Practice Questions

Q12.In the given circuit, the terminal potential difference of the cell is : (1) 2 V (2) 3 V (3) 4 V (4) 1.5 V

Q12.A current of 200 μA deflects the coil of a moving coil galvanometer through 60°. The current to cause 𝜋 deflection through radian is 10 (1) 30 μA (2) 120 μA (3) 60 μA (4) 180 μA

Q13.In a coil, the current changes from −2 A to +2 A in 0.2 s and induces an emf of 0.1 V. The self inductance of the coil is : (1) 4mH (2) 1mH (3) 5mH (4) 2.5mH

Q13.A galvanometer has a resistance of 50 𝛺 and it allows maximum current of 5 mA. It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance. (1) 5975 𝛺 (2) 20050 𝛺 (3) 19950 𝛺 (4) 19500 𝛺

Q13.The electric current through a wire varies with time as I = I0 + βt, where I0 = 20 A and β = 3 A s−1 . The amount of electric charge crossed through a section of the wire in 20 s is: JEE Main 2024 (29 Jan Shift 1) JEE Main Previous Year Paper (1) 80 C (2) 1000 C (3) 800 C (4) 1600 C

Q13.The electrostatic force → and magnetic force acting on a charge q moving with velocity v can be (→F2) (F1) written : (1) →F1 = q →E, →F2 = q(→V × →B) (2) →F1 = q →B, →F2 = q(→B × →V ) (3) →F1 = q →E, →F2 = q(→B × →V ) (4) →F1 = q →V ⋅→E, →F2 = q(→B ⋅→V )

Q13.Match List I with List II List I List II 1 A. Gauss’s law of magnetostatics i. ∮ →𝐸· d→𝑎= ∫𝜌d𝑉 𝜀0 B. Faraday’s law of electro magnetic induction ii. ∮ →𝐵· d→𝑎= 0 → → -d → C. Ampere’s law iii. ∮ 𝐸· d 𝑙= 𝐵· d →𝑎 d𝑡∫ D. Gauss’s law of electrostatics iv. ∮ →𝐵· d →𝑙= 𝜇0𝐼 Choose the correct answer from the options given below: (1) A-I, B-III, C-IV, D-II (2) A-III, B-IV, C-I, D-II (3) A-IV, B-II, C-III, D-I (4) A-II, B-III, C-IV, D-I

Q13.Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are 𝑉1, 𝑉2 and 𝑉3. Choose the correct option. JEE Main 2024 (27 Jan Shift 2) JEE Main Previous Year Paper (1) 𝑉1 = 𝑉2 (2) 𝑉1 ≠𝑉3 - 𝑉2 (3) 𝑉1 + 𝑉2 > 𝑉3 (4) 𝑉1 + 𝑉2 = 𝑉3

Q14.An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then : (1) the electron will continue to move with uniform (2) the electron will be accelerated along the axis. velocity along the axis of the solenoid. (3) the electron path will be circular about the axis. (4) the electron will experience a force at 45∘ to the axis and execute a helical path.

Q14.A LCR circuit is at resonance for a capacitor C , inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now: JEE Main 2024 (08 Apr Shift 1) JEE Main Previous Year Paper (1) Zero (2) same (3) halved (4) double

Q14.An AC voltage 𝑉= 20sin200𝜋𝑡 is applied to a series LCR circuit which drives a current 𝐼= 10sin200π𝑡+ 3 . The average power dissipated is: JEE Main 2024 (31 Jan Shift 2) JEE Main Previous Year Paper (1) 21 . 6 W (2) 200 W (3) 173 . 2 W (4) 50 W

Q14.A beam of unpolarised light of intensity 𝐼0 is passed through a polaroid 𝐴 and then through another polaroid 𝐵 which is oriented so that its principal plane makes an angle of 45° relative to that of 𝐴. The intensity of emergent light is : (1) 𝐼0 (2) 𝐼0 4 𝐼0 𝐼0 (3) (4) 2 8

