Important Questions for Class 12 Physics

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Very Short Answer Questions (1 Marks Questions)

Ques. In which orientation, a dipole placed in uniform electric is in stable and unstable equilibrium?

Ans. If a dipole is placed parallel to the electric field, then it is in stable equilibrium. And if a dipole is placed anti-parallel to the electric field, then it is in unstable equilibrium.

Ques. Why do the electric field lines not form closed loops?

Ans. Electric field lines always start from positive charge and end on negative charge.

To form a close loop, an electric field line can start from either positive or negative charge and also can end on either positive or negative charge. Hence electric field lines do not form closed loops.

Ques. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?

Ans. At stopping potential, the maximum kinetic energy of photoelectrons must be equal to the energy acquired by an electron while passing through that potential.

(K.E)_max = eV₀ where, V₀ is stopping potential.

Hence, maximum kinetic energy, (K.E)_max = 1.5 eV

Ques. An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field?

Ans. The force experience a charge q moving in uniform magnetic field B with velocity v, is give by

F_m = qvB sinθ

Where, θ is the angle between direction of velocity and uniform magnetic field.

Since, the charge does not suffer any deflection, therefore, F_m = 0

⇒ qvB sinθ = 0 ⇒ sinθ = 0 ⇒ θ = 0^o or 180^o

Hence, the direction of the magnetic field is parallel or antiparallel to the direction of velocity of the electron.

Ques. Two thin lenses of power + 4D and – 2D are in contact. What is the focal length of the combination?

Ans. Given, P₁ = +4 D and P₂ = -2 D

Equivalent power for the combination of lenses, P = P₁ + P₂

⇒ P = +4 - 2 = 2 D

Equivalent focal length of the combination, f = \(\frac{1}{P} = \frac{1}{2} \)= 0.5 m = 50 cm

Ques. Define the term ‘stopping potential’ in relation to photoelectric effect.

Ans. The minimum negative potential for which the photoelectric current becomes zero is called stopping potential for the given frequency of the incident radiation.

It is also called Cut-off potential.

Ques. State the conditions for the phenomenon of total internal reflection to occur.

Ans. There are two conditions for the phenomenon of total internal reflection to occur:

• The light must travel from denser to rarer medium.
• The angle of incidence must be greater than the critical angle for the given pair of media.

Ques. Why should the spring/suspension wire in a moving coil galvanometer have a low torsional constant?

Ans. Since, the current sensitivity of a moving coil galvanometer is inversely proportional to the torsional constant. Hence, to increase the current sensitivity, a moving coil galvanometer have low torsional constant

Ques. Define self-inductance of a coil. Write its S.I. unit.

Ans. Self inductance is defined as the magnetic flux linked with the coil, when current flows through it.

Self inductance, L = \(\frac{\Phi}{I}\)

SI unit of self inductance is Henry (H)

Ques. Define ionization energy. What is its value for a hydrogen atom?

Ans. The energy required to knock out an electron from an atom is called ionization energy of the atom.

Ionization energy of hydrogen atom is, E = 13.6 eV.

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Short Answer Questions (2 Marks Questions)

Ques. A certain radioactive element disintegrates for an interval of time equal to its mean life.

• What fraction of the element remains undecayed?
• What fraction of elements has disintegrated?

Ans. We know, means life, \(\tau = \frac{1}{\lambda}\)

According to decay law, N=N₀e^-λt

Given, t = \(\tau\)

⇒ N = \(N_0e^{-\lambda \times \frac{1}{\lambda}} = N_0e^{-1}\)

⇒ N =\( \frac{N_0}{e}\) = 0.3679 N₀

(i) Fraction of element remains undecayed = \(\frac{N}{N_0}\) = 0.3679

(ii) Fraction of element disintegrated = 1 - 0.3679 = 0.6321

Ques. The charge +q is lying at center C of a circle. What is the work done in carrying another charge Q from point X to Y on the circumference of the circle.

Ans. Work done in carrying a charge Q from one point to another point is given by

dW = QdV

Circular path around a point charge (+q) is actually a path on an equipotential surface.

Since, the value of potential at point X and Y is same, i.e. dV = 0, hence work done, dW = 0.

Therefore, no work has to be done in carrying charge Q from point X to Y on the circumference of the circle.

Ques. Can an object have a charge of 2.8 x 10^-18 C?

Ans. Given, charge Q = 2.8 x 10^-18 C

Charge of an electron is, e = 1.6 x 10^-19 C

According to the principle of Quantization of charge, we have

Q = ne

Where, n is the number of electrons and e is the charge of an electron.

Therefore, number of electrons, n = \(\frac{Q}{e}\)

n = \(\frac{2.8 \times 10^{-18}}{1.6 \times 10^{-19}}\) =17.5

As per quantization of electric charges, the number of elementary charges must be an integer. Since n is not an integer, the object cannot have the given charge.

Ques. Calculate the frequency associated with a photon of energy 3.3 X 10^-20 J.

Ans. We know, energy of a photon is given by

E = hf

Where, h is Planck’s constant and f is frequency associated with a particle

f = \(\frac{E}{h} = \frac{3.3 \times 10^{-20}}{6.6 \times 10^{-34}}\) = 0.5 x 10¹⁴ Hz

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Long Answer Questions (3 Marks Questions)

Ques. Why do reflected and refracted light have the same frequency as that of monochromatic incident light?

