BPSC TRE 3.0 Physics Question Paper (Class 11–12, 22 July 2024) with Official Answers

BPSC TRE 3.0 Physics question paper (Class 11–12, 22 July 2024): all 80 Physics questions from Part III with BPSC’s official revised final answers and a short explanation for each.

BPSC TRE 4.0 is now open for 33,320 teacher posts, 16,540 of them for Class 11–12. Your subject part is 80 of the 150 marks, so solving the last paper is the best way to see the level BPSC sets for Physics.

Table of Contents

• Exam overview
• Physics questions with answers (Q71–150)
• How to prepare
• FAQs

BPSC TRE 3.0 Physics (Class 11–12): Exam Overview

Exam	BPSC School Teacher (Re-)Examination, TRE 3.0 (Advt. No. 22/2024)
Subject	Physics, Class 11–12
Exam date	22 July 2024, 1st sitting
Paper	150 questions, 150 marks, 2 hours 30 minutes
Part I	Language (qualifying): Q1–30
Part II	General Studies: Q31–70 (solve here)
Part III	Physics: Q71–150 (on this page)
Answers	Revised Final Answer Key, notice dated 6 November 2024 (Set-E column), revised: yes. Differs from the Final Answer Key of 17 October 2024 only at Q104 (D → C). Q102 deleted in both., booklet Series E
Deleted by BPSC	Q102

BPSC TRE 3.0 Physics Questions with Answers (Q71–150)

Question numbers are as in the original booklet. Answers follow BPSC’s revised final answer key; explanations are by the ExamYaari team.

Q71. Dimensions of elasticity constant is

(A) MLT⁻¹
(B) ML⁻¹T⁻²
(C) ML²T⁻¹
(D) More than one of the above
(E) None of the above

Answer: (B) ML⁻¹T⁻²
Explanation: An elastic constant (modulus) is stress ÷ strain; strain has no unit, so it has the dimensions of stress = force/area = MLT⁻²/L² = ML⁻¹T⁻².

Q72. The electric current in a charging RC circuit is given by i = i₀e^−t/RC, where i₀, R and C are constant parameters of the circuit and t is the time. The rate of change of current at t = 0 is

(A) i₀/RC
(B) i₀
(C) −i₀/RC
(D) More than one of the above
(E) None of the above

Answer: (C) −i₀/RC
Explanation: di/dt = −(i₀/RC)e^−t/RC. At t = 0, e⁰ = 1, so di/dt = −i₀/RC.

Q73. A projectile is fired with a speed u at an angle θ, with the horizontal. Its speed when its direction of motion makes an angle α with the horizontal will be

(A) u cos θ / cos α
(B) u cos α / cos θ
(C) u cos θ
(D) More than one of the above
(E) None of the above

Answer: (A) u cos θ / cos α
Explanation: The horizontal velocity never changes: v cos α = u cos θ, so v = u cos θ / cos α.

Q74. Which of the following exerts nuclear force only

(A) neutron on proton
(B) proton on proton
(C) neutron on neutron
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: A neutron has no charge, so between a neutron and a proton, or two neutrons, only the nuclear force acts. Two protons also repel electrically. So (A) and (C) are both right, which makes the answer (D). (The Hindi version of option C is misprinted; the English version is the standard.)

Q75. 1 micrometer / 1 femtometer =

(A) 10⁹
(B) 10⁻⁹
(C) 10⁶
(D) More than one of the above
(E) None of the above

Answer: (A) 10⁹
Explanation: 1 μm = 10⁻⁶ m and 1 fm = 10⁻¹⁵ m, so the ratio is 10⁻⁶/10⁻¹⁵ = 10⁹.

Q76. Percentage error in (250 ± 0.5) cm distance is

(A) 4%
(B) 0.04%
(C) 5%
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Percentage error = (0.5/250) × 100 = 0.2%. That is not listed, so the answer is (E).

