिस्टर पर क्लिक करें।
• दिशा-निर्देशों को ठीक से पूरा करें। रजिस्टर करने के लिए बंद करें।
• पंजीकृत होने के बाद, छात्र आवेदन आईडी पर ध्यान दें और आवेदन करना जारी रखें।
• आवेदक करेंगे उनके पंजीकृत मोबाइल नंबर पर एक ओटीपी प्राप्त करें। OTP की पुष्टि करें। पासवर्ड को बदलें।
• अपना पंजीकरण नंबर, जन्म तिथि और पासवर्ड के साथ दर्ज करें।
• आपको फॉर्म भरने के बारे में महत्वपूर्ण निर्देश देने वाले वेबपृष्ठ पर निर्देशित किया जाएगा।
• सभी निर्देशों को अच्छी तरह से पढ़ें और पृष्ठ के अंत में दिए गए बॉक्स पर टिक करें। और “आगे बढ़ें” पर क्लिक करें।
• जिस मिनट पर आप “आगे बढ़ें” बटन पर क्लिक करते हैं, आपको उपयोगकर्ता डैशबोर्ड पर निर्देशित किया जाएगा।
• “आवेदन पत्र भरें” भाग पर क्लिक करें। छात्रवृत्ति आवेदन पत्र में अतिरिक्त विवरण भरें।
• इस पर क्लिक करें। सबमिट बटन पर क्लिक करें।
• यदि आप आवेदन पत्र को पूरा करने के बाद सबमिट पर क्लिक करते हैं, तो आपको अपनी तस्वीर और अन्य सहायक दस्तावेजों को अपलोड करना होगा।
• आवेदन पत्र के अंतिम जमा करने की दिशा में आगे बढ़ते हुए, उम्मीदवारों को किसी भी से बचने के लिए सावधानीपूर्वक भरे गए प्रत्येक विवरण से गुजरने का सुझाव दिया जाता है।
• बाद में त्रुटि की तरह।
• इसके अलावा, उम्मीदवार द्वारा भरी गई जानकारी में बदलाव करने की आवश्यकता नहीं है, एक बार अंत में आवेदन पत्र जमा करें।
• आवेदन पत्र के अंतिम रूप से जमा करने के बाद, उम्मीदवारों को फॉर्म का प्रिंट आउट लेने और इसे जमा करने के लिए कहा जाता है।
  अन्य सहायक दस्तावेज अपने संबंधित शिक्षण संस्थानों में।

Bihar Scholarship Documents Required for Application

• 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

बिहार छात्रवृत्ति आवेदन के लिए आवश्यक दस्तावेज

• अंतिम योग्यता परीक्षा मार्क शीट
• कैटेगिरी सर्टिफिकेट
• प्रमाण पत्र आय प्रमाण पत्र
• पासबुकफाइ
• रसीद नंबरए नॉन
• नॉन रिफंडेबल अमाउंट
• ईयररोलमेंट नंबर
• स्टूडेंट आईडी प्रूफडोमाइल
• सर्टिफिकेट
• आधार कार्ड नंबरलैट
• पासपोर्ट साइज फोटो

Bihar Scholarship Reward Details

Get the details of the scholarship amount that each Bihar scholarship will offer and the number of students who will be benefitted through these scholarships. The prize money or reward amount of scholarship and the number of students receiving benefits from them differs from scholarship to scholarship. Moreover, the government scholarships transfer the amount directly into the bank account of the receivers through Direct Benefit Transfer (DBT).

Post Matric Scholarship (PMS) For OBC/SC/ST Students

• There are 3,00,000 scholarships available in this category
• Maintenance allowance of up to Rs.1200 per month (for hostellers) and INR 550 per month (for day scholars) for 10 months
• An additional allowance of up to Rs.240 per month for students with disabilities
• Reimbursement of compulsory non-refundable fees
• Study tour charges of up to Rs.1600 per annum
• Thesis typing and printing charges of up to Rs.1600 for research scholars