Ans. When monochromatic light falls on a reflecting surface, the electrons of the atom of the surface start vibrating with the same frequency as that of incident light. These vibrating electrons emit light having the same frequency as that of the incident light.

When monochromatic light goes from one medium to another medium (i.e. when refraction takes place), the electrons of the atom of the second medium are set into forced vibrations of frequency equal to the frequency of incident light. These forced oscillations of electrons emit light of the same frequency as that of the incident light.

Ques. Explain the energy losses in a transformer. How are they minimized?

Ans. Various types of energy losses in the transformer and ways to minimize it, are given below.

(i) Copper Loss

1. Due to resistance of the copper coils used in the windings of a transformer, some energy is lost in the form of heat, which is known as copper loss.
2. Copper loss can be minimized by using thick wire for the windings.

(ii) Hysteresis Loss

1. Loss of energy due to continuous magnetization and demagnetization of the transformer is called hysteresis loss.
2. Hysteresis loss can be minimized by using suitable material having a narrow hysteresis loop for the core of a transformer.

(iii) Flux Loss

1. If magnetic flux linked with the primary coil is not equal to the magnetic flux linked with the secondary coil. So a certain amount of energy supplied to the primary winding is wasted, which is known as Flux loss.
2. Flux loss in a transformer can be reduced by winding the primary and secondary coils one over the other.

(iv) Eddy current Loss

1. Energy loss in a metallic plate when kept in a time-varying magnetic field causes eddy current loss.
2. This loss can be minimized by using a laminated iron core in the transformer.

Ques. Explain the formation of n-type of semiconductor from silicon.

Ans. When suitable pentavalent impurity is added to pure silicon, we get an extrinsic semiconductor, known as n- type semiconductor.

• When an atom of pentavalent impurity (say arsenic) is added to silicon, it replaces the silicon atom.
• The arsenic atom makes four covalent bonds with the silicon atom. The fifth valence electron of arsenic remains un-accomodated.
• This electron is loosely bound to its parent nucleus and easily detached even at room temperature.
• Now, it becomes a free electron and moves randomly through the crystal.
• In this way a number of free electrons are available when a small amount of arsenic is added.

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Very Long Answer Questions (5 Marks Questions)

Ques. A coil of inductance 0.50 H and resistance 100 ohm is connected to 200 volt, 50 Hz a.c. supply. Find the maximum current in the coil? Also find the time lag between the maximum voltage and maximum current.

Ans. We know, V_rms=\( \frac{V_0}{2}\) or V₀=V_rms√2

Given, V_rms = 200 volt

Maximum voltage, V₀ = 200 x 1.414 = 282.8 volt

Impedance of LR circuit is, Z_L = \(\sqrt{R^2+ \omega ^2 L^2}\) = \(\sqrt{R^2+ (2 \pi f^2) L^2}\)

Given, f = 50 Hz, R = 100 ohm and L = 0.50 H

Z_L= =\(\sqrt{100^2+ (2\pi \times 50)^2 (0.50)^2}\) = 186.1 ohm

Therefore, maximum current, I₀ = \(\frac{V_0}{Z_L}\) = \(\frac{282.8}{186.1}\) = 1.52 A

In LR circuit phase difference between current and voltage is Φ, given by

tan Φ = \(\frac{L \omega}{R}= \frac{L \times \pi 2f}{R}= \frac{0.50 \times 23.14 \times 50}{100}\)

⇒ tan Φ = 1.57 ⇒ Φ = 57.5^o = 0.3194π radians

Here, V = V₀ sin ωt and I = I₀ sin (ωt - Φ)

Time lag between V and I is given by ω=t

t = \(\frac{\phi}{\omega} = \frac{\phi}{2\pi f} = \frac{0.3194 \times \pi}{2 \times \pi \times 50}\)= 3.194 x 10^-3 seconds

Ques. Using Gauss’ law, deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point.

• Outside
• Inside the shell.

Plot a graph showing the variation of the electric field as a function of r > R and r < R (r is the distance from the center of a shell)

Ans. 1) Electric field intensity at a point outside the uniformly charged spherical conducting shell 

Consider a positive charge q distributed uniformly on the surface of a spherical shell of radius R.

According to Gauss’s Theorem,

2) Electric field intensity at a point inside the shell

According to Gauss’s theorem,

There is no electric field inside a uniformly charged thin spherical shell.

The Graph showing the variation of electric field intensity (E) with distance (r) from the center of the charged spherical shell is shown below.

Ques. Draw the plot of binding energy per nucleon as a function of mass number. Write two important conclusions that can be drawn regarding the nature of nuclear force.

Use this graph to explain the release of energy in both the process of nuclear fusion and fission.

Ans. The variation of binding energy per nucleon with mass number is shown in figure below.

Conclusions regarding the nature of nuclear force

• For nuclei of middle mass number 30 < A > 170, the binding energy per nucleon is almost constant. Maximum of about 8.75 MeV for A = 56 and has a value of 7.6 MeV for A = 238.
• For both light nuclei (A < 30) and heavy nuclei (A > 170) binding energy per nucleon is low.

Nuclear Fission:

There will be a gain in the overall binding energy when we move from the heavy nuclei region to the middle class region, hence when a heavy nucleus splits into two roughly equal fragments, energy can be released. This process is called Nuclear Fission.

Nuclear Fusion:

There will be gain in the overall binding energy, when we move from lighter nuclei to heavier nuclei. Hence, energy is released, when two or more lighter nuclei fuse together. This process is called Nuclear Fusion.

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