Bhotik Academy | JEE Advanced Physics Mock Test Set 4

Bhotik Academy — JEE Advanced Physics Mock Test Set 4

Level: Advanced | Questions: 30 | Type: Single-correct MCQ | Suggested Time: 75 minutes

Multi-concept numericals, conceptual traps, graphs, circuits, optics, mechanics and modern physics.

Pulley + frictionRotational dynamicsMoving rod induction Capacitor slabThermo cycleOptics diagramsCross-checked

Question 1 · Pulley + friction

A block of mass 4 kg lies on a rough horizontal table, coefficient of friction μ = 0.25. It is connected by a light string over a pulley to a hanging mass 3 kg. Take g = 10 m/s². The acceleration of the system is:

2.0 m/s² 2.86 m/s² 5.0 m/s² 6.0 m/s²

Answer: B
Friction on 4 kg block = μmg = 0.25×4×10 = 10 N. Driving force = 30 − 10 = 20 N. Total mass = 7 kg. Hence a = 20/7 = 2.86 m/s².

Question 2 · Moving rod induction

A conducting rod of length 0.5 m moves with speed 6 m/s on smooth rails in a uniform magnetic field B = 2 T into the page. If circuit resistance is 3 Ω, the induced current is:

0.5 A 1 A 2 A 4 A

Answer: C
Induced emf e = Blv = 2×0.5×6 = 6 V. Current I = e/R = 6/3 = 2 A.

Question 3 · Capacitor slab insertion

A parallel plate capacitor has capacitance C. A dielectric slab of thickness d/2 and dielectric constant K = 4 is inserted, filling the full area. New capacitance is:

8C/5 5C/8 2C 4C/5

Answer: A
Equivalent separation = d/2 + (d/2)/K = d/2 + d/8 = 5d/8. C' = ε₀A/(5d/8)=8C/5.

Question 4 · Rotational dynamics trap

A solid sphere rolls without slipping down an incline of angle 30°. Its acceleration is:

g/2 2g/5 5g/7 5g/14

Answer: D
For rolling sphere, a = g sinθ /(1+I/mR²)=g(1/2)/(1+2/5)=g/2 × 5/7 = 5g/14.

Question 5 · COM + collision

A 2 kg block moving at 6 m/s collides inelastically with a stationary 4 kg block. They stick together. Loss of kinetic energy is:

12 J 24 J 36 J 48 J

Answer: B
Final velocity = (2×6)/6 = 2 m/s. Initial KE = 36 J. Final KE = 1/2×6×4 = 12 J. Loss = 24 J.

Question 6 · Magnetic field superposition

Two long straight wires carry equal currents I in opposite directions and are separated by distance 2a. Magnetic field at midpoint is:

0 μ₀I/(2πa) μ₀I/(πa) 2μ₀I/(πa)

Answer: C
At midpoint, distance from each wire is a. Field due to one wire = μ₀I/(2πa). For opposite currents, fields at midpoint are in same direction. Net = μ₀I/πa.

Question 7 · Thermodynamics cyclic process

In a cyclic process, heat absorbed by gas is 500 J and heat rejected is 300 J. Work done by gas in one cycle is:

200 J 300 J 500 J 800 J

Answer: A
For complete cycle ΔU=0. Net work = net heat = 500 − 300 = 200 J.

Question 8 · Optics ray diagram

A ray enters a glass slab obliquely. Which quantity remains unchanged after emergence?

Speed Wavelength Direction inside slab Frequency

Answer: D
Frequency of light remains unchanged during refraction. Speed and wavelength change in different media.

Question 9 · Modern physics numerical

A photon of energy 10 eV falls on metal of work function 4 eV. Maximum kinetic energy of photoelectron is:

4 eV 6 eV 10 eV 14 eV

Answer: B
Kmax = hν − φ = 10 − 4 = 6 eV.

Question 10 · Graph interpretation

For a particle, v-t graph is a straight line from v = 2 m/s at t = 0 to v = 10 m/s at t = 4 s. Distance covered is:

16 m 20 m 24 m 32 m

Answer: C
Distance = area under v-t graph = average velocity × time = (2+10)/2 ×4 = 24 m.

Question 11 · Assertion-Reason

Assertion: In pure rolling, friction may do zero work. Reason: The point of contact is instantaneously at rest.

Both true and Reason explains Assertion Both true but Reason does not explain Assertion true, Reason false Assertion false, Reason true

Answer: A
For ideal pure rolling on fixed ground, contact point has zero instantaneous velocity, so static friction can do zero work.

Question 12 · SHM energy

In SHM, when displacement is half of amplitude, fraction of total energy that is kinetic is:

1/4 1/2 2/3 3/4

Answer: D
PE/E = x²/A² = 1/4. Hence KE/E = 3/4.

