CBSE Class 12

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

Question 1.
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of
(a) reflected, and
(b) refracted light ? Refractive index of water is 1.33. (C.B.S.E. Sample Paper 1991)
Answer:
(a) For reflected light wavelength is unchanged i. e.
X = 589 x 10^-9 m = 589 nm
Also, speed of light in air c = 3 x 10⁸ m s ^-1

Question 2.
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Answer:
(a) Spherical shape
(b) Plane wavefront
(c) Plane wavefront

Question 3.
(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in a vacuum is 3.0 x 10⁸ m s^-1)
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism ?
Answer:
(a) Here ,n=105,c=3.0 x 10⁸ ms^-1

speed of light when passing through glass depends on colour of light. λ_r > λ_υ , therefore the speed of violet light is less than the red light.

Question 4.
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 Determine the wavelength of light used in the experiment.
Answer:

Question 5.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is K units. What is the intensity of light at a point where path difference is λ/2 ?
Answer:

Question 6.
A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide ?
The distance between two slits is 2 mm and distance between the plane of the slits and the screen is 1.2 m.
Answer:
(a) λ = 650 nm = 650 x 10^-9 m,
d = 2 mm = 2 x 10^-3 m,
D = 1.2 m
Distance of mth bright fringe from the central maximum is given by

Question 7.
In a double-slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experiment apparatus is immersed in water ? Take
refractive index of water to be \(\frac { 4 }{ 3 } \)
Answer:


Question 8.
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Answer:
µ = 1.5
According to Brewster’s law,
µ = tan p
tan p = 1.5
⇒ p = 56.31º

Question 9.
Light of wavelength 5000 A falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what angle of incidence is the reflected ray normal to the incident ray ?
Answer:
Here X = 5000 A = 5000 X 10^-10 m,
c =3 x 10⁸ m s^-1
Wave length of reflected light
= Wavelength of incident light = 5000 Å


Question 10.
Estimate the distance for which ray optics is good approximation  for an aperture  of
4 mm and wavelength 400 nm.
Answer:
Here X = 400 nm = 400 x 10^-9 m, Aperture, a = 4 mm = 4 x 10^-3 m
.’. Distance for which  ray optics is a good approximation is Fresnel’s distance


Question 11.
The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15Å. Estimate the speed with which the star is receding from the Earth.
Answer:

Negative sign shows that the star is receding away from the earth.

Question 12.
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water ? If not, which alternative picture of light is consistent with experiment ?
Answer:
According to Corpuscular theory, when light in the form of particles enters into denser medium from a rarer medium, a force of attraction comes into play on the particles normal to the surface. Thus, the component of velocity normal to the surface of water increases whereas the component of velocity parallel to surface does not change. Therefore,

velocity in water is greater than velocity of light in air. However, in actual case, c > υ. Huygen’s wave theory of light is consistent with the experiment.

Question 13.
You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Answer:
Let there be a point object A at a distance y from a plane mirror. Treating this point to be a point source of light, we can assume spherical wavefronts progressing from A of radius y. Let there be no mirror then after time t, the wavefront will reach A’ as wavefront I. If a mirror is placed as shown in the figure then image will be formed at A’ represented by II.

It is seen that OA’ = OA i. e. virtual image is formed at a distance equal to the distance of object from the mirror.

Question 14.
Let us list some of the factors, which could possibly influence the speed of wave propagation:
(1) nature of the source,
(2) direction of propagation.
(3) the motion of the source and/or observer.
(4) wavelength
(5) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, glass or water), depend ?
Answer:
(a) Speed of light in a vacuum is an absolute constant (universal constant). It is independent of any factor. It is independent of the relative motion between source and observer even.
(b)

1. Speed of light in a medium depends upon wavelength.
2. It is independent of the nature of the source and motion of the source relative to the medium.
3. It depends upon the properties of the medium of propagation and motion of the observer relative to the medium,
4. It is independent of the direction of propagation for isotropic medium,
5. It is independent of the intensity of the wave.

Question 15.
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations :
(1) a source at rest; observer moving, and
(2) source moving ; observer at rest. The exact Doppler formulae for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulae to be strictly identical for the two situations in case of light travelling in a medium ?
Answer:
Sound requires material medium for propagation. Though situations
(1) and (2) may correspond to the same relative motion, yet they are not identical physically as the motion of observer relative to medium may be different in both situations. Hence, Doppler effect for sound cannot be same in both situations. Light when passing through material medium is also governed by different Doppler formulae for
(1) a source at rest; observer moving and
(2) source moving; observer at rest.
But when light passes through vacuum the formulae become exactly same for the two different situations because speed of light and frequency/wavelength of light remain unchanged in vacuum.

Question 16.
In a double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between two slits?
Answer:

Question 17.
Answer the following questions :
(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction
band ?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? (C.B.S.E. 2013, 2013 )
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle.
Explain why? (C. B. S. E. 2013 )
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily ?   (C.B.S.E. 1990 )
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?   (C.B.S.E. 1990)
Answer:
(a) The width of central maxima = 2λD/d.
When the width (d) of slit is doubled, then the width of central diffraction maxima reduces to half and the intensity of the central band increases four times as amplitude of light wave is doubled.
(b) Intensity of fringes produced in the double-slit experiment is changed due to diffraction pattern superposing due to each slit.
(c) Light waves diffract at the edges of the circular obstacle. These diffracted waves interfere constructively and give rise to the bright spot at the center of the geometrical shadow.
(d) Diffraction is observed when the wavelength of the wave is of the order of the size of the obstacle. The wavelength of sound wave (≈ 0.33 m) is larger than the light wave (≈10^-7 m) and is also comparable to wall, so diffraction of sound waves takes place and hence the students can converse easily. On the other hand, the wavelength of light is very small as compared to the obstacle e. 1 m high wall so the diffraction of light waves does not take place.
(e) In optical instruments, size of apertures are much larger than the wavelength of light. So diffraction of light is negligible. Hence, the assumption that light can travel in straight line is used in optical instruments.