Q77. Velocity of a boat with respect to land is (3î + 4ĵ) m s⁻¹ and velocity of water with respect to land is (−3î − 4ĵ) m s⁻¹, then velocity of boat with respect to water is

(A) 8ĵ m s⁻¹
(B) −6î − 8ĵ m s⁻¹
(C) 6î + 8ĵ m s⁻¹
(D) More than one of the above
(E) None of the above

Answer: (C) 6î + 8ĵ m s⁻¹
Explanation: Velocity of boat relative to water = v(boat) − v(water) = (3î + 4ĵ) − (−3î − 4ĵ) = 6î + 8ĵ m s⁻¹.

Q78. A body of mass m₁ moving with velocity v, collides with another stationary body of mass m₂. They get embedded. The velocity of the resultant body after the collision will be

(A) (m₁ + m₂)v / m₂
(B) m₂v / (m₁ + m₂)
(C) m₁v / m₂
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Momentum is conserved: m₁v = (m₁ + m₂)V, so V = m₁v/(m₁ + m₂). None of the options matches, so the answer is (E).

Q79. Vulcanized rubber is an example of

(A) Ductile
(B) Brittle
(C) Elastomer
(D) More than one of the above
(E) None of the above

Answer: (C) Elastomer
Explanation: Vulcanized rubber can be stretched a lot and still return to its shape, so it is an elastomer.

Q80. What should be the velocity of a person of weight 60 Kg, so that his kinetic energy becomes 270 J?

(A) 10 m s⁻¹
(B) 20 m s⁻¹
(C) 3 m s⁻¹
(D) More than one of the above
(E) None of the above

Answer: (C) 3 m s⁻¹
Explanation: ½mv² = 270 ⇒ v² = 2 × 270/60 = 9 ⇒ v = 3 m s⁻¹.

Q81. A piece of ice is floating in the water in a container. When the ice is completely melted in the water, height of the water will

(A) Remains same
(B) Increase
(C) Decrease
(D) More than one of the above
(E) None of the above

Answer: (A) Remains same
Explanation: Floating ice displaces water equal to its own weight. When it melts it becomes exactly that much water, so the level stays the same.

Q82. Value of Reynolds number is less for the fluid whose

(A) Velocity is less
(B) Density is less
(C) Coefficient of viscosity is high
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: Reynolds number = ρvD/η. It is smaller when velocity or density is lower, or when viscosity is higher, so all three options are right and the answer is (D).

Q83. Weight of a metal in air is 210 g and in water it is 140 g. Density of water is 1 g cm⁻³. The specific density of the metal is

(A) 7/4
(B) 7/6
(C) 7/2
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Relative density = weight in air ÷ loss of weight in water = 210/(210 − 140) = 210/70 = 3. That is not listed, so the answer is (E).

Q84. A sphere of mass m and radius r is falling freely in a viscous medium. Its terminal velocity is

(A) Directly proportional to m
(B) Inversely proportional to r
(C) Inversely proportional to √r
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: At terminal velocity, weight balances Stokes’ drag (ignoring buoyancy): mg = 6πηrv, so v = mg/(6πηr). For a given m it is ∝ m and ∝ 1/r, so (A) and (B) are both right and BPSC’s answer is (D).

Q85. A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is

(A) 76 cm
(B) < 76 cm
(C) > 76 cm
(D) More than one of the above
(E) None of the above

Answer: (C) > 76 cm
Explanation: In a lift accelerating upward the mercury feels a larger effective g, so P = hρ(g + a). A 76 cm column therefore means a pressure greater than 76 cm of mercury under normal g.

Q86. Working of Venturi meter is based on

(A) Bernoulli’s principle
(B) Torricelli’s law
(C) Stock’s law
(D) More than one of the above
(E) None of the above

Answer: (A) Bernoulli’s principle
Explanation: A Venturi meter measures flow speed from the pressure drop at a narrow section, which follows from Bernoulli’s principle.

Q87. Observations on Celsius and Fahrenheit scale becomes equal at

(A) −273.15°C
(B) −100°C
(C) 0°C
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Put F = C in F = 9C/5 + 32: C = 9C/5 + 32 gives C = −40. So the readings match at −40°, which is not listed; the answer is (E).

