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Light Reflection and Refraction Class 10 Notes Science Chapter 10

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CBSE Class 10 Science Notes Chapter 10 Light Reflection and Refraction Pdf free download is part of Class 10 Science Notes for Quick Revision. Here we have given NCERT Class 10 Science Notes Chapter 10 Light Reflection and Refraction.

According to new CBSE Exam Pattern, MCQ Questions for Class 10 Science pdf Carries 20 Marks.

CBSE Class 10 Science Notes Chapter 10 Light Reflection and Refraction

REFLECTION

Reflection of Light: The phenomenon of bouncing back of light into the same medium by the smooth surface is called reflection.

Incident light: Light which falls on the surface is called incident light.

Reflected light: Light which goes back after reflection is called reflected light.

The angle of incidence: The angle between the incident ray and the normal.

An angle of reflection: The angle between the reflected ray and the normal.

Mirror: The surface which can reflect the light is a mirror.

Plane Mirror: If the reflecting surface is a plane then the mirror is plane.

Spherical Mirror: If the reflecting surface is part of the hollow sphere then the mirror is a spherical mirror.
The spherical mirror is of two types:

  • Convex mirror: In this mirror reflecting surface is convex. It diverges the light so it is also called a diverging mirror.
  • Concave mirror: In this mirror reflecting surface is concave. It converges the light so it is also called converging mirror.

Parameters of Mirror:

  • Center of Curvature: The centre of hollow sphere of which mirror is a part.
  • The radius of curvature: The radius of hollow sphere of which mirror is a part.
  • Pole: The centre of mirror (middle point) is pole.
  • Principal axis: The line joining the pole and center of curvature is called principal axis.
  • Aperture: Size of mirror is called aperture of mirror.
  • Principal Focus: The point on the principal axis, where all the incident rays parallel to principal axis converge or diverge after reflection through mirror.
  • Focal Length: The distance between pole and focus point is focal length.

Special Rays for Formation of Image:

  • A ray of light which is parallel to the principal axis of a spherical mirror, after reflection converges or diverges from focus.
  • A ray of light passing through or appearing from the center of curvature of spherical mirror is reflected back along the same path.
  • A ray of light passing through or appearing from the focus of spherical mirror becomes parallel to the principal axis.
  • A ray of light which is incident at the pole of a spherical mirror is reflected back making same angle with principal axis.

Use of Concave Mirror: It is used as a makeup mirror, the reflector in torches, in headlights of cars and searchlights, doctor’s head-mirrors, solar furnace, etc.

Sign Conventions of Spherical Mirror

  • All the distances are measured from the pole of the mirror as the origin.
  • Distances measured in the direction of incident rays are taken as positive.
  • Distances measured opposite to the direction of incident rays are taken as negative.
  • Distances measured upward and perpendicular to the principal axis are taken as positive.
  • Distances measured downward and perpendicular to the principal axis are taken as negative.
    \frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } …where f, v and u are focal length, image distance, object distance

Linear Magnification: This is the ratio of the height of the image to the height of the object.
m=\frac { { h }^{ ' } }{ h } …where m = magnification, h = height of image, h’ = height of object

Use of Convex Mirror: Convex mirror used as rear view mirror in vehicles, as shop security mirrors, etc.

REFRACTION

Refraction of Light: The bending of light at the interface of two different mediums is called Refraction of light.

  • If the velocity of light in medium is more, then medium is called optical rarer.
    Example, air or vacuum is more optical rarer.
  • If the velocity of light in medium is less, then medium is called optical denser.
    Example, glass is more denser than air.

Refractive Index: It represents the amount or extent of bending of light when it passes from one medium to another.
There are two types of refractive index

  • Relative refractive index and
  • Absolute refractive index.

Refractive index of medium with respect to other medium is called Relative Refractive Index.
Refractive index of medium 1 with respect to medium 2 = \frac { Speed\quad of\quad light\quad in\quad medium\quad 2(V2) }{ Speed\quad of\quad light\quad in\quad medium\quad 1(V1) }

Refractive index of medium with respect to air or vacuum is called Absolute Refractive Index.
Absolute refractive index of medium (m) = \frac { Speed of light in air(c) }{ Speed of light in medium (Vm) }

Incident ray: It is incoming ray on the refracting surface.

Refracted ray: It is an outgoing ray from the refracting surface.

An angle of incidence (i): It is the angle between incident rays and perpendicular line (normal) at the point of incidence.

An angle of refraction (r): It is the angle between refracted rays and perpendicular line (normal) at the point of incidence.

Law of Refraction: According to this law

  • “The incident ray, refracted ray and normal at the point of incidence all lie in the same plane.”
  • “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.”
    \frac { sin\quad i }{ sin\quad r } = constant (µ)

Lens: The transparent refracting medium bounded by two surfaces in which at least one surface is curved is called lens.
Lenses are mainly two type

  • Convex lens and
  • Concave lens.

Center of Curvature: The centres of two spheres, of which lens is part is called the centre of curvature.

Radii of Curvature: The radii of spheres, of which lens is part is called radius of curvature.

Principal Axis: The line joining the centres of curvature of two surfaces of lens is called principal axis.

Optical Center: It is a special point on the principal axis. Light incident on the optical centre passes through the lens without deviation.

Principal Focus: The point on the principal axis at which all incident rays parallel to the principal axis converge or appear to diverge after refraction through the lens.

Special Rays for Image Formation by Lens:

  • An incident ray, parallel to the principal axis, after refraction passes through (or appears to come from), second focus of the lens.
  • An incident ray, passing through the optical center of the lens, goes undeviated from the lens.
  • An incident ray, passing through the (first) principal focus of the lens, or directed toward it, becomes parallel to the principal axis after refraction through lens.

Use of Lens: In photographic cameras, magnifying glass, microscope, telescope, the human eye.

1. Light travels in a straight line.

2. Light gets reflected when it falls on polished surfaces; like mirrors.

3. Light suffers refraction when it travels from one medium to another.

4. There is a change in the wavelengths!light when it moves from one medium into another.

5. The bouncing back of light when it strikes a smooth or polished surface is called reflection of light. Reflection is of two types; Specular or regular and Diffuse or irregular reflection.

6. The angle of incidence is equal to the angle of reflection. Mathematically, we have ∠i = ∠r.

7. The image is as far behind the mirror as the object is in front.

8. The image is unmagnified, virtual and erect.

9. The image has right-left reversal.

10. Focal length of a plane mirror is infinity.

11. Power of a plane mirror is zero.

12. If a plane mirror is turned by an angle, the reflected ray turns by 2θ.

13. The least size of a plane mirror to view an object is equal to half the size of the object.

14. Pole (Vertex): The central point of a mirror is called its pole.

15. Centre of curvature : The centre of the sphere of which the mirror is a part is called the centre of curvature. It is denoted by C.

16. Radius of curvature : The radius of the sphere of which the mirror is a part is called the radius of curvature. It is denoted by R.

17. Principal axis : The straight line passing through the pole and the centre of curvature of the mirror is called the principal axis.

18. Principal focus : It is a point on the principal axis at which the rays parallel to the principal axis meet after reflection or seem to come from. For a concave mirror, the focus lies in front of the mirror and for a convex mirror, it lies behind the mirror. In short, a concave mirror has a real focus while aconvex mirror has a virtual focus.

19. Focal plane : A plane, drawn perpendicular to the principal axis and passing through the principal focus.

20. Focal length : The distance between the pole and the focus is called the focal length. It is represented by f. The focal length is half the radius of curvature.

21. Aperture: The size of the mirror is called its aperture. It is also defined as the effective diameter of the light reflecting area of the mirror.

22. Real image : When the rays of light, after reflection from a mirror, actually meet at a point, then the image formed by these rays is said to be real. Real images can be obtained on a screen.

23. Virtual image: When the rays of light, after reflection from a mirror, appear to meet at a point, then the image formed by these rays is said to be virtual. Virtual images can’t be obtained on a screen.

24. The following rays are used while drawing ray diagrams to find the position of an image :

  • A ray of light parallel to the principal axis after reflection passes through the focus. (1)
  • A ray of light passing through the focus after reflection becomes parallel to the principal axis. (2)
  • A ray of light incident on the centre of curvature retraces its path after reflection form the mirror.
    light-reflection-and-refraction-cbse-notes-for-class-10-science-1

25. For mirrors, the following results hold :
u is – ve, if the object is in front of the mirror.
(Real object)
u is + ve, if the object is behind the mirror.
(Virtual object)
v is – ve, if the image is in front of the mirror.
(Real image)
vis +ve, if the image is behind the mirror.
(Virtual image)
Focal length of a concave mirror is taken as – ve. Focal length of a convex mirror is taken as +ve.

26. When the image formed by a spherical mirror is real, it is also inverted and is on the same side of the mirror as the object. Since both v and u are negative, the magnification is negative.

27. When the image formed by a spherical mirror is virtual, it is also erect and is on the other side of the mirror as the object. In this case, u is – ve and v is + ve , therefore, m is positive.

28. The expression for the mirror formula is 1/u+1/v = 1/f

29. Linear magnification is given by the expression
light-reflection-and-refraction-cbse-notes-for-class-10-science-2

30. If m is positive, the image is erect w.r.t the object and if m is negative, the image is inverted w.r.t. the object.

31.The position of the image for various positions of the object for a concave mirror is as shown in the table below. The table also shows the use of the mirror for different positions of the object.
light-reflection-and-refraction-cbse-notes-for-class-10-science-3
The position of the image for various positions of the object for a convex mirror is as shown in the table below. The table also shows the use of the mirror for different positions of the object.
light-reflection-and-refraction-cbse-notes-for-class-10-science-4

32. The bending of light when it travels from one medium into another is called refraction of light

33.
light-reflection-and-refraction-cbse-notes-for-class-10-science-5

34. As light travels from ,one medium to another, the frequency of light does not change.

35. Light refracts because it has different speeds in different media.

36. The refraction of light obeys the following two laws :

  • The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
  • The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.This
    constant is called the index of refraction or refractive index.
    light-reflection-and-refraction-cbse-notes-for-class-10-science-6

37. If wng is the refractive index of glass w.r.t. water, ang  be the refractive index of glass w.r.t. air and anw  be the refractive index of water w.r.t. air ,then
light-reflection-and-refraction-cbse-notes-for-class-10-science-7

38. The most familiar and widely used optical device is the lens. A lens is an optical system with two refracting surfaces. The simplest lens has two spherical surfaces close enough together that we can neglect the distance between them. Such a lens is called a thin lens. The two common types of lenses are Converging lens or Convex lens, Diverging lens or Concave lens.

39. It should be noted that, if the above lenses are surrounded by .a material with a refractive index greater than that of the lens, the convex lens gets converted into a concave lens and vice-versa.

40. Any lens that is thicker at its centre than at its edges is a converging lens with positive f, and any lens that is thicker at its edges than at the centre is a diverging lens with negative f.

41. Optical centre : The central point C in the lens is called the optical centre. If a ray is incident towards the optical centre, it passes undeviated .through the lens.
light-reflection-and-refraction-cbse-notes-for-class-10-science-8

42.Principal axis: Since the lens contains two spherical surfaces, therefore, it has two centres of curvatures.
The line joining these centres and passing through the optical centre is called principal axis.

43. Aperture: The effective width of a lens through which refraction takes place is called the aperture.

44. Focus and Focal Length : If a beam of light moving parallel to the principal axis of a convex lens is incident on it, the rays converge or meet at a point on the principal axis. This point F is called the focus. The distance CF is called the focal length. If a beam of light moving parallel to the principal axis is incident on a concave lens, the beam of light diverges. If these diverged rays are produced backward, they meet at a point F on the principal . axis. The transmitted rays appear to come from this point. This point F is called the focus and distance CF is called the focal length.

45. For drawing the ray diagrams, we note the following :
light-reflection-and-refraction-cbse-notes-for-class-10-science-9
light-reflection-and-refraction-cbse-notes-for-class-10-science-10

  • All rays parallel to the principal axis after refraction pass through the principal focus or seem to come from it.
  • A ray of light passing through the focus after refraction becomes parallel to the principal axis.
  • A ray of light passing through the optical centre of the lens after refraction passes undeviated.

