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Important Question of Physics Class 12 with Answers

Important Questions for Class 12 Physics with Answers

As Central Board of Secondary Education has announced the CBSE Term 2 Exam Date 2022. The students have to practice Class 12 Physics Term 2 Important Questions given on this page to get good marks in the exam. On this page, we have given Class 12 Physics Term 2 Important Questions as per the latest exam pattern of CBSE Term 2. We have segregated 2 marks, 3 marks, and 5 marks questions from each chapter of Class 12 Physics and given them on this page to help students in their Class 12 Term 2 Exam preparation. The students appearing in CBSE Class 12 Term 2 exam can bookmark this page to get the latest updates from the Central Board of Secondary Education. 

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Physics Important Questions for Class 12 Term 2 – Exam pattern

Preparing for any exam requires familiarity with the exam pattern. If required, students will be provided a log table, and the use of calculators in exams will be restricted. The CBSE Term 2 Physics question paper is divided into three sections: Section A, Section B, and Section C. Check the details given below: 

Section A: In this section, each question carries 2 marks. 

Section B:  In this section, each question carries 3 marks. 

Section C:   In this section, each question carries 5 marks.

Class 12 Term 2 Physics Important Questions with Answers – 2 Marks

Unit – Electromagnetic Waves

Q. The charging current for a capacitor is 0.25Ω0.25Ω. What is the displacement current across its plates?

Ans: Displacement current is equivalent to and is the same as charging current 025A025A.

Q. Write the following radiations in a descending order of frequencies: red light, x-rays, microwaves, radio waves

Ans: The descending order of frequencies: red light, x-rays, microwaves, radio waves are given as: X-rays > Red light >Microwaves >Radio Waves

Q. How does the frequency of a beam of ultraviolet light change, when it goes from air into glass?

Ans: We know that the frequency of UV light is unaffected by the frequency of a beam of ultraviolet light change and when it goes from air into glass.

Q. What is the ratio of speed of gamma rays and radio waves in vacuum?

Ans: As we know, the ratio of speed of gamma rays and radio waves in vacuum is one

Q. It is necessary to use satellites for long distance TV transmission. Why?

Ans: Because television signals are not reflected by the ionosphere layer of the atmosphere, TV broadcasts from air earth stations must be reflected back to the ground via an artificial satellite.

Q. What is the role of ozone layer in the atmosphere?

Ans: The main role of the ozone layer in the atmosphere is that it absorbs all harmful UV rays and shields us from their detrimental effects.

Q. What is the nature of waves used in radar?

Ans: The characteristics of the waves utilized in radar employs microwaves.

Q. What physical quantity is the same for XX rays of wavelength 10−10m10−10m, red light of wavelength 6800 A and radio waves of wavelength 500m500m ?

Ans: As we know that the speed of light (3×108m/s)(3×108m/s) all wavelengths are the same in a vacuum. In the vacuum and they are unaffected by the wavelength.

Unit – Optics

Q. How do the focal lengths of a lens change with increase in the wavelength of the light?

Ans: We know that

1f=(μ1)(1R11R2)1f=(μ−1)(1R1−1R2)

Also,

μ1λμ∝1λ

Clearly, when wavelength (λ)(λ) increases, refractive index

(μ)(μ)

decreases.

Similarly, μfμ∝f.

Clearly, as refractive index

(μ)(μ)

decreases, focal length (f)(f) increases.

Q. Show with a ray diagram, how an image is produced in a total reflecting prism?

Ans: Consider the two rays from the object PQ, as shown below.

They undergo total internal reflection firstly at the face AB and then at BC forming the final image

PQP′Q′

(real and inverted image).

Q. The radii of the curvature of the two spherical surfaces which is a lens of required focal length are not the same. It forms an image of an object. The surfaces of the lens facing the object and the image are interchanged. Will the position of the image change?

Ans: As we know that,

1f=(μ1)(1R11R2)1f=(μ−1)(1R1−1R2)

When the radii of curvature

R1R1

and

R2R2

are interchanged, the focal length of the lens also changes. Hence, the position of the image will reduce gradually.

Q. A thin converging lens has focal length (f) when illuminated by violet light. State with reason how the focal length of the lens will change if violet light is replaced by red light.

Ans: We know that,

1f=(n1)(1R11R2)1f=(n−1)(1R1−1R2)

Since, nn for violet is more that nn for red colour, and since mfm∝f, we can say that the focal length of the lens will decrease when violet light is replaced by red light.

Q. What is the shape of the wave front when light is diverging from a point source?

Ans: When light diverges from a point source the wave front would show spherical shape.

Q. State the conditions that must be satisfied for two light sources to be coherent. 

Ans: The conditions to be satisfied are the following:

a) They must emit waves continuously of same wavelengths.

b) The phase difference between the waves must be zero or constant.

Q. You read a newspaper, because of the light it reflects. Then why do you not see even a faint image of yourself in the newspaper?

Ans: We know that image is formed due to the regular reflection of light.

However, when we read a newspaper, there is diffused (irregular) reflection of light, thus we are not able to see even a faint image of ourselves on the newspaper.