Question 18.
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects ?
Answer:
If A and B are two hills and C is the hill peak midway

Question 19.
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2-5 mm from the center of the screen. Find the width of the slit.( C.B.S.E. 2013)
Answer:

Question 20.
Answer the following questions :
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TY screen. Suggest a possible explanation.
(b) As you have learned in the text, the principle of linear superposition of wave displacement is basic to understanding distributions in diffraction and interference patterns. What is the justification of this principle?
Answer:
(a) When a low flying aircraft passes overhead, the metallic body of the aircraft reflects the TV signal. A slight shaking of the picture on the TV screen takes place due to the interference of the reflected signal from the aircraft and the direct signal received by the antenna.
(b) The linear combination of wave equations is also a wave equation. This is the very basis of the superposition principle.

Question 21.
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angle of nλ/α. Justify this by suitably dividing the slit to bring out the cancellation.
Answer:
Let us suppose that we have n slits each of width

Therefore, each of the n slits of width d’ each sends zero intensity in the direction 9. As a result, the net resultant of intensity due to n such slits is zero.

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NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Question 1.
A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of a radius of curvature of 36 cm. At what distance from the mirror should a screen be placed in order to receive a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer:
Here, h = 2.5 cm, u = – 27 cm, R = – 36 cm.

Nature of image : Real, inverted and magnified. When the position of the object (i.e. candle) is moved closer to the concave mirror, the distance of the image moves away from the screen till the distance of the candle from the concave mirror is less than 18 cm. Hence, the screen has to be moved away from the concave mirror. When the distance of the candle is less than 18 cm from the concave mirror, a virtual and magnified image of the candle is formed behind the mirror. This image is not obtained on the screen.

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and magnification. Describe what happens as the needle is moved farther from the mirror.
Answer:
Here, h₁ = 4.5 cm, μ = – 12 cm, f= 15 cm

Thus, the image is virtual, erect and diminished. As we move the needle away from the mirror, the image goes on decreasing in size and moves towards the principal focus on the other side.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of the needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water ? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again ? (C.B.S.E. 2009 )
Answer:

Question 4.
The following figures (a) and (b) show refraction of an incident ray in air at 60° with the normal to glass-air and water-air interface, respectively. Predict the angle of refraction of an incident ray in water at 45° with the normal to a water-glass interface [Fig. (c)]

Answer:

Question 5.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out ? Refractive index of water is 1.33 (consider the bulb to be a point source).
Answer:

The light rays from the bulb B, which fall on the surface of water at an angle equal to critical angle (θ_C), grazes on the surface of water and the rays of light which fall on the surface of water at an angle greater than θ_C are totally internally reflected back into the water. The rays of light images emerges out of water through a circular patch of radius r.

Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. By rotating the prism, the angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism ? If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. The refracting angle of the prism is 60°.
Answer:

Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm ?
Answer:
Here, n = 1.55, R₁ = R and R₂ = – R, f= 20 cm Using lens maker formula, we get

Question 8.
A beam of light converges to a point P. A lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm?
Answer:
(a) When a convex lens is placed in the path of light converging at P, the beam converges at P_t. Thus, point P acts as virtual object for the convex lens.
Now, u = 12 cm, f= 20 cm.

Question 9.
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved farther from the lens?
Answer:

Thus, the image is virtual, erect, diminished and is formed on the same side of the lens at a distance of 8.4 cm from the lens. If the object is moved away from the lens, the image moves towards the principal focus and goes on decreasing in size.

Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer:
Here, f₁ = 30 cm and f₂ = -20 cm
For the combination of two thin lenses, the focal length of the combination is given by

Since the focal length of the system of lens is negative, therefore, the combination behaves as a diverging lens.

Question 11.
A compound microscope consists of an objective lens of focal length 2-0 cm and an eye-piece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct   vision (25 cm), (b) infinity ? What is the magnifying power of the microscope in each case? (C.B.S.E. 2008)
Answer:
Here, f_n =2.0 cm, f = 6.25 cm,
Distance between object lens and eye piece = 15 cm (a) For the formation of image at the least distance of distinct vision,

Question 12.
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eye-piece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses ? How much is the magnifying power of the microscope?
Answer:

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?
Answer:
Here, focal length of objective lens, f₀ = 144 cm Focal

Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eye-piece of focal length 1.0 cm is used, what is the angular magnification of the telescope ?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 x 10⁶m and the radius of the lunar orbit is 3.8 x 10⁸m. (C.B.S.E. 2008, 2011)
Answer:

Question 15.
Use the mirror equation to deduce that
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. (C.B.S.E. 2011)
Answer:

Question 16.
A small pin fixed on the table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table?
Refractive index of glass = 1.5. Does the answer depend upon the location of the slab?
Answer:
Here, t = 15 cm, n = 1.5
The lateral displacement

For small angles of incidence, the answer does not depend upon the location of the slab.

Question 17.
(a) Following figure shows a cross-section of a ‘light pipe’ made of glass fibre of refractive index 1.68. The outer covering of the pipe is made of material of refractive index 1.44. What is the range of the angles of incident rays with the axes of the pipe for which the total internal reflection inside the pipe take place as shown in the figure ?
(b) What is the answer if there is no outer covering of pipe ?

Answer:


Therefore all incident rays in the range 0 to 90°suffer total internal refletion.

Question 18.
Answer the following questions:
(a) You have learned that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake, would the fisherman look taller or shorter to the diver than what he actually is ?
(d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decreases
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter ?
Answer:
(a) Yes. They can produce real images if the object is a virtual object as shown in figure.

(b) here is no contradiction in this case. The virtual image of the object acts as an object for the convex lens of our eye and the lens of our eye make a real image of this object on the ratina.
(c) Let AB be the fisherman standing on the bank of the lake. The rays of light from the head of the fisherman bends towards the normal on refraction at the interface separating water and air. The refracted rays appear to come from point B’ instead of point B for the fish. Thus, for a diver the height of the fisherman is AB’ which is greater than his actual height AB.

(d) The apparent depth of a pond of water decreases when viewed obliquely. This is due to the refraction of light from the surface of water.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose ?
Answer:
The minimum distance between a real object and its real image formed by a convex lens of focal length/is given by L = 4f

Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer:
Let O be the position of object and I is the position of image when lens is at L₁ and then at L₂


Question 21.
(a) Determine the effective focal length of the combination of the two lenses in the question 910 if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangements. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answer:

Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face ? The refractive index of the material of the prism is 1.524.
Answer:
Let the ray of light be incident on the face AB at angle i so that it is totally internally reflected at face AC.

