CUET Physics Practice Quiz 3
Welcome to this CUET Physics Practice Quiz, the third in a series of three unique variations designed to help you master key concepts for the Common University Entrance Test. This quiz focuses on application-based problems and conceptual understanding across core topics like electrostatics, magnetism, optics, modern physics, and semiconductor devices.
Each of the 20 questions is crafted to mirror the style and difficulty of the actual CUET exam. Read each question carefully, choose the best answer, and use the detailed explanations provided to solidify your understanding. This is an excellent way to identify your strengths and pinpoint areas that need more revision.
Good luck, and let’s get started!
Q1. A charged particle enters a uniform magnetic field perpendicular to its velocity. If the speed of the particle is doubled, the radius of its circular path will:
The radius of a charged particle moving in a magnetic field is given by r = mv/(qB). If the speed (v) is doubled, the radius (r) also doubles, assuming mass, charge, and magnetic field remain constant.
Q2. In a series LCR circuit, at resonance, the impedance of the circuit is equal to:
At resonance in a series LCR circuit, the inductive reactance (XL) equals the capacitive reactance (XC). They cancel each other out, leaving the impedance equal to the resistance (R) of the circuit.
Q3. The work function of a metal is 2.5 eV. What is the maximum kinetic energy of the photoelectrons emitted when light of energy 4.5 eV is incident on it?
The maximum kinetic energy of emitted photoelectrons is given by Einstein's photoelectric equation: KEmax = hν - φ, where hν is the incident photon energy and φ is the work function. KEmax = 4.5 eV - 2.5 eV = 2.0 eV.
Q4. Two capacitors of capacitances 2 µF and 4 µF are connected in series. The equivalent capacitance of the combination is:
For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances. 1/Ceq = 1/2 + 1/4 = 3/4. Therefore, Ceq = 4/3 µF ≈ 1.33 µF.
Q5. A convex lens has a focal length of 20 cm. The power of the lens is:
The power of a lens is the reciprocal of its focal length in meters. P = 1/f = 1/0.20 m = +5 Diopters (D). The sign is positive for a convex (converging) lens.
Q6. In a Young's double-slit experiment, the fringe width is 0.5 mm. If the distance between the slits is doubled, the new fringe width will be:
Fringe width (β) is given by β = λD/d, where d is the distance between slits. If the distance between slits (d) is doubled, the fringe width (β) is halved. New fringe width = 0.5 mm / 2 = 0.25 mm.
Q7. The half-life of a radioactive substance is 10 days. The time taken for 75% of the sample to decay is:
After one half-life (10 days), 50% of the sample remains. After two half-lives (20 days), 25% of the sample remains, meaning 75% has decayed. Therefore, the time taken is 20 days.
Q8. In a p-n junction diode, the depletion region is formed due to the:
The depletion region is formed due to the diffusion of electrons from the n-side to the p-side and holes from the p-side to the n-side across the junction. This diffusion leaves behind immobile ions, creating a region devoid of free charge carriers.
Q9. The electric potential at a point due to a point charge is 100 V. If the distance from the charge is doubled, the new potential will be:
The electric potential (V) due to a point charge is given by V = kQ/r. If the distance (r) is doubled, the potential (V) is halved. New potential = 100 V / 2 = 50 V.
Q10. A current of 2 A flows through a conductor of resistance 10 Ω for 5 minutes. The heat generated in the conductor is:
The heat generated is given by Joule's law: H = I²Rt. Here, I = 2 A, R = 10 Ω, and t = 5 minutes = 300 seconds. H = (2)² × 10 × 300 = 4 × 10 × 300 = 12000 J.
Q11. The de Broglie wavelength of a particle is inversely proportional to its:
The de Broglie wavelength (λ) is given by λ = h/p, where p is the momentum of the particle. Therefore, the wavelength is inversely proportional to the momentum.
Q12. In an electromagnetic wave, the ratio of the magnitudes of the electric field (E) to the magnetic field (B) is equal to:
In an electromagnetic wave, the magnitudes of the electric and magnetic fields are related by E = cB, where c is the speed of light. Therefore, the ratio E/B is equal to the speed of light (c).
Q13. A wire of length 1 m is moving with a velocity of 5 m/s perpendicular to a magnetic field of 0.2 T. The induced EMF across the ends of the wire is:
The induced EMF in a moving conductor is given by ε = Bvl, where B is the magnetic field, v is the velocity, and l is the length of the conductor. ε = 0.2 T × 5 m/s × 1 m = 1 V.
Q14. The energy of a photon of frequency 5 × 10^14 Hz is approximately: (Use h = 6.63 × 10^-34 Js)
The energy of a photon is given by E = hν, where h is Planck's constant and ν is the frequency. E = 6.63 × 10^-34 Js × 5 × 10^14 Hz = 3.315 × 10^-19 J ≈ 3.3 × 10^-19 J.
Q15. In a common emitter transistor amplifier, the current gain (β) is defined as the ratio of:
The current gain (β) in a common emitter configuration is defined as the ratio of the change in collector current (ΔIC) to the change in base current (ΔIB). It is a key parameter for amplifier applications.
Q16. The Brewster angle for a transparent medium is 60°. The refractive index of the medium is:
According to Brewster's law, the refractive index (n) is given by n = tan(ip), where ip is the polarizing angle (Brewster angle). n = tan(60°) = √3 ≈ 1.73.
Q17. A body is projected vertically upward with a velocity of 20 m/s. The maximum height reached by the body is: (Take g = 10 m/s²)
The maximum height (H) reached by a projectile is given by H = u²/(2g), where u is the initial velocity. H = (20)² / (2 × 10) = 400 / 20 = 20 m.
Q18. The potential difference across a 2 µF capacitor is 100 V. The energy stored in the capacitor is:
The energy stored in a capacitor is given by U = (1/2)CV². U = 0.5 × 2 × 10^-6 F × (100 V)² = 0.5 × 2 × 10^-6 × 10000 = 0.01 J.
Q19. In a nuclear fission reaction, the total mass of the products is:
In nuclear fission, a heavy nucleus splits into lighter nuclei. The total mass of the products is slightly less than the mass of the original nucleus. This mass defect is converted into energy according to Einstein's equation, E = mc².
Q20. The mutual inductance between two coils depends on:
Mutual inductance depends on the number of turns in each coil, their geometry (size, shape, and relative orientation), and the magnetic properties (permeability) of the core material between them.