46. A convex and a concave lens can be supposed to be made-up of prisms.

47. Image formation by a concave lens.
light-reflection-and-refraction-cbse-notes-for-class-10-science-11

48. Image formation by a convex lens.
light-reflection-and-refraction-cbse-notes-for-class-10-science-12

49. New Cartesian sign conventions :

  • All distances, object distance (u), image distance (v) and focal length f are measured from the optical centre.
  • The distances measured in the direction of incident ray are taken as positive and distances measured against the direction of incident ray are taken as negative.
  • All distances (heights) of objects and images above principal axis are taken as positive and those below the principal axis are taken as negative.

50. For the two lenses, the sign conventions take the form

  • u is- ve, if the object is in front of the lens. (Real object)
  • u is +ve, if the object is virtual.
  • v is – ve, if the image is on the same side as that of the object. (Virtual image )
  • v is +ve, if the image is real.
  • Focal length of a concave lens is taken as – ve.
  • Focal length of a convex lens is taken as +ve.

51. Lens formula for convex lens 1/v-1/u = 1/f

52. The linear magnification produced by a lens is defined as the ratio of the size of the image (h’) to the size of the object (h). It is represented by m i.e.,
light-reflection-and-refraction-cbse-notes-for-class-10-science-13

53. If the magnification of a lens is negative, then the image formed is inverted and real.

54. If the magnification of a lens is positive, then the image formed is erect and virtual.

55. Power is defined as the reciprocal of the focal length. Power is measured in dioptre.

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Heredity and Evolution Class 10 Notes Science Chapter 9

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CBSE Class 10 Science Notes Chapter 9 Heredity and Evolution Pdf free download is part of Class 10 Science Notes for Quick Revision. Here we have given NCERT Class 10 Science Notes Chapter 9 Heredity and Evolution.

According to new CBSE Exam Pattern, MCQ Questions for Class 10 Science pdf Carries 20 Marks.

CBSE Class 10 Science Notes Chapter 9 Heredity and Evolution

Heredity and Inherited Traits: Mendel’s Experiment; Sex determination.
Heredity refers to the transmission of characters from parents to offsprings. An inherited trait is a particular genetically determined feature that distinguishes a person from the others for example; attached or free ear lobes in human beings.

Rules for the inheritance of traits:
Mendel’s contribution: The rules for inheritance of traits in human beings are related to the fact that both mother and father contribute an equal amount of genetic material i.e. DNA to their offspring. So an offspring will get two versions of that trait from the two parents. Mendel worked out rules for inheritance of these traits. Gregor Johann Mendel regarded as the ‘Father of Genetics’ performed his experiments with garden peas (Pisum sativum) in the garden behind his monastery. He observed a number of contrasting characters in garden peas and observed their inheritance.

Some important terms
1. Chromosomes are long thread-like structures present in the nucleus of a cell which contain hereditary information of the cell in the form of genes.

2. DNA is a chemical in the chromosome which carries the traits in a coded form.

3. Gene is the part of a chromosome which controls a specific biological function.

4. Contrasting characters: A pair of visible charactes such as tall and dwarf, white and violet flowers, round and wrinkled seeds, green and yellow seeds etc.

5. Dominant trait: The character which expresses itself in a (Ft) generation is dominant trait. Example : Tallness is a dominant character in pea plant.

6. Recessive trait: The character which does not express itself but is present in a generation is recessive trait. Ex. dwarfism in the pea plant.

7. Homozygous: A condition in which both the genes of same type are present for example; an organism has both the genes for tallness it is expressed as TT and genes for dwarfness are written as tt.

8. Heterozygous: A condition in which both the genes are of different types for example; an organism has genes Tt it means it has a gene for tallness and the other for dwarfness only tall character is expressed.

9. Genotype: It is genetic make up of an individual for example; A pure tall plant is expressed as TT and hybrid tall as Tt.

10. Phenotype: It is external appearance of the organism for example; a plant having Tt composition will appear tall although it has gene for dwarfness.

11. Homologous pair of characters are those in which one member is contributed by the father and the other member by the mother and both have genes for the same character at the same position.

Mendel’s Experiment: Mendel started his experiment on the pea plants. He conducted first monohybrid and then dihybrid crosses.

Monohybrid Cross: The cross in which Mendel showed inheritance of dominant and recessive characters is monohybrid cross. To observe inheritance of single pair of contrasting characters
Heredity and Evolution Class 10 Notes Science Chapter 9 1
he took pure tall (genotype TT) and pure dwarf (genotype tt) pea plants and cross pollinated them to obtain first generation or first filial generation. In this figuration (F1 generation) he obtained only tall plants. This meant that only one of the parental traits was seen, not the mixture of the two. The plants of F generation or progeny are then self pollinated to obtain F2 generation or progeny. Now all plants were not tall. He obtained 75% tall plants and 25% dwarf plants i.e. the phenotypic ratio was 3:1. This indicates that in the F, generation both tall and dwarf traits were inherited but tallness expressed it self. Tallness is a dominant trait and dwarfness is a recessive trait. F2 generation has a genotypic ratio of 1 : 2 : 1 of three types of plants represented by TT, Tt and tt as shown in the cross.

Conclusion: Phenotypic ratio—Tall : Dwarf 3 : 1
Genotype ratio—Pure Tall : Hybrid Tall : Pure Dwarf 1 : 2 : 1

Law of Dominance: When parents having pure contrasting characters are crossed then only one character expresses itself in the Ft generation. This character is the dominant character and the character/factor which cannot express itself is called the recessive character.

Dihybrid Cross: Mendel also carried out experiments to observe inheritance of two pairs of contrasting characters, which is called dihybrid cross. He cross breed pea plants bearing round green seed with plants bearing wrinkled and yellow seeds. In the Fx generation he obtained all round and yellow seeds it means round and yellow traits of seeds are dominant features while wrinkled and green are recessive. He self-pollinated the plants of F: generation to obtain F2 generation, he obtained four different types of seeds round yellow, round green, wrinkled yellow and wrinkled green in the ratio of 9 : 3 : 3 : 1. He concluded that traits are independently inherited

Conclusion

  • Round and yellow seeds-9.
  • Round and green seeds-3.
  • Wrinkled and yellow seeds-3.
  • Wrinkled and green seeds-1.

How do traits get expressed?
Cellular DNA is the information source for making proteins in the cell.
A part of DNA that provides information for one particular protein is called a gene for that protein for example; the height of a plant depends upon the growth hormone which is in turn controlled by the gene. If the gene is efficient and more growth hormone is secreted the plant will grow tall. If the gene for that particular protein gets altered and less of it is secreted when the plant will remain short. Both the parents contribute equally to the DNA of next generation during sexual reproduction. They actually contribute a copy of the same gene for example; when tall plant is crossed with short plant the gametes will have single gene either for tallness or for shortness. F1 generation will get one gene for tallness and other for shortness also.
Heredity and Evolution Class 10 Notes Science Chapter 9 2

How do germ cells i.e. gametes get single set of genes from parents who have two copies in them ?
Each gene set is present, not as a single long thread of DNA, but as separate independent pieces each called a chromosome. Each cell gets two copies of the chromosome, one from each parent. Each germ cell or gamete has one copy of it because there is reductional division in the sex organs at the time of formation of gametes. When fertilization takes place normal number of chromosomes is restored in the progeny ensuring the stability of DNA of the species.

How is the sex of a newborn individual determined?
It is the process by which sex of a newborn can be determined.

Different species use different strategies for this :

  • In some animals the temperature at which fertilized eggs are kept determines whether the developing animals will be males or females.
  • Some animals like snails can change sex indicating that sex is not genetically determined.
  • In human beings sex of the individual is determined genetically; means genes inherited from the parents decide the sex of the offspring.

Sex determination in human beings: In human beings, all chromosomes are not paired. 22 chromosomes are paired but one pair called sex chromosome is odd in not having a perfect pair in males. Females have a perfect pair both represented by XX. On the other hand males have a normal sized X but the other is short called Y so it is shown as XY. All gametes or ova formed by the homogenetic female are similar i.e. have X chromosome. Males heterogenetic form two types of sperms i.e. half with X chromosome and the other half with Y chromosome. Sex of the baby will depend on fertilization. There are two possibilities :
Heredity and Evolution Class 10 Notes Science Chapter 9 3

Autosomes: Those chromosomes which do not play any role in sex determination.

Sex chromosomes: Those chromosomes which play a role in determining sex of the newborn.

  • If the sperm having X chromosome fertilizes with ovum with X chromosome then the baby will have XX chromosome and it will be female.
  • If the sperm having Y chromosome fertilizes with ovum with X chromosome then the baby will have XY chromosomes and it will be male.

Evolution: Acquired and inherited traits, Speciation, Evolution and classification, Evolution by stages, Human evolution.

Evolution: It is the sequence, of gradual, irreversible changes which took place in the primitive organisms over millions of years to form new present-day species. Variations that resulted in formation of new species occurred basically due to errors in DNA copying as well as due to sexual reproduction.

An Illustration to show variations in a population: A group of twelve red beetles live in green bushes and reproduce sexually so are likely to develop variations. There are the following possibilities

First situation: Crows eat these beetles as they can easily pick up red ones in the green bushes There is a colour variation during sexual reproduction and green beetles appears, it reproduces and its population increases. Crows are not able to see green beetles so their population continues to increase but that of red beetles decreases. This type of variation gives a survival advantage.

Second situation: Due to a colour variation few blue beetle appear forming blue population. Crows can see both red and blue and eat them. Initially there are more of red beetles and less of blue. There is sudden calamity, an elephant kills red beetles by stamping on bush, blue beetles survive reproduce and increase in number. In this case there is no survival advantage but provides diversity without any adaptation.

Third situation: As the population of beetles increases, the bushes suffer from a disease and the availability of food for beetles decreases. The size of beetles decrease but after a few years as the plant disease is eliminated and enough food is available for the beetles they come back to their normal size. This type of change is not inherited.
Heredity and Evolution Class 10 Notes Science Chapter 9 4

Acquired Traits: Acquired traits are those which are not inherited over generations as they are caused due to change in the non-reproductive tissue and are not passed on the DNA of the germ cells for example; the size of the beetles in the population decreased due to scarcity of food.

Inherited Trait: Inherited traits are caused due to changes in the DNA of germ cells which are inherited from generation to generation, for example; formation of green beetles in the population of red beetles.

Acquired Traits and Inherited Traits

Acquired TraitsInherited Traits
(i) These are the traits which are developed in an individual due to special conditions.(i) These are the traits which are passed from one generation to the next.
(ii) They cannot be transferred to the progeny.(ii) They get transferred to the progeny.
(iii) They cannot direct evolution, e.g. low weight of starving beetles.(iii) They cannot direct evolution, e.g. low weight of starving beetles.

Charles Darwin’s Idea of Evolution: His concept of evolution was based on the idea that new species were formed due to variations that occurred in the organisms Nature played an important role in selecting the organisms having suitable variations.

Speciation: It means the development of one or more species from an existing species The factors that could lead to rise of a new species are :

Gene flow: It means the exchange of genetic material by interbreeding between populations of the same species or between individuals within a population. It increases the variation in the genetic composition of a population.

Genetic drift: It is random change in the frequency of alleles in a populate over successive generation due to errors in the gametes. The process is rapid in smaller population. Genetic drift can lead to accumulation of changes in the generations.

Natural selection: According to Darwin, natural selection also plays an important role in bringing about evolution of new species of plants and animals. According to him variations existed between the individuals of a population and some natural phenomena eliminated those individuals which were less adapted. The surviving population would pass the hereditary advantageous features to their offsprings. With time this process would give rise to organisms different from the original population and new species are formed.