Q. 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. Motion of the source and/or observer

  4. Wave length

  5. Intensity of the wave. On which of these factors, if any, does

  6. The speed of light in vacuum

Ans: We know that the speed of light in vacuum, 3×108 m/s3×108 m/s  (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. So, the given factor does not affect the speed of light in a vacuum.

Q. The speed of light in a medium (say, glass or water), depends.

Ans: Out of the listed factors, the speed of light in a medium would depend on the wavelength of light in that medium.

Q. 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?

Ans: Myopia and hypermetropia are common eye defects.

A myopic or hypermetropic person need not necessarily suffer a partial loss in their eyes’ ability of accommodation.

Myopia occurs when the eye-balls engage in elongation from the front to the back whereas hypermetropia occurs when the eye-balls shorten themselves.

On the other hand, when the eye-lens completely loses its ability of adjusting itself, then the defect is called presbyopia.

Unit – Dual Nature of Radiation and Matter

Q. Mention one physical process for the release of electrons from the surface of a metal.

Ans: Photoelectric emission.

The phenomenon in which the electrons from the surface of a metal are given energy in form of electromagnetic waves and they are ejected out, this phenomenon is called the photoelectric emission.

Q. The maximum kinetic energy of photoelectron is?

2.8 eV2.8 eV

Q. What is the value of stopping potential?

Ans: Given an electron that is moving with a kinetic energy. For it to be not ejected, it has to be held back using a stopping potential

V0V0

The relation between the two is:

KE=eVo=2.8eVKE=eVo=2.8eV

Vo=2.8V

Q. Ultraviolet light is incident on two photosensitive materials having work functionϕ1ϕ1 and  ϕ2ϕ2 (ϕ1>ϕ2ϕ1>ϕ2). In which of the case will K.E. of emitted electrons be greater? Why?

Ans: According to the energy balance equation of the photoelectric effecthv=ϕo+K.Ehv=ϕo+K.E

If  ϕ1>ϕ2ϕ1>ϕ2 thus K.E. will be more for second surface whose work function is less.
Q. Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of incident radiations.

Ans: Suppose

νoνo is the threshold of frequency or cut off frequency;

VoVo is the corresponding stopping potential

Q. How does the stopping potential applied to a photocell change if the distance between the light source and the cathode of the cell is doubled?

Ans: Intensity of light drops quadratically with distance. However, the stopping potential does not depend on the intensity of the light. Hence it is independent of distance as well.

Q. On what factor does the retarding potential of a photocell depend? 

Ans: The retarding photocell depends upon the frequency of the incident light

Q. Electron and proton are moving with same speed, which will have more wavelength?

Ans: Since the wavelength is inversely proportional to the square root of the mass of the body, λ1mλ∝1m. So, electrons being lighter will have more wavelengths.

Unit – Atoms and Nuclei

Q. Complete the following nuclear reactions:

a) 4Be9+1H1→3Li6+……….

Ans: 4Be9+1H1→3Li6+2He4

b) 5B10+2He4→7N13+……….

Ans:  5B10+2He4→7N13+0n1

Q. What is the Q-value of a nuclear reaction?

Ans:  Qvalue=(Mass of reactantsMass of products)  

Q. The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 9546∘A,  6463∘A and 1216∘A. Which one of these wavelengths belongs to the Lyman series?

Ans: 1216∘A belong to the Lyman series.

Q. Name the series of hydrogen spectrum lying in ultraviolet and visible   region.

Ans: Lyman series lies in ultraviolet region while Balmer series lies in visible region.

Q. State the limitations of Bohr’s atomic model.

Ans: The limitations of Bohr’s atomic model are:

(1) It does not give any indication regarding the arrangement and distribution of electrons in an atom.

(2) It could not account for the wave nature of electrons.

Unit – Electronic Devices

Q. What type of impurity is added to obtain n-type semiconductor?

Ans: Pentavalent atoms (group –1515) like Phosphorus (P), Arsenic (As), etc.

Q. Doping of silicon with indium leads to which type of semiconductor?

Ans: Doping of Silicon with Indium produces a p-type semiconductor as Indium is a trivalent impurity.

Q. How does the energy gap of an intrinsic semiconductor vary, when doped with a trivalent impurity?

Ans: An acceptor energy level is formed in the forbidden energy gap above the valence band when an intrinsic semiconductor is doped with a trivalent impurity. Due to this, electrons quickly jump to the acceptor energy level.

Q. How does the width of the depletion layer of p-n-junction diode change with decrease in reverse bias?

Ans: The width of the depletion layer will decrease with decrease in reverse bias.

Q. Under what condition does a junction diode work as an open switch?

Ans: A junction diode works as an open switch when it is connected under reverse bias conditions.

Q. Which type of biasing gives a semiconductor diode very high resistance?

Ans: Reverse biasing gives a semiconductor diode very high resistance.

Q. Define current amplification factor in a common – emitter mode of transistor.

Ans: Current amplification factor is the fraction of a small change in collector current to the slight difference in base current at constant collector-emitter junction voltage.

Q.  Why is a semiconductor damaged by a strong current?

Ans: When a strong current traverses through a semiconductor, a large amount of heat is generated, which breaks the covalent links in the semiconductor. This causes the semiconductor to get damaged.’