Q88. In an adiabatic process relationship between pressure and temperature for an ideal gas is

(A) P^1−γ T^γ = constant
(B) P^γ−1 T^γ = constant
(C) P^γ T^1−γ = constant
(D) More than one of the above
(E) None of the above

Answer: (A) P^1−γ T^γ = constant
Explanation: Combine PV^γ = constant with V ∝ T/P: P(T/P)^γ = constant, i.e. P^1−γ T^γ = constant.

Q89. For a process in an ideal gas dW = 0 and dQ < 0, then for that gas

(A) Temperature will increase
(B) Temperature will decrease
(C) Pressure remains constant
(D) More than one of the above
(E) None of the above

Answer: (B) Temperature will decrease
Explanation: First law: dU = dQ − dW = dQ < 0. The internal energy of an ideal gas falls only when its temperature falls, so the temperature decreases.

Q90. A Carnot engine does works between 1000 K and 500 K. The engine does 210 J mechanical work per cycle. The amount of heat required by the engine per cycle from the heat source will be

(A) 500 Cal
(B) 200 Cal
(C) 100 Cal
(D) More than one of the above
(E) None of the above

Answer: (C) 100 Cal
Explanation: Efficiency η = 1 − 500/1000 = 0.5, so heat taken Q = W/η = 210/0.5 = 420 J = 420/4.2 = 100 cal.

Q91. Performance coefficient of a heat pump

(A) Is greater than 1
(B) Is less than 1
(C) Can not be infinity
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: COP of a heat pump = Q₁/W = Q₁/(Q₁ − Q₂), which is always more than 1; it would be infinite only if no work were needed, which the second law forbids. So (A) and (C) are right: answer (D).

Q92. In order to make volume of a gas four times

(A) Its temperature should be made four times
(B) Its pressure should be made four times
(C) Its temperature should be made four times at a constant pressure
(D) More than one of the above
(E) None of the above

Answer: (C) Its temperature should be made four times at a constant pressure
Explanation: By Charles’s law V ∝ T only when pressure is constant, so the (absolute) temperature must be made four times at constant pressure. Raising pressure would shrink the volume.

Q93. In a small closed room, if the fan is kept on, then the temperature of the room

(A) Decreases
(B) Increases
(C) Remains same
(D) More than one of the above
(E) None of the above

Answer: (B) Increases
Explanation: A fan does not cool air; the electrical energy it uses ends up as heat in the closed room, so the room temperature slowly rises.

Q94. Mean free path of gas molecules is

(A) Directly proportional to the diameter of the molecule
(B) Inversely proportional to the square of the diameter
(C) Inversely proportional to the square root of the diameter
(D) More than one of the above
(E) None of the above

Answer: (B) Inversely proportional to the square of the diameter
Explanation: Mean free path λ = 1/(√2 π d² n), so it is inversely proportional to the square of the molecular diameter d.

Q95. A particle moves in a circular path with uniform speed. Its motion is

(A) Simple harmonic
(B) Oscillatory
(C) Periodic
(D) More than one of the above
(E) None of the above

Answer: (C) Periodic
Explanation: Uniform circular motion repeats after every revolution, so it is periodic. It is not to-and-fro about a mean position, so it is neither oscillatory nor SHM.

Q96. A pendulum clock that keeps correct time on the earth is taken to the moon. It will run

(A) 6 times faster
(B) √6 times faster
(C) √6 times slower
(D) More than one of the above
(E) None of the above

Answer: (C) √6 times slower
Explanation: T = 2π√(l/g). On the moon g is about g/6, so T becomes √6 times larger and the clock runs √6 times slower.

Q97. The resultant force acting on a particle executing simple harmonic motion is 4 N when it is 10 cm away from the centre of oscillation. The spring constant will be

(A) 40 N m⁻¹
(B) 80 N m⁻¹
(C) 0.4 N m⁻¹
(D) More than one of the above
(E) None of the above

Answer: (A) 40 N m⁻¹
Explanation: F = kx ⇒ k = 4 N / 0.10 m = 40 N m⁻¹.