Isolation: When a population of a species splits into two, it cannot reproduce with each other and forms a new species, for example; when a population of beetles feed on bushes on a mountain range, some may start feeding on nearby bushes finding entry into a new subpopulation. They reproduce with them so genes enter in a new population. Ultimately the two groups will be incapable of reproducing with each other and new species will be formed.

Evolution and Classification: The organisms show certain features, like appearance and behaviour which are called characteristics for example; Plants can perform photosynthesis. The basic characteristics are shared by a large number of organisms. More characteristics which two species have in common more closely are related, if they are more closely related then they have common ancestors (explain the example of brother sister and cousins). Evolutionary relationships can be traced with the help of the following :

Homologous organs: Those organs which have the same basic structural design and developmental origin but perform different functions and appearance, for example; Forelimbs of frog, lizard, bird, bat and human beings. They have same design of bones but they perform different functions.
Heredity and Evolution Class 10 Notes Science Chapter 9 5

Analogous organs: Those organs which have different basic design and developmental origin but have similar appearance and perform a similar function, for example; wings of bat and bird. Wings of bat are folds of skin attached between fingers. But wing of birds are modified forelimbs.
Heredity and Evolution Class 10 Notes Science Chapter 9 6

Study of Fossils: Fossils are preserved remains of living organisms that lived in the past. When living organisms die their bodies decompose but some parts of their body may be in such an environment that they do not decompose for example; if a dead insect gets caught in hot mud it will not decompose quickly but the mud will harden and retain impressions of the body parts of the insects. These impressions are also called fossils: The age of fossil can be estimated in two ways :
The fossils that occur closer to earth surface are more recent to those found in deeper layers.
The second method is isotope dating i.e. detecting the ratio of different isotopes of the same element in the fossil material.
Heredity and Evolution Class 10 Notes Science Chapter 9 7

Significance of fossils: Fossils are formed layer by layer in the earths crust. The animals and plants which existed earlier are buried in the deeper layer which ones found in the upper layer. It is found that, deeper fossils have simpler structure than found than upper layer. Complete fossil record of animals like horse, camel, man has helped us to study the stages of evolution.

Evolution by stages: Evolution is a continuous and gradual process, complicated organs did not evolve by a single DNA change but were formed by bit by bit change over generations for example; complex organs like eyes were created by bit by bit changes, in between the rudimentary eye in some insects also provided a fitness advantage. The structure of eye in all organisms is different enough to have evolutionary origins. Some organs even developed for one particular function but later become useful for quite a different function, e.g Feathers developed to provide warmth to the animal but later helped in flight.

Some dinosaurs had feathers although they could not fly, this shows that birds are closely related to reptiles, since dinosaurs were reptiles Some dissimilar looking structures also evolved from common ancestors. The current example of such a process is wild cabbage plant from which different vegetables are generated by artificial selection rather than natural selection
Heredity and Evolution Class 10 Notes Science Chapter 9 8

  • Selection of short distance between the leaves has led to formation of cabbage that, we eat.
  • Selection for arrested flower development had led to broccoli,
  • Selection for sterile flowers had made cauliflower,
  • Selection for swollen-stem had formed kohlrabi.
  • Selection for large leaves had formed leafy vegetable kale,
  • Selection for colored leaves formed red cabbage.

To sum up we can say that evolutionary relationships can be established by

  • Study of Homologous organs
  • Study of Analogous organs
  • Study of fossils
  • Changes in DNA during reproduction

Evolution versus Progress: Evolution can not be called progress from lower forms to higher forms. It is basically forming more complex designs while the simpler once also keep growing. Evolution is generation of diversity with the help of environmental selection. Bacteria which were formed first have the capacity to live in diverse conditions and are still flourishing; on the other hand human beings which are highly evolved species can not be called the pinnacle of evolution but yet another species in the evolving life forms.

Human Evolution: Human evolution has been studied with the help of excavation; time dating and fossil study All human beings belong to single species i.e. Homo sapiens. Human species have come from Africa. Some of our ancestors left Africa while others stayed on. These migrants slowly spread across the planet i.e. West Asia, Central Asia, Eurasia, South Asia and East Asia They traveled to Indonesia, the Philippines, Australia and America They traveled forward and backward sometimes separating and sometimes coming back to mix with each other. They had come into being as an accident of evolution.
Heredity and Evolution Class 10 Notes Science Chapter 9 9

Although there is a great diversity of human forms all over the world get all humans are single species.

  • They didn’t go in a single line.
  • They went forward and backward.
  • Moved in and out of Africa.
  • Sometimes come back to mix with each other.

Heredity and Evolution Class 10 Notes Science Chapter 9 10
Genetics: Branch of science that deals with heredity and variation.

Heredity: It means the transmission of features/characters/traits from one generation to the next generation.

Variation: The differences among the individuals of a species/population are called variations.

Mendel and his work on Inheritance.
Gregor Johann Mendel started his experiments on plant breeding and hybridisation. Mendel was known as Father of Genetics.
The plant selected by Mendel was Pisutn sativum (garden pea). Mendel used a number of contrasting characters for garden pea.

Sex Determination: Phenomenon of decision or determination of sex of an offspring.

Factors Responsible for Sex Determination:

  • Environmental: In some animals, the temperature at which the fertilised eggs are kept decides the gender. Example, in turtle.
  • Genetic: In some animals like humans gender or individual is determined by a pair of chromosomes called sex chromosomes (XX – female; XY – male).

Sex Chromosomes: In human beings, there are 23 pairs of chromosomes. Out of these 22 chromosome pairs are called autosomes and the last pair of chromosomes that help in deciding the gender of that individual are called sex chromosome.
XX – female; XY – male
The cross done shows that half the children will be boys and half will be girls. All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus sex of children will be determined by what they inherit from their father, and not from their mother.

Acquired Traits:

  • These are the traits which are developed in an individual due to special conditions.
  • They cannot be transferred to the progeny.
  • They cannot direct evolution, for example, the low weight of starving beetles.

Inherited Traits:

  • These are the traits which are passed from one generation to the next.
  • They get transferred to the progeny.
  • They are helpful in evolution, for example, the colour of eyes and hair.

Microevolution: It is the evolution which takes place on a small scale. Example, change in body colour of beetles.

Speciation: It is the process of formation of new species. A species is a group of similar individuals that belong to a population that can interbreed and produce fertile offspring. Speciation takes place when the variation is combined with geographical isolation.

Gene flow: It is the exchange of genetic material by interbreeding between populations of the same species or individuals. Gene flow occurs between populations that are partly but not completely separated.

Genetic Drift: It is the random change in the frequency of alleles (gene pair) in a population over successive generations.
Genetic drift takes place due to:

  • severe changes in the DNA.
  • change in the number of chromosomes.

Natural Selection: The process by which nature selects and consolidates those organisms which are more suitably adapted and possesses favourable variations.

Evolution and classification. Both evolution and classification are interlinked.

  • Classification of species is a reflection of their evolutionary relationship.
  • The more characteristics two species have in common the more closely they are related.
  • The more closely they are related, the more recently they have a common ancestor.
  • Similarities among organisms allow us to group them together and to study their characteristics.

Tracing Evolutionary Relationships:

  • Homologous Organs: Morphological and anatomical evidences. These are the organs that have same basic structural plan and origin but different functions.
    Example, forelimb of a horse (running), wings of bat (flying), paw of a cat (walk/ scratch/ attack) — same basic structure but different functions.
  • Analogous Organs: These are the organs that have different origin and structural plan but same functions.
    Example, wings of a bat (elongated fingers with skin folds), wings of bird (feathery covering along the arm) — different structures but same functions.
  • Fossils: The remains and relics of dead organisms that lived in the remote past. Fossils provide evidence of evolution. Example, a fossil called Archaeopteryx has feathered wings like birds but teeth and tail like reptiles hence suggesting that birds and reptiles had a common ancestor.

Artificial Selection: Humans have been a powerful agent in modifying wild species to suit their own requirement throughout ages by using artificial selection. Example, wheat (many varieties obtained due to artificial selection).

1. Heredity : It refers to the transmission of characters or traits from the parents to their offspring. Heredity is the continuity of features from one generation to another which are present in fertilised egg or zygote. The zygote develops into an organism of a particular type only.

2. Genetics : It is the branch of biology which deals with heredity and variation. Genetics is to help our understanding of heredity by knowing how offspring inherit characteristics from their parents.

3. Variation : It means the differences in the characters or traits among the individuals of a species. Variations occur during reproduction both because of error in DNA copying and as a result of sexual reproduction. Variations contribute to evolution.
Causes of variations:

  • Different combinations of genetic material.
  • Some positive gene mutations.
  • Interaction of genes with environmental changes (adaptations).

Importance of variations:

  • It forms, the. basis of heredity.
  • It causes adaptations due to which organism can easily adjust to its changing environment.
  • Accumulation of variations forms the basis of evolution.

Remember!
Variations are produced both in sexual and asexual reproduction but amount of variations produced in asexual reproduction are subtle (so little) that they are hardly noticeable as compared to variations caused due to sexual reproduction.

4. Genotype : The genetic constitution of an organism e.g., Genotype of human male is 44 + XY and
genotype of human female is 44 + XX

5. Phenotype : The appearance of the organism, i.e., the way in which genotype is expressed. Phenotype is the result of interaction of genes with the environment.
e.g., Red colour may be controlled by a pair of genes RR. Now if genotype is RR phenotype will be red only but if genotype is Rr then also phenotype will be red since R is a dominant gene.

6. Gene : It is the basic unit of inheritance by which characters are transferred from parents to their offspring. Gene consists of a specific length of DNA on a chromosome. A specific Segment of DNA that provides the information for one protein is called gene for that protein.
According to Mendel, both parents must contribute equally to the DNA of the progeny during sexual reproduction. As both parents determine the trait in the progeny, so both parents must be contributing a copy of the same gene.

7. Chromosomes : These are the long threads present in the nucleus of every cell. Chromosomes are made- up of DNA and protein. Each chromosome contains very long molecule of DNA.
Remember!
Each gene set is present as separate independent pieces each called a chromosome. Each cell have two copies of each chromosome, one each from male and female parents. Every germ cell will take one chromosome from each pair and these may be of either maternal or paternal origin. When two germ cells combine, they will restore the normal number of chromosomes in the progeny, ensuring the stability of the DNA of the species. Such mechanism of inheritance is used by all sexually and asexually reproducing organisms.

8. Allele: It is an alternative form of a gene occupying the same position on a chromosome and affecting the same characters but in two alternative ways, e.g., the free and attached ear lobe are the alleles of ear lobe character.
Expressing allele of a gene :

  • Homozygous dominant in capital letters, e.g., tallness(TT)
  • Homozygous recessive in smalMetters, e.g., shortness or dwarfness (tt)
  • Heterozygous (Tt)-lt will be called hybrid tall.

9. Dominant allele: An allele that affects the phenotype of an organism both in heterozygous and homozygous condition. It is denoted by a capital letter, e.g., tallness in pea plant is denoted by ‘T.

10. Recessive allele: An allele that affects the phenotype of the organism in absence of a dominant allele, i.e., in homozygous recessive individuals. It is denoted by a small alphabet, e.g., dwarfness in pea plant is denoted by’t’.

11. Homozygous: When both alleles of a particular gene are the same, e.g., TT

12. Heterozygous : When both alleles of a particular gene are different, e.g., Tt

13. Diploid : Cells or organism containing two sets of genes, e.g., human body cells. Diploid cells have genetic constitution of 2n.

14. Haploid : Cells or organism containing one set of genes, e.g., human reproductive cells (sperms and ova). Haploid cells have genetic constitution of n.

15. Monohybrid cross : A cross between two parents taking the alternative traits of one single character, e.g., A cross between tall and dwarf pea plants.
Monohybrid Ratio :

  • In F1 generation : 100% hybrid
  • In F2 generation : phenotypic ratio is 3 : 1 and genotypic ratio is 1 : 2 : 1

16. Dihybrid cross: A cross between two parents taking into consideration alternative traits of two different characters, e.g., A cross between two pea plants one having round, green seeds and the other having wrinkled, yellow seeds.
Dihybrid Ratio :

  • F1 ratio is 100% Hybrid type.
  • F2 ratio : Phenotypic is 9 : 3 : 3 : 1 and Genotypic . ratio is very complex.