Class 12 Physics Term 2 Important Questions with Answers

Class 12 Term 2 Physics Important Questions with Answers– 3 Marks

Unit – Electromagnetic Waves

Q. Write the characteristics of emf waves? Write the expression for velocity of electromagnetic waves in terms of permittivity and permeability of the medium?

Ans: The Characteristics of emf waves are as follows :

(1) It travels in free space with speed of light c=3×106m/sc=3×106m/s,

(2) Electromagnetic waves are transverse in nature.

Velocity of emf waves in vacuum C =1μ06−−−−√=1μ0∈6

Q. The electric field of a plane electromagnetic wave in vacuum is represented byby

Ex=0,Ey=0.5cos(2π×108(tx/c))Ex=0,Ey=0.5cos⁡(2π×108(t−x/c)) and Ez=0Ez=0

a) What is the direction of propagation of electromagnetic wave?

Ans: For the given equation  It represents the wave is propagating along +x¯+x¯ – axis

b) Determine the wavelength of the wave?

Ans: Now, by Comparing equation with the standard one

Ey=E0cosw(tx/c)Ey=E0cos⁡w(t−x/c)

w=2π×102w=2π×102

zt⃗ t=2z×108z′t→t=2′z×108

v=108v=108

λ=cv=3×108108⇒λ=cv=3×108108

λ=3mλ=3m

c) Compute the component of associated magnetic field

Ans: As it is associated magnetic field to electric field and the direction of propagation. Since wave is propagating along x=x= axis, electric field is along, y=y= axis

Thus, magnetic field is along z=z= axis

Bx=0,By=0⇒Bx=0,By=0

Bz=B0cos(2π×101(tx/x))Bz=B0cos⁡(2π×101(t−x/x))

Bz=E0Ccos(2π×101(tx/x))Bz=E0Ccos⁡(2π×101(t−x/x))

Q. Find the wavelength of electromagnetic waves of frequency 5×1018Hzin free space5×1018Hzin free space.  Give its two applications.

Ans: By Using C=λvC=λv

v=5×1015Hzv=5×1015Hz

λ=3×1015×106λ=3×1015×10∘6

λ=0.6×1011=6×1012mλ=0.6×10−11=6×10−12m

These are Gamma Rays.

Applications for gamma rays :

(1) These rays are used to get information regarding atomic structure.

(2) They have very high penetrating power so they are used for detection purpose

Q. State the condition under which a microwave oven heats up food items containing water molecules most efficiently?

Ans: (1) Microwaves must have a frequency that is equal to the resonant frequency of the water molecules in the food item.

(2) Name the radiations which are next to these radiations in emf. Spectrum having (a) Shorter wavelength (b) Longer wavelength

(a) visible light is the radiation are the radiation next to shorter wavelength.

(b) Microwaves light is the radiation are the radiation next to longer wavelength.

Q.  Long distance radio broadcasts use short-wave bands. Why?

Ans: Shortwave bands are used for long-distance radio broadcasts because only these bands can be refracted by the ionosphere.

Q. It is necessary to use satellites for long distance TV transmission. Why?

Ans: Because television signals have high frequencies and energy, satellites are required for long-distance TV transmissions. As a result, the ionosphere does not reflect these signals. As a result, satellites aid in the reflection of television transmissions.

Q. Optical and radio telescopes are built on the ground but   -ray astronomy is possible only from satellites orbiting the earth. Why? 

Ans: As far as -ray astronomy is concerned, -rays are absorbed by the atmosphere. Visible and radio waves, on the other hand, can pass through it. As a result, optical and radio telescopes are constructed on the ground, but X-ray astronomy is only conceivable thanks to satellites orbiting the Earth.

Q. The small ozone layer on top of the stratosphere is crucial for human survival. Why?

Ans: The ozone layer at the top of the atmosphere is critical for human survival because it absorbs damaging ultraviolet radiation from the sun and keeps it from reaching the Earth’s surface.

Q. If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

Ans: There would be no greenhouse effect on the Earth’s surface if it did not have an atmosphere. As a result, the Earth’s temperature would swiftly drop, making it cold and uncomfortable to live on and difficult for human survival.

Q. Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

Ans: On the surface of the Earth, a global nuclear war would be catastrophic. The Earth will experience a harsh winter following a nuclear war because the war will produce clouds of smoke that will cover the majority of the sky, preventing solar light from reaching the atmosphere. It will also contribute to the ozone layer’s depletion.

Unit – Optics

Q. 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. Motion of the source and/or observer

  4. Wave length

  5. Intensity of the wave. On which of these factors, if any, does

  1. The speed of light in vacuum

Ans: We know that the speed of light in vacuum, 3×108 m/s3×108 m/s  (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. So, the given factor does not affect the speed of light in a vacuum.

  1. The speed of light in a medium (say, glass or water), depends.

Ans: Out of the listed factors, the speed of light in a medium would depend on the wavelength of light in that medium.

Q. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations given and explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium? 

  1. source at rest; observer moving, and 

  2.  source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations.

Ans: We know that sound waves can propagate only through a medium. The two given situations are not scientifically identical as the motion of an observer relative to a medium is different in the given two situations. So, the Doppler formulas for the given two situations cannot be the same.