Q14.In an a.c. circuit, voltage and current are given by: V = 100 sin(100t) V and I = 100 sin(100t + π3 ) mA respectively. The average power dissipated in one cycle is : (1) 5 W (2) 10 W (3) 2. 5 W (4) 25 W

Q15.Critical angle of incidence for a pair of optical media is 45∘ . The refractive indices of first and second media are in the ratio: (1) 1 : √2 (2) √2 : 1 (3) 2 : 1 (4) 1 : 2

Q15.UV light of 4.13eV is incident on a photosensitive metal surface having work function 3.13eV. The maximum kinetic energy of ejected photoelectrons will be: (1) 4.13eV (2) 3.13eV (3) 1eV (4) 7.26eV

Q15.In series LCR circuit, the capacitance is changed from 𝐶 to 4𝐶. To keep the resonance frequency unchanged, the new inductance should be : 1 (1) reduced by 4𝐿 (2) increased by 2𝐿 (3) reduced by 3 (4) increased to 4𝐿 4𝐿 o

Q15.Arrange the following in the ascending order of wavelength: A. Gamma rays (λ1) B. x - rays (λ2) C. Infrared waves (λ3) D. Microwaves (λ4) Choose the most appropriate answer from the options given below (1) λ4 < λ3 < λ1 < λ2 (2) λ2 < λ1 < λ4 < λ3 (3) λ1 < λ2 < λ3 < λ4 (4) λ4 < λ3 < λ2 < λ1

Q15.A plane electromagnetic wave of frequency 35 MHz travels in free space along the X -direction. At a → particular point (in space and time) E = 9. 6ˆj V m−1 . The value of magnetic field at this point is: (1) 3. 2 × 10−8ˆk T (2) 3. 2 × 10−8ˆi (3) 9. 6ˆj T (4) 9. 6 × 10−8ˆk T

Q15.Given below are two statements: Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields. Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface. In the light of the above statements, choose the most appropriate answer from the options given below: (1) Statement I is incorrect but Statement II is (2) Both Statement I and Statement II are correct correct (3) Both Statement I and Statement II are incorrect (4) Statement I is correct but Statement II is incorrect

Q15. List-I List-II EM-Wave Wavelength Range (A) Infra-red (I) < 10−3 nm Match List-I with List-II : (B) Ultraviolet (II) 400 nm to 1 nm (C) X-rays S (III) 1 mm to 700 nm (D) Gamma rays (IV) 1 nm to 10−3 nm (1) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (2) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (3) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (4) (A)-(I), (B)-(III), (C)-(II), (D)-(IV)

Q16.A proton and an electron have the same de Broglie wavelength. If Kp and Ke be the kinetic energies of proton and electron respectively, then choose the correct relation : (1) Kp > Ke (2) Kp < Ke (3) Kp = Ke (4) Kp = Ke2

Q16.When unpolarized light is incident at an angle of 60° on a transparent medium from air. The reflected ray is completely polarized. The angle of refraction in the medium is (1) 30° (2) 60° (3) 90° (4) 45°

Q16.The electric field of an electromagnetic wave in free space is represented as →E = E0 cos(ωt −kz)ˆi corresponding magnetic induction vector will be : (1) → (2) → E0 B = E0C B = C cos(ωt −kz)ˆj cos(ωt −kz)ˆj (3) → (4) → E0 B = E0C + B = C + cos(ωt kz)ˆj cos(ωt kz)ˆj JEE Main 2024 (30 Jan Shift 1) JEE Main Previous Year Paper

Q16.In photoelectric experiment energy of 2.48eV irradiates a photo sensitive material. The stopping potential was measured to be 0.5 V. Work function of the photo sensitive material is : (1) 1.68eV (2) 2.48eV (3) 1.98eV (4) 0.5eV

Q16.For the photoelectric effect, the maximum kinetic energy 𝐸𝑘 of the photoelectrons is plotted against the frequency ( 𝑣) of the incident photons as shown in figure. The slope of the graph gives (1) Ratio of Planck’s constant to electric charge (2) Work function of the metal (3) Charge of electron (4) Planck’s constant