17. Human Blood Groups: There are four types of blood groups A, B, AB or O. These are controlled by a gene which is denoted by symbols IA, IB and IO (sometimes also denoted as i). The genes IA and IB show no dominance over each other (they are codominant, i.e., both expresses themselves independently). But these both genes are dominant over the gene IO. Therefore, blood group of a person depends on the type of genes present, e.g., (i) Blood group A has the following gene types :
heredity-and-evolution-cbse-notes-for-class-10-science-1

18. Determining sex of a newborn individual genetically:

  • In human beings the sex of the individual is ” determined genetically.
  • There are 23 pairs of chromosomes of which 22 are similar in male and female and are known as autosomes.
  • The remaining one is sex chromosome which is XY in males and XX in females.
  • Males produce two types of sperms X and Y, while female produces one type of egg X.
  • If a X type of sperm fertilizers the egg then the sex of baby will be female (XX).
  • If Y type of sperm fertilizers the egg then the sex of the baby will be male (XY).
    heredity-and-evolution-cbse-notes-for-class-10-science-2

19. Mendel’s experiment to show that traits may be dominant or recessive:

  • Mendel conducted breeding experiments in garden pea.
  • selected pure plant of a tall/short plant.
  • produced first generation plants by crossing them.
  • found that all plants were tall.
  • produced the second generation by self-fertilization of hybrids.
  • found that three-quarter of the plants was tall and one quarter was short.
    heredity-and-evolution-cbse-notes-for-class-10-science-3

20. Homologous chromosomes: A pair of corresponding chromosomes of the same shape and size, one from each parent.

21. Autosomes and Sex chromosomes : The identical » chromosome pairs are called autosomes. The
chromosome pair which is different are called sex chromosomes. Humans have 23 pairs of chromosomes. 1-22 pairs are autosomes while 23rd pair (XX in females and XY in males) which are designated as X and Y are sex chromosomes.

22. Molecular Phylogeny: It is the study of evolutionary relationships by comparing DNA of different species.

23. Natural selection : Natural selection is one of the basic mechanisms of evolution, along with mutation, migration and genetic drift. Natural selection means the environmental conditions prevailing around an organism against which organism adapts itself, grows – and reproduces further. This leads to a change in the composition of genes within a population further causing evolution. Thus, it can be said that,
Natural selection results in adaptation in population to fit their environment better. Thus, natural selection direct evolution in the population of a particular species.

24. Fossils of the information which they provide regarding evolution: Fossils are the remains of ancient life forms, which got preserved somehow in the layers of earth, snow or oil.
Information given by fossils:

  • They reveal that the life forms which existed earlier do not exist today which indicate that the living forms are ever changing (evolving).
  • They are used to guess the time when a particular organism existed on earth. It is done through carbon dating.

25. Genetic drift: The change in the frequency of some genes in a population which provides diversity without any survival advantage is called genetic drift.

26. The various ways in which individuals with a particular trait may increase in a population : Differences in population are responsible for the diversity such as, colour of eyes, hair, shape of ear lobes. This occurs due to : (i) Sexual reproduction (ii) Inaccuracies during DNA replication (iii) Due to environmental changes. This diversity will increase with time as these variations can be passed on only through DNA/genes during reproduction through reproductive tissue (germ cells or gametes).

  • If these variations give survival advantage, then such traits are selected in nature and such traits increase in a population.
  • Due to genetic drift. This occurs due to geographical or reproductive isolation. It results in the change in gene frequency in a particular: population.
  • Migration which leads to gene flow in and out of the population.
  • The mutation caused due to particular type of environment. ,
  • Acquired traits due to particular type of environment.

27. Evidence of evolution: Errors in DNA copying (mutation) and sexual reproduction lead to variations which form the basis of evolution. Characteristics that
are common in different kinds of living organisms provide evidence in favour of evolution.

28.Evolution : Evolution can be defined as a naturally occurring slow, continuous and irreversible process of change. The gradual change of living organisms from pre-existing organisms since the beginning of life is called organic evolution. Whereas, gradual change in elements from one form to another with time is termed as inorganic evolution, i.

29.Inherited traits : are those traits which are passed from one generation to another through specific genes. Any change in DNA of the germ cells will be passed.
30. Acquired traits : are those traits which are acquired by the organism in its lifetime, e.g., removal of tail cannot change the genes of the germ cells of the mice thus cannot be passed to next generation.

31. Speciation : It means the origin of new species from the existing ones. It happens when different populations of the same species evolve along different lines.
How speciation occurs ?

  • It occurs when two populations are isolated (both geographically and reproductively) leading to almost no gene flow between the two populations.
  • Over generations, genetic drift will accumulate different changes in each sub-population.
  • Natural selection may also operate differently in these different locations.
  • Together natural selection and genetic drift will cause such changes (severe changes in the DNA) that these two groups will not be able to reproduce with each other even if they happen to meet.
  • When DNA changes occur to larger extent, it may lead to change in the number of chromosomes or gene expression, eventually the germ cells of the two groups cannot fuse with each other. This leads to emergence of new species.

32. Estimating Age of Fossil: There are 2 methods :

  • Relative method : On digging, the fossils which are closer to the surface are more recent than the fossils found in deeper layers.
  • Dating fossils (carbon dating method): It is done by detecting the ratios of different isotopes of the same element (i.e., isotope of C-14 which is radioactive) in the fossil material.

33. Evolution by stages :

Complex organs like eye has evolved from rudimentary organs, (e.g., rudimentary eye in flatworm might be useful enough to give only a fitness advantage and the structure of eye in different organisms is different indicating them to have different evolutionary origins) not by a single DNA change but created bit-by-bit over generations.

A change that is useful for one property to start with can become useful later for quite a different
function (e.g., Feathers might start as providing insulation in cold weather. But later, they might be useful for flight. Some heavy birds and reptiles also have feathers but they do not fly.

Some very dissimilar looking structures evolve from a common ancestral design, e.g., wild cabbage was cultivated as a food plant and many different vegetables were generated by selection over last two thousand years, (a) Selection of very small distances between the leaves gave rise to cabbage we eat. (b) Selection for arrested flower development gave rise to broccoli, (c) Selection for sterile flowers gave rise to cauliflower (d) Selection for swollen parts gave rise to kohlrabi.(e) Selection for larger leaves gave rise to leafy vegetable kale. It suggests that, if these selections were not done then there would have been only wild cabbage.

34. Homologous organs are organs having same origin and basic structure but they appear different and perform different functions in various organisms,
e.g.,

  • Forelimbs of horse and arms of man.
  • Wings of birds and flippers of whale.

Similarities in basic structure of (homologous) organs in different organisms, indifferent groups indicate common ancestry.

35. Analogous organs are organs, which look similar because they perform same function, but they do not have same origin and basic structure.
e.g.,

  • Wings of birds and wings of insects.
  • Fins of fish and flippers of the whale.

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Nationalism in India Class 10 Notes History Chapter 3

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Nationalism in India Class 10 Notes Social Science History Chapter 3 SST Pdf free download is part of Class 10 Social Science Notes for Quick Revision. Here we have given Nationalism in India Class 10 History Chapter 3 Notes. According to new CBSE Exam Pattern, MCQ Questions For Class 10 Social Science with Answers Carries 20 Marks.

Formulae Handbook for Class 10 Maths and Science

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectSocial Science Notes
ChapterHistory Chapter 3
Chapter NameNationalism in India
CategoryCBSE Revision Notes

Nationalism in India Class 10 Notes Social Science History Chapter 3

Mahatma Gandhi and the idea of Satyagraha:
Mahatma Gandhi returned to India in 1915 from South Africa. Gandhiji’s novel method of mass agitation is know as ‘Satyagraha’. Satyagraha emphasized truth. Gandhiji believed that if the cause is true, if the struggle is against injustice, then physical force was not necessary to fight the oppressor. A satyagrahi can win the battle through non-violence. People, including oppressors, had to be persuaded to see the truth. Truth was bound to ultimately triumph.

In India the first was at Champaran in 1916 to inspire plantation workers to struggle against oppressive plantation system. In 1917 Satyagraha at Kheda to support peasants.

In 1918 Satyagraha at Ahmadabad:
Among the cotton mill workers.

‘Hind Swaraj’:
The famous book written by Mahatma Gandhi, which emphasized non-cooperation to British rule in India.

New economic situation created in India by the First World War:

  1. Manchester imports into India declined as the British mills were busy with war production to meet the needs of the army paving the way for the Indian mills to supply for the huge home market
  2. As the war prolonged, Indian factories were called upon to supply war needs. As a result new factories were set up, new workers were employed and everyone was made to work longer hrs.
  3. Cotton production collapsed and exports of cotton cloth from Britain fell dramatically after the war, as it was unable to modernize and compete with US, Germany, Japan. Hence within colonies like India, local industrialists gradually consolidated their position capturing the home market.

The Rowlatt Act of 1919:
It gave the British government enormous power to repress political activities and allowed detention of political prisoners without trial for two years.

Jallianwala Bagh incident:
On 13th April 1919, a crowd of villagers who had come to attend a Baisakhi fair, gathered in the enclosed ground of Jallianwala Bagh. Being from outside the city, many were not aware of the martial law that had been imposed as a repressive measure. General Dyer with his British troops entered the park and closed the only exit point without giving any warning to the assembled people and ordered the troops to fire at the crowds, killing hundreds. This brutal act of General Dyer provoked unparalleled indignation. As the news of Jallianwala Bagh spread, crowds took to the streets in many North Indian towns. There were hartals, clashes and attacks on government buildings.

Non-cooperation programme was adopted at Nagpur in Dec. 1920.

Effects of the Non-cooperation Movement on the economy of India:
Foreign goods were boycotted, liquor shops were picketed and foreign cloth was burnt. The import of foreign cloth halved between 1921-1922. Its value dropped from Rs 102 crore to Rs 57 crore. Many merchants and traders refused to trade in foreign goods or finance foreign trade. People began discarding imported clothes and wearing Indian ones. The production of Indian textile mills and hand looms went up. Use of khadi was popularized.

Non-cooperation Movement in the countryside:

  • In Awadh, the peasants’ movement led by Baba Ramchandra was against talukdars and landlords who demanded extremely high rents and a variety of other ceases from the peasants. Peasants were forced to work in landlords’ farms without any payment (beggar). Peasants had no security of tenure, thus being regularly evicted so that they could acquire no right over the leased land. The demands of the peasants were— reduction of revenue, abolition of beggar and social boycott of oppressive landlords.
  • In the Gudem Hills of Andhra Pradesh a militant guerrilla movement spread in the early 1920s against the closure of forest areas by the colonial government, preventing people from entering the forests to graze their cattle, or to collect fuel wood and fruits. They felt that their traditional rights were being denied.
  • For plantation workers in Assam, freedom meant the right to move freely in and out of the confined space in which they were enclosed. It meant retaining a link with the village from which they had come. Under the Inland Emigration Act of 1859, plantation workers were not permitted to leave tea gardens without permission. In fact the permission was hardly granted. When they heard of the Non-Cooperation Movement, thousands of workers defied the authorities and left for their homes.

Slowing down of Non-cooperation Movement in cities:

  • Khadi cloth was more expensive than mill cloth and poor people could not afford to buy it. As a result they could not boycott mill cloth for too long.
  • Alternative Indian institutions were not there which could be used in place of the British ones.
    These were slow to come up.
  • So students and teachers began trickling back to government schools and lawyers joined back work in government courts.

Khilafat movement:
Khilafat movement was started by Mahatma Gandhi and the Ali Brothers, Muhammad Ali and Shaukat Ali in response to the harsh treatment given to the Caliph of Ottoman empire and the dismemberment of the Ottoman empire by the British.