Q98. In a stationary wave

(A) All the particles of the medium vibrate in phase
(B) All the particles between consecutive nodes vibrate in phase
(C) The alternate antinodes vibrate in phase
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: In a standing wave all particles between two neighbouring nodes move in phase, and each loop is opposite in phase to the next, so alternate antinodes are in phase. (B) and (C) are right, so the answer is (D).

Q99. Velocity of sound in air is 332 m s⁻¹. Its velocity in vacuum will be

(A) > 332 m s⁻¹
(B) < 332 m s⁻¹
(C) = 332 m s⁻¹
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Sound needs a medium and cannot travel through vacuum at all, so none of the given speeds is right; the answer is (E).

Q100. Which of the following is a mechanical wave?

(A) Radio wave
(B) Microwave
(C) Light wave
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Radio waves, microwaves and light are all electromagnetic waves that need no medium. None of them is a mechanical wave, so the answer is (E).

Q101. Mechanical waves transfers

(A) Energy
(B) Matter
(C) Frequency
(D) More than one of the above
(E) None of the above

Answer: (A) Energy
Explanation: In a mechanical wave the particles only vibrate about their places; what travels forward is energy, not matter.

Q102. In a stationary wave distance between two successive nodes and antinodes is

(A) λ
(B) λ/4
(C) λ/2
(D) More than one of the above
(E) None of the above

Answer: Question deleted (BPSC removed it in the final answer key)
Explanation: BPSC deleted this question in the final answer key. The wording is unclear: two successive nodes (or two successive antinodes) are λ/2 apart, while a node and the next antinode are λ/4 apart.

Q103. Electric field at a distance r from an infinite thin plane sheet of uniform surface charge density σ is

(A) Inversely proportional to r
(B) Is independent of r
(C) Is independent of σ
(D) More than one of the above
(E) None of the above

Answer: (B) Is independent of r
Explanation: For an infinite sheet E = σ/(2ε₀). It depends on σ but not on the distance r.

Q104. The electric displacement vector D

(A) Relates the electric field to free charge in vacuum
(B) Relates the electric field to free charge in dielectric medium
(C) Is related to electric field in vacuum by relation, D = ε₀E
(D) More than one of the above
(E) None of the above

Answer: (C) Is related to electric field in vacuum by relation, D = ε₀E
Explanation: In general D = ε₀E + P, where P comes from bound charges in a dielectric. In vacuum P = 0, so D = ε₀E. BPSC’s revised key (6 Nov 2024) changed the answer from (D) to (C).

Q105. No current flows between two charged bodies when connected by a conducting wire, if they have same

(A) Capacity
(B) Charge
(C) Potential
(D) More than one of the above
(E) None of the above

Answer: (C) Potential
Explanation: Charge flows only because of a potential difference. If both bodies are at the same potential, no current flows, whatever their charge or capacity.

Q106. The direction of the magnetic field of a long straight wire carrying current is

(A) In the direction of the current
(B) Radially outward
(C) Radially inward
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: The field lines are circles around the wire, so the field is tangential to these circles (right-hand thumb rule). It is neither along the current nor radial, so the answer is (E).

Q107. The permeability of a paramagnetic substance is

(A) Slightly less than that of vacuum
(B) Slightly more than that of vacuum
(C) Much more than that of vacuum
(D) More than one of the above
(E) None of the above

Answer: (B) Slightly more than that of vacuum
Explanation: A paramagnetic material has a small positive susceptibility, so μᵣ is just above 1: its permeability is slightly more than μ₀. (Diamagnetic is slightly less; ferromagnetic is much more.)

Q108. If a charged particle q enters parallel to the magnetic field B, with velocity v, then the magnetic force on the charge particle is

(A) 0
(B) qvB
(C) qB
(D) More than one of the above
(E) None of the above

Answer: (A) 0
Explanation: F = qvB sin θ. Moving parallel to the field, θ = 0 and sin 0 = 0, so the force is zero.

Q109. Soft iron is used to manufacture electromagnets because its

(A) Area of hysteresis is large
(B) Retentivity is high
(C) Coercive force is high
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Soft iron suits electromagnets because it has high permeability, low coercivity and a narrow (small-area) hysteresis loop, so it magnetises and demagnetises easily. All three options state the opposite, so the answer is (E).

