CBSE Class 12 Pre-Board Exam 2024-25: Physics Most Important Questions with Answers

The CBSE Class 12 pre-board exams for 2024-25 are an important step in your exam preparation. These exams help you check how well you’re prepared for the final board exams.

To help you study better, we’ve created a list of the most important physics questions with answers. This Physics question focuses on essential topics to help you perform well.

The CBSE Pre-board Exam Physics questions cover key topics. They include objective, short/long answer, and competency-based questions, all matching the latest exam pattern.

CBSE 12 Pre-board Exam 2024-25 Physics Most Important Questions

1. A thin plastic rod is bent into a circular ring of radius R. It is uniformly charged with charge density λ. The magnitude of the electric field at its centre is:

Ans. (B) Zero

2. Ten capacitors, each of capacitance 1 μF, are connected in parallel to a source of 100 V. The total energy stored in the system is equal to:

(A) 10^-2 J

(B) 10^-3 J

(D) 5.0 × 10^-2 J

Ans. (D) 5.0 × 10^-2 J

3. Consider the circuit shown in the figure. The potential difference between points A and B is:

(A) 6 V

(B) 8 V

(D) 12 V

Ans. (B) 8 V

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4. A loop carrying a current I clockwise is placed in x-y plane, in a uniform magnetic field directed along z-axis. The tendency of the loop will be to:

(A) move along x-axis

(B) move along y-axis

(D) expand

Ans. (C) shrink

Question 5

Ans. (B)

6. A galvanometer of resistance G Ω is converted into an ammeter of range 0 to IA. If the current through the galvanometer is 0.1% of I A, the resistance of the ammeter is:

(A) G/999 Ω

(B) G/1000 Ω

(D) G/100.1 Ω

Ans. (B) G/1000 Ω

7. The reactance of a capacitor of capacitance C connected to an ac source of frequency ω is 'X'. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become:

(A) X/6

(B) 6X

(D) 3X/2

Ans. (A) X/6

8. In the four regions, I, II, III and IV, the electric fields are described as:

Region I: E_x = E₀ sin(kz - ωt)

Region II: E_x = E₀

Region III: E_x = E₀ sin kz

Region IV: E_x = E₀ cos kz

The displacement current will exist in the region:

(A) I

(B) IV

(D) III

Ans. (A) I

9. The transition of electron that gives rise to the formation of the second spectral line of the Balmer series in the spectrum of hydrogen atom corresponds to:

(A) n_f = 2 and n_i = 3

(B) n_f = 3 and n_i = 4

(D) n_f = 2 and n_i = ∞

Ans. (C) n_f = 2 and n_i = 4

10. Ge is doped with As. Due to doping,

(A) the structure of Ge lattice is distorted.

(B) the number of conduction electrons increases.

(D) the number of conduction electrons decreases.

Ans. (B) the number of conduction electrons increases.

11. Two beams, A and B whose photon energies are 3.3 eV and 11.3 eV respectively, illuminate a metallic surface (work function 2.3 eV) successively. The ratio of maximum speed of electrons emitted due to beam A to that due to beam B is:

(A) 3

(B) 9

(D) 1/9

Ans. (C) 1/3

12. The waves associated with a moving electron and a moving proton have the same wavelength λ. It implies that they have the same:

(A) momentum

(B) angular momentum

(D) energy

Ans. (A) momentum

Questions number 13 to 16 are Assertion (A) and Reason (R) type questions. Two statements are given one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is notthe correct explanation of the Assertion (A).

(D) Assertion (A) is false and Reason (R) is also false.

13. Assertion (A): In photoelectric effect, the kinetic energy of the emitted photoelectrons increases with increase in the intensity of the incident light.

Reason (R): Photoelectric current depends on the wavelength of the incident light.

Ans. (D) Assertion (A) is false and Reason (R) is also false.

14. Assertion (A): The mutual inductance between two coils is maximum when the coils are wound on each other.

Reason (R): The flux linkage between two coils is maximum when they are wound on each other.

Ans. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

15. Assertion (A): Two long parallel wires, freely suspended and connected in series to a battery, move apart.

Reason (R): Two wires carrying current in opposite directions repel each other.

Ans. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

16. Assertion (A): Plane and convex mirrors cannot produce real images under any circumstance.

Reason (R): A virtual image cannot serve as an object to produce a real image.

Ans. (D) Assertion (A) is false and Reason (R) is also false.

17. Find the temperature at which the resistance of a wire made of silver will be twice its resistance at 20°C. Take 20°C as the reference temperature and temperature coefficient of resistance of silver at 20°C = 4.0 × 10^-3 K^-1.