Question 23.
You are given prism made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion (b) disperse (and displace) a pencil of white light without much deviation.
Answer:
(a) Angular dispersion produced by two prisms i.e. crown glass and flint glass should be zero in this case


In the combination of prisms, flint glass prism of greater angle may be tried but in any case still this angle will be smaller than the angle of the crown glass prism in opposite order as shown in figure.

Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 2.5 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation {i.e., the range of converging power of the eye-lens) of a normal eye.
Answer:
When the object is placed at infinity, the eye makes use of the least converging power, Therefore, total converging power of cornea and the eye lens = 40 + 20 = 60 dioptre.


Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation ? If not, what might cause these defects of vision ?
Answer:
No. Myopia may arise due to the elongation of the eye ball and hypermetropia may arise due to the decrease in the size of the eye ball even when the eye has the normal ability of accommodation. There is another defect in the eye called presbyopia similar to hypermetropia. However, the causes of presbyopia and hypermetropia are different. Presbyopia arises in elderly persons and is corrected by using a bi-focal lens.

Question 26.
A myopic person has been using spectacles of power – 1.0 dioptres for distant vision. During old age he also needs to use separate reading glass of power +2 dioptres. Explain what may have happened.
Answer:
For -1.0 dioptre, the far point for eyes is 1 m i.e. 100 cm. The near point is 25 cm. The objects lying at infinity are brought at 100 cm from his eyes using the concave lens and the objects lying in between 25 cm and 100 cm are brought to focus using the ability of accommodation of the eye lens. In the old age, this ability of accommodation is reduced and the near point reaches 50 cm from his eyes.


Question 27.
A person looking at a mesh of crossed wires is able to see the vertical wires more distinctly than the horizontal wires. What is this defect due to ? How is such a defect of vision corrected ?
Answer:
This is due to the defect of lenses called astigmatism. The defect arises because of the fact that curvature of the eye-lens and the cornea is not same in different planes. This defect is removed by using cylindrical lens with vertical axis.

Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifing glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass ?
(b) What is the maximum and minimum angular magnification (magnifying power) possible using the above simple microscope ?
Answer:
(a) To see the object at a closest distance, the image of object should be formed at the least distance of distinct vision.


Question 29.
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to eye.
(a) What is the magnification (image size/object size) produced by the lens ? How much is the area of each square in the virtual object ?
(b) What is the angular magnification (magnifying powers) of the lens ?
(c) Is the magnification in
(1) equal to magnifying power in
(2) ? Explain.
Answer:

(c) Clearly magnification and power magnification are not equal to each other unless the image is located near the least distance of distinct vision, e. v = D.

Question 30.
(a) At what distance should the lens be held from the figure in the above exercise in order to view the squares distinctly with maximum possible magnifying power ?
(b) What is the magnification (image size/object size) in this case ?
(c) Is the magnification equal to magnifying power in this case ? Explain.
Answer:
(a) The magnifying power is maximum if the image is formed at the least distance of distinct point from the eye, i.e., if υ = -25 cm ; Also, f = 10 cm

=> the linear magnification and magnifying power is equal in this case.

Question 31.
What should be the distance between the object in the previous exercise and the magnifying glass if the virtual image of each square in the figure is to have an area 6.25 mm²? Would you be able to see the squares distinctly with your eyes very close to the magnifier?
Answer:


Question 32.
Answer the following questions :
(а)  The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virutal image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eye-piece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eye-piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer:
(a) When magnifying glass is not used, object to be seen clearly is to be placed at 25 cm. However, while using magnifying glass, object can be placed closer to eye than at 25 cm. The closer object has large angular size than the same object placed at 25 cm. It is in this sense that magnifying glass provides angular magnification.
(b) Yes. The angular magnification decreases slightly because angle subtended at eye is somewhat less than the angle subtended at the lens.
(c) The aberrations like spherical and chromatic aberrations start croping up if the convex lens of smaller and smaller focal length is made.
(d) Angular magnification of eye piece is given by (1+D/f_e)  and angular magnification of objective is approximately given by υ/f₀. Clearly for better magnification focal length of eye piece f_e and focal length of objective f_e should be small.
(e) If we position our eye very close to the eyepiece, the whole light will not fall on our eye and the field of view will decrease. So we place our eye a short distance away from the eye-piece to collect the large amount of light refracted through the eyepiece to increase the field of view.

Question 33.
An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and eye piece of focal length 5 cm. How will you set up the compound microscope ?
Answer:
For the image formed at the least distance of distinct vision, the magnifying power is given by



Question 34.
A small telescope has an objective lens of focal length 140 cm and an eye piece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e.,when the final image is formed at infinity) ?
(b) the final image is formed at the least distance of distinct vision (25 cm) ?
Answer:

Question 35.
For the telescope described in the last exercise, in 9.34
(a) what is the separation between the objective lens and the eye-piece ?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of image of the tower formed by objective lens ?
(c) What is the height of final image of the tower if it is formed at 25 cm ?
Answer:
(a) Since the final image is formed at infinity, the distance between the object lens and the eye-piece is f₀ + f_e = 140 + 5 = 145 cm


Question 36.
A cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with mirrors 20 mm apart. If the radius of curvature of large mirror is 220 mm and the small mirror is 140 mm, where will be the final image of an object at infinity be ?

Answer:

Question 37.
The adjoining figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of liquid ?