Q110. Two long straight parallel conductors carrying equal steady currents in same direction

(A) Attract each other
(B) Repel each other
(C) Neither attract nor repel each other
(D) More than one of the above
(E) None of the above

Answer: (A) Attract each other
Explanation: Parallel wires with currents in the same direction attract each other; opposite currents repel. The force per unit length is μ₀I₁I₂/(2πd).

Q111. Which is the incorrect statement about the electromagnetic wave?

(A) The electromagnetic field vectors E and B are mutually perpendicular.
(B) The electromagnetic field vectors E and B are in same phase.
(C) The electromagnetic waves are transverse.
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: All three statements are true for an EM wave in free space: E ⊥ B, E and B oscillate in phase, and the wave is transverse. No statement is incorrect, so the answer is (E).

Q112. The electric field intensity on the surface of a charged conductor is

(A) Directed normally to the surface
(B) Directed tangentially to the surface
(C) Zero
(D) More than one of the above
(E) None of the above

Answer: (A) Directed normally to the surface
Explanation: The surface of a conductor is an equipotential, so the field there has no tangential part and points normal to the surface (magnitude σ/ε₀).

Q113. For a given dielectric, the electronic polarizability αₑ

(A) Increases with temperature
(B) Decreases with temperature
(C) Independent of temperature
(D) More than one of the above
(E) None of the above

Answer: (C) Independent of temperature
Explanation: Electronic polarizability comes from the shift of the electron cloud relative to the nucleus, αₑ = 4πε₀R³ (R = atomic radius). It does not depend on temperature, unlike orientational polarizability.

Q114. Phenomenon responsible for working of a transformer is

(A) Doppler effect
(B) Induction heating
(C) Mutual induction
(D) More than one of the above
(E) None of the above

Answer: (C) Mutual induction
Explanation: A changing current in the primary coil changes the flux through the secondary coil and induces an emf there; this is mutual induction.

Q115. When speed of a dc motor increases, current flowing through the armchair [armature]

(A) Decreases
(B) Increases
(C) Remains same
(D) More than one of the above
(E) None of the above

Answer: (A) Decreases
Explanation: As the motor speeds up, its back emf e grows, so the armature current I = (V − e)/R falls. (The English booklet prints ‘armchair’, a misprint for armature.)

Q116. In optical fibre, the refractive index of the material of the core is

(A) Equal to that of the cladding
(B) Less than that of the cladding
(C) Higher than that of the cladding
(D) More than one of the above
(E) None of the above

Answer: (C) Higher than that of the cladding
Explanation: Total internal reflection needs light to go from a denser to a rarer medium, so the core’s refractive index is higher than the cladding’s.

Q117. Which of the following do not have zero average value in plane electromagnetic wave?

(A) Electric energy
(B) Magnetic energy
(C) Magnetic field
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: The fields E and B swing positive and negative, so they average to zero. Energy densities ½ε₀E² and B²/(2μ₀) are never negative and have non-zero averages. So (A) and (B) are right: answer (D).

Q118. Maxwell’s equations represents fundamental laws of

(A) Electricity
(B) Magnetism
(C) Mechanics
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: Maxwell’s four equations are the basic laws of electricity and magnetism (electromagnetism), not mechanics. So (A) and (B) are right: answer (D).

Q119. A dc voltage of 10 volts is applied to an RC series circuit, in which R is 50 Ω and C is 2000 μF. The time constant of the circuit is

(A) 0.1 s
(B) 10 s
(C) .01 s
(D) More than one of the above
(E) None of the above

Answer: (A) 0.1 s
Explanation: τ = RC = 50 × 2000 × 10⁻⁶ = 0.1 s. The applied voltage does not matter.