Now, in case of light waves, sound can travel in a vacuum. But in vacuum, the above two cases are found to be identical as the speed of light is independent of the motion of the observer and the motion of the source. While, when light travels in a medium, the above two cases are not identical as the speed of light would now depend on the wavelength of the medium.

Q. In double-slit experiment using light of wavelength 600nm600nm, the angular width of a fringe formed on a distant screen is  0.10.1∘. What is the spacing between the two slits?

Ans: We are given:

Wavelength of light used, λ=6000 nm =600×109 mλ=6000 nm =600×10−9 m

Angular width of fringe, θ=0.1=0.1×λ180=3.141800 radθ=0.1∘=0.1×λ180=3.141800 rad

Angular width of a fringe is related to slit spacing (d)(d) as:

θ=λdθ=λd
d=λθ⇒d=λθ
d=600×1093.141800=3.44×104 m∴d=600×10−93.141800=3.44×10−4 m

Therefore, the spacing between the slits is found to be 3.44×104 m3.44×10−4 m.

Q.  In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/anλ/a . Justify this by suitably dividing the slit to bring out the cancellation. 

Ans: Consider that a single slit of width dd is divided into nn smaller slits.

That is, width of each slit,  d=dnd=dn

Angle of diffraction is given by the relation,

θ=ddλd=λdθ=ddλd=λd

Now, each of these infinitesimally small slits send zero intensity in direction θθ. Thus, the combination of these slits will give zero intensity.

Q. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Ans: In the given case, the person is able to see vertical lines more distinctly than horizontal lines.

This means that the refracting system (cornea and eye-lens) of the eye is not working effectively in different planes. This defect is called astigmatism.

The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient.

Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

Q. Answer the following questions:

  1. When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation. 

Ans: We know that weak radar signals are sent by a low flying aircraft and this can interfere with the TV signals received by the antenna. As a result of this, the TV signals might get distorted. Hence, when a low flying aircraft passes overhead, we could sometimes notice a slight shaking of the picture on our TV screen.

  1. As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

Ans: For our understanding of intensity distributions and interference patterns, the principle of linear superposition of wave displacement is essential. This is because superposition follows from the linear character of a differential equation that is known to govern wave motion. Let y1y1 and y2y2 are the solutions of the second order wave equation, then any linear combination of y1y1 and y2y2 might also be the solution of the wave equation.

Q. What is polarization of light? What type of waves show the property of polarization? Name any two methods to produce plane polarized light.

Ans: The phenomenon by virtue of which the vibrations of a light vector is restricted in a particular direction in a plane perpendicular to the direction of propagation of light is called polarisation of light. Transverse waves are known to show the property of polarisation. Two methods to produce plane polarised light are:

  1. Polarisation by Reflection and

  2. Polarization by scattering.

Q. Draw the curve depicting variation of intensity in the interference pattern in young’s double slit experiment. State conditions for obtaining sustained interference of light? 

Ans:  The curve could be drawn as,

Conditions for sustained interference of light are:

  1. Two sources causing the interference must be coherent sources of light.

  2. Two sources causing the interference should have nearly equal amplitudes and intensities and should be monochromatic.

Q. What is the shape of the wave front in each of the following cases:

  1. Light diverging from a point source. 

Ans: The shape of the wave front in case of a light that is diverging from a point source is spherical. The wave front emanating from a point source would be as shown in the below figure.

(Image will be uploaded soon)

  1. Light emerging out of a convex lens when a point source is placed at its focus.

Ans: The shape of the wave front in case of a light emerging out of a convex lens when a point source is placed at its focus would be a parallel grid. This would be as shown in the below figure.

  1. The portion of the wave front of light from a distant star intercepted by the Earth.

Ans: The portion of the wavefront of light from a distant star intercepted by the Earth would be a plane.

Unit – Dual Nature of Radiation and Matter

Q. Show on a plot the nature of variation of photoelectric current with the intensity of radiation incident on a photosensitive surface.
Answer:
Important Question of Physics Class 12 with Answers_40.1

Q. Figure shows a plot of 1V, where V is the accelerating potential, vs. the de-Broglie wavelength ‘λ’ in the case of two particles having same charge ‘q’ but different masses m1 and m2. Which line (A or B) represents a particle of larger mass? 
Important Question of Physics Class 12 with Answers_50.1
Answer:
B line represents particle of larger mass because slope 1m.

Q. Find the ratio of de-Broglie wavelengths associated with two electrons accelerated through 25 V and 36 V. 
Answer:
Important Question of Physics Class 12 with Answers_60.1

Q. Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit. 
Answer:
It is the number of photo-electrons emitted per second per unit area.
SI unit : m-2S-1

Q. The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work- function? Justify your answer. 
Important Question of Physics Class 12 with Answers_70.1
Answer:
Metal ‘A’, because of higher threshold frequency for it.

Q.  The graph shows variation of stopping potential V0 versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why?

Important Question of Physics Class 12 with Answers_80.1
Answer:
Metal ‘A’, because of higher threshold frequency for it.

Q. An electron is revolving around the nucleus with a constant speed of 2.2 × 108 m/s. Find the de-Broglie wavelength associated with it
Answer:
Important Question of Physics Class 12 with Answers_90.1

Q. Draw a plot showing the variation of de Broglie wavelength of electron as a function of its K.E.
Answer:
Important Question of Physics Class 12 with Answers_100.1

Q. Name the phenomenon which shows the quantum nature of electromagnetic radiation.

Answer:
Photoelectric Effect is the phenomenon which shows the quantum nature of electro-magnetic radiation.

