JEE Main Practice — Physics Set 4

JEE Main · 2026 · with answers and explanations

questions
25
Total marks
100
Duration
60 minutes
Negative marking
−1

Subjects: Physics

Question 1

Physics · 4 marks · −1

A 2 kg block slides down a rough incline of inclination 37∘37^\circ with coefficient of kinetic friction 0.5. Its acceleration is (sin⁡37∘=0.6\sin37^\circ=0.6, g=10 m/s2g=10\ \mathrm{m/s^2})

  1. A.6 m/s2^2
  2. B.4 m/s2^2
  3. C.2 m/s2^2
  4. D.1 m/s2^2

Correct answer: C

Explanation: a=g(sin⁡θ−μcos⁡θ)=10(0.6−0.5×0.8)=2a=g(\sin\theta-\mu\cos\theta)=10(0.6-0.5\times0.8)=2 m/s2^2 (independent of mass).

Question 2

Physics · 4 marks · −1

A 2 kg block is held at rest against a rough vertical wall by a horizontal force FF. If the coefficient of static friction is 0.4, the minimum value of FF is (g=10 m/s2g=10\ \mathrm{m/s^2})

  1. A.8 N
  2. B.20 N
  3. C.80 N
  4. D.50 N

Correct answer: D

Explanation: Normal reaction N=FN=F. Friction must balance weight: μF≥mg⇒Fmin=200.4=50\mu F\ge mg\Rightarrow F_{min}=\frac{20}{0.4}=50 N.

Question 3

Physics · 4 marks · −1

A 2 kg block rests on a 4 kg block that lies on a smooth horizontal floor. The coefficient of friction (static and kinetic) between the blocks is 0.3. A horizontal force of 30 N is applied to the lower block. The acceleration of the lower block is (g=10 m/s2g=10\ \mathrm{m/s^2})

  1. A.5 m/s2^2
  2. B.7.5 m/s2^2
  3. C.3 m/s2^2
  4. D.6 m/s2^2

Correct answer: D

Explanation: Maximum acceleration friction can give the upper block: μg=3\mu g=3 m/s2^2. Without slipping, a=30/6=5>3a=30/6=5>3, so the blocks slip. Kinetic friction =0.3×20=6=0.3\times20=6 N. Lower block: a=30−64=6a=\frac{30-6}{4}=6 m/s2^2 (upper block: 3 m/s2^2).

Question 4

Physics · 4 marks · −1

A 0.2 kg ball strikes a wall perpendicularly at 20 m/s and rebounds along the same line at 15 m/s. If the contact lasts 0.05 s, the magnitude of the average force exerted by the wall on the ball (in newton) is ______.

Correct answer: 140

Explanation: Δp=0.2×(15−(−20))=7\Delta p=0.2\times(15-(-20))=7 kg m/s. F=70.05=140F=\frac{7}{0.05}=140 N.

Question 5

Physics · 4 marks · −1

A road of radius of curvature 100 m is banked at an angle θ\theta with tan⁡θ=0.4\tan\theta=0.4. The speed (in m/s) at which a car can take the turn with no tendency to slip, even on a frictionless road, is ______. (g=10 m/s2g=10\ \mathrm{m/s^2})

Correct answer: 20

Explanation: Without friction: Nsin⁡θ=mv2rN\sin\theta=\frac{mv^2}{r}, Ncos⁡θ=mg⇒v=rgtan⁡θ=100×10×0.4=20N\cos\theta=mg\Rightarrow v=\sqrt{rg\tan\theta}=\sqrt{100\times10\times0.4}=20 m/s.

Question 6

Physics · 4 marks · −1

A proton moves with a speed of 2×1062\times10^6 m/s perpendicular to a uniform magnetic field of 0.5 T. The magnitude of the magnetic force on it is (e=1.6×10−19e=1.6\times10^{-19} C)

  1. A.3.2×10−133.2\times10^{-13} N
  2. B.1.6×10−131.6\times10^{-13} N
  3. C.1.6×10−141.6\times10^{-14} N
  4. D.0.8×10−130.8\times10^{-13} N

Correct answer: B

Explanation: F=qvBsin⁡90∘=1.6×10−19×2×106×0.5=1.6×10−13F=qvB\sin90^\circ=1.6\times10^{-19}\times2\times10^6\times0.5=1.6\times10^{-13} N.