Pre-Matric Scholarship for OBC Students

• There are 3,00,000 scholarships available in this category
• For day-scholars of class 1 to 4 – Rs.50 per month
• For day scholars of class 5 to 6 – Rs.100 per month
• For day scholars of class 7 to 10 – Rs.150 per month
• For hostellers of class 1 to 10 – Rs.250 per month

Chief Minister Medhavi Yojana for EBC/BC Students

• There are 1,55,000 scholarships available in this category
• One-time financial assistance of Rs.10,000 to each scholar

BTSE Bihar Talent Search Examination

• Rs.20,000 and a laptop for students scoring from 91% to 100% in BTSE
• Rs.15,000 and a notepad for students scoring from 81% to 90% in BTSE
• Rs.10,000 and a tab for students scoring from 71% to 80% in BTSE
• Rs.5,000 and a smartwatch for students scoring from 61% to 70% in BTSE
• Gift hampers for students scoring from 51% to 60% in BTSE

Combined Counselling Board (CCB) Scholarship

• Scholarship from Rs.1 Lakh to Rs.3 Lakh

Bihar Scholarship Amount

Important Instructions for Bihar Scholars

• Fields provided with a red asterisk (*) mark are mandatory fields.
• Candidates should separately inform the mistakes detected by them to the Institute/District/Region/State. The software provides the facility at the level of the Institute & State to edit & correct limited information.
• The Fields which can be edited are, Gender, Religion, Category, Profession, Annual Income, Aadhar Number, Disability, Day Scholar/Hostler, Mode of Study, IFSC Code, Account No., Admission Fees, and Tuition Fees. However, corrections made by the Institute/State, if any, would be conveyed instantly to the student through SMS/email.
• Candidates can fill up the online application in as many sittings as he/she wishes until he/she are satisfied that you have entered all desirable fields correctly. The software provides the facility to save their application at every stage.
• Aadhaar No. is not Mandatory for the Students in order to Register and fills up the application form online. Students can apply for Scholarship without entering the Aadhaar no. but in that case, they have to enter Aadhaar Enrollment Id.
• An Application ID (Permanent ID) will be provided to the candidate once his/her Registration is done. It will be conveyed to candidates through SMS and e-mail. Students should memorize their Application ID as it will be required while applying for a Fresh/renewal scholarship.
• Student cannot apply as a fresh if he/she is a Renewal candidate. Their application will be rejected in that case.
• The student should immediately approach the institute to contact the nodal officer of the State where the institute is located. Students can also approach the Nodal Officer of that State directly through e-mail under intimation to the Ministry. If their institute is an eligible institution, the State Government concerned would enter it into the database and then they can apply.
• The name and contact details of the Nodal Officer/State Department of all States/UTs are available in the “Services->Know your State Nodal Officer” option.
• Students can check the status of Online Application by submitting his/her Permanent id and Date Of Birth and open the link “Check your Status”.
• Documents are required to be uploaded only for the Scholarship Amount more than ₹50000 per annum

FAQ’s on Bihar Scholarship 2023

Question 1.
What is the Bihar Scholarship 2023 Online last date for submitting applications online?

Answer:
Closure dates for acceptance of various scholarship applications are available in National Scholarships Portal.

Question 2.
How can a candidate apply online for scholarships?

Answer:
In order to apply online, please visit the website through URL www.scholarships.gov.in

Question 3.
Who are eligible to apply for Scholarship Schemes?

Answer:
Students fulfilling the Scheme guidelines of various Ministries are eligible to apply for these scholarships. These are available on the Home Page of the Portal.

Question 4.
Can an applicant, edit the information already saved and up-to what time?

Answer:
All the information can be edited until the closure of the application form. After final submission, your application will be forwarded to the next level and application hereby cannot be edited.

Question 5.
Do I have to fill up the online application in one sitting?

Answer:
No, you can easily fill up your Bihar Scholarship 2023 Online in as many sittings as you wish, until you have provided all desirable fields correctly.

Question 6.
How to check the Bihar Scholarship Status of my application?

Answer:
Candidates can check the status of Online Application by submitting his/ her Permanent ID and Date Of Birth and open the link “Check your Status”.

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

The post Bihar Scholarship 2023 | List, Eligibility, Application Procedure, Timeline appeared first on Learn CBSE.