Q. State one factor which determines the intensity of light in the photon picture of light.
Answer:
The factor determining the intensity of light is number of electrons emitted per second.

Q. State one reason to explain why wave theory of light does not support photoelectric effect. (Comptt. Delhi 2014)
Answer:
One reason why wave theory of light does not support photoelectric effect is that the kinetic energy of photo electrons does not depend on the intensity of incident light.

Q.  Write three basic properties of photons which are used to obtain Einstein’s photoelectric equation. Use this equation to draw a plot of maximum kinetic energy of the electrons emitted versus the frequency of incident radiation.

Answer:
Properties.
Einstein’s photoelectric equation is Kmax = hv – ϕ0
(i) We find Kmax depends linearly on V only. It is independent of intensity of radiation.
Important Question of Physics Class 12 with Answers_110.1
(iii) Greater the number of energy quanta, greater is the number of photoelectrons. So, photoelectric current is proportional to intensity.

Plot of maximum kinetic energy vs. frequency
Important Question of Physics Class 12 with Answers_120.1

Q. (i) Define the term ‘threshold frequency’ as used in photoelectric effect.
(ii) Plot a graph showing the variation of photoelectric current as a function of anode potential for two light beams having the same frequency but different intensities I1 and I2 (I1 > I2). 
Answer:
(i) Threshold frequency. The minimum frequency v0 which the incident light must possess so as to eject photoelectrons from a metal surface, is called threshold frequency of the metal.
Important Question of Physics Class 12 with Answers_130.1

Q. A proton and an a-particle have the same de- roglie wavelength. Determine the ratio of
(i) their accelerating potentials
(ii) their speeds.
Answer:
The de-Broglie wavelength for a proton is,
Important Question of Physics Class 12 with Answers_140.1

Q. Using the graph shown in the figure for stopping potential v/s the incident frequency of photons, calculate Planck’s constant. 
Important Question of Physics Class 12 with Answers_150.1
Answer:
Important Question of Physics Class 12 with Answers_160.1

Unit – Atoms and Nuclei

Q. (a) The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain.
(b) Plot a graph showing the variation of potential energy of a pair of nucleons as a function of their separation.
Answer:
(a) When nucleons approach each other to form a nucleus, they strongly attract each other. Their potential energy decreases and becomes negative. It is this potential energy which holds the nucleons together in the nucleus. The decrease in’ potential energy results in the decrease in the mass of the nucleons inside the nucleus.
Important Question of Physics Class 12 with Answers_170.1

Q. A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV. 
Answer:
Important Question of Physics Class 12 with Answers_180.1
∴ Gain in binding energy for nucleon = 8.5 – 7.6 = 0.9 MeV
Hence total gain in binding energy per nucleus fission = 240 × 0.9 = 216 MeV

Q. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.
Answer:
Two important conclusions :
(i) Nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This explains constancy of the binding energy per nucleon for large-size nucleus.
Important Question of Physics Class 12 with Answers_190.1
(ii) Graph explains that force is attractive for distances larger than 0.8 fin and repulsive for distances less than 0.8 fm.

Q. Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A ≤ 240. How do you explain the constancy of binding energy per nucleon in the range 30 < A < 170 using the property that nuclear force is short-ranged? 
Answer:
(a) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short ranged.
If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is approximately equal to pk. The property that a given nucleon influences only nucleons close to it, is referred to as saturation property of the nuclear force.
Important Question of Physics Class 12 with Answers_200.1

(b) Nuclear force is short-ranged for a sufficiently large nucleus. A nucleon is under the influence of only some of its neighbours, which come within the range of the nuclear force. If a nucleon can have maximum of P neighbours within the range of nuclear force, its binding energy would be proportional to ‘P’ Thus on increasing ‘A’ by adding nucleons binding energy will remain constant.

Q. Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release of energy in the processes of nuclear fission and nuclear fusion can be explained.
Answer:
1. Nuclear fission : Binding energy per nucleon is smaller for heavier nuclei than the middle ones i.e. heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, the B.E./nucleon changes (increases) from about 7.6 MeV to 8.4 MeV. Greater binding energy of the product nuclei results in the liberation of energy. This is what happens in nuclear fission which is the basis of the atom bomb.
Important Question of Physics Class 12 with Answers_210.1
2. Nuclear fusion : The binding energy per nucleon is small for light nuclei, i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy.

Q. Complete the following nuclear reactions :
Important Question of Physics Class 12 with Answers_220.1
Important Question of Physics Class 12 with Answers_230.1

Answer:
Important Question of Physics Class 12 with Answers_240.1

Q. If both the number of protons and neutrons in a nuclear reaction is conserved, in what way is mass converted into energy (or vice verse)? Explain giving one example. 
Answer:
Explanation for release of energy in a nuclear reaction : Since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of the nuclear reaction.
But total binding energy of nuclei on the left side need not be the same as that on the right hand side. The difference in binding energy causes a release of energy in the reaction.
Examples :
Important Question of Physics Class 12 with Answers_250.1

Q. 
Important Question of Physics Class 12 with Answers_260.1
Answer:
Important Question of Physics Class 12 with Answers_270.1

Q. Define ionization energy.
How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge?
Answer:
Definition of ionization energy : “The minimum energy, required to free the electron from the ground state of the hydrogen atom, is known as Ionization Energy.”
The ionization energy is given by :
Important Question of Physics Class 12 with Answers_280.1
∴Ionization Energy will become 200 times,
∵ the mass of given particle is 200 times.