Chauri Chaura incident:
In February 1922, Gandhiji decided to launch a no tax movement. The police opened fire at the people who were taking part in a demonstration, without any provocation. The people turned violent in their anger and attacked the police station and set fire to it. The incident took place at Chauri Chaura in Uttar Pradesh.

When the news reached Gandhiji, he decided to call off the Non-cooperation movement as he felt that it was turning violent and that the satyagrahis were not properly trained for mass struggle.

Swaraj Party was founded by C.R. Das and Moti Lai Nehru for return to council Politics. Simon Commission 1928 and boycott. Lahore Congress session and demand for Puma Swaraj in 1929. Dandi march and the beginning of civil Disobedience movement.
Features of Civil Disobedience Movement:

  • People were now asked not only to refuse cooperation with the British but also to break colonial laws.
  • Foreign cloth was boycotted and people were asked to picket liquor shops.
  • Peasants were asked not to pay revenue and chaukidari taxes.
  • Students, lawyers and village officials were asked not to attend English medium schools, colleges, courts and offices.

‘Salt March’:
On 31st January, 1930 Mahatma Gandhi sent a letter to Viceroy Irwin stating eleven demands, one of which was the demand to abolish Salt Tax. Salt was one of the most essential food items consumed by the rich and poor alike and a tax on it was considered an oppression on the people by the British Government. Mahatma Gandhi’s letter was an ultimatum and if his demands were not fulfilled by March 11, he had threatened to launch a civil disobedience campaign. So, Mahatma Gandhi started his famous Salt March accompanied by 78 of his trusted volunteers. The march was over 240 miles, from Gandhiji’s ashram in Sabarmati to the Gujarati coastal town of Dandi. The volunteers walked for 24 days, about 10 miles a day. Thousands came to hear Mahatma Gandhi wherever he stopped, and he told them what he meant by Swaraj and urged them to peace-fully defy the British. On 6th April, he reached Dandi, and ceremonially violated the law, manufacturing salt by boiling sea water. This marked the beginning of the Civil Disobedience Movement.

Who participated in the movement?
Civil Disobedience Movement came into force in various parts of the country. Gandhiji led the salt march from Sabarmati Ashram to Dandi with his followers starting the Civil Disobedience Movement. In the countryside, the rich Patidars of Gujarat and Jats of Uttar Pradesh were active in the movement. As rich communities were very hard hit by the trade depression and falling prices, they became enthusiastic supporters of the Civil Disobedience Movement. Merchants and industrialists supported the movement by giving financial assistance and also by refusing to buy and sell the imported goods. The industrial working class of Nagpur region also participated in the Civil Disobedience Movement. Railway workers, dock workers, mineral of Chhota Nagpur, etc. participated in protest rallies and boycott campaigns.

Limits of the movement
less participation by untouchables—Ambedker for separate electorate and Poona pact of 1932, Luke warm response by some Muslim Political Organization.

Provisions of Poona pact of 1932:
Signed between Dr. Ambedkar and Gandhiji. It gave depressed classes reserved seats in central provincial councils but they were to be voted by the general electorate.

The sense of collective belonging:
Though nationalism spread through the experience of united struggle but a variety of cultural processes captured the imagination of Indians and promoted a sense of collective belonging:

  1. Use of figures or images: The identity of India came to be visually associated with the image of Bharat Mata. Devotion to the mother figure came to be seen as an evidence of one’s nationalism
  2. Indian folklore: Nationalists started recording and using folklore’s and tales, which they believed, gave a true picture of traditional culture that had been corrupted and damaged by outside forces. So preservation of these became a way to discover one’s national identity and restore a sense of price in one’s past.
  3. Use of icons and symbols in the form of flags: Carrying the tricolor flag and holding it aloft during marches became a symbol of defiance and promoted a sense of collective belonging.
  4. Reinterpretation of history: Indians began looking into the past to rediscover the glorious developments in ancient times in the field of art, science, mathematics, religion and culture, etc. This glorious time was followed by a history of decline when India got colonized, as Indian history was miserably written by the colonizers.

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NTSE Uttar Pradesh 2020 for Class X Admit Card (Released) | Exam Date, Eligibility, Exam Pattern

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NTSE Uttar Pradesh 2019: NTSE Uttar Pradesh exam board has released the admit card for the stage 2 exam. Candidates who want to appear for the Uttar Pradesh NTSE exam 2019-20 can refer the downloading process of admit card from the article given below the NTSE Uttar Pradesh exam is going to be conducted in 2 stages. The result for stage 1 is going to be declared in February 2019 while the result for stage 2 is going to be conducted in August 2019. Candidates that qualify the stage 1 exam are the only ones eligible for the stage 2 exam. The final merit of the NTSE exam will be prepared based on the overall performance of the candidates.

NTSE Uttar Pradesh Exam 2019

Candidates can their main exam result by going to the official website of NTSE Uttar Pradesh. The NTSE UP result contains details like roll number, candidate name, date of birth, marks scored by the candidate in CAT, MAT, school name, and language test. The final result of NTSE UP exam comprises of 2 results will be declared in August.

NTSE or national talent search exam is a scholarship exam that is conducted by NCERT. This exam is conducted to nurture and identify young talent in India. Students that are currently studying in class 10 can appear for NTSE 2019 exam in order to get the scholarship till the doctorate level of their academics.

Click Here to download UP NTSE Admit Card

NTSE Uttar Pradesh Exam Dates

NTSE UP Exam EventsDates
Application form releasing dateAugust 2019
Application concludes30 September 2019
Admit cards24 October 2019
NTSE Stage 1 Uttar Pradesh examNovember 3, 2019
NTSE stage 1 answer keyThe second week of November 2019
NTSE Uttar Pradesh Result & CutoffLast week of February 2020
NTSE admit card for stage 2April 2020
NTSE Stage 2 examMay 10, 2020

NTSE UP Exam Application Form

Below is the important information related to the NTSE UP exam application form for the candidates

  • The candidates that appearing for the NTSE exam and applying through online mode are instructed to fill the application form and submit the form.
  • However, for filling the application form, candidates need to register themselves on the official NTSE exam website.
  • After registration is done successfully, candidates need to download the application form from the main website of the university.
  • Then the candidates need to keep all the important documents ready to fill in with the application form.
  • During the filling of the application form, candidates will be required to enter the personal details, contact details, course-related details, and more additional details will be required.
  • Candidates are advised to fill all the details in the application form properly and correctly.
  • Once the details are entered in the application form completely, candidates are instructed to upload the relevant documents with the application form in the size and format mentioned on the main website.
  • Once the procedure for filling the application form is done applicants are requested to take a printout of the application form.
  • After that applicant has to pay the application fee as mentioned below.
  • After the application form is duly signed, submit it before the due date and complete your application process.

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NTSE UP Exam Fee

  • For the application process to be over, it is important that the candidates pay the application fee and after that appear for the exam.
  • However, candidates are required to pay Rs. 50 as the application fee if he/she or belongs to an OBC or general category.
  • Candidates belonging to reserved categories are required to pay Rs. 30 as the application fee for the exam.
  • Candidates need to pay the application fee through a bank account along with the duly signed application form.
  • Candidates should note that the bank slip submitted should be in favor of the exam and it is payable at the secretary exam regulatory authority, UP Allahabad.
  • Here are the bank details for the candidate’s reference
    • AC no. 519701010028014
    • IFSC code – UBIN0551970
    • Union Bank of India, LIC Colony, Allahabad

Online Application for UP ITI has been started from 15th July 2019.

NTSE UP Question Paper Pattern

The NTSE UP exam stage 1 divided into 3 sections. These are the mental ability test, language comprehension test, and scholastic ability test. Below is the table where the distribution details of maximum marks, questions, and time allotted for every section is provided.

SectionsNumber of questionsMaximum MarksTime Allotted
Language Comprehension Test (LCT)505045 minutes
Mental Ability Test (MAT)505045 minutes
Scholastic Ability Test (SAT)10010090 minutes
  1. Each right answer in MAT, as well as SAT stage 1 exam, carries 1 mark. Also, there are no negative marks in any of this section.
  2. SAT and MAT sections in the NTSE stage 1 exam have MCQs with 4 options. Candidates need to mark the correct answer to each question given on the OMR sheet.
  3. The LCT section given is just for qualifying the exam. The marks scored in this section will not be added in the final score of the candidate. Thus, it is not considered as a decisive factor in the last stage of the state-level merit list.
  4. The SAT section given evaluates candidates’ knowledge and most likely covers subjects like science, mathematics, and social science for class 10 and 12 levels.
  5. The MAT section evaluates candidates’ ability to think, reasoning skills, and spatial orientation, etc. Questions that are asked in the MAR section are based on series, analogies, coding-decoding, patterns, problem-solving, etc.

NTSE UP Exam Answer Key

The answer key for the NTSE UP exam for stage 1 will be released on the official NTSE UP exam website which is scertup.co.in. once the exam is over. The answer key for the NTSE exam helps the candidates to check their answers and get an idea of what their score might be. Below is the step by step process for checking the NTSE UP exam answer key:

  1. Go to the official website of the NTSE UP exam and click on the answer key option.
  2. After the answer key is downloaded, save it and download your OMR sheets for evaluating.
  3. Candidates can also download their individual OMR sheets published on the main website and can compare the answers which they have submitted in the main exam.
  4. Also, candidates can challenge the answer key if they have substantial proof. Based on this proof, a revised answer key set will be released on the official website.
  5. The final answer key will be used for the NTSE exam results.

How to Check NTSE UP Exam Result?

  • Go to the official NTSE UP website.
  • There will be an NTSE result tab on the homepage, click on it.
  • Enter your details like date of birth and roll number.
  • The result will be visible on your desktop screen.

Details Mentioned On NTSE UP Exam Result

NTSE Uttar Pradesh exam result will have the following details

  • School name
  • Candidates name
  • Candidates roll number
  • Marks scored by candidates in MAT, LCT & SAT
  • Candidates category, gender, and physical disability status
  • Rank secured
  • Grand total

NTSE UP Exam Cutoff Scores

Cutoff scores are the minimum scores that one needs to score in the main exam to qualify for the NTSE stage 2 exam. The cutoff scores every year are changed on the basis of

  • Regulation of the conducting authority of the exam
  • Exam difficulty level
  • Previous year’s cutoff trends
  • Number of students that are appearing for the exam

The qualifying marks for stage 1 category students is given in the table below

CategorySectionQualifying marks
GeneralLAT20
MAT20
SAT40
SC/ST/PHCLAT16
MAT16
SAT32

The overall cutoff marks for the NTSE UP exam for each category have been released. Below is the table that contains category wise cutoff for NTSE UP stage 1 exam:

CategoryCut Off 2017-18
General133
SC115
ST109
PH81

State-level National Talent Search Exam

The post NTSE Uttar Pradesh 2020 for Class X Admit Card (Released) | Exam Date, Eligibility, Exam Pattern appeared first on Learn CBSE.