Q120. An ac voltage v = vₘ sin ωt is applied to an inductor of inductive reactance X_L. The current, i in the inductor will be

(A) i = Vₘ/X_L
(B) i = (Vₘ/X_L) sin(ωt + π/2)
(C) i = (Vₘ/X_L) sin(ωt − π/2)
(D) More than one of the above
(E) None of the above

Answer: (C) i = (Vₘ/X_L) sin(ωt − π/2)
Explanation: In a pure inductor the current lags the voltage by π/2, with peak value Vₘ/X_L, so i = (Vₘ/X_L) sin(ωt − π/2).

Q121. If an ac electric current i = Iₘ sin ωt flows through a resistor R

(A) Power loss does not take place in R
(B) The average power loss will be Iₘ² × R / 2
(C) Power loss changes from instant to instant
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: Instantaneous power i²R = Iₘ²R sin²ωt keeps changing, and its average is Iₘ²R/2 because the average of sin²ωt is ½. So (B) and (C) are right: answer (D).

Q122. In an LCR circuit the discharge will be oscillatory if

(A) R² = 4L/C
(B) R² > 4L/C
(C) R² < 4L/C
(D) More than one of the above
(E) None of the above

Answer: (C) R² < 4L/C
Explanation: The discharge oscillates (underdamped) when R² < 4L/C; R² = 4L/C is critical damping and R² > 4L/C is overdamped.

Q123. In a simple circuit containing an inductance L and resistance R, and having a steady value of the current I₀, the rate of increase of current is

(A) Constant
(B) Proportional to R/L
(C) Proportional to L/R
(D) More than one of the above
(E) None of the above

Answer: (B) Proportional to R/L
Explanation: During growth i = I₀(1 − e^−Rt/L), so di/dt = (R/L)(I₀ − i) = (R/L)I₀e^−Rt/L. The rate is proportional to R/L, as BPSC’s key says.

Q124. Alternating current or voltage can be generated by

(A) Rotating a magnetic field in a stationary coil
(B) Rotating a coil in a steady magnetic field
(C) Rectifier circuit
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: An AC emf appears whenever the flux through a coil changes periodically, either by rotating the field (alternators) or by rotating the coil. A rectifier does the opposite (AC to DC). So (A) and (B) are right: answer (D).

Q125. A parallel LC circuit works as a rejector for a frequency fᵣ, then fᵣ =

(A) 1/(2π√(LC))
(B) 1/√(LC)
(C) 2π√(LC)
(D) More than one of the above
(E) None of the above

Answer: (A) 1/(2π√(LC))
Explanation: A parallel LC circuit has maximum impedance (rejects current) at resonance, where fᵣ = 1/(2π√(LC)).

Q126. The form factor in an ac circuit is

(A) The ratio of rms to average value
(B) The ratio of average to rms value
(C) The ratio of rms to maximum value
(D) More than one of the above
(E) None of the above

Answer: (A) The ratio of rms to average value
Explanation: Form factor = rms value ÷ average value (about 1.11 for a sine wave).

Q127. For a convex mirror

(A) Image distance is ≤ focal length
(B) Image distance is ≥ focal length
(C) Image is imaginary
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: A convex mirror always forms a virtual (imaginary) image between the pole and the focus, so the image distance is never more than the focal length. (A) and (C) are right: answer (D).

Q128. If v₁ and v₂ are the velocities of light in medium 1 and medium 2 respectively, then refractive index of medium 2 with respect to medium 1 is

(A) v₁/v₂
(B) v₂/v₁
(C) v₁v₂
(D) More than one of the above
(E) None of the above

Answer: (A) v₁/v₂
Explanation: ₁n₂ = n₂/n₁ = (c/v₂)/(c/v₁) = v₁/v₂.

Q129. Laws of reflection are true for

(A) Plane mirror
(B) Curved reflective surface
(C) Total internal reflection
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: The laws of reflection hold at every point of any reflecting surface, plane or curved, and also in total internal reflection. More than one option is right, so the answer is (D).

Q130. If a prism is arranged in a position of minimum deviation angle, then

(A) Angle of incidence < angle of emergence
(B) Angle of incidence > angle of emergence
(C) Angle of incidence = angle of emergence
(D) More than one of the above
(E) None of the above

Answer: (C) Angle of incidence = angle of emergence
Explanation: At minimum deviation the ray passes symmetrically through the prism, so the angle of incidence equals the angle of emergence (i = e).