E-District Scholarship 2023-24: Stipends are a boon for those students and aspirants who are in severe need of financial support and who aspire to higher education. Looking at India’s economic crisis and financial instability, scholarships are the savior for such candidates. E-District Scholarships are a reward for candidates from across the country from the Government of India to pursue their studies without any lack of funding problems. Each candidate who receives the Scholarship is awarded an E–District Certificate that proves their excellence. The student must fill in the E-Scholarship Application Form to qualify for this scholarship.

Applicants applying for these scholarships can check their status as an E–Scholarship Application. By visiting the official websites (the link below) candidates can find out about e-Application Status for District Revenue. To find out all the necessary details about EDistrictScholarships, read the full article.

E-District Scholarship Requirements

In order to fulfill those requirements, aspirants who are trying to obtain the award of E–scholarship are needed. If the candidate fails to follow and meet the eligibility criteria for the E–District Scholarship, no scholarship will be awarded. The appropriate requirements are set out below:

• The candidate must belong to a reserved category acceptable in the country of India in question.
• The nominee must have completed their last test with a score of at least 50 points.
• The candidate’s parents ‘ annual income must be lower than Two Lakh Rupees per year.
• The applicant from the economically backward class must apply online for the E–District Scholarship. Below is the process explained.

E-District Scholarship 2023 – Details

E-District Scholarship Availability

The E–District Scholarship can be used by candidates who need financial support and who genuinely belong to the categories that can not provide for their Higher Level Education. The E-Scholarship is awarded to eligible candidates for the 11th, 12th, ITI, ITC affiliated course, XI & XII NCVT classes, Polytechnic, Nursing Diploma, Teacher Training, Undergraduate, Postgraduate Courses, M.Phil, Ph.D. Level Study in State Government Aided and Recognized Private Institutions.

E-Scholarships are also awarded as a Top Class Scholarship Program to the students qualifying to India’s top colleges such as IIT and IIM.

E-District Scholarship Eligibility Criteria

Eligible Students  from Delhi and Tamil Nadu States

• Must be a Resident of India
• Belonging to Scheduled Castes (SC), Scheduled Tribes (ST), Other Backward Classes (OBC) and Economically Weaker Sections (EWS) and possess a government-affiliated caste certificate.
• Candidates must be enrolled in a course while applying for the scholarship.
• The annual family income of candidates must not exceed INR 2 lakhs from all sources.

E-District Scholarship Application Procedure

Until submitting the E District Scholarship Form, applicants qualifying for the E–District Scholarship login must follow the eligibility criteria. Applicants should visit the official website and follow the steps to apply for the EDistrict Scholarship. Before completing the application form, it is advisable that the candidate read the official notification and the instructions given on the official website.

• Visit E-Scholarship’s official website scholarships.gov.in New applicants pick the New Registration page.
• Read the instructions on the page Instructions for Applications After reading all the instructions, click on the Continue button.
• Fill in the registration form with all the necessary details needed.
• Give all the reports you want.
• Review all the information you’ve filled in.
• Review the statement at the end of the article.
• Then press the Record button.
• Candidates can use valid DISE or AISHE code on the portal itself to check the link to Add School/Institute.
• Once you have registered, change your password for security purposes provided by NSP.
• When the whole process is over. Taking the verification page’s printout.

Documents Required for E-District Scholarship

During their New Registration, the candidates who apply for the first time for the E–District Scholarship must provide the following list of documents. Save all documents incompatible file size in PDF format and upload them to the Application e-district scholarship portal when requested. Below is a list of all the documents required.

• Recent passport size photo Mark-sheet(s) (Consolidated Mark-sheet only)
• Community Certificate
• Income Certificate for self-employed parents/guardian
• An affidavit in a non-judicial stamp paper with the value of Rs.10/-(original)
• Address Proof (Ration Card, Voter ID, PAN Card, etc)
• Fee receipts charged by IFSC Code (11 Digits) Student Evidence of Bank Account Number (Core Banking Service)

FAQ’s on E-District Scholarship

Question 1.
Which students are available under the E district scholarship?