Answer:
In the presence of the liquid, the distance of the needle from the lens is equal to the focal length f of the combination of the convex lens and the piano concave lens formed by the liquid below it i.e. f = 45 cm. Also n = 1.5
In the absence of the liquid, the distance of the needle and the lens is equal to the focal length of the convex lens only i.e. f = 30 cm
.’.  If f₂ is the focal length of plane concave lens formed

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NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

Question 1.
(a) Two stable isotopes of lithium \(_{ 3 }^{ 6 }{ Li } \) and \(_{ 3 }^{ 7 }{ Li } \) have respective abundance of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic weight of lithium.
(b) Boron has two stable isotopes \(_{ 5}^{ 10 }{ Li } \) and \(_{5 }^{ 11 }{ Li } \) ^.Their respective masses are 10.01294 u and 11.00931 u and the atomic weight of boron is 10.811 u. Find the abundances of \(_{ 5 }^{ 10 }{ Li } \) and
\(_{5 }^{ 11 }{ Li } \)
Answer:
(a) Atomic weight of lithium


Question 2.
The three stable isotopes of neon :\(_{ 20}^{ 10 }{ Ne } \) and \(_{ 22}^{ 10 }{ Ne } \)  have respective abundance of 90.51%, 0.27% and 9.22%. The atomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
Answer:

Question 3.
Obtain the binding energy of a nitrogen nucleus (\(_{ 7}^{ 14 }{ N } \)) from the following data :
m_H = 1.00783 u
m_n = 1.00867 u
_mn = 14.00307 u
Give your answer in MeV.
Answer:

Question 4.
Obtain the binding energy of the nuclei \(_{ 26 }^{ 56 }{ Fe } \) and
in units of \(_{ 83 }^{ 209 }{ Bi} \) from the following data:
m_H =1007825u
m_n =1008665u
m (\(_{ 26 }^{ 56 }{ Fe } \))= 55.934939 u
m (\(_{83}^{209 }{ Bi} \))
Which nucleus has greater binding energy per nucleon?
Answer:

Question 5.
A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of \(_{ 29 }^{ 63 }{ Cu } \) atoms (of mass 62.92960 u). The masses of proton and neutron are 1.00783 u and 1.00867 u, respectively.
Answer:

Question 6.
Write nuclear equations for :
(a) the α-decay of \(_{ 86 }^{226 }{ Ra} \)
(b) the β^–-decay of \(_{ 15 }^{ 32 }{ p } \)
(c) the β⁺-decay of \(_{ 6 }^{ 11 }{ p } \)
Answer:

Question 7.
A radioactive isotope has a half-life of T years. After how much time is its activity reduced to 3.125% of its original activity (b) 1% of original value ?
Answer:

Question 8.
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive \(_{ 6 }^{ 14 }{ C } \) present with the stable carbon isotope
\(_{ 12 }^{ 6 }{ C } \) When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of \(_{ 6 }^{ 14 }{ C } \) , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of \(_{ 6 }^{ 14 }{ C } \) dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Answer:

Question 9.
Obtain the amount of \(_{ 27 }^{ 60 }{ Co } \) necessary to provide a radioactive source of 8.0 mCi strength. The half­ life of \(_{ 27 }^{ 60 }{ Co } \) is 5.3 years.
Answer:

Question 10.
The half-life of \(_{ 38 }^{ 90 }{ Sr } \) is 28 years. What is the disintegration rate of 15 mg of this isotope?
Answer:

Question 11.
Obtain approximately the ratio of the nuclear radii of the gold isotope \(_{ 79 }^{ 197}{ Au } \) and silver isotope \(_{47}^{107 }{ Au } \).
Answer:

Question 12.

Answer:

Question 13.

Answer:

Here m_N stands for the nuclear mass of the element or particle. In order to express the Q value in terms of the atomic masses, 6 m_e mass has to be subtracted from the atomic mass of \(_{6}^{11 }{ Au } \) and 5 m_e mass has to be

Question 14.
The nucleus \(_{ 10 }^{ 23 }{ Ne} \) decays by β~ emission. Write down the p-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m(\(_{ 10 }^{ 23 }{ Sr } \)) = 22.994466 u
m(\(_{ 11 }^{ 23 }{ Sr } \)) = 22.989770 u.
Answer:

Question 15.
The Q value of a nuclear reaction A + b ⇒ C + d is defined by [Q = m_A + m_b-m_c– m_d] c² where the masses refer to nuclear rest masses. Determine from the given data whether the following reactions are exothermic or endothermic.

Answer:

Question 16.
Suppose, we think of fission of a \(_{ 26}^{ 56 }{ Fe} \) nucleus into two equal fragments, if \(_{ 13}^{ 28 }{ Al } \). Is the fission energetically possible ? Argue by working out Q of the process. Given, m (\(_{ 26}^{ 56 }{ Fe} \)) = 55.93494 u and m (\(_{ 13}^{ 28 }{ Al } \))= 27.98191
Answer:


Question 17.
The fission properties of \(_{ 94}^{ 239 }{ Pu} \) are very similar to those of \(_{ 92}^{ 235 }{ u} \)u. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure
\(_{94}^{ 239 }{ Pu} \) undergo fission ?
Answer:

Question 18.
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much
\(_{ 92}^{ 235 }{ u} \) did it contain initially ? Assume that all the energy generated arises from the fission of \(_{92}^{ 235 }{ u} \) and that this nuclide is consumed by the fission process.
Answer:

Question 19.
How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium ? The fusion reaction can be taken as

Answer:

Question 20.
Calculate the height of Coulomb barrier for the head on collision of two deuterons. The effective radius of deuteron can be taken to be 2.0 fm.
Answer:
The initial mechanical energy E of the two deutrons before collision is given by
E = 2 K.E.

Question 21.
From the relation R = R₀ A^1/3, where R₀  is a constant and A is the mass number of a nucleus, show that nuclear matter density is nearly constant (i.e. independent of A)
Answer:

Question 22.
For the β⁺ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K- shell, is captured by the nucleus and a neutrino is emitted).

Show that if β⁺ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa
Answer:

Question 23.
In a Periodic Table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on Earth. The three isotopes and their masses are \(_{ 12}^{ 24 }{ Mg} \) (23.98504u), ? \(_{12}^{ 25 }{ Mg} \) (24.98584) and \(_{ 12}^{ 26 }{ Mg} \) (25.98259u). The natural abundance of \(_{ 12}^{ 24 }{ Mg} \) is 78.99% by mass. Calculate the abundances of the other two isotopes.
Answer:

Question 24.
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei \(_{ 12}^{ 24 }{ Ca} \) and \(_{ 13}^{ 27 }{ Al} \) from the following data :

Answer:

Question 25.