Physics MCQs for Class 12 Chapter Wise with Answers Pdf Download

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  • Chapter 1 Electric Charges and Fields
  • Chapter 2 Electrostatic Potential and Capacitance
  • Chapter 3 Current Electricity
  • Chapter 4 Moving Charges and Magnetism
  • Chapter 5 Magnetism and Matter
  • Chapter 6 Electromagnetic Induction
  • Chapter 7 Alternating Current
  • Chapter 8 Electromagnetic Waves
  • Chapter 9 Ray Optics and Optical Instruments
  • Chapter 10 Wave Optics
  • Chapter 11 Dual Nature of Radiation and Matter
  • Chapter 12 Atoms
  • Chapter 13 Nuclei
  • Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
  • Chapter 15 Communication Systems

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Chemistry MCQs for Class 12 Chapter Wise with Answers Pdf Download

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  • Chapter 1 The Solid State
  • Chapter 2 Solutions
  • Chapter 3 Electrochemistry
  • Chapter 4 Chemical Kinetics
  • Chapter 5 Surface Chemistry
  • Chapter 6 General Principles and Processes of Isolation of Elements
  • Chapter 7 The p-Block Elements
  • Chapter 8 The d-and f-Block Elements
  • Chapter 9 Coordination Compounds
  • Chapter 10 Haloalkanes and Haloarenes
  • Chapter 11 Alcohols, Phenols and Ethers
  • Chapter 12 Aldehydes, Ketones and Carboxylic Acids
  • Chapter 13 Amines
  • Chapter 14 Biomolecules
  • Chapter 15 Polymers
  • Chapter 16 Chemistry in Everyday Life

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  • Chapter 1 Reproduction in Organisms
  • Chapter 2 Sexual Reproduction in Flowering Plants
  • Chapter 3 Human Reproduction
  • Chapter 4 Reproductive Health
  • Chapter 5 Principles of Inheritance and Variation
  • Chapter 6 Molecular Basis of Inheritance
  • Chapter 7 Evolution
  • Chapter 8 Human Health and Disease
  • Chapter 9 Strategies for Enhancement in Food Production
  • Chapter 10 Microbes in Human Welfare
  • Chapter 11 Biotechnology: Principles and Processes
  • Chapter 12 Biotechnology and its Applications
  • Chapter 13 Organisms and Populations
  • Chapter 14 Ecosystem
  • Chapter 15 Biodiversity and Conservation
  • Chapter 16 Environmental Issues

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  • Chapter 1 Physical World
  • Chapter 2 Units and Measurements
  • Chapter 3 Motion in a Straight Line
  • Chapter 4 Motion in a Plane
  • Chapter 5 Laws of Motion
  • Chapter 6 Work, Energy and Power
  • Chapter 7 System of Particles and Rotational Motion
  • Chapter 8 Gravitation
  • Chapter 9 Mechanical Properties of Solids
  • Chapter 10 Mechanical Properties of Fluids
  • Chapter 11 Thermal Properties of Matter
  • Chapter 12 Thermodynamics
  • Chapter 13 Kinetic Theory
  • Chapter 14 Oscillations
  • Chapter 15 Waves

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  • Chapter 1 Some Basic Concepts of Chemistry
  • Chapter 2 Structure of Atom
  • Chapter 3 Classification of Elements and Periodicity in Properties
  • Chapter 4 Chemical Bonding and Molecular Structure
  • Chapter 5 States of Matter
  • Chapter 6 Thermodynamics
  • Chapter 7 Equilibrium
  • Chapter 8 Redox Reactions
  • Chapter 9 Hydrogen
  • Chapter 10 The s-Block Elements
  • Chapter 11 The p-Block Elements
  • Chapter 12 Organic Chemistry: Some Basic Principles and Techniques
  • Chapter 13 Hydrocarbons
  • Chapter 14 Environmental Chemistry

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  • Chapter 1 The Living World
  • Chapter 2 Biological Classification
  • Chapter 3 Plant Kingdom
  • Chapter 4 Animal Kingdom
  • Chapter 5 Morphology of Flowering Plants
  • Chapter 6 Anatomy of Flowering Plants
  • Chapter 7 Structural Organisation in Animals
  • Chapter 8 Cell: The Unit of Life
  • Chapter 9 Biomolecules
  • Chapter 10 Cell Cycle and Cell Division
  • Chapter 11 Transport in Plants
  • Chapter 12 Mineral Nutrition
  • Chapter 13 Photosynthesis in Higher Plants
  • Chapter 14 Respiration in Plants
  • Chapter 15 Plant Growth and Development
  • Chapter 16 Digestion and Absorption
  • Chapter 17 Breathing and Exchange of Gases
  • Chapter 18 Body Fluids and Circulation
  • Chapter 19 Excretory Products and their Elimination
  • Chapter 20 Locomotion and Movement
  • Chapter 21 Neural Control and Coordination
  • Chapter 22 Chemical Coordination and Integration

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Chemistry MCQs for Class 12 with Answers Chapter 2 Solutions

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 2 Solutions. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry Solutions MCQs Pdf with Answers to know their preparation level.

Solutions Class 12 Chemistry MCQs Pdf

Question 1. Mole fraction of glycerine C3H5(OH)3 in solution containing 36 g of water and 46 g of glycerine is
(a) 0.46
(b) 0.40
(c) 0.20
(d) 0.36

Answer

Answer: c


Question 2. Out of molality (m), molarity (M), formality (F) and mole fraction (x), those which are independent of temperature are
(a) M, m
(b) F, x
(c) m, x
(d) M, x

Answer

Answer: c


Question 3. Which of the following condition is not satisfied by an ideal solution?
(a) ΔHmixing = 0
(b) ΔVmixing = 0
(c) Raoult’s Law is obeyed
(d) Formation of an azeotropic mixture

Answer

Answer: d


Question 4. The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol. The mixture shows
(a) no deviation from Raoult’s Law.
(b) positive deviation from Raoult’s Law.
(c) negative deviation from Raoult’s Law.
(d) that the solution is unsaturated.

Answer

Answer: b


Question 5. Which has the lowest boiling point at 1 atm pressure?
(a) 0.1 M KCl
(b) 0.1 M Urea
(c) 0.1 M CaCl2
(d) 0.1 M A1Cl3

Answer

Answer: b


Question 6. Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in moles/litre will be
(a) 0.33
(b) 0.666
(c) 0.3 × 10-2
(d) 3

Answer

Answer: c


Question 7. People add sodium chloride to water while boiling eggs. This is to
(a) decrease the boiling point.
(b) increase the boiling point.
(c) prevent the breaking of eggs.
(d) make eggs tasty.

Answer

Answer: b


Question 8. The van’t Hoff factor (i) accounts for
(a) degree of solubilisation of solute.
(b) the extent of dissociation of solute.
(c) the extent of dissolution of solute.
(d) the degree of decomposition of solution.

Answer

Answer: b


Question 9. Which relationship is not correct?
Chemistry MCQs for Class 12 with Answers Chapter 2 Solutions

Answer

Answer: b


Question 10. The molal elevation constant depends upon
(a) nature of solute.
(b) nature of the solvent.
(c) vapour pressure of the solution.
(d) enthalpy change.

Answer

Answer: b


We hope the given Chemistry MCQs for Class 12 with Answers Chapter 2 Solutions will help you. If you have any query regarding CBSE Class 12 Chemistry Solutions MCQs Pdf, drop a comment below and we will get back to you at the earliest.

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Chemistry MCQs for Class 12 with Answers Chapter 3 Electrochemistry

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 3 Electrochemistry. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry Electrochemistry MCQs Pdf with Answers to know their preparation level.

Electrochemistry Class 12 Chemistry MCQs Pdf

Question 1. The charge required for the reduction of 1 mol of MnO2 to MnO2 is
(a) 1 F
(b) 3 F
(c) 5 F
(d) 6 F

Answer

Answer: b


Question 2. The cell reaction of the galvanic cell.
Chemistry MCQs for Class 12 with Answers Chapter 3 Electrochemistry 1

Answer

Answer: d


Question 3. Which of the following reaction is used to make fuel cell?
Chemistry MCQs for Class 12 with Answers Chapter 3 Electrochemistry 2

Answer

Answer: c


Question 4. If limiting molar conductivity of Ca2+ and Cl are 119.0 and 76.3 S cm2 mol-1, then the value of limiting molar conductivity of CaCl2 will be
(a) 195.3 S cm2 mol-1
(b) 271.6 S cm2 mol-1
(c) 43.3 S cm2 mol-1
(d) 314.3 S cm2 mol-1.

Answer

Answer: b


Question 5. NH4NC>3 is used in salt bridge because
(a) it forms a jelly like material with agar-agar.
(b) it is a weak electrolyte.
(c) it is a good conductor of electricity.
(d) the transport number of NH4+ and NO3 ions are almost equal.

Answer

Answer: d


Question 6.
Chemistry MCQs for Class 12 with Answers Chapter 3 Electrochemistry 3

Answer

Answer: b


Question 7. The reaction, 3ClO (aq) → ClO3 (aq) + 2Cl (aq) is an example of
(a) Oxidation reaction
(b) Reduction reaction
(c) Disproportionation reaction
(d) Decomposition reaction

Answer

Answer: c


Question 8. The emf of the cell:
Ni / Ni2+ (1.0 M) // Au3+ (1.0 M) / Au (E° = -0.25 V for Ni2+/Ni; E° = 1.5 V for Au3+/Au) is
(a) 1.25 V
(b) -1.25 V
(c) 1.75 V
(d) 2.0 V

Answer

Answer: c


Question 9. The standard emf of a galvanic cell involving cell reaction with n = 2 is formed to be 0.295 V at 25° C. The equilibrium constant of the reaction would be
(a) 1.0 × 1010
(b) 2.0 × 1011
(c) 4.0 × 1012
(d) 1.0 × 102
[Given F = 96500 (mol-1); R = 8.314 JK-1 mol-1]

Answer

Answer: a


Question 10. If E°Fe2+/Fe = -0.441 V and E°Fe2+/Fe2+ = 0.771 V, the standard EMF of the reaction,
Fe + 2Fe3+ → 3Fe2+ will be
(a) 1.212 V
(b) 0.111 V
(C) 0.330 V
(d) 1.653 V

Answer

Answer: a


We hope the given Chemistry MCQs for Class 12 with Answers Chapter 3 Electrochemistry will help you. If you have any query regarding CBSE Class 12 Chemistry Electrochemistry MCQs Pdf, drop a comment below and we will get back to you at the earliest.

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Chemistry MCQs for Class 12 with Answers Chapter 4 Chemical Kinetics

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 4 Chemical Kinetics. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry Chemical Kinetics MCQs Pdf with Answers to know their preparation level.

Chemical Kinetics Class 12 Chemistry MCQs Pdf

Question 1. For the reaction N2 + 3H2 → 2NH3 if \frac{\Delta\left[\mathrm{NH}_{3}\right]}{\Delta t} = 2 × 10-4 mol L-1s-1, the value of \frac{-\Delta\left[\mathrm{H}_{2}\right]}{\Delta t} would be
(a) 1 × 10-4 mol L-1s-1
(b) 3 × 10-4 mol L-1s-1
(c) 4 × 10-4 mol L-1s-1
(d) 6 × 10-4 mol L-1s-1

Answer

Answer: b


Question 2. The rate of a certain hypothetical reaction
A + B + C → products
is given by r = \frac{-d[\mathbf{A}]}{d t} \mathbf{K}[\mathbf{A}]^{1 / 2}[\mathbf{B}]^{1 / 3}[\mathbf{C}]^{1 / 4}. The order of the reaction is
(a) 13/11
(b) 13/14
(c) 12/13
(d) 13/12

Answer

Answer: d


Question 3. In the formation of S02 by contact process;
2SO2 + O2 → 2SO3, the rate of reaction was measured as \frac{-d\left[\mathrm{O}_{2}\right]}{d t} = 2.5 × 10-4 mol L-1s-1. at
The rate of formation of of S03 will be
(a) -5.0 × 10-4 mol L-1s-1
(b) -1.25 × 10-4 mol L-1s-1
(c) 3.75 × 10-4 mol L-1s-1
(d) 5.00 × 10-4 mol L-1s-1

Answer

Answer: d


Question 4. For a chemical reaction A→B, it is found that the rate of reaction doubles when the concentration of A is increased four times. The order of reaction is
(a) Two
(b) One
(c) Half
(d) Zero

Answer

Answer: c


Question 5. The half life of the first order reaction having rate constant K = 1.7 x 10-5s-1 is
(a) 12.1 h
(b) 9.7 h
(c) 11.3 h
(d) 1.8 h

Answer

Answer: c


Question 6. What will be the fraction of molecules having energy equal to or greater than activation energy, Ea?
(a) K
(b) A
(c) Ae-Ea/Rt
(d) e-Ea/Rt

Answer

Answer: d


Question 7. Chemistry MCQs for Class 12 with Answers Chapter 4 Chemical Kinetics 1
What type of reaction is this?
(a) Second order
(b) Unimolecular
(c) Pseudo-unimolecular
(d) Third order

Answer

Answer: c


Question 8. Which among the following is a false statement?
(a) Rate of zero order reaction is independent of initial concentration of reactant.
(b) Half life of a third order reaction is inversely proportional to square of initial concentration of the reactant.
(c) Molecularity of a reaction may be zero or fraction.
(d) For a first order reaction, t_{1 / 2}=\frac{0.693}{\mathrm{K}}

Answer

Answer: c


Question 9. Which of the following statements about the catalyst is true?
(a) A catalyst accelerates the rate of reaction by bringing down the activation energy.
(b) A catalyst does not participate in reaction mechanism.
(c) A catalyst makes the reaction feasible by making ∆G more negative.
(d) A catalyst makes equilibrium constant more favourable for forward reaction.