Answer:
The E-Scholarship is awarded to eligible candidates for the 11th, 12th, ITI, ITC affiliated course, XI & XII NCVT classes, Polytechnic, Nursing Diploma, Teacher Training, Undergraduate, Postgraduate Courses, M.Phil, Ph.D. Level Study in State Government Aided and Recognized Private Institutions.

Question 2.
Which institution is responsible for rewarding the E district scholarship 2023 to the students?

Answer:
E-District Scholarships are a reward for candidates from across the country from the Government of India to pursue their studies without any lack of funding problems.

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

The post E-District Scholarship 2023 | Application Procedure, Documents Required, Benefits appeared first on Learn CBSE.

National Accounting Talent Search (NATS): National Accounting Talent Search welcomes applications from undergraduate students who are seeking B.Com, BBA, BBM or any other degree course and are below 25 years of age. It is a national level competition administered at a junior and senior level to examine the accounting knowledge of the students. The key purpose of the competition is to stimulate young talents to apprehend the cause and subject of accounting.

The National Accounting Talent Search (NATS) competition is a sole initiative of the Indian Accounting Association. It was during the succeeding part of 2008 that the organization in its National Executive conference accepted the plan to organize such a competition at the National level. The purpose was primarily to drive young talent to understand the subject and cause of accounting. Since 2009 the competition is being arranged in all parts of India every year. During the initial years, NATS was arranged for undergraduate students only. Later on, the scope of the competition was increased to be organized at the senior level also in which any postgraduate or professional eligibility holder up to the age of 25 years can participate.

NATS Eligibility Conditions

• For Junior Level Competition: Any student doing B.com, BBA, BBM or any other undergraduate course.
• For Senior Level Competition: Any person below the age of 25 years as on July 1, 2023 (born after June 30, 1998)

NATS Awards

NATS Application Procedure

• Go to the official website and click on the registration link.
• Now, select the mode of payment, level and enter your date of birth.
• You can pay the fee either through credit card/debit card or net banking.
• For the junior level, the applicable fee is INR 250/- and for senior-level, the registration fee is INR 500/-.
• Click on the ‘Next’ button.
• Now, fill all the details in the application form.
• Upload all the supporting documents and submit the form.

Documents Required for NATS

• Attested mark sheet of class 12th.
• 1st and 2nd year of a graduation certificate from institution head stating that the candidate has not completed graduation.

NATS Important Dates

• Registration open from October 4, 2022 to December 31, 2022.
  
• Examination on Sunday, February 5, 2023 shall be conducted online and candidate can participate from home.

NATS Exam Pattern

NATS Syllabus of the Courses

Junior Level

• Financial Accounting Cycle including Journal, Day Books, Ledger, Trial Balance, Adjustment Entries, Final Accounts, Rectification of Errors,
• Bank Reconciliation Statement, Depreciation Accounting, Accounting for Non-Profit Making Organizations, Accounting for consignment and joint venture, insurance claims.
• Partnership Accounts, Accounting for issue, reissue, and forfeiture of shares, issue, and redemption of preference shares and debentures.
• Costing Concepts, elements, and Classification of Cost, Unit Costing.
• Elementary knowledge of IND AS

Senior Level

• Financial & Management Accounting
• Basic Accounting concepts, Capital and Revenue, Financial statements
• Partnership Accounts: Admission, Retirement, Death, Dissolution and Cash Distribution.
• Advanced Company Accounts: Issue, forfeiture, Purchase of Business, Liquidation, Valuation of shares, Amalgamation, Absorption, and Reconstruction, Holding Company Accounts.
• Cost and Management Accounting: Ratio Analysis, Funds Flow Analysis, Cash Flow Analysis, Marginal costing and Break-even analysis, Standard costing, Budgetary control, Costing for decision-making.
• Responsibility accounting

Business Statistics

• Data types, Data collection and analysis, sampling need, errors and methods of sampling, Normal distribution, Hypothesis testing, Analysis and Interpretation of Data.
• Correlation and Regression, small sample tests-t-test, F-test and chi-square test.
• Financial Management
• Capital Structure, Financial and Operating leverage
• Cost of capital, Capital budgeting
• Working capital management
• Dividend Policy