Question 7

Physics · 4 marks · −1

A circular loop of radius 10 cm carries a current of 5 A. The magnetic field at its centre is

  1. A.π×10−5\pi\times10^{-5} T
  2. B.2π×10−52\pi\times10^{-5} T
  3. C.π×10−6\pi\times10^{-6} T
  4. D.π2×10−5\frac{\pi}{2}\times10^{-5} T

Correct answer: A

Explanation: B=μ0I2R=4π×10−7×50.2=π×10−5B=\frac{\mu_0I}{2R}=\frac{4\pi\times10^{-7}\times5}{0.2}=\pi\times10^{-5} T.

Question 8

Physics · 4 marks · −1

A proton and an α\alpha-particle having the same kinetic energy enter a uniform magnetic field perpendicular to it. The ratio of the radius of the α\alpha-particle's path to that of the proton's path is (mα=4mpm_\alpha=4m_p, qα=2eq_\alpha=2e)

  1. A.2
  2. B.12\frac12
  3. C.1
  4. D.2\sqrt2

Correct answer: C

Explanation: r=mvqB=2mKqBr=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB}. rαrp=42=1\frac{r_\alpha}{r_p}=\frac{\sqrt{4}}{2}=1.

Question 9

Physics · 4 marks · −1

Two long straight parallel wires 10 cm apart carry currents of 10 A and 20 A in the same direction. The force per unit length between them is

  1. A.4×10−44\times10^{-4} N/m, repulsive
  2. B.2×10−42\times10^{-4} N/m, attractive
  3. C.8×10−48\times10^{-4} N/m, attractive
  4. D.4×10−44\times10^{-4} N/m, attractive

Correct answer: D

Explanation: Fl=μ0I1I22πd=2×10−7×10×200.1=4×10−4\frac{F}{l}=\frac{\mu_0I_1I_2}{2\pi d}=\frac{2\times10^{-7}\times10\times20}{0.1}=4\times10^{-4} N/m. Parallel currents in the same direction attract.

Question 10

Physics · 4 marks · −1

A galvanometer of resistance 50 Ω50\,\Omega gives full-scale deflection for a current of 2 mA. To convert it into a voltmeter of range 0–10 V, one must connect

  1. A.4950 Ω4950\,\Omega in series
  2. B.5000 Ω5000\,\Omega in series
  3. C.5050 Ω5050\,\Omega in series
  4. D.4950 Ω4950\,\Omega in parallel

Correct answer: A

Explanation: Ig(G+R)=V⇒R=102×10−3−50=5000−50=4950 ΩI_g(G+R)=V\Rightarrow R=\frac{10}{2\times10^{-3}}-50=5000-50=4950\,\Omega, in series.

Question 11

Physics · 4 marks · −1

A proton enters a uniform magnetic field of 0.2 T with speed 4×1054\times10^5 m/s at an angle of 60∘60^\circ to the field. The pitch of its helical path is approximately (mp=1.67×10−27m_p=1.67\times10^{-27} kg, e=1.6×10−19e=1.6\times10^{-19} C)

  1. A.6.6 cm
  2. B.11.4 cm
  3. C.3.3 cm
  4. D.1.8 cm

Correct answer: A

Explanation: Period T=2πmqB=2π×1.67×10−271.6×10−19×0.2=3.28×10−7T=\frac{2\pi m}{qB}=\frac{2\pi\times1.67\times10^{-27}}{1.6\times10^{-19}\times0.2}=3.28\times10^{-7} s. Pitch =vcos⁡60∘⋅T=2×105×3.28×10−7≈6.6×10−2=v\cos60^\circ\cdot T=2\times10^5\times3.28\times10^{-7}\approx6.6\times10^{-2} m. (11.4 cm uses vsin⁡60∘v\sin60^\circ; 1.8 cm is the radius of the helix.)

Question 12

Physics · 4 marks · −1

A square coil of side 10 cm has 50 turns and carries a current of 2 A. It is placed in a uniform magnetic field of 0.5 T. The maximum torque on the coil is x×10−2x\times10^{-2} N m. The value of xx is ______.

Correct answer: 50

Explanation: τmax=NIAB=50×2×(0.1)2×0.5=0.5\tau_{max}=NIAB=50\times2\times(0.1)^2\times0.5=0.5 N m =50×10−2=50\times10^{-2} N m.

