JEE Main Practice — Physics Set 2

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

An alternating voltage is given by v=311sin⁡(100πt)v=311\sin(100\pi t) volt. Its frequency and rms value are respectively

  1. A.50 Hz and 220 V
  2. B.100 Hz and 311 V
  3. C.50 Hz and 311 V
  4. D.100 Hz and 220 V

Correct answer: A

Explanation: ω=100π=2πf⇒f=50\omega=100\pi=2\pi f\Rightarrow f=50 Hz. Vrms=3112≈220V_{rms}=\frac{311}{\sqrt2}\approx220 V.

Question 2

Physics · 4 marks · −1

A series LCR circuit has L=0.5L=0.5 H, C=8 μC=8\ \muF and R=10 ΩR=10\,\Omega. The quality factor of the circuit at resonance is

  1. A.50
  2. B.25
  3. C.250
  4. D.12.5

Correct answer: B

Explanation: ω0=1LC=10.5×8×10−6=12×10−3=500\omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.5\times8\times10^{-6}}}=\frac{1}{2\times10^{-3}}=500 rad/s. Q=ω0LR=500×0.510=25Q=\frac{\omega_0L}{R}=\frac{500\times0.5}{10}=25.

Question 3

Physics · 4 marks · −1

A conducting rod of length 0.5 m slides with constant speed 4 m/s on two parallel frictionless rails in a uniform magnetic field of 0.6 T perpendicular to the plane of the rails. The rails are joined at one end, and the total resistance of the circuit is 2 Ω2\,\Omega. The mechanical power needed to keep the rod moving is

  1. A.0.72 W
  2. B.0.18 W
  3. C.1.2 W
  4. D.0.36 W

Correct answer: A

Explanation: ε=BLv=0.6×0.5×4=1.2\varepsilon=BLv=0.6\times0.5\times4=1.2 V; I=ε/R=0.6I=\varepsilon/R=0.6 A. Retarding force F=BIL=0.6×0.6×0.5=0.18F=BIL=0.6\times0.6\times0.5=0.18 N. Power =Fv=0.72=Fv=0.72 W (equal to ε2/R=1.44/2\varepsilon^2/R=1.44/2).

Question 4

Physics · 4 marks · −1

An inductor of 2 H and a resistor of 4 Ω4\,\Omega are connected in series to a 12 V ideal battery at t=0t=0. The energy stored in the inductor at t=0.5ln⁡2t=0.5\ln2 s is

  1. A.9 J
  2. B.4.5 J
  3. C.2.25 J
  4. D.1.125 J

Correct answer: C

Explanation: τ=L/R=0.5\tau=L/R=0.5 s and I0=12/4=3I_0=12/4=3 A. At t=τln⁡2t=\tau\ln2: I=I0(1−e−ln⁡2)=1.5I=I_0(1-e^{-\ln2})=1.5 A. Energy =12LI2=12×2×2.25=2.25=\frac12LI^2=\frac12\times2\times2.25=2.25 J.

Question 5

Physics · 4 marks · −1

A series circuit has R=30 ΩR=30\,\Omega, XL=80 ΩX_L=80\,\Omega and XC=40 ΩX_C=40\,\Omega across a 200 V (rms) supply. A capacitor is now added in series so that the power factor becomes unity, with the supply unchanged. The ratio of the new average power to the original average power is

  1. A.5/35/3
  2. B.9/259/25
  3. C.3/53/5
  4. D.25/925/9

Correct answer: D

Explanation: Originally Z=302+(80−40)2=50 ΩZ=\sqrt{30^2+(80-40)^2}=50\,\Omega, I=4I=4 A, P1=I2R=480P_1=I^2R=480 W. At unity power factor Z=R=30 ΩZ=R=30\,\Omega, I=203I=\frac{20}{3} A, P2=200230=40003P_2=\frac{200^2}{30}=\frac{4000}{3} W. P2P1=4000/3480=259\frac{P_2}{P_1}=\frac{4000/3}{480}=\frac{25}{9}. (Equivalently P∝cos⁡2ϕP\propto\cos^2\phi at fixed V,RV,R: 1/0.621/0.6^2.)

Question 6

Physics · 4 marks · −1

A step-down transformer of efficiency 80% works on a 2200 V line and delivers 8.8 kW to a load at 440 V. The current in the primary coil (in ampere) is ______.

Correct answer: 5

Explanation: Input power =8.80.8=11=\frac{8.8}{0.8}=11 kW. Primary current =110002200=5=\frac{11000}{2200}=5 A.

Question 7

Physics · 4 marks · −1

The magnetic flux linked with a coil of resistance 5 Ω5\,\Omega changes from 0 to 0.5 Wb (flux linkage, all turns included) in 0.1 s. The charge that flows through the coil (in mC) is ______.

Correct answer: 100

Explanation: Q=∫I dt=1R∫ε dt=ΔΦR=0.55=0.1Q=\int I\,dt=\frac{1}{R}\int\varepsilon\,dt=\frac{\Delta\Phi}{R}=\frac{0.5}{5}=0.1 C =100=100 mC. (Independent of how fast the flux changes.)

Question 8

Physics · 4 marks · −1

Which of the following sequences of electromagnetic radiations is in order of increasing frequency?

  1. A.Radio waves, microwaves, infrared, X-rays
  2. B.Microwaves, radio waves, X-rays, infrared
  3. C.Infrared, microwaves, radio waves, X-rays
  4. D.X-rays, infrared, microwaves, radio waves

Correct answer: A

Explanation: In order of increasing frequency (decreasing wavelength): radio << microwave << infrared << visible << UV << X-rays << γ\gamma-rays. Only option A preserves this order.

Question 9

Physics · 4 marks · −1

In a plane electromagnetic wave travelling in vacuum, the amplitude of the magnetic field is 2×10−82\times10^{-8} T. The amplitude of the electric field is

  1. A.60 V/m
  2. B.6 V/m
  3. C.0.6 V/m
  4. D.1.5 V/m

Correct answer: B

Explanation: E0=cB0=3×108×2×10−8=6E_0=cB_0=3\times10^8\times2\times10^{-8}=6 V/m.

Question 10

Physics · 4 marks · −1

A plane electromagnetic wave in vacuum has an electric field amplitude of 50 V/m. Its average intensity is approximately (ε0=8.