Ans. Finding the temperature

R = R₀[1 + α(T - T₀)]

R = 2R₀ [Given]

2R₀ = R₀[1 + a(T - T₀)]

On solving

T = T₀ + 250

T = 270°C or 543K

18. (a) Monochromatic light of frequency 5.0 × 10¹⁴ Hz travels from air into a medium with refractive index 1.5. Find the wavelength of (i) reflected light, and (ii) refracted light at the interface of the two media.

Ans. Finding the wavelength of

(i) Reflected Light

(ii) Refracted Light

(i) v = υ λ

3 × 10⁸ = 5 × 10¹⁴ × λ

λ = 600 nm or 6 × 10^-7m

18. (a) (ii)

(b) A plano-convex lens of focal length 16 cm is made of a material of refractive index 1.4. Calculate the radius of the curved surface of the lens.

Ans. Calculating the radius of the curved surface

19. An object is placed 30 cm in front of a concave mirror of radius of curvature 40 cm. Find the (i) position of the image formed and (ii) magnification of the image.

Ans. Finding the

(i) position of the image formed

(ii) magnification of the image

20. Consider a neutron (mass m) of kinetic energy E and a photon of the same energy. Let λ_n and λ_p be the de Broglie wavelength of neutron and the wavelength of photon respectively. Obtain an expression for λ_n/λ_p.

Ans. Obtaining an expression for λ_n/λ_p

21. Plot a graph showing the variation of current with voltage for the material GaAs. On the graph, mark the region where:

(a) resistance is negative, and

(b) Ohm's law is obeyed.

Ans. Plotting the graph

Marking the region where:

(a) resistance is negative

(b) Ohm’s law is obeyed

Question 22

Calculate:

(a) the flux passing through the cube, and

(b) the charge within the cube.

Ans. Calculating

(a) the flux passing through the cube

(b) the charge within the cube

Question 23

Ans. (a)

Current density is the amount of charge flowing per second per unit area normal to the flow.

Alternatively:

𝑗 = I/A

It is a vector quantity.

(b) What is a Wheatstone bridge ? Obtain the necessary conditions under which the Wheatstone bridge is balanced.

Ans. Defining Wheatstone bridge

Obtaining balancing conditions

Alternatively:

If the figure is explained in words full credit to be given.

For loop ADBA:

–I₁ R₁ + I₂ R₂ + I_g G = 0 (1)

For loop CBDC:

I₄R₄ - I₃ R₃ - I_g G = 0 (2)

For balanced wheatstone bridge, Ig = 0

And by applying Kirchoff’s junction rule to junction D and B,

I₁ = I₃ & I₂ = I₄

From eqn (1) and (2)

24. A proton with kinetic energy 1·3384 10^-14 J moving horizontally from north to south, enters a uniform magnetic field B of 2·0 mT directed eastward.

Calculate :

(a) the speed of the proton

(b) the magnitude of acceleration of the proton

[Take (q/m) for proton = 1·0 10⁸ C/kg]

Ans. Calculating

(a) the speed of the proton

(b) the magnitude of the acceleration of the proton

25. An inductor, a capacitor and a resistor are connected in series with an ac source v = v_m sin ωt. Derive an expression for the average power dissipated in the circuit. Also obtain the expression for the resonant frequency of the circuit.

Ans. Deriving an expression for the average power dissipated in series LCR circuit

Obtaining expression for the resonant frequency

The average power over a cycle is given by the average of the two terms in RHS of eqn (1). It is only the 2^nd term which is time dependent. It’s average is zero. Therefore,

26. (a) "The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates." Give two examples to justify this statement.

(b) (i) Long distance radio broadcasts use short-wave bands. Why?

(ii) Optical and radio telescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the Earth. Why?

Ans. (a) Two examples

(b) (i) Reason for use of short waves bands

(ii) Reason for x-ray astronomy from satellites

(a) (Any Two)

Gamma radiation having wavelength of 10^–14 m to 10^–15 m, typically originate from an atomic nucleus.
X-rays are emitted from heavy atoms.
Radio waves are produced by accelerating electrons in a circuit. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna.

(b) (i) Ionosphere reflects waves in these bands

(ii) Atmosphere absorbs x-rays, while visible and radio waves can penetrate it

Note: Full credit to be given for part (b) for mere attempt.

27. Write the drawbacks of Rutherford's atomic model. How did Bohr remove them? Show that different orbits in Bohr's atom are not equally spaced.