Advanced Accounting and Finance

• Accounting standards: IND AS and IFRS, Responsibility Accounting, Social Accounting.
• Money and Capital market, Working of stock exchanges in India, NSE, OTCEI, NASDAQ, Derivatives, and Options.
• Regulatory Authorities: SEBI, Ratting Agencies; New Instruments: GDRs, ADRs
• Venture Capital Funds, Mergers, and Acquisitions, Mutual Funds, Lease Financing, Factoring, Measurement of risk and returns, securities and portfolios

Income Tax and Tax Planning

• Basic concepts, Residential status, and tax incidence, exempted incomes, computation of taxable income under various heads.
• Computation of taxable income of individuals and firms.
• Deduction of tax, filing of returns, different types of assessment; Defaults and penalties.
• Tax planning: Concepts, significance and problems of tax planning, Tax evasion, and tax avoidance, methods of tax planning.
• Tax considerations in specific business decisions, viz., make or buy; own or lease, retain or replace; export or domestic sales; shut-down or closure; expand or contract; invest or disinvest
• Computer Application in Income Tax and Tax Planning.

NATS Exam Centers

• Agra
• Ahmedabad
• Agartala
• Aizawl
• Ajmer
• Alappuzha
• Allahabad
• Amritsar
• Amarkantak
• Bangluru
• Banasthali
• Barrailey
• Baru Sahib(Sirmour)
• Bhilwara
• Bhavnagar
• Bikaner
• Bhopal
• Bhuvaneshwar
• Burdwan
• Chandigarh
• Chennai
• Coimbatore
• Delhi
• Dehradun
• Dharwad
• Gorakhpur
• Gulbarga
• Guwahati
• Gwalior
• Hyderabad
• Indore
• Jabalpur
• Jaipur
• Jalgaon
• Jammu
• Jodhpur
• Jorhat
• Kanpur
• Kolkata
• Kota
• Lakshmangarh
• Lucknow
• Midnapore
• Mumbai
• Nathdwara
• Panjim
• Patiala
• Patna
• Pondicherry
• Pune
• Raipur
• Rajkot
• Ranchi
• Ratlam
• Rohtak
• Sagar
• Salem
• Shillong
• Shimla
• Silchar
• Srinagar
• Surat
• Tezpur
• Trivandrum
• Tiruchirappalli
• Udaipur
• Vallabh
• Vidya Nagar
• Vadodara
• Varanasi
• Vishakhapatnam
• Wardha

FAQ’s on Natio0nal Accounting Talent Search

Question 1.
What is the National Accounting Talent Search Competition?

Answer:
National Accounting Talent Search (NATS) competition is a sole initiative of the Indian Accounting Association for undergraduate students to examine the accounting knowledge of the students at a junior and senior level.

Question 2.
What are the important dates for NATS?

Answer:

• Registration opens – October 4, 2022
• Registration closes – December 31, 2022
• Date of Examination – February 5, 2023

Question 3.
How to Apply for NATS 2023-24?

Answer:

• Go to the official website.
• Under the Exam icon at the topmost bar, click on registration.
• Make the required payment based on Junior and senior-level through debit/credit card.
• Keep your scanned copy of the photo and signature.
• After payment is done, click on the Next button.
• Fill the application form and upload the required documents.
• Submit the form.

Question 4.
How to make Online payment?

Answer:

1. Deposit Rs Amount in any branch of State Bank of Bikaner and Jaipur in the account of Coordinator, National Accounting Talent Search (Account No 61060736681), No bank charges are to be paid.
   or
2. Send a Demand Draft in Favour of Coordinator, National Accounting Talent Search  payable at Udaipur Rajasthan and send it to Prof. G. Soral, Coordinator, National Accounting Talent Search, Department of Accountancy and Statistics, UCCMS, Mohanlal Sukhadia University, Udaipur- 313 001 Rajasthan
3. You can make payment online via Debit card/Credit Card and NetBanking, We accept all bank cards.

Scholarships for Students

The post National Accounting Talent Search (NATS) 2023-24 | Eligibility, Awards, Application Procedure appeared first on Learn CBSE.