Answer

Answer: a


Question 10. An endothermic reaction with high activation energy for the forward reaction is given by the diagram.
Chemistry MCQs for Class 12 with Answers Chapter 4 Chemical Kinetics 2

Answer

Answer: c


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Chemistry MCQs for Class 12 with Answers Chapter 5 Surface Chemistry

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 5 Surface Chemistry. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry Surface Chemistry MCQs Pdf with Answers to know their preparation level.

Surface Chemistry Class 12 Chemistry MCQs Pdf

Question 1. In Freundlich adsorption isotherm x/m = Kp1/n, the value of ‘n’ at low pressure is
(a) more than one.
(b) less than one.
(c) equal to one.
(d) from zero to one.

Answer

Answer: c


Question 2. According to adsorption theory of catalysis, the speed of the reaction increases because
(a) the concentration of the reactant molecules at the active centres of the catalyst becomes high due to adsorption.
(b) in the process of adsoption, the activation energy of the molecules becomes large.
(c) adsorption produces heat which increases the speed of the reaction.
(d) adsorption lowers the activation energy of the reaction.

Answer

Answer: d


Question 3. Which shape selective catalyst is used to convert alcohol to gasoline?
(a) Trpsin
(b) Calgon
(c) ZSM-5
(d) Zeigler-Natta catalyst

Answer

Answer: c


Question 4. Which of the following is an example of heterogenous catalyst?
Chemistry MCQs for Class 12 with Answers Chapter 5 Surface Chemistry

Answer

Answer: c


Question 5. When a small amount of FeCl3 is added to a freshly precipitated Fe(OH)3, b reddish brown colloidal solution is obtained. This pheno¬menon is known as
(a) dialysis
(b) peptization
(c) protection
(d) dissolution

Answer

Answer: c


Question 6. Lyophillic colloids are stable due to
(a) charge on the particles.
(b) large size of the particles.
(c) small size of the particles.
(d) layer of dispersion of medium on the particles.

Answer

Answer: d


Question 7. Cottrell precipitator is used to
(a) precipitate mud from muddy water.
(b) precipitate carbon particles from smoke.
(c) purify the ordinary drinking water.
(d) precipitate salts in qualitative analysis.

Answer

Answer: b


Question 8. The potential difference between the fixed charged layer and the diffused layer having opposite charge is called
(a) Zeta potential
(b) Electrokinetic potential
(c) Both (a) and (b)
(d) Streaming potential

Answer

Answer: a


Question 9. Peptization is a process of
(a) precipitation of colloidal particles.
(b) purification of colloids.
(c) dispersing precipitate into colloidal solution.
(d) movement of colloidal particles in the electric field.

Answer

Answer: c


Question 10. An emulsifier is a substance which
(a) stabilises the emulsion.
(b) homogenises the emulsion.
(c) Coagulates the emulsion.
(d) Accelerates the disperson of liquid in liquid.

Answer

Answer: a


We hope the given Chemistry MCQs for Class 12 with Answers Chapter 5 Surface Chemistry will help you. If you have any query regarding CBSE Class 12 Chemistry Surface Chemistry MCQs Pdf, drop a comment below and we will get back to you at the earliest.

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Chemistry MCQs for Class 12 with Answers Chapter 6 General Principles and Processes of Isolation of Elements

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 6 General Principles and Processes of Isolation of Elements. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry General Principles and Processes of Isolation of Elements MCQs Pdf with Answers to know their preparation level.

General Principles and Processes of Isolation of Elements Class 12 Chemistry MCQs Pdf

Question 1. The electrolytic reduction technique is used in the extraction of
(a) Highly electronegative elements.
(b) Highly electropostive elements.
(c) Metalloids.
(d) Transition metals.

Answer

Answer: b


Question 2. In the commercial electrochemical process for aluminium extraction, electrolyte used is
(a) Al(OH)3 is NaOH solution.
(b) An aqueous solution of Al2 (SO4)3.
(c) A molten mixture of Al2O3 and Na3AlF6.
(d) A molten mixture of Al2O3 and Al(OH)3.

Answer

Answer: c


Question 3. Which ore can be best concentrated by froth floatation process?
(a) Malachite
(b) Cassiterite
(c) Galena
(d) Magnetite

Answer

Answer: c


Question 4. Electrolytic reduction of Al2O3 to Al by Hall- Herault process is carried out
(a) in presence of NaCl.
(b) in presence of fluorite.
(c) in presence of cryolite which forms a melt with lower melting point.
(d) in presence of cryolite which forms a melt with high melting point.

Answer

Answer: c


Question 5. The chemical composition of ‘slag’ formed during the melting process in the extraction of copper is
(a) Cu2O + FeS
(b) FeSiO3
(c) CuFeS2
(d) Cu2S + FeO

Answer

Answer: b


Question 6. Bessemer converter is used in the manufacture of
(a) Pig iron
(b) Steel
(c) Wrought iron
(d) Cast iron

Answer

Answer: b


Question 7. The method of zone refining of metals is based on the principle of
(a) greater mobility of the pure metal than that of the impurity.
(b) higher melting point of the impurity than that of the pure metal.
(c) greater noble character of the solid metal than that of impurity.
(d) greater solubility of the impurity in the molten state than in the solid.

Answer

Answer: d


Question 8. In the leaching of Ag2S with NaCN, a stream of air is also passed. It is because
(a) The reaction between Ag2S and NaCN is reversible.
(b) to oxidise Na2S formed in the reaction to Na2SO4.
(c) to oxidise Ag2S to Ag2O.
(d) Both (a) and (b).

Answer

Answer: d


Question 9. Purest form of iron is
(a) Cast iron
(b) Hard Steel
(c) Stainless steel
(d) Wrought iron

Answer

Answer: d


Question 10. Consider the following reaction at 1000° C
Chemistry MCQs for Class 12 with Answers Chapter 6 General Principles and Processes of Isolation of Elements
Choose the correct statement at 1000°C
(a) Zinc can be oxidised by carbon monoxide.
(b) Zinc oxide can be reduced by graphite.
(c) Both statements (a) and (b) are correct.
(d) Both statements (a) and (b) are false.

Answer

Answer: b


We hope the given Chemistry MCQs for Class 12 with Answers Chapter 6 General Principles and Processes of Isolation of Elements will help you. If you have any query regarding CBSE Class 12 Chemistry General Principles and Processes of Isolation of Elements MCQs Pdf, drop a comment below and we will get back to you at the earliest.

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Chemistry MCQs for Class 12 with Answers Chapter 7 The p-Block Elements

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 7 The p-Block Elements. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry The p-Block Elements MCQs Pdf with Answers to know their preparation level.

The p-Block Elements Class 12 Chemistry MCQs Pdf

Question 1. H2S is more acidic than H2O because
(a) oxygen is more electronegative than sulphur.
(b) atomic number of sulphur is higher than oxygen.
(c) H — S bond dissociation energy is less as compared to H — O bond.
(d) H — O bond dissociation energy is less also compared to H — S bond.

Answer

Answer: b


Question 2. The boiling points of hydrides of group 16 are in the order
(a) H2O > H2Te > H2S > H2Se
(b) H2O > H2S > H2Se > H2Te
(c) H2O > H2Te > H2Se > H2S
(d) None of these

Answer

Answer: b


Question 3. In the manufacture of sulphuric acid by contact process Tyndall box is used to
(a) convert SO2 and SO3
(b) test the presence of dust particles
(c) filter dust particles
(d) remove impurities

Answer

Answer: b


Question 4. Fluorine differs from rest of the halogens in some of its properties. This is due to
(a) its smaller size and high electronegativity.
(b) lack of d-orbitals.
(c) low bond dissociation energy.
(d) All of the these.

Answer

Answer: b


Question 5. The set with correct order of acidity is
(a) HClO < HClO2 < HClO3 < HClO4
(b) HClO4 < HClO3 < HClO2 < HClO
(c) HClO < HClO4 < HClO3 < HClO2
(d) HClO4 < HClO2 < HClO3 < HClO

Answer

Answer: b


Question 6. When chlorine reacts with cold and dilute solution of sodium hydroxide, it forms
(a) Cl and ClO
(b) Cl and ClO2
(c) Cl and ClO3
(d) Cl and ClO4

Answer

Answer: a


Question 7. The formation of O2+ [PtF6] is the basis for the formation of first xenon compound. This is because
(a) O2 and Xe have different sizes.
(b) both O2 and Xe are gases.
(c) O2 and Xe have comparable electro-negativities.
(d) O2 and Xe have comparable ionisation enthalpies.

Answer

Answer: d


Question 8. Partial hydrolysis of XeF4 gives
(a) XeO3
(b) XeOF2
(c) XeOF4
(d) XeF2

Answer

Answer: b


Question 9. Helium is preferred to be used in balloons instead of hydrogen because it is
(a) incombustible
(b) lighter than hydrogen
(c) more abundant than hydrogen
(d) non polarizable

Answer

Answer: a


Question 10. The increasing order of reducing power of the halogen acids is
(a) HF < HCl < HBr < HI
(b) HI < HBr < HCl < HF
(c) HBr < HCl < HF < HI
(d) HCl < HBr < HF < HI

Answer

Answer: a


We hope the given Chemistry MCQs for Class 12 with Answers Chapter 7 The p-Block Elements will help you. If you have any query regarding CBSE Class 12 Chemistry The p-Block Elements MCQs Pdf, drop a comment below and we will get back to you at the earliest.

The post Chemistry MCQs for Class 12 with Answers Chapter 7 The p-Block Elements appeared first on Learn CBSE.

Chemistry MCQs for Class 12 with Answers Chapter 8 The d-and f-Block Elements

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 8 The d-and f-Block Elements. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry The d-and f-Block Elements MCQs Pdf with Answers to know their preparation level.

The d-and f-Block Elements Class 12 Chemistry MCQs Pdf

Question 1. Anomalous electronic configuration in the 3d series are of
(a) Cr and Fe
(b) Cu and Zn
(c) Fe and Cu
(d) Cr and Cu

Answer

Answer: d


Question 2. Which of the following are d-block elements but not regarded as transistion elements?
(a) Cu, Ag, Au
(b) Zn, Cd, Hg
(c) Fe, Co, Ni
(d) Ru, Rh, Pd

Answer

Answer: b


Question 3. CuSO4. 5H2O is blue is colour because
(a) It contains water of crystallization.
(b) SO42- ions absorb red light.
(c) Cu2+ ions absorb orange red light.
(d) Cu2+ ions absorb all colours except red from the white light.

Answer

Answer: c


Question 4. Transistion elements form alloys easily because they have
(a) Same atomic number
(b) Same electronic configuration
(c) Nearly same atomic size
(d) None of the above

Answer

Answer: c


Question 5. Which one of the following characteristics of the transistion metals is associated with higher catalytic activity?
(a) High enthalpy of atomisation
(b) Paramagnetic behaviour
(c) Colour of hydrate ions
(d) Variable oxidation states

Answer

Answer: d


Question 6. Which of the following has the maximum number of unpaired electrons?
(a) Mg2+
(b) Ti3+
(c) V3+
(d) Fe2+

Answer

Answer: d


Question 7. The property which is not characteristic of transistion metals is
(a) variable oxidation states.
(b) tendency to form complexes.
(c) formation of coloured compounds.
(d) natural radioactivity.