Q131. In Rayleigh scattering wavelength of the incident light wave is

(A) Less than the dimension of the particle
(B) Greater than the dimension of the particle
(C) Equal to the dimension of the particle
(D) More than one of the above
(E) None of the above

Answer: (B) Greater than the dimension of the particle
Explanation: Rayleigh scattering (intensity ∝ 1/λ⁴) applies when the scattering particles are much smaller than the wavelength of light.

Q132. In Young’s double-slit experiment, distance between two successive fringes is

(A) Proportional to the frequency of the light wave
(B) Dependent on intensity of light
(C) Equal
(D) More than one of the above
(E) None of the above

Answer: (C) Equal
Explanation: Fringe width β = λD/d is the same for all fringes, so successive fringes are equally spaced. It is proportional to λ (not frequency) and does not depend on intensity.

Q133. In polarization of light, which property of light changes?

(A) Frequency
(B) Wavelength
(C) Phase
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Polarization restricts the direction of vibration of the electric field (and reduces intensity); frequency, wavelength and phase are not what it changes. So the answer is (E).

Q134. From the Brewster’s law of polarization it can be said that the polarization angle depends on

(A) Wavelength of light
(B) Refractive index of the medium
(C) Rotation of plane of polarization
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: Brewster’s law: tan θₚ = n. The angle depends on the refractive index, and since n changes with wavelength, it also depends on wavelength. (A) and (B) are right: answer (D).

Q135. Human body at its normal temperature

(A) Radiates most strongly in the visible part of the spectrum
(B) Radiates most strongly in the infrared part of the spectrum
(C) Radiates most strongly in the radio wave part of the spectrum
(D) More than one of the above
(E) None of the above

Answer: (B) Radiates most strongly in the infrared part of the spectrum
Explanation: Wien’s law: λₘₐₓ = b/T ≈ 2.9 × 10⁻³ / 310 ≈ 9.4 μm, which lies in the infrared.

Q136. Work function of a metal depends on

(A) Type of metal
(B) Type of the surface of the metal
(C) Type of radiation incident on the metal
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: Work function is a property of the metal and of its surface condition (cleanliness, coatings, crystal face). It does not depend on the incident radiation. (A) and (B) are right: answer (D).

Q137. Momentum of a photon of frequency f is given by

(A) hf/c
(B) c/hf
(C) hf
(D) More than one of the above
(E) None of the above

Answer: (A) hf/c
Explanation: p = E/c = hf/c (equivalently h/λ).

Q138. Bohr’s model can be applied to

(A) Hydrogen atom
(B) He⁺ ion
(C) All ions with unit electric charge
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: Bohr’s model works only for one-electron systems such as H, He⁺ and Li²⁺. Ions with unit charge in general (e.g. Li⁺) have more electrons. (A) and (B) are right: answer (D).

Q139. Principle quantum number decides

(A) Energy of electron in an atom
(B) Distance of electron from nucleus
(C) Total angular momentum
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: The principal quantum number n fixes the main energy level (Eₙ ∝ −1/n²) and the shell size or distance from the nucleus (rₙ ∝ n²). Total angular momentum depends on l and s. (A) and (B) are right: answer (D).

Q140. In a hydrogen atom radius of second Bohr orbit in terms of Bohr’s radius a₀ is given by

(A) √2 a₀
(B) 8 a₀
(C) 4 a₀
(D) More than one of the above
(E) None of the above

Answer: (C) 4 a₀
Explanation: rₙ = n²a₀, so for n = 2, r = 4a₀.

Q141. Plutonium (₉₄Pu²³⁹) is

(A) Highly stable
(B) Highly radio active
(C) Fissionable
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: Pu-239 is radioactive (an alpha emitter) and is fissile, used in reactors and weapons. So (B) and (C) are right and BPSC’s answer is (D).