Answer:

Question 26.
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes :

(a) Calculate the Q values for these decays and determine that both are energetically possible.
(b) The Coulomb barrier height for α-particle

Answer:

Question 27.
Consider the fission of \(_{ 92}^{ 239}{u} \) by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products, after the beta-decay of the primary fragments, are \(_{58}^{ 140}{Ce} \) and \(_{44}^{ 99}{Ru} \). Calculate Q for this fission process. The relevant atomic and particle masses

Answer:

Question 28.
Consider the D-T reaction (deuterium-tritium-fusion) given in eqn. :

(b) Consider the radius of both deuterium and tritium to be approximately 1.5 fm. What is the kinetic energy needed to overcome the Coulomb repulsion? To what temperature must the gases be heated to initiate the reaction?
Answer:
From the equation given in the question,

m_N refers to the nuclear mass of the element given in the brackets and m_n = mass of the neutron. If in represents the atomic mass, then


Question 29.
Obtain the maximum kinetic energy of p-particles and the radiation frequencies to y decay in the following decay scheme. You are given that
m (¹⁹⁸Au) = 197.968233 u
m (¹⁹⁸Hg) = 197.966760 u
Answer:
The total energy released for the transformation of \(_{79}^{ 198}{Au} \) to \(_{80}^{ 198}{u} \) can be found by considering the energies of ϒ-rays. We first find the frequencies of the ϒ-rays emitted.




Question 30.
Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the sun and (b) the fission of 1.0 kg of ²³⁵U in a fission reactor.
Answer:

Question 31.
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plant. Suppose we are given that, on average, the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium did our country need per year by 2000 ? Take the heat energy per fission of ²³⁵U to be about 200 MeV. Avogadro’s number = 6.023 x 10²³ mol^-1.
Answer:

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NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

Question 1.
What is the force between two small charged spheres having charges of 2 x 10^-7 C and 3 x 10^-7 C placed 30 cm apart in the air?
Answer:

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
Answer:
(a) Force on charge 1 due to charge 2 is given by the relation

Question 3.
Check that the ratio ke²/G m_em_p. is dimensionless. Lookup a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Answer:

Thus, the given ratio is a number and is dimensionless. This ratio signifies that electrostatic force between electron and proton is very-very large as compared to the gravitational force between them.

Question 4.
(a) Explain the meaning of the statement ‘electric charge of a body is quantised.’
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large-scale charges?
Answer:
(a) Charge occurs in nature as a discrete entity. One packet of charge (least quantity) is called quantum charge. It is represented by ‘e‘. Generally, charge of an electron is represented by ‘e’ and charge of a proton is represented by ‘+ e’. Therefore, the charge possessed by any charged body will be an integral multiple of ± e, ie., ne where n = 1, 2, 3,
∴ q = ± ne
The fact that electric charge collections are integral multiples of the fundamental electronic charge was proved experimentally by Millikan.

(b) While dealing with the large-scale electrical phenomenon, we ignore the quantization of charge because the magnitude of charge of proton and electron is so small. For continuous charge distribution, charge can be accounted in terms of charge density such as linear charge density λ etc. We need not go for individual charges.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:
Before rubbing, both the glass rod and the silk cloth are electrically neutral. In other words, the net charge on the glass rod and the silk cloth is zero. When the glass rod is rubbed with silk cloth, a few electrons from the glass rod get transferred to the silk cloth. As a result, the glass rod becomes positively charged and the silk cloth negatively charged. Since the magnitude of positive charge on the glass rod is the same as that of negative charge on the silk, the net charge on the system is zero. Thus the appearances of charge on the glass rod and the silk cloth is in accordance with the law of conservation of charges.

Question 6.
Four-point charges q_A = 2 μC, q_B = -5 μC, or q_c = -2 μC and q_D = -5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1μC placed at the center of the Square?
Answer:
The symmetry of the figure clearly indicates that 1μC charge will experience equal and opposite forces due to equal charges of 2μC placed at A and C. Similarly, 1μC charge will experience equal and opposite forces due to -5 μC charges placed at D and B.

Thus, net force = zero.

Question 7.
(a) An electrostatic field line is a continuous curve.
That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?
Answer:
(a) Electric lines of force exist throughout the region of an electric field. The electric field of a charge decreases gradually with increasing distance from it and becomes zero at infinity (i)e., electric field can’t vanish abruptly. So a line of force can’t have sudden breaks, it must be a continuous curve.

(b) If two lines of force intersect, then there would be two tangents and hence two directions of electric fields at the point of intersection, which is not possible.

Question 8.
Two-point charges q_A = 3μC and q_B = -3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 x 10^-9 C is placed at this point, what is the force experienced by the test charge? (C.B.S.E. 2003)
Answer:
(a) Electric field at the midpoint of the separation between two equal and opposite charges is given by

(b) Force experienced by test charge = q₀^E
= (1.5 x 10^-9) (5.4 x 10⁶)
= 8.1 x 10^-3 N along BA

Question 9.
A system has two charges q_A = 2.5 x 10^-7 C and q_B = -2.5 x 10^-7 C located at points
A : (0,0, -15 cm) and B : (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moments of the system?
Answer:
Clearly, the given points are lying on the z-axis.
Distance between charges, 21

= 15 + 15 = 30 cm = 0.3 m
Total charge = (2.5 x 10-⁷) – (2.5 x 10-⁷) = 0
Dipole moment
= q x 21 = 2.5 x 10-⁷ x 0.3
= 7.5 x 10^-8 C m along negative z-axis.

Question 10.
An electric dipole with dipole moment 4 x 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 x 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole.
Answer:
Using τ = pE sin 0, we get
= (4 x 10-⁹) (5x 10-⁴ sin 30°)
= 2 x 10-⁴ x \(\frac { 1 }{ 2 } \) = 10-⁴ N m 2

Question 11.
A polythene piece rubbed with wool is found to have a negative charge of 3.2 x 10-⁴ C.
(a) (Estimate the number of electrons transferred (from which to which ?)
(b) Is there a transfer of mass from wool to polythene?
Answer:
(a) Using q = ne, we get

(b) Yes, but of negligible amount because mass of an electron is very-very small (mass transferred = m_e x n = 91 x 10-³¹ x 2 x 10¹² = 1.82 x 10^-18 kg).