Q. Calculate the shortest wavelength of the spectral lines emitted in Balmer series.
[Given Rydberg constant, R = 107 m-1]

Answer:
Important Question of Physics Class 12 with Answers_290.1

Q. The electron, in a hydrogen atom, is in its second excited state.
Calculate the wavelength of the lines in the Lyman series, that can be emitted through the permissible transitions of this electron.
(Given the value of Rydberg constant, R = 1.1 × 107 m-1
Answer:
Important Question of Physics Class 12 with Answers_300.1
Important Question of Physics Class 12 with Answers_310.1

Question 23.
An α-particle moving with initial kinetic energy K towards a nucleus of atomic number z approaches a distance ‘d’ at which it reverses its direction. Obtain the expression for the distance of closest approach ‘d’ in terms of the kinetic energy of α-particle K. (Comptt. All India 2016)
Answer:
Important Question of Physics Class 12 with Answers_320.1

Question 24.
Find the ratio between the wavelengths of the ‘most energetic’ spectral lines in the Balmer and Paschen series of the hydrogen spectrum. (Comptt. All India 2016)
Answer:
Important Question of Physics Class 12 with Answers_330.1

Q.Define the distance of closest approach. An a-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is V. What will be the distance of closest approach for an a-particle of double the kinetic energy? 
Answer:
The distance of closest approach is defined as “the distance of charged particle from the centre of the nucleus, at which the whole of the initial kinetic energy of the (far off) charged particle gets converted into the electric potential energy of the system”.
Distance of closest approach (rc) is given by
Important Question of Physics Class 12 with Answers_340.1
Important Question of Physics Class 12 with Answers_350.1

Q. Write two important limitations of Rutherford nuclear model of the atom. 
Answer:
Important limitations of Rutherford Model :

  1. According to Rutherford model, electron orbiting around the nucleus, continuously radiates energy due to the acceleration; hence the atom will not remain stable.
  2. As electron spirals inwards; its angular velocity and frequency change continuously; therefore it. will emit a continuous spectrum.

Question 27.
Find out the wavelength of the electron orbiting in the ground state of hydrogen atom. (Delhi 2016)
Answer:
Important Question of Physics Class 12 with Answers_360.1

Q. 
Find the wavelength of the electron orbiting in the first excited state in hydrogen atom. (Delhi 2016)
Answer:
Important Question of Physics Class 12 with Answers_370.1

Q. A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted. 
Answer:
Important Question of Physics Class 12 with Answers_380.1
Important Question of Physics Class 12 with Answers_390.1

Q. The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 A Calculate the short wavelength limit for Balmer series of the hydrogen spectrum.

Answer:
Important Question of Physics Class 12 with Answers_400.1

 

Class 12 Term 2 Physics Important Questions with Answers– 5 Marks

Unit – Electromagnetic Waves

Q. In a plane electromagnetic wave, the electric field oscillates sinusoid ally at a frequency of 2.0×1010Hz2.0×1010Hz and amplitude 48v1m48vm−1.

(a) What is the wavelength of the wave?

(b) What is the amplitude of the oscillating magnetic field?

(c) Show that the average energy density of the E field equals the average energy density of the B field. (c =3×108ms1=3×108ms−1. )

Ans: As it is given that,

Frequency of the electromagnetic wave, v=2.0×1010Hzv=2.0×1010Hz

Electric field amplitude, E0=48Vm1E0=48Vm−1

Speed of light, c=3×103m/sc=3×103m/s

(a) What is the wavelength of the wave?

Ans: Now, Wavelength of a wave is given as:

λ=cvλ=cv

=3×1082×1012=0.015m=3×1082×1012=0.015m

(b) What is the amplitude of the oscillating magnetic field?

Ans: Where,

Magnetic field strength is given as:

B0=E0cB0=E0c

=483×102=1.6×107T=483×102=1.6×10−7T

(c) Show that the average energy density of the E field equals the average energy density of the B field. (c =3×108ms1=3×108ms−1. )

Ans: Then,  Energy density of the electric field is given as:

UE=120E2(1)UE=12∈0E2…(1)

And, energy density of the magnetic field is given as:

UB=12μ0B2(2)UB=12μ0B2…(2)

Where,

ε0ε0

– Permittivity of free space

μ0=μ0= Permeability of free space

We have the relation connecting EE and BB as:

E=cB(3)E=cB…(3)

Where,

c=1μ0ε0−−−−√c=1μ0ε0

Now, Substituting equation (3) in equation (1), we get

Squaring both sides, we get

UE=UB⇒UE=UB

Q. Suppose that the electric field amplitude of an electromagnetic wave is E0=120N/CE0=120N/C and that its frequency is v=50.0MHzv=50.0MHz. (a) Determine, B0,ω,kB0,ω,k, and λλ. (b) Find expressions for EE and BB.