Question 13

Physics · 4 marks · −1

Two long straight parallel wires 16 cm apart carry currents of 3 A and 5 A in opposite directions. The net magnetic field is zero at a point in their plane at a distance dd (in cm) from the 3 A wire. The value of dd is ______.

Correct answer: 24

Explanation: For antiparallel currents the fields oppose only outside the pair, and the null point lies nearer the smaller current, on its far side. 3d=5d+16⇒3d+48=5d⇒d=24\frac{3}{d}=\frac{5}{d+16}\Rightarrow3d+48=5d\Rightarrow d=24 cm.

Question 14

Physics · 4 marks · −1

A 0.5 kg mass attached to a spring of force constant 200 N/m oscillates on a smooth horizontal surface. Its time period is

  1. A.π10\frac{\pi}{10} s
  2. B.π5\frac{\pi}{5} s
  3. C.π20\frac{\pi}{20} s
  4. D.10π10\pi s

Correct answer: A

Explanation: T=2πmk=2π0.5200=2π×0.05=π10T=2\pi\sqrt{\frac mk}=2\pi\sqrt{\frac{0.5}{200}}=2\pi\times0.05=\frac{\pi}{10} s.

Question 15

Physics · 4 marks · −1

The displacement of a particle is x=5sin⁡(πt+π3)x=5\sin\left(\pi t+\frac\pi3\right) cm. The magnitude of its maximum acceleration is

  1. A.5π5\pi cm/s2^2
  2. B.25π225\pi^2 cm/s2^2
  3. C.5π25\pi^2 cm/s2^2
  4. D.π25\frac{\pi^2}{5} cm/s2^2

Correct answer: C

Explanation: amax=ω2A=π2×5=5π2a_{max}=\omega^2A=\pi^2\times5=5\pi^2 cm/s2^2.

Question 16

Physics · 4 marks · −1

A 2 kg block is attached to two springs of force constants 100 N/m and 300 N/m. The ratio of the time period when the springs are joined in series to that when they are joined in parallel is

  1. A.34\frac{\sqrt3}{4}
  2. B.43\frac{4}{\sqrt3}
  3. C.163\frac{16}{3}
  4. D.22

Correct answer: B

Explanation: ks=100×300400=75k_s=\frac{100\times300}{400}=75 N/m, kp=400k_p=400 N/m. T∝k−1/2T\propto k^{-1/2}: TsTp=40075=163=43\frac{T_s}{T_p}=\sqrt{\frac{400}{75}}=\sqrt{\frac{16}{3}}=\frac{4}{\sqrt3}.

Question 17

Physics · 4 marks · −1

A particle executes SHM of amplitude AA. When its displacement is A/2A/2, the ratio of its kinetic energy to potential energy is

  1. A.1:31:3
  2. B.1:11:1
  3. C.2:12:1
  4. D.3:13:1

Correct answer: D

Explanation: U=12kx2=18kA2U=\frac12kx^2=\frac18kA^2, total E=12kA2E=\frac12kA^2, so K=38kA2K=\frac38kA^2. K:U=3:1K:U=3:1.

Question 18

Physics · 4 marks · −1

A simple pendulum has a period of 2 s in a stationary lift. If the lift accelerates downward at g/4g/4, the new period is approximately

  1. A.1.73 s
  2. B.2.31 s
  3. C.2.83 s
  4. D.1.79 s

Correct answer: B

Explanation: Effective gravity geff=g−g4=3g4g_{eff}=g-\frac g4=\frac{3g}{4}. T′=Tggeff=2×23≈2.31T'=T\sqrt{\frac{g}{g_{eff}}}=2\times\frac{2}{\sqrt3}\approx2.31 s. (1.79 s would be for upward acceleration g/4g/4.)