Question 12.
(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10–⁷ C ? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Answer:

Question 13.
Suppose the spheres A and B in Q 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Answer:
Charge on sphere A on contact with the third sphere (say C) having no charge is given by

When third sphere, how having charge 3.25 10^-7 C is brought in contact with sphere B, the charge left on sphere B is given by,

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Answer:
Unlike charges attract each other, therefore particle 1 and 2 are negatively charged whereas particle 3 has a positive charge. Particle 3 gets maximum deflection so it has the highest charge (e) to mass (m) ratio because deflection, y α e/m

Question 15.
Consider a uniform electric field \(\overrightarrow { E } \)
= 3 x 10³ \(\hat { i } \)N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Answer:
(a) Electric flux through the square,

Question 16.
What is the net flux of the uniform electric field of Problem 1.15 through a cube of the side 20 cm oriented so that its faces are parallel to the co­ordinate planes?
Answer:
Zero, because the number of field lines entering the cube is equal to the number of field lines coming out of the cube.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 10³ Nm2/C.
Answer:
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?
Ans.
(a) Using φ = φ /ε₀ we get q =φ ε₀
= (8 x 10³) (8.854 x 10^-12)
= 70.8 x 10-⁹ C = 0.07 μC
(b) No, it cannot be said so because there may be an equal number of positive and negative elementary changes inside the box. It can only be said that the net charge inside the box is zero.

Question 18.
A point charge + 10 μC is a distance 5 cm directly above the center of a square of side 10 cm, as shown in the given figure. What is the magnitude of the electric flux through the square? [Hint. Think of the square as one fact of a cube with edge 10 cm.]

Answer:
The charge can be assumed to be placed as shown in the figure.

Question 19.
A point charge of 2.0 μC is at the center of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Answer:

Question 20.
A point charge causes an electric flux of -1.0 x 10³ Nm²/C to pass through a spherical Gaussian surface of a 10.0 cm radius centered on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
Answer:
(a) The electric flux depends only on the charge enclosed by the Gaussian surface and independent of the size of the Gaussian surface. The electric flux through new Gaussian surface remains same
i.e. -1 x 10³ Nm² C-¹ because the charge enclosed remains same in this case also.
(b) Using φ = q/ε₀, we get q = ε₀ φ = (8.85 x 10-¹²) (-1 x 10³)
i.e. q = -8.85 x 10^-9 C.

Question 21.
a conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the center of the sphere is 1.5 x 10³ N/C and points radially inward, what is the net charge on the sphere?
Answer:

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.4 μC /m².
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere? (B.S.E. 2009 C)
Answer:
(a) Charge on the sphere is given by

Question 23.
An infinite line charge produces a field of 9 X 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density
Answer:

Question 24.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10^-22 C/m². What is E?

(a) in the outer region of the first plate,
(b) in the outer region of the second plate, and
(c) between the plates?
Answer:

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 x 10⁴ NC^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop, (g = 9.81 ms^-2 ; e = 1.60 x 10^-19 C.)
Answer:
Charge on the drop,

Question 26.
Which among the curves shown in the figure cannot possibly represent electrostatic field lines?



Answer:
(a) Incorrect, because the field lines should be normal to the surface of a conductor.
(b) Incorrect, because the field lines cannot start from a negative charge.
(c) Correct
(c) Incorrect, because electric field lines cannot intersect with each other.
(d) Incorrect, because electrostatic field lines cannot form a closed loop.

Question 27.
In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ N C^-1 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 C m in the negative z-direction?
Answer:
Suppose the dipole is along the z-axis.


Question 28.
(a) A conductor A with a cavity as shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Figure (b)].
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Answer:
Select a Gaussian surface lying wholly inside the conductor but very near to the surface of the conductor.
(a) There is no electric field inside the conductor so electric flux through Gaussian surface is zero or in other words, net charge inside the Gaussian surface is zero. Then it can be said that the charge lies outside the Gaussian surface e. on the outer surface of the conductor.
(b) Charge q inside the cavity will induce a charge -q on the inner side of the cavity and thus +q will appear on the outer surface. Thus total charge will be (q + Q).
(c) The instrument should be enclosed in a metallic shell so that the effect of the electrostatic field is cancelled out.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε₀) \(\hat { n } \), where \(\hat { n } \) is the unit vector in the outward normal direction, and a is the surface charge density near the hole.
Answer:
Let the tiny hole of the conductor be considered as filled up. The field inside the conductor is zero, whereas outside it is given by

This field is infact due to
(1) field (E₁) due to plugged hole and
(2) field E₂ due to rest of the charged conductor. Inside the conductor, these fields are equal but opposite, whereas outside they are exactly same. i.e.

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density X without using Gauss’s law.
[Hint. Use Coulomb’s law directly and evaluate the necessary integral.]
Answer:
Consider a long thin wire of uniform linear charge density X placed along the X-axis. Let P be a point lying on the y-axis






Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves build-out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3)e, and the ‘down’ quark (denoted by d) of charge
(-l/3)e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.)
Answer:

Question 32.
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Answer:
(a) To prove the result, let us assume that the test charge placed at the null point is in stable equilibrium. If it is so, then on being displaced slightly away from the null point, the test charge should return to its position., It implies that if a closed surface is drawn around the test charge, there will be a net inward flux of the electric field through its surface. According to Gauss law, there cannot be any electric flux through its surface as it does not enclose any charge. Hence our assumption is wrong and the test

(b) For the configuration of the two charges of the same magnitude and sign, the null point is the midpoint of the line joining the two charges. If the test charge is displaced slightly from the null point along the line, it will return back due to the restoring force that comes into the day. But if the charge is displaced slightly from the null – point along normal to the line it will not return. This is because the resultant force due to the configuration of two charges will take it away from the null point. For the test charge to be in stable equilibrium restoring force must come into play, when it is displaced in any direction. Hence the test charge cannot be in stable equilibrium.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed σ_x, (like particle 1 in figure.) The length of plate is L and a uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2m υ_x²).
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10. of Class XI Textbook of Physics.
Answer:
Consider a uniform electric field \(\overrightarrow { E } \) set up between two oppositely charged parallel plates (Figure). Let a positively charged particle having charge +q and mass m enters the region of electric field E at O with velocity E along X-direction.
Step 1.