Answer

Answer: d


Question 8. Which of the following is incorrect for KMnO4 to be used as an oxidising agent?
(a) HCl cannot be used because some KMnO4 is consumed in the reaction.
(b) Nitric acid is not used for the above purpose because it itself acts as a self oxidising agent and will react with the reducing agent.
(c) The equivalent weight of KMnO4 in basic medium is 158.
(d) The number of electrons involved in oxidation of KMnO4 in acidic medium is 3.

Answer

Answer: d


Question 9. Transistion metals, despite high E° oxidation, are poor reducing agents. The incorrect reason is
(a) high heat of vaporization.
(b) high ionization energies.
(c) low heats of hydration.
(d) complex forming nature.

Answer

Answer: d


Question 10. Which of the following has magnetic moment value of 5.9?
(a) Fe2+
(b) Fe3+
(c) Ni2+
(d) Cu2+

Answer

Answer: b


We hope the given Chemistry MCQs for Class 12 with Answers Chapter 8 The d-and f-Block Elements will help you. If you have any query regarding CBSE Class 12 Chemistry The d-and f-Block Elements MCQs Pdf, drop a comment below and we will get back to you at the earliest.

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Chemistry MCQs for Class 12 with Answers Chapter 9 Coordination Compounds

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Free PDF Download of CBSE Chemistry Multiple Choice Questions for Class 12 with Answers Chapter 9 Coordination Compounds. Chemistry MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Chemistry Coordination Compounds MCQs Pdf with Answers to know their preparation level.

Coordination Compounds Class 12 Chemistry MCQs Pdf

Question 1. The solution of the complex [Cu(NH3)4] SO4 in water will
(a) give the tests of Cu2+ ion
(b) give the tests of NH3
(c) give the tests of SO42- ions
(d) not give the tests of any of the above

Answer

Answer: c


Question 2. IUPAC name of [Pt(NH3)3 Br (NO2) Cl] Cl isw
(a) triamminechlorodibromidoplatinum (IV) chloride
(b) triamminechloridobromidonitrochloride- platinum (IV) chloride
(c) triamminebromidochloridonitroplatinum (IV) chloride
(d) triamminenitrochlorobromoplatinum (IV) chloride

Answer

Answer: c


Question 3. Trunbull’s blue is
(a) Ferricyanide
(b) Ferrous ferricyanide
(c) Ferrous cyanide
(d) Fe3[Fe(CN)6]4

Answer

Answer: b


Question 4. Primary and secondary valency of Pt in [Pt(en)2Cl2] are
(a) 4, 4
(b) 4, 6
(c) 6, 4
(d) 2, 6

Answer

Answer: d


Question 5. The complex ions [Co(NH3)5(NO2)]2+ and [Co(NH3)5 (ONO)]2+ are called
(a) Ionization isomers
(b) Linkage isomers
(c) Co-ordination isomers
(d) Geometrical isomers

Answer

Answer: b


Question 6. Which of the following has square planar structure?
(a) [NiCl4]2-
(b) [Ni(CO)4]
(c) [Ni(CN)4]2-
(d) None of these

Answer

Answer: c


Question 7. Which of the following has magnesium?
(a) Chlorophll
(b) Haemocyanin
(c) Carbonic anhydrate
(d) Vitamin B12

Answer

Answer: a


Question 8. Mohr’s salt is
(a) Fe2(SO4) 3 . (NH4)2SO4 . 6H2O
(b) FeSO4 . (NH4)2 . SO4 . 6H2O
(c) MgSO4 . 7H2O
(d) FeSO4 . 7H2O

Answer

Answer: b


Question 9. Which of the following shall form an octahedral complex?
(a) d4 (low spin)
(b) d8 (high spin)
(c) d6 (low spin)
(d) All of these

Answer

Answer: b


Question 10. EDTA is used for the estimation of
(a) Na+ and K+ ions
(b) Cl and Br ions
(c) Cu2+ and Cs+ ions
(d) Ca2+ and Mg2+ ions

Answer

Answer: d


We hope the given Chemistry MCQs for Class 12 with Answers Chapter 9 Coordination Compounds will help you. If you have any query regarding CBSE Class 12 Chemistry Coordination Compounds MCQs Pdf, drop a comment below and we will get back to you at the earliest.

The post Chemistry MCQs for Class 12 with Answers Chapter 9 Coordination Compounds appeared first on Learn CBSE.

Physics MCQs for Class 12 with Answers Chapter 1 Electric Charges and Fields

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Free PDF Download of CBSE Physics Multiple Choice Questions for Class 12 with Answers Chapter 1 Electric Charges and Fields. Physics MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Electric Charges and Fields MCQs Pdf with Answers to know their preparation level.

Electric Charges and Fields Class 12 Physics MCQs Pdf

1. SI unit of permittivity of free space is
(a) Farad
(b) Weber
(c) C2N-1 m-2
(d) C2N-1 m-2

Answer

Answer: c


2. A charge Q is placed at the centre of the line joining two point charges +q and +q as shown in the figure. The ratio of charges Q and q is
Physics MCQs for Class 12 with Answers Chapter 1 Electric Charges and Fields 1
(a) 4
(b) 1/4
(c) -4
(d) -1/4

Answer

Answer: d


3. The force per unit charge is known as
(a) electric flux
(b) electric field
(c) electric potential
(d) electric current

Answer

Answer: b


4. Electric field lines provide information about
(a) field strength
(b) direction
(c) nature of charge
(d) all of these

Answer

Answer: d


5. Which of the following figures represent the electric field lines due to a single negative charge?
Physics MCQs for Class 12 with Answers Chapter 1 Electric Charges and Fields 2

Answer

Answer: b


6. The SI unit of electric flux is
(a) N C-1 m-2
(b) N C m-2
(c) N C-2 m2
(d) N C-1 m2

Answer

Answer: d


7. The unit of electric dipole moment is
(a) newton
(b) coulomb
(c) farad
(d) debye

Answer

Answer: d


8. Consider a region inside which, there are various types of charges but the total charge is zero. At points outside the region
(a) the electric field is necessarily zero.
(b) the electric field is due to the dipole moment of the charge distribution only.
(c) the dominant electric field is inversely pro-portional to r3, for large r (distance from ori-gin).
(d) the work done to move a charged particle along a closed path, away from the region will not be zero.

Answer

Answer: c


9. The surface considered for Gauss’s law is called
(a) Closed surface
(b) Spherical surface
(c) Gaussian surface
(d) Plane surface

Answer

Answer: c


10. The total flux through the faces of the cube with side of length a if a charge q is placed at corner A of the cube is
Physics MCQs for Class 12 with Answers Chapter 1 Electric Charges and Fields 3
Physics MCQs for Class 12 with Answers Chapter 1 Electric Charges and Fields 4

Answer

Answer: a


11. Which of the following statements is not true about Gauss’s law?
(a) Gauss’s law is true for any closed surface.
(b) The term q on the right side side of Gauss’s law includes the sum of all charges enclosed by the surface.
(c) Gauss’s law is not much useful in calculating electrostatic field when the system has some symmetry.
(d) Gauss’s law is based on the inverse square dependence on distance contained in the coulomb’s law

Answer

Answer: c


We hope the given Physics MCQs for Class 12 with Answers Chapter 1 Electric Charges and Fields will help you. If you have any query regarding CBSE Class 12 Physics Electric Charges and Fields MCQs Pdf, drop a comment below and we will get back to you at the earliest.

The post Physics MCQs for Class 12 with Answers Chapter 1 Electric Charges and Fields appeared first on Learn CBSE.

Physics MCQs for Class 12 with Answers Chapter 2 Electrostatic Potential and Capacitance

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Free PDF Download of CBSE Physics Multiple Choice Questions for Class 12 with Answers Chapter 2 Electrostatic Potential and Capacitance. Physics MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Electrostatic Potential and Capacitance MCQs Pdf with Answers to know their preparation level.

Electrostatic Potential and Capacitance Class 12 Physics MCQs Pdf

1. Which of the following statement is true?
(a) Electrostatic force is a conservative force.
(b) Potential at a point is the work done per unit charge in bringing a charge from any point to infinity.
(c) Electrostatic force is non-conservative
(d) Potential is the productof charge and work.

Answer

Answer: a


2. 1 volt is equivalent to
Physics MCQs for Class 12 with Answers Chapter 2 Electrostatic Potential and Capacitance 1

Answer

Answer: c


3. The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential at that point is
Physics MCQs for Class 12 with Answers Chapter 2 Electrostatic Potential and Capacitance 2

Answer

Answer: b


4. Consider a uniform electric field in the z-direction. The potential is a constant
(a) for any x for a given z
(b) for any y for a given z
(c) on the x-y plane for a given z
(d) all of these

Answer

Answer: d


5. Equipotential surfaces
(a) are closer in regions of large electric fields compared to regions of lower electric fields.
(b) will be more crowded near sharp edges of a conductor.
(c) will always be equally spaced.
(d) both (a) and (b) are correct.

Answer

Answer: d


6. In a region of constant potential
(a) the electric field is uniform.
(b) the electric field is zero.
(c) there can be no charge inside the region.
(d) both (b) and (c) are correct.

Answer

Answer: d


7. A test charge is moved from lower potential point to a higher potential point. The potential energy of test charge will
(a) remain the same
(b) increase
(c) decrease
(d) become zero

Answer

Answer: c


8. An electric dipole of moment \vec{p} is placed in a uniform electric field \vec{E} . Then
(i) the torque on the dipole is \vec{p} \times \overrightarrow{\mathrm{E}}
(ii) the potential energy of the system is \vec{p} . \overrightarrow{\mathrm{E}}
(iii) the resultant force on the dipole is zero. Choose the correct option.
(a) (i), (ii) and (iii) are correct
(b) (i) and (iii) are correct and (ii) is wrong
(c) only (i) is correct
(d) (i) and (ii) are correct and (iii) is wrong

Answer

Answer: b


9. If a conductor has a potential V ≠ 0 and there are no charges anywhere else outside, then
(a) there must be charges on the surface or in¬side itself.
(b) there cannot be any charge in the body of the conductor.
(c) there must be charges only on the surface.
(d) both (a) and (b) are correct.

Answer

Answer: c


10. Which of the following statements is false for a perfect conductor?
(a) The surface of the conductor is an equipoten-tial surface.
(b) The electric field just outside the surface of a conductor is perpendicular to the surface.
(c) The charge carried by a conductor is always uniformly distributed over the surface of the conductor.
(d) None of these.

Answer

Answer: d


11. Dielectric constant for a metal is
(a) zero
(b) infinite
(c) 1
(d) 10

Answer

Answer: b


12. When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance
(a) increases K times
(b) remains unchanged
(c) decreases K times
(d) increases K-1 times

Answer

Answer: c


13. In a parallel plate capacitor, the capacity increases if
(a) area of the plate is decreased.
(b) distance between the plates increases.
(c) area of the plate is increased.
(d) dielectric constantly decreases.

Answer

Answer: c


14. A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
(a) increases
(b) decreases
(c) does not change
(d) becomes zero

Answer

Answer: a


15. Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
(a) The charges on the free plated connected to-gether are destroyed.
(b) The energy stored in ths system increases.
(c) The potential difference between the free plates is 2V.
(d) The potential difference remains constant.

Answer

Answer: c


16. A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then
(a) capacitance will increase.
(b) energy stored will decrease.
(c) electric field will increase.
(d) voltage will decrease.

Answer

Answer: c


17. Two spherical conductors each of capacity C are charged to potential V and -V. These are then connected by means of a fine wire. The loss of energy is
(a) zero
(b) \frac{1}{2}CV2
(c) CV2
(d) 2 CV2

Answer

Answer: c


We hope the given Physics MCQs for Class 12 with Answers Chapter 2 Electrostatic Potential and Capacitance will help you. If you have any query regarding CBSE Class 12 Physics Electrostatic Potential and Capacitance MCQs Pdf, drop a comment below and we will get back to you at the earliest.

The post Physics MCQs for Class 12 with Answers Chapter 2 Electrostatic Potential and Capacitance appeared first on Learn CBSE.

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