Question 13 · Current electricity

A cell of emf 12 V and internal resistance 1 Ω is connected to external resistance 5 Ω. Terminal voltage is:

8 V 10 V 12 V 2 V

Answer: B
Current I = 12/(5+1)=2 A. Terminal voltage V = IR = 2×5 = 10 V.

Question 14 · Electrostatics

Two equal charges q are placed at distance 2a. Electric potential at midpoint is:

0 kq/a 2kq/a kq/2a

Answer: C
Potential is scalar. Each charge contributes kq/a. Net = 2kq/a.

Question 15 · NLM trap

A force F pulls a block of mass m on smooth horizontal surface. If F acts at angle θ above horizontal, acceleration is:

F cosθ/m F/m F sinθ/m F tanθ/m

Answer: A
Only horizontal component causes horizontal acceleration on smooth surface: ma = F cosθ.

Question 16 · Gravitation

Orbital speed of a satellite close to earth surface is approximately:

√gR/2 gR √(2gR) √(gR)

Answer: D
For near earth orbit, GM/R² = g and v = √(GM/R)=√(gR).

Question 17 · AC circuit

In a series LCR circuit at resonance, impedance is:

Maximum Minimum and equal to R Zero Infinite

Answer: B
At resonance XL=XC, so net reactance zero and Z=R, minimum.

Question 18 · Wave optics

In YDSE, slit separation is doubled and screen distance is halved. Fringe width becomes:

β β/2 β/4 4β

Answer: C
β = λD/d. New β' = λ(D/2)/(2d)=β/4.

Question 19 · Ray optics

For a concave mirror, object at centre of curvature gives image:

At C, real, inverted, same size At F, highly diminished Behind mirror, virtual At infinity

Answer: A
Standard concave mirror result: object at C forms real, inverted, same-size image at C.

Question 20 · Fluid mechanics

Water flows through a pipe whose radius becomes half. If flow is steady and incompressible, speed becomes:

v/2 2v v/4 4v

Answer: D
Continuity Av = constant. Radius half makes area one-fourth, so speed becomes 4v.

Question 21 · EMI

Magnetic flux through a coil changes from 5 Wb to 1 Wb in 2 s. Magnitude of induced emf is:

1 V 2 V 4 V 8 V

Answer: B
|e| = |Δφ/Δt| = |1−5|/2 = 2 V.

Question 22 · Nuclear physics

In alpha decay, mass number and atomic number change as:

A−2, Z−4 A−4, Z−4 A−4, Z−2 A+4, Z+2

Answer: C
Alpha particle is helium nucleus ⁴₂He, so daughter has A−4 and Z−2.

Question 23 · Dimensional analysis

Dimension of angular momentum is:

ML²T⁻¹ MLT⁻¹ ML²T⁻² MT⁻¹

Answer: A
Angular momentum = r × p. Dimension = L × MLT⁻¹ = ML²T⁻¹.

Question 24 · Rotation

Torque on a body is zero. Which quantity must remain constant?

Linear momentum Kinetic energy Angular velocity always Angular momentum

Answer: D
Net external torque τ = dL/dt. If τ = 0, angular momentum remains constant.

Question 25 · Semiconductor

In a p-n junction diode under forward bias, barrier potential:

Increases Decreases Remains same Becomes infinite

Answer: B
Forward bias opposes the built-in field, reducing barrier potential and allowing current.

Question 26 · Projectile motion

For maximum range on horizontal ground, angle of projection is:

30° 60° 45° 90°

Answer: C
Range R = u²sin2θ/g is maximum when sin2θ = 1, so 2θ = 90°, θ = 45°.

Question 27 · Work-energy

A force F = 3x² acts along x-axis. Work done from x = 0 to x = 2 m is:

8 J 6 J 12 J 16 J

Answer: A
W = ∫₀² 3x² dx = [x³]₀² = 8 J.

Question 28 · Heat transfer

Two rods of same material and length have radii r and 2r. Ratio of heat currents is:

1:2 2:1 1:1 1:4

Answer: D
Heat current H ∝ A ∝ r². Ratio = r²:(2r)² = 1:4.

Question 29 · Magnetic force

A charged particle enters uniform magnetic field perpendicular to velocity. Its speed:

Increases Remains constant Decreases First increases then decreases

Answer: B
Magnetic force is perpendicular to velocity, so it does no work. Kinetic energy and speed remain constant.

Question 30 · Advanced mixed concept

A particle moves in a circle with constant speed. Which statement is correct?

Velocity is constant Acceleration is zero Speed is constant but velocity changes Net force is zero

Answer: C
In uniform circular motion, speed is constant but direction of velocity continuously changes; centripetal acceleration exists.