Ans.

Drawbacks of Rutherford’s atomic model
Bohr’s explanation
Showing different orbits are not equally spaced

Drawbacks:

(i) According to classical electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, continuously decrease. The electron would spiral inward and eventually fall into the nucleus. Thus, such an atom cannot be stable.

(ii) As the electrons spiral inwards, their angular velocities and hence their frequencies would change continuously. Thus, they would emit a continuous spectrum, in contradiction to the line spectrum actually observed.

Bohr postulated stable orbits in which electrons do not radiate energy Alternatively:

Bohr’s postulates (Any ONE of the three)

(i) An electron in an atom could revolve in certain stable orbits without the emission of radiant energy.

(ii) The electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π

(iii) An electron might make a transition from one of its specified nonradiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states.

The radius of the n^th orbit is found as

r_n α n²

Alternatively:

Difference in radius of consecutive orbits is

r_n+1 – r_n = k [(n+1)² – n²)]

= k (2n + 1) which depends on n, and is not a constant

28. (a) State any two properties of a nucleus.

(b) Why is the density of a nucleus much more than that of an atom?

(c) Show that the density of the nuclear matter is the same for all nuclei.

Ans. (a) Stating two properties of a nucleus

(b) Why density of a nucleus is much more than that of an atom

(a) (Any TWO)

(i) The nucleus is positively charged

(ii) The nucleus consists of protons and neutrons

(iii) The nuclear density is independent of mass number

(iv) The radius of the nucleus, R = Ro A^1/3

(b) Atoms have large amount of empty spaces. Mass is concentrated in nucleus.

So, density is independent of mass number

29. A lens is a transparent medium bounded by two surfaces, with one or both surfaces being spherical. The focal length of a lens is determined by the radii of curvature of its two surfaces and the refractive index of its medium with respect to that of the surrounding medium. The power of a lens is reciprocal of its focal length. If a number of lenses are kept in contact, the power of the combination is the algebraic sum of the powers of the individual lenses.

(i) A double-convex lens, with each face having same radius of curvature R, is made of glass of refractive index n. Its power is :

(A) 2 (n - 1)/R

(B) (2n - 1)/R

(D) (2n - 1)/2R

Ans. (A) 2 (n - 1)/R

(ii) A double-convex lens of power P, with each face having same radius of curvature, is cut into two equal parts perpendicular to its principal axis. The power of one part of the lens will be :

(A) 2P

(B) P

(D) P/2

Ans. (D) P/2

(iii) The above two parts are kept in contact with each other as shown in the figure. The power of the combination will be :

(A) P/2

(B) P

(D) P/4

Ans. (B) P

(iv) (a) A double-convex lens of power P, with each face having same radius of curvature, is cut along its principal axis. The two parts are arranged as shown in the figure. The power of the combination will be :

(A) Zero

(B) P

(D) P/2

Ans. (C) 2P

(b) Two convex lenses of focal lengths 60 cm and 20 cm are held coaxially in contact with each other. The power of the combination is :

(A) 6·6 D

(B) 15 D

(D) 1/80 D

Ans. (A) 6·6 D

30. Junction Diode as a Rectifier :

The process of conversion of an ac voltage into a dc voltage is called rectification and the device which performs this conversion is called a rectifier. The characteristics of a p-n junction diode reveal that when a p-n junction diode is forward biased, it offers a low resistance and when it is reverse biased, it offers a high resistance. Hence, a p-n junction diode conducts only when it is forward biased. This property of a p-n junction diode makes it suitable for its use as a rectifier.

Thus, when an ac voltage is applied across a p-n junction, it conducts only during those alternate half cycles for which it is forward biased. A rectifier which rectifies only half cycle of an ac voltage is called a half-wave rectifier and one that rectifies both the half cycles is known as a full-wave rectifier.

Ans.

(ii) In a full-wave rectifier, the current in each of the diodes flows for :

(A) Complete cycle of the input signal

(B) Half cycle of the input signal

(D) Only for the positive half cycle of the input signal

Ans. (B) Half cycle of the input signal

(iii) In a full-wave rectifier :

(A) Both diodes are forward biased at the same time.

(B) Both diodes are reverse biased at the same time.

(D) Both are forward biased in the first half of the cycle and reverse biased in the second half of the cycle.

Ans. (C) One is forward biased and the other is reverse biased at the same time.