Question 19

Physics · 4 marks · −1

A block of mass mm attached to a horizontal spring oscillates on a smooth floor with amplitude AA. As it passes through the equilibrium position, a lump of putty of mass mm is dropped vertically onto it and sticks. The new amplitude of oscillation is

  1. A.A2\frac A2
  2. B.A2\frac{A}{\sqrt2}
  3. C.2A\sqrt2A
  4. D.AA

Correct answer: B

Explanation: Horizontal momentum is conserved: mv=2mv′⇒v′=v2mv=2mv'\Rightarrow v'=\frac v2, where v=ωAv=\omega A. New angular frequency ω′=k2m=ω2\omega'=\sqrt{\frac{k}{2m}}=\frac{\omega}{\sqrt2}. The collision occurs at equilibrium, so A′=v′ω′=ωA/2ω/2=A2A'=\frac{v'}{\omega'}=\frac{\omega A/2}{\omega/\sqrt2}=\frac{A}{\sqrt2}.

Question 20

Physics · 4 marks · −1

A particle in SHM has speeds 8 cm/s and 6 cm/s at displacements 6 cm and 8 cm from the mean position respectively. Its amplitude (in cm) is ______.

Correct answer: 10

Explanation: v2=ω2(A2−x2)v^2=\omega^2(A^2-x^2): 64=ω2(A2−36)64=\omega^2(A^2-36) and 36=ω2(A2−64)36=\omega^2(A^2-64). Subtracting, 28=28ω2⇒ω=128=28\omega^2\Rightarrow\omega=1 rad/s; then A2=100A^2=100, A=10A=10 cm.

Question 21

Physics · 4 marks · −1

A uniform rod of length 6 m is pivoted at one end and oscillates as a compound pendulum in a vertical plane with small amplitude. Taking g=π2g=\pi^2 m/s2^2, its time period (in second) is ______.

Correct answer: 4

Explanation: I=mL23I=\frac{mL^2}{3}, distance of centre of mass from pivot =L2=\frac L2. T=2πImg(L/2)=2π2L3g=2π4π2=4T=2\pi\sqrt{\frac{I}{mg(L/2)}}=2\pi\sqrt{\frac{2L}{3g}}=2\pi\sqrt{\frac{4}{\pi^2}}=4 s.

Question 22

Physics · 4 marks · −1

A wire 2 m long with cross-sectional area 1 mm2^2 stretches by 1 mm under a load of 100 N. The Young's modulus of the material is

  1. A.2×10102\times10^{10} Pa
  2. B.1×10111\times10^{11} Pa
  3. C.2×1082\times10^{8} Pa
  4. D.2×10112\times10^{11} Pa

Correct answer: D

Explanation: Y=FLA ΔL=100×210−6×10−3=2×1011Y=\frac{FL}{A\,\Delta L}=\frac{100\times2}{10^{-6}\times10^{-3}}=2\times10^{11} Pa.

Question 23

Physics · 4 marks · −1

The absolute pressure at a depth of 10 m below the surface of a lake is (atmospheric pressure =1×105=1\times10^5 Pa, ρwater=1000\rho_{water}=1000 kg/m3^3, g=10g=10 m/s2^2)

  1. A.1×1051\times10^5 Pa
  2. B.2×1052\times10^5 Pa
  3. C.1.1×1051.1\times10^5 Pa
  4. D.3×1053\times10^5 Pa

Correct answer: B

Explanation: P=P0+ρgh=105+1000×10×10=2×105P=P_0+\rho gh=10^5+1000\times10\times10=2\times10^5 Pa.

Question 24

Physics · 4 marks · −1

Eight identical spherical raindrops, each falling with terminal speed vv, coalesce into a single drop. The terminal speed of the combined drop is

  1. A.2v2v
  2. B.8v8v
  3. C.4v4v
  4. D.16v16v

Correct answer: C

Explanation: Volume conservation: R3=8r3⇒R=2rR^3=8r^3\Rightarrow R=2r. Terminal speed (Stokes) vt=2r2(ρ−σ)g9η∝r2v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2, so v′=4vv'=4v.

Question 25

Physics · 4 marks · −1

Water rises in a clean glass capillary tube of radius 0.1 mm (angle of contact 0∘0^\circ). If the surface tension of water is 0.07 N/m, the height of rise is (g=10g=10 m/s2^2)

  1. A.14 cm
  2. B.7 cm
  3. C.1.4 cm
  4. D.28 cm

Correct answer: A

Explanation: h=2Tcos⁡θρgr=2×0.071000×10×10−4=0.14h=\frac{2T\cos\theta}{\rho gr}=\frac{2\times0.07}{1000\times10\times10^{-4}}=0.14 m =14=14 cm.

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