Q142. Complete the nuclear process : ₆C¹¹ → ₅B¹¹ + β⁺ + ______

(A) Electron
(B) Proton
(C) Anti-neutrino
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: In β⁺ decay a proton becomes a neutron, emitting a positron and a neutrino (ν), not an anti-neutrino. The neutrino is not listed, so the answer is (E).

Q143. Which element can be used to make P-type semiconductor?

(A) Phosphorous
(B) Antimony
(C) Arsenic
(D) More than one of the above
(E) None of the above

Answer: (E) None of the above
Explanation: Phosphorus, antimony and arsenic are pentavalent and make n-type semiconductors. P-type needs a trivalent dopant such as boron, aluminium, gallium or indium, none of which is listed; the answer is (E).

Q144. With increase in temperature resistivity of metals

(A) Increases
(B) Decreases
(C) Remains same
(D) More than one of the above
(E) None of the above

Answer: (A) Increases
Explanation: In metals, heating makes the ions vibrate more, so electrons collide more often and resistivity increases: ρ = ρ₀(1 + αΔT).

Q145. In rectification process

(A) ac voltage (or current) is converted into dc voltage (or current)
(B) dc voltage (or current) is converted into ac voltage (or current)
(C) ac voltage (or current) is amplified
(D) More than one of the above
(E) None of the above

Answer: (A) ac voltage (or current) is converted into dc voltage (or current)
Explanation: A rectifier converts AC into DC. Converting DC into AC is done by an inverter, and amplifying is done by an amplifier.

Q146. Which of the following indicates forward bias of a diode?

(A) [Diode symbol, arrow pointing right: anode (left) at +2 V, cathode (right) at 0 V]
(B) [Diode symbol, arrow pointing right: anode (left) at −3 V, cathode (right) at 0 V]
(C) [Diode symbol, arrow pointing right: anode (left) at −2 V, cathode (right) at −3 V]
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: A diode is forward biased when its anode (p side) is at a higher potential than its cathode. In (A) +2 V > 0 V and in (C) −2 V > −3 V, so both are forward biased; (B) is reverse biased. Answer (D).

Q147. In common base amplifier the phase difference between the input and the output voltages is

(A) π
(B) π/2
(C) 0
(D) More than one of the above
(E) None of the above

Answer: (C) 0
Explanation: In a common-base amplifier the output voltage is in phase with the input, so the phase difference is 0. (In common-emitter it is π.)

Q148. In a common emitter transistor amplifier, current gain of the transistor is 100, input resistance rᵢ = 1 kΩ, and load resistance is 10 kΩ. What will be the voltage gain of the amplifier?

(A) 500
(B) 1000
(C) 100
(D) More than one of the above
(E) None of the above

Answer: (B) 1000
Explanation: Voltage gain = β × (R_L/rᵢ) = 100 × (10 kΩ / 1 kΩ) = 1000.

Q149. A NOR gate has two inputs A and B. Its output Y can be given as

(A) Y = A + B
(B) Y = A + B
(C) Y = A · B
(D) More than one of the above
(E) None of the above

Answer: (D) More than one of the above
Explanation: NOR output is the complement of OR: Y = A + B. By De Morgan’s theorem this equals A · B. So (B) and (C) are both right: answer (D).

Q150. In a logic gate when all the inputs are one, the output is zero. The logic gate is

(A) a NAND gate
(B) an OR gate
(C) an AND gate
(D) More than one of the above
(E) None of the above

Answer: (A) a NAND gate
Explanation: A NAND gate gives 0 only when all inputs are 1. OR and AND would both give 1 for all-1 inputs.

How to Prepare for BPSC TRE Physics

• Solve this paper in one sitting with a timer: about 1 minute per question.
• For every wrong answer, read the explanation and revise that chapter from your NCERT and graduation books.
• Watch for (D) More than one and (E) None of the above: check each option, don’t stop at the first one that looks right.
• Practise the General Studies part too: BPSC TRE 3.0 GS paper (Class 11–12).
• Take timed mock tests on the ExamYaari app.

More BPSC TRE Resources

• यही प्रश्न पत्र हिंदी में देखें
• All BPSC TRE General Studies question papers
• BPSC TRE 3.0 General Studies (Class 11–12)

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