(iv) (a) An alternating voltage of frequency of 50 Hz is applied to a half-wave rectifier. Then the ripple frequency of the output will be :

(A) 100 Hz

(B) 50 Hz

(D) 150 Hz

Ans. (B) 50 Hz

(b) A signal, as shown in the figure, is applied to a p-n junction diode. Identify the output across resistance R_L :

Ans.

31.

Ans. (i)

Deriving the expression for potential energy
Maximum & Minimum value of potential energy

(ii) Finding the torque.

31.

Ans. (i) Deriving expression for potential

(ii) New charge on Sphere S₁

Alternatively:

By geometry

Using equations (i) ,(ii) & (iii) & p = 2qa

When connected by a thin wire they acquire a common potential V and the charge remains conserved.

Alternatively:

When connected by a thin wire they acquire a common potential V and the charge remains conserved.

32. (a) (i) A resistor and a capacitor are connected in series to an ac source v = v_m sin ωt. Derive an expression for the impedance of the circuit.

(ii) When does an inductor act as a conductor in a circuit ? Give reason for it.

(iii) An electric lamp is designed to operate at 110 V dc and 11 A current. If the lamp is operated on 220 V, 50 Hz ac source with a coil in series, then find the inductance of the coil.

Ans.

(i) Deriving expression for impedance

(ii) Reason

(iii) Inductance of coil

(ii) For direct current (dc), an inductor behaves as a conductor.

As X_L = ωL = 2π ν L

For dc ν = 0 ⇒ X_L = 0

Alternatively: -

Induced emf (ε) = -LdI/dt

For dc; dI = 0 ⇒ ε = 0

(iii) R = 110/11 = 10 Ω

Squaring both sides:

(b) (i) Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes for energy losses in a real transformer.

(ii) A step-up transformer converts a low voltage into high voltage. Does it violate the principle of conservation of energy ? Explain.

(iii) A step-up transformer has 200 and 3000 turns in its primary and secondary coils respectively. The input voltage given to the primary coil is 90 V. Calculate :

(1) The output voltage across the secondary coil

(2) The current in the primary coil if the current in the secondary coil is 2·0 A.

Ans. (i) Labelled diagram of step – up transformer

Describing working principle

Three causes

(ii) Explanation

(iii) (1) Output voltage across secondary coil

(2) Current in primary coil

The working principle of transformer is mutual induction.

When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it.

Causes of energy losses (Any three)

(a) Flux leakage

(b) Resistance of the windings

(d) Hysteresis

(ii) No

Current changes correspondingly. So, the input power is equal to the output power.

(iii)

33. (a) (i) A ray of light passes through a triangular prism. Show graphically, how the angle of deviation varies with the angle of incidence ? Hence define the angle of minimum deviation.

(ii) A ray of light is incident normally on a refracting face of a prism of prism angle A and suffers a deviation of angle δ. Prove that the refractive index n of the material of the prism is given by sin A sin (A + δ )/sin A.

(iii) The refractive index of the material of a prism is √2. If the refracting angle of the prism is 60° , find the

(1) Angle of minimum deviation, and

(2) Angle of incidence.

Ans.

Minimum deviation angle is defined as the angle at which angle of incidence is equal to the angle of emergence.

Alternatively

At minimum deviation refracted ray inside the prism becomes parallel to the base of the prism.

(b) (i) State Huygens principle. A plane wave is incident at an angle i on a reflecting surface. Construct the corresponding reflected wavefront. Using this diagram, prove that the angle of reflection is equal to the angle of incidence.

(ii) What are the coherent sources of light ? Can two independent sodium lamps act like coherent sources ? Explain.

(iii) A beam of light consisting of a known wavelength 520 nm and an unknown wavelength λ, used in Young's double slit experiment produces two interference patterns such that the fourth bright fringe of unknown wavelength coincides with the fifth bright fringe of known wavelength. Find the value of λ.

Ans.

(i) Statement of Huygens’ Principle

Construction of reflected wave front

Proof of angle of reflection is equal to angle of incidence

(ii) Definition of coherent sources

Explanation

(iii) Finding the unknown wavelength

(i) Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the spread of the wave. Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time.

ΔEAC is congruent to ΔBAC; so ∠i = ∠r

(ii) Two sources are said to be coherent if the phase difference between them does not change with time.

No, two independent sodium lamps cannot be coherent.

Two independent sodium lamps cannot be coherent as the phase between them does not remain constant with time.

(iii)

👉 CBSE Class 12 Study Materials

CBSE Class 12 Syllabus 2024-25	CBSE Class 12 Previous Year Papers
NCERT Books For Class 12 Books	NCERT Class 12 Solutions
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