Physics Class 12 CBSE Solved Question Paper – Section A

Section A – Multiple Choice Questions (1 mark each)

1. A battery supplies 0.9 A current through a 2 Ω resistor and 0.3 A current through a 7 Ω resistor when connected one by one. The internal resistance of the battery is:
   • (A) 2 Ω
   • (B) 1.2 Ω
   • (C) 1 Ω
   • (D) 0.5 Ω
   SHOW ANSWER

   Answer: (D) 0.5 Ω
   Solution: Let R_int be the internal resistance of the battery, and V be the electromotive force (emf) of the battery.
   For the 2 Ω resistor: V = 0.9(2 + R_int)
   For the 7 Ω resistor: V = 0.3(7 + R_int)
   Equating the two expressions for V:
   0.9(2 + R_int) = 0.3(7 + R_int)
   Expanding both sides: 1.8 + 0.9R_int = 2.1 + 0.3R_int
   Rearranging terms: 0.9R_int − 0.3R_int = 2.1 − 1.8
   0.6R_int = 0.3
   Solving for R_int: R_int = 0.3 / 0.6 = 0.5 Ω
   
2. A particle of mass m and charge q describes a circular path of radius R in a magnetic field. If its mass and charge were 2m and q/2 respectively, the radius of its path would be:
   • (A) R/4
   • (B) R/2
   • (C) 2R
   • (D) 4R
   SHOW ANSWER

   Answer: (D) 4R
   Solution: The radius R of the circular path of a charged particle in a magnetic field is given by: R = (mv) / (qB).
   For the new mass 2m and charge q/2, the radius becomes:
   R’ = (2mv) / ((q/2)B) = 4R.
   Thus, the new radius is 4R.
   
3. Which of the following pairs is that of paramagnetic materials?
   • (A) Copper and Aluminium
   • (B) Sodium and Calcium
   • (C) Lead and Iron
   • (D) Nickel and Cobalt
   SHOW ANSWER

   Answer: (B) Sodium and Calcium

   Solution:

   Paramagnetic materials are those that have unpaired electrons and are weakly attracted to magnetic fields.

   • Copper is diamagnetic, Aluminium is paramagnetic.
   • Sodium and Calcium are paramagnetic.
   • Lead is diamagnetic, Iron is ferromagnetic.
   • Nickel and Cobalt are ferromagnetic, not paramagnetic.

   Thus, Sodium and Calcium are paramagnetic.
   

4. A galvanometer of resistance 50 Ω is converted into a voltmeter of range (0–2 V) using a resistor of 1.0 kΩ. If it is to be converted into a voltmeter of range (0–10 V), the resistance required will be:
   • (A) 4.8 kΩ
   • (B) 5.0 kΩ
   • (C) 5.2 kΩ
   • (D) 5.4 kΩ
   SHOW ANSWER

   Answer: (C) 5.2 kΩ

   Solution:

   The total resistance of the voltmeter is given by: R_v = R_g + R_s.
   The ratio of the ranges is proportional to the ratio of total resistances. For extending the range from 2 V to 10 V:
   (R_v2) / (R_v1) = 10 / 2
   R_v2 = 5 × (1 kΩ + 50 Ω) = 5250 Ω.
   Thus, the required resistance is R_s = 5250 Ω – 50 Ω = 5200 Ω = 5.2 kΩ.
   

5. Two coils are placed near each other. When the current in one coil is changed at the rate of 5 A/s, an emf of 2 mV is induced in the other. The mutual inductance of the two coils is:
   • (A) 0.4 mH
   • (B) 2.5 mH
   • (C) 10 mH
   • (D) 2.5 H
   SHOW ANSWER

   Answer: (A) 0.4 mH

   Solution:

   The emf induced in the second coil is given by: ε = M (dI/dt).
   Given: ε = 2 mV = 2 × 10^-3 V, dI/dt = 5 A/s.
   Rearranging the formula to solve for M:
   M = ε / (dI/dt) = (2 × 10^-3) / 5 = 0.4 mH.
   

6. The electromagnetic waves used to purify water are:
   • (A) Infrared rays
   • (B) Ultraviolet rays
   • (C) X-rays
   • (D) Gamma rays
   SHOW ANSWER

   Answer: (B) Ultraviolet rays
   Solution: Ultraviolet rays are used to purify water by killing harmful microorganisms. Infrared, X-rays, and Gamma rays are not commonly used for this purpose.
   
7. The focal lengths of the objective and the eyepiece of a compound microscope are 1 cm and 2 cm respectively. If the tube length of the microscope is 10 cm, the magnification obtained by the microscope for most suitable viewing by a relaxed eye is:
   • (A) 250
   • (B) 200
   • (C) 150
   • (D) 125
   SHOW ANSWER

   Answer: (D) 125

   Solution:

   The total magnification of a compound microscope is M = M_o × M_e, where M_o = L / f_o and M_e = 25 / f_e.
   Here, L = 10 cm, f_o = 1 cm, f_e = 2 cm.
   Hence, M = (10/1) × (25/2) = 125.
   

8. The variation of the stopping potential (V₀) with the frequency (ν) of the incident radiation for four metals A, B, C, and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for metal:
   
   • (A) A
   • (B) B
   • (C) C
   • (D) D
   SHOW ANSWER

   Answer: (A) A

   Solution:

   The kinetic energy of photoelectrons is given by K.E. = hν – φ, where φ is the work function.
   The stopping potential is related to the kinetic energy, and metals with lower stopping potentials have higher kinetic energy for the same frequency of incident light.
   