Force acting on the charge +q due to electric field E is given by
\(\overrightarrow { F } \) =Q \(\overrightarrow { E } \)
The direction of the force is along the direction of
\(\overrightarrow { E } \)and hence the charged particle is deflected accordingly.
Acceleration produced in the charged particle is given by

Step 2.
The charged particle will accelerate in the direction of E . As soon as the particle leaves the region of electric field, it travels due to inertia of motion and hits the screen at point P. Let t be the time taken by the charged particle to traverse the region of electric field of length L. Let y be the distance travelled by the particle along y-direction (i.e. direction of electric field). Using a standard equation of motion,
S = ut + \(\frac { 1 }{ 2 } \) at².
For horizontal motion. S = L, u = υ_x and a = 0.
(∴ no force acts on the particle along x-direction)
From equation (ii), we have

Equation (i) is the equation of a parabola.
Hence a charged particle moving in a uniform electric field follows a parabolic path.

Question 34.
Suppose that the particle in Q 1.33 is an electron projected with velocity
υ_x = 2.0 x 10⁶ ms^-1. If E between the plates separated by 0.5 cm is 9.1 x 10² N/C, where
will the electron strike the upper plate ? ( |e| = 1.6 x 10^-19 C, m_e = 9.1 x 10-³¹ kg.)
Answer:

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NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts.

NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts

Question 1.
Explain with examples what historians mean by the integration of cults.
Solution :
The integration of cults means that there was composition, compilation and preservation of Puranic texts in simple Sanskrit verse, explicitly meant to be accessible to women and Shudras, who were generally excluded from Vedic learning. The Brahmanas also accepted and reworked the beliefs and practices of these and other social categories. The example of this integration of cults is at Puri, Orissa where the principal deity was identified as Jagannatha (literally, the lord of the world), a form of Vishnu. Here the deity is represented in a very different way. In this case, a local deity, whose image was and continues to be made of wood by local tribal specialists, was recognised as a form of Vishnu. At the same time, Vishnu was visualised in a way, that was very different from that in other parts of the country. Such instances of integration were evident amongst goddess cults as well where the local deities were provided an identity as the wife of the principal male deities – Lakshmi, the wife of Vishnu or Parvati, the wife of Shiva.

Question 2.
To what extent do you think the architecture of mosques in the subcontinent reflects a combination of universal ideals and local traditions?
Solution :
With the arrival of Islam in the Medieval ages, the architecture of Islam also came to India. However, the Arab-cum-Islamic architecture got impacted by the local traditions and rites too. Hence, we see a fusion of the two. This can be further elaborated by the examples of architecture mainly the constructions of the mosques of those days.

Some features of the architecture of mosques are universal. All mosques have orientation towards Mecca. This is manifested in the placement of Mehrab and Minar within a mosque. But at the same time we have influences that can be described only as local influences. A 13th Century mosque in Kerala has a shikhar like roof unlike a normal mosque where it is dome. The Shah Hamdan Mosque in Kashmir is made of Kashmiri woods and its facade is like that of a temple. The Atia Mosque in Bangladesh is made of bricks, though its roof is round. Thus, we can see that the architecture of Mosques is that of fusion.

Question 3.
What were the similarities and differences between the be-shari‘a and ba- shari‘a sufi traditions ?
Solution :
(a) Similarities :

• Both be-sharia and ba-shari‘a sufis turned to asceticism and mysticism and were against the growing materialism of the Caliphate as a religious and political institution.
• They are critical of the dogmatic definitions and scholastic methods of interpreting the Quran and sunna adopted by the theologians.
• Both laid emphasis on seeking salvation through intense devotion and love for God.
• Both sought an interpretation of the Quran on the basis of their personal experience.

(b) Differences :

 Question 4.
Discuss the ways in which the Alvars, Nayanars and Virashaivas expressed critiques of the caste system.
Solution :
The Alvars, Nayanars and Virashaivas expressed critiques of the caste system in the following ways :
(a) Alvars and Nayanars :

1. The Alvars and Nayanars protested against the caste system or at least attempted to reform the system. Their bhaktas hailed from diverse social backgrounds ranging from Brahmanas to artisans and cultivators and even from castes considered “untouchable”. Thus, people from all walks of society were welcomed by them.
2. They stated that their compositions were as important as the Vedas. For example, one of the major anthologies of compositions by the Alvars, the Nalayira Divyaprabandham, was described as the Tamil Veda.
3. Tondaradippodi, a Brahmana Alvar, opposed the cast system in the following way :
   “You (Vishnu) manifestly like those “servants” who express their love for your feet, though
   they may be bom outcastes, more than the Chaturvedins who are strangers and without allegiance to your service.”
4. Another saint, a Nayanar, composed the following verse in protest of the caste system : “O rogues who quote the law books, Of what use are your gotra and kula, Just bow to Marperu’s lord (Shiva who resides in Marperu, in Thanjavur, Tamil Nadu) as your sole refuge.”

(b) Virashaiva:

1. Virashaiva or Lingayats challenged the idea of caste and the “pollution” attributed to certain groups by Brahmanas. As a result of it, marginalised people within the Brahmanical order joined this tradition.
2. They questioned the theory of rebirth.
3. The Lingayats encouraged certain practices disapproved in the Dharmashastras, such as post-puberty marriage and remarriage of widows.