Ans: Given that ,

Electric field amplitude, E0=120N/CE0=120N/C

Frequency of source, v=50.0MHz=50×105Hzv=50.0MHz=50×105Hz

Speed of light, c=3×108m/sc=3×108m/s

(a) Magnitude of magnetic field strength is given as:

B0E0cB0E0c

=1203×102=1203×102

=4×107T=400nT=4×10−7T=400nT

Angular frequency of source is given as:

ω=2m=2π×50×106ω=2m=2π×50×106

=3.14×108rad/s=3.14×108rad/s

Propagation constant is given as:

k=(i)ck=(i)c

=3.14×1083×108=1.05rad/m=3.14×1083×108=1.05rad/m

Now,

Wavelength of wave is given as:

λ=cvλ=cv

=3×10850×102=6.0m=3×10850×102=6.0m

(b) Assume the wave is travelling in a positive direction. The electric field vector will then point in the positive direction, and the magnetic field vector will also point in the positive direction. This is due to the fact that all three vectors are perpendicular to one another.

Equation of electric field vector is given as:

E¯=E0sin(kc(i)t)jE¯=E0sin⁡(kc−(i)t)j

=120sin[1.05x3.14×101t]j=120sin⁡[1.05x−3.14×101t]j

And, magnetic field vector is given as:

B¯=B0sin(kx(t)t)k^B¯=B0sin⁡(kx−(t)t)k^

B⃗ =(4×101)sin[1.05x3.14×109t]k^B→=(4×10−1)sin⁡[1.05x−3.14×109t]k^

Q. A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R=6.0cmR=6.0cm has a capacitance C=100pFC=100pF. The capacitor is connected to a 230V230V ac supply with a (angular) frequency of 300rads1300rads−1

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

(c) Determine the amplitude of BB at a point 3.0cm3.0cm from the axis between the plates.

Ans: Given that ,

Radius of each circular plate, R=6.0cm=0.06mR=6.0cm=0.06m

Capacitance of a parallel plate capacitor, C=100pF=100×1012FC=100pF=100×10−12F

Supply voltage, V=230VV=230V

Angular frequency, ω=300rads1ω=300rads−1

(a) Rms value of conduction current, I=VXI=VXℓ

Where,

=1ωC=1ωC

I=V×ωC∴I=V×ωC

=230×300×100×1012=230×300×100×10−12

=6.9×106A=6.9×10−6A

=6.9HA=6.9HA

Hence, the rms value of conduction current is 6.9 μAμA

(b) Yes, the displacement current equal to conduction current

(c) Where Magnetic field is given as:

B=μerr2πR2IeB=μerr2πR2Ie

Where,

μ0=μ0= Free space permeability =4π×107NA1=4π×10−7NA−1

10=10= Maximum value of current =2–√l=2l

r=r= Distance between the plates from the axis =3.0cm=0.03m=3.0cm=0.03m

B=4π×107×0.03×2–√×6.9×1052π>0(06)2∴B=4π×10−7×0.03×2×6.9×10−52π>0(06)2

=1.63×1011T=1.63×10−11T Hence, the magnetic field at that point is 1.63×1011T1.63×10−11T.

Q. Figure 8.6 shows a capacitor made of two circular plates each of radius 12cm12cm, and separated by 5.0cm5.0cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A0.15A

(a) Calculate the capacitance and the rate of charge of potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Ans: Given data,

Radius of each circular plate, r=12cm=0.12mr=12cm=0.12m

Distance between the plates, d=5cm=0.05md=5cm=0.05m

Charging current, I=0.15AI=0.15A

Permittivity of free space, ε0=8.85×1012C2N1m2ε0=8.85×10−12C2N−1m−2

(a) Capacitance between the two plates is given by the relation.

Ans: C=ε0Adλ=cv=3×10830×106=10mC=ε0Ad′λ=cv=3×10830×106=10m

Where,

A=A= Area of each plate =πr2=πr2

C=ε0πr2dC=ε0πr2d

=8.85×10π12×0k2()20.05=8.85×10π12×0k2()20.05

8.0032×1012F=80.032pF−8.0032×10−12F=80.032pF

Charge on each plate, q=CVq=CV

Where,

VV = Potential difference across the plates

Differentiation on both sides with respect to time (t)(t) gives:

dqdt=CdVdtdqdt=CdVdt

But dqat=dqat= current (I)(I)

dVdt=1C∴dVdt=1C

0.1580.032×101.2=1.87×109V/s⇒0.1580.032×10−1.2=1.87×109V/s

Therefore, the change in potential difference between the plates is 1.87×109V/s1.87×109V/s.

(b) Obtain the displacement current across the plates.

Ans: The displacement current

(id)(id)

will be same as the conduction current or equal to the conduction current

(ic)(ic)
(id)=(ic)=0.15amp∴(id)=(ic)=0.15amp

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Ans: Yes, Kirchhoff’s first law of electrical circuits will be valid at each plate of the given capacitor, unless we consider the total sum of the both conduction and displacement currents.