9. The energy of an electron in the ground state of a hydrogen atom is -13.6 eV. The kinetic and potential energy of the electron in the first excited state will be:
   • (A) -13.6 eV, 27.2 eV
   • (B) -6.8 eV, 13.6 eV
   • (C) 3.4 eV, -6.8 eV
   • (D) 6.8 eV, -3.4 eV
   SHOW ANSWER

   Answer: (C) 3.4 eV, -6.8 eV
   Solution: In the first excited state (n = 2), the total energy of the electron is -3.4 eV.
   The kinetic energy is equal to -E, so K.E. = 3.4 eV.
   The potential energy is twice the total energy with a negative sign, so P.E. = -6.8 eV.
   
10. A Young’s double-slit experimental setup is kept in a medium of refractive index 4/3. Which maximum in this case will coincide with the 6th maximum obtained if the medium is replaced by air?
    • (A) 4th
    • (B) 6th
    • (C) 8th
    • (D) 10th
    SHOW ANSWER

    Answer: (C) 8th

    Solution:

    The wavelength of light decreases in a medium with a refractive index greater than 1. The order of the maxima is inversely proportional to the wavelength.
    Thus, the 6th maximum in air coincides with the 8th maximum in the medium with a refractive index of 4/3.
    

11. The potential energy between two nucleons inside a nucleus is minimum at a distance of about:
    • (A) 0.8 fm
    • (B) 1.6 fm
    • (C) 2.0 fm
    • (D) 2.8 fm
    SHOW ANSWER

    Answer: (B) 1.6 fm
    Solution: The potential energy between nucleons is minimum at a distance of about 1.6 femtometers (fm), which is the typical range of the strong nuclear force that binds protons and neutrons together.
    
12. A pure Si crystal having 5 × 10²⁸ atoms/m³ is doped with 1 ppm concentration of antimony. If the concentration of holes in the doped crystal is found to be 4.5 × 10⁹ m^-3, the concentration (in m^-3) of intrinsic charge carriers in Si crystal is about:
    • (A) 1.2 × 10¹⁵
    • (B) 1.5 × 10¹⁶
    • (C) 3.0 × 10¹⁵
    • (D) 2.0 × 10¹⁶
    SHOW ANSWER

    Answer: (B) 1.5 × 10¹⁶

    Solution:

    The intrinsic carrier concentration n_i in a semiconductor is related to the electron and hole concentrations by n_i² = n_e × n_h.
    Using the given values, n_i = √(5 × 10²² × 4.5 × 10⁹) = 1.5 × 10¹⁶ m^-3.
    

13. Assertion (A): Equal amount of positive and negative charges are distributed uniformly on two halves of a thin circular ring as shown in figure. The resultant electric field at the center O of the ring is along OC.
    Reason (R): It is so because the net potential at O is not zero.
    • (A) Both A and R are true, and R is the correct explanation of A.
    • (B) Both A and R are true, but R is not the correct explanation of A.
    • (C) A is true, but R is false.
    • (D) Both A and R are false.
      
    SHOW ANSWER

    Answer: (C) A is true, but R is false.
    Solution: The electric field at the center is along OC due to the asymmetry in charge distribution. However, the net potential at the center is zero, so the reason is incorrect.
    
14. Assertion (A): The energy of a charged particle moving in a magnetic field does not change.
    Reason (R): It is because the work done by the magnetic force on the charge moving in a magnetic field is zero.
    • (A) Both A and R are true, and R is the correct explanation of A.
    • (B) Both A and R are true, but R is not the correct explanation of A.
    • (C) A is true, but R is false.
    • (D) Both A and R are false.
    SHOW ANSWER

    Answer: (A) Both A and R are true, and R is the correct explanation of A.

    Solution:

    The magnetic force does no work because it is always perpendicular to the velocity of the particle, so the energy of the particle does not change.
    

15. Assertion (A): In a Young’s double-slit experiment, interference pattern is not observed. When two coherent sources are infinitely close to each other.
    Reason (R): The fringe width is proportional to the separation between the two sources.
    • (A) Both A and R are true, and R is the correct explanation of A.
    • (B) Both A and R are true, but R is not the correct explanation of A.
    • (C) A is true, but R is false.
    • (D) Both A and R are false.
    SHOW ANSWER

    Answer: (A) Both A and R are true, and R is the correct explanation of A.
    Solution: If the slits are infinitely close, the path difference between the waves at any point is very small, leading to no observable interference. Fringe width is proportional to slit separation.
    
16. Assertion (A): An alpha particle is moving towards a gold nucleus. The impact parameter is maximum for the scattering angle of 180°.
    Reason (R): The impact parameter in an alpha particle scattering experiment does not depend upon the atomic number of the target nucleus.
    • (A) Both A and R are true, and R is the correct explanation of A.
    • (B) Both A and R are true, but R is not the correct explanation of A.
    • (C) A is true, but R is false.
    • (D) Both A and R are false.
    SHOW ANSWER

    Answer: (D) Both A and R are false.

    Solution:

    In an alpha particle scattering experiment, the scattering angle is related to the impact parameter. For a scattering angle of 180°, the impact parameter is zero (not maximum). Therefore, both the assertion and the reason are false.