Question 5.
Describe the major teachings of either Kabir or Baba Guru Nanak, and the ways in which these have been transmitted.
Solution :

1. The major teachings of Kabir and Baba Guru Nanak are as given below :
   • Teachings of Kabir :
     
     • He described the Ultimate Reality as Allah, Khuda, Hazrat and Pir. He also used terms drawn on Vedantic traditions, alakh (the unseen), nirakar (formless), Brahman and Atman. Terms such as shabda (sound) or shunya (emptiness) were drawn from yogic traditions.
     • There is only one God. He was against the distinction made between gods of different communities.
     • Kabir was against idol worship and Hindu polytheism.
     • He was in favour of the Hindu practice of nam-simaran (remembrance of God’s name).
   • Teachings of Guru Nanak : The message of Baba Guru Nanak is spelt out in his hymns and teachings which are as given below :
     • Baba Guru Nanak believed in nirguna bhakti.
     • He was against the external practices of religions.
     • He rejected sacrifices, ritual baths, image worship, austerities and the scriptures of both Hindus and Muslims.
     • For him, the Absolute or “arab” had no gender or form.
     • He favoured a simple way to connect to the Divine by remembering and repeating the Divine Name.
2. The ways in which the major teachings of Kabir and Baba Guru Nanak were transmitted are as given below –
   • Compositions of the sants were sung by roadside musicians.
   • Baba Guru Nanak would sing his compositions in various ragas while his attendant Mardana played the rabab.
   • Baba Guru Nanak organised his followers into a community. He set up rules for congregational worship (sangat) involving collective recitation.

Question 6.
Discuss the major beliefs and practices that characterised Sufism.
Solution :
After the advent of Islam in the early, middle ages , it saw a new movement in later part. The movement has had great impact and reach in the Indian subcontinent. It is called Sufi movement. The Sufi saints were mystics. Their preachings included:
1. Sufi saints did not subscribe to the theological and rigid interpretations of religious scriptures of Islam. They believed that the interpretation have to be based on individual experiences. This way the theological interpretations became flexible. Further the control of the orthodox religious leaders got weakened. This was a people centric move.
2. They rejected the high sounding rituals. They also emphasised on simplicity in religious traditions and rites.
3. Sufi saints prescribed devotion to Almighty as path to salvation. They even approved of singing and dancing as part of devotion. It is notable that classical Islam has forbidden singing, dancing and any music.
4. The most important theme of Sufi philosophy was that serving people is the true religion. With the objective of serving the poor people they also held Langar. Today also one can go to Ajmer and can partake in the Langar organised on the tomb of Nijammudin Auliya, the great Sufi saint.
5. Sufi saints also emphasised on the equality among people and oneness among all.

Question 7.
Examine how and why rulers tried to establish connections with the traditions of the Nayanars and the sufis.
Solution :
The rulers tried to establish connections with the traditions of the Nayanars and the sufis to claim divine support and proclaim their own power and status.
(a) Rulars and Sufis :
(i) The Turks who had set up Delhi Sultanate also required legitimation from the sufis because they resisted the insistence of the ulama on imposing sharva as state law because they anticipated opposition from their non-Muslim subjects. The sultans, therefore, sought the support of the Sufis who derived their authority directly from God and did not depend on jurists to interpret the sharva.

(ii) The sultans wanted to have the blessings of Sufi saints because it was believed that the Auliya could intercede with God in order to improve the material and spiritual conditions of ordinary human beings. The Sultans of Delhi set up charitable trusts as endowments for hospices and granted tax-free land.

(iii) Akbar maintained connections to get their blessings for new conquests, fulfilment of vows and birth of sons. He used to visit the Dargah of Muinuddin Chishti. He went there 14 times and gave generous gifts, which were recorded in imperial documents. For example, in 1568, he offered a huge cauldron (degh) to facilitate cooking for pilgrims. He also had a mosque constructed within the compound of the dargah.

(b) Rulers and Nayanars :
(i) The Chola kings tried to establish connections with Nayanars and Alvars by building splendid temples that were adorned with stone and metal sculpture to recreate the visions of these popular saints who sang in the language of the people.

(iii) Those kings also introduced the singing of Tamil Shaiva hymns in the temples under royal patronage taking initiative to collect and organise them into a text.

(iii) An inscription suggests that Chola ruler Parantaka I had consecrated metal images of Appar, Sambandar and Sundarar in a Shiva temple. These were carried in processions during the festivals of these saints.

Question 8.
Analyse, with illustrations, why bhakti and sufi thinkers adopted a variety of languages in which to express their opinions.
Solution :
The bhakti and sufi thinkers adopted a variety of languages to express their opinions due to the following reasons :

1. The local languages were used so that people might understand their verses easily.
2. The language of the Vedas was Sanskrit. The Bhakti traditions were against the beliefs and practices of the Vedas. So, it was necessary to use local languages which people might understand and follow them in practice. That is why Kabir’s poems have survived in several languages. Similarly, Baba Guru Nanak expressed his ideas through hymns called “shabad” in Punjabi, the language of the region.
3. The Chishtis too adopted local languages such as Hindavi because poetic compositions were recited in hospices, usually during ‘sama’. In Bijapur, sufi poetry was composed in Dakhani because these were probably sung by women while performing household chores like grinding grain and spinning.

Question 9.
Read any five of the sources included in this chapter and discuss the social and religious ideas that are expressed in them.
Solution :
The period of the Bhakti Movement and Sufi Movement also has many sources that contribute to the history of those days. Some of the major social and religious ideas expressed in the various sources of history are as follows:
1. The first is the architecture. The different types of stupas, temple, monasteries all symbolise different types of religious belief system and practices. Some of them exist as it is and enable us to look into the annals of history of those days. Some of them are in the form of ruins but they also throw light on the, religion and society of those days alike.
2. The next important source of history is the composition of the saints both Bhakti
and Sufi. In terms of content they are religious but they are not the divine textbooks of religion that are sacrosanct. The compilation throws light on the life of common men and village lifestyle. They also impact the music and art of those days.
3. Another very important source of the history of those days is the biographies of the Saints. The biographies include the description of the society and prevalent beliefs and practices. It is notable that such biographies may not be in the written form still they can give insight into the life of those days. It is the story prevalent that I when Kabirdas died, both Hindus and Muslims fought for his dead body later on his body turned into flowers. Some were taken by Muslims and others by Hindus.
This represents that there conflict and collaboration between both Hindus and Muslims of those days.
4. This was also the period of rise of religious leaders who were intermediaries between common men and God. Earlier it was only the Brahmins who got this role. Now many people from other background also joined in. To some extent it acted as the the force that idolised equality and fraternity.
5. The other source is the folklore. They are described in our art forms. It may be dance, paintings, and sculpture and so on. They all talk about the universal brotherhood of mankind and love for one and all.

We hope the NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts, drop a comment below and we will get back to you at the earliest.