Unit – Optics

Q. State Huygens’s principle for constructing wave fronts. 

Ans: According to Huygens’s principle:

  1. Each source of light would spread waves in all directions.

  2. Each point on the wave front would give rise to new disturbance which in-turn produces secondary wavelets which travel with the speed of light.

  3. Only the forward envelope which encloses the tangent would give the new position of wave front.

  4. Rays are always found to be perpendicular to the wave front.

  1. Using Huygens’s principle, deduce the laws of reflection of light.

Ans: A plane wave front ABAB incident at AA hence every point on ABAB gives rise to new waves. Time taken by the ray to reach from PP to RR

t=PQv+QRvt=PQv+QRv ……………….. (1)

In ΔPAQΔPAQ , sini=PQAQsin⁡i=PQAQ

PQ=AQsini⇒PQ=AQsin⁡i

In ΔRCQΔRCQ , sinr=QRQCsin⁡r=QRQC

QR=QCsinr⇒QR=QCsin⁡r

Substituting in equation (1),

t=AQsiniv+(ACAQ)sinrv⇒t=AQsin⁡iv+(AC−AQ)sin⁡rv

t=AQsiniv+ACsinrvAQsinrv⇒t=AQsin⁡iv+ACsin⁡rv−AQsin⁡rv

t=AQ(sinisinr)v+ACsinrv⇒t=AQ(sin⁡i−sin⁡r)v+ACsin⁡rv

Since all the secondary wavelets takes the same time to go from the incident wave front to the reflected wave front hence it must be independent of QQ

 i.e.,

sinisinr=0sin⁡isin⁡r=0

sini=sinr∴sin⁡i=sin⁡r

or i=ri=r (Law of Reflection of light)

Q. What changes in diffraction pattern of a single slit will you observe when the monochromatic source of light is replaced by a source of white light? 

Ans: The changes would be:

1) The diffracted light consists of different colours.

2) It results in overlapping of different colours.

Q. Coloured spectrum is seen when we look through a muslin cloth. Why?

Ans: Muslin cloth is known to consist of very fine threads which acts as fine slits and when light passes through it, light would get diffracted giving rise to a coloured spectrum.

Q. What changes in diffraction pattern of a single slit will you observe when the monochromatic source of light is replaced by a source of white light? 

Ans: The changes are:

  1. Diffracted lights consist of different colours.

  2. It would result in overlapping of different colours.

Q. A slit of width a′a′  is illuminated by light of wavelength 6000A6000A◯ . For what value of a′a′ will the:

  1. First maximum fall at an angle of diffraction of  3030∘

Ans: We are given:

λ=6000A=6000×1010mλ=6000A◯=6000×10−10m

θ1=30,m=1θ1=30∘,m=1

For first maximum

sinQm=(m+12)λasin⁡Qm=(m+12)λa

sinQ1=3λ2a⇒sin⁡Q1=3λ2a

or a=3λ2sinθ1=3×6×1072×sin30a=3λ2sin⁡θ1=3×6×10−72×sin⁡30∘

Q. First minimum fall at an angle of diffraction 3030∘

Ans: For first minimum

sinQm=mλasin⁡Qm=mλa

Or sinQ1=λasin⁡Q1=λa

a=λsinθ1⇒a=λsin⁡θ1

a=6×107sin30⇒a=6×10−7sin⁡30∘

a=1.2×105m∴a=1.2×10−5m

Q. Derive all expressions for the fringe width in Young’s double slit experiment. 

Ans: Path difference between

S1PS1P  and S2PS2P

Δx=S2PS1PΔx=S2P−S1P…………………………… (A)

In ΔS2PΔS2P,

(S2P)=[(S2B)2+(PB2)]12⇒(S2P)=[(S2B)2+(PB2)]12…………………………… (1)

S2P=D[1+(y+d2)D2]12S2P=D[1+(y+d2)D2]12

Using Binomial theorem expand equation(1)(1) and then neglect higher terms to get,

S2P=D+(y+d2)22DS2P=D+(y+d2)22D

Similarly,  S1P=D+(yd2)22D(2)S1P=D+(y−d2)22D⋯(2)

Substituting equation (1) and (2) in equation (A)(A),

Δx=(y+d2)2(yd2)22D⇒Δx=(y+d2)2−(y−d2)22D

Δx=y2+d42+ydy2d42+yd2D⇒Δx=y2+d42+yd−y2−d42+yd2D

Δx=2yd2DΔx=2yd2D

Δx=ydDΔx=ydD

For bright fringes

Path difference =xλ=xλ

xλ=ydDxλ=ydD

y=xλDd⇒y=xλDd

For m=1, y1=λDdm=1, y1=λDd

n=2, y2=λDdn=2, y2=λDd

For fringe width,

β=y2y1β=y2−y1

β=λdd∴β=λdd

Q. If the two slits in Young’s double slit experiment have width ratio 4:14:1 deduce the ratio of intensity of maxima and minima in the interference pattern. 

Ans: From the given ratio, a21a22=w1w2=41a12a22=w1w2=41

a1a2=21a1a2=21

Or

a1=2a2a1=2a2

Using

ImaxImin=(a1+a2)2(a1a2)2ImaxImin=(a1+a2)2(a1−a2)2
ImaxImin=(2a2+a2)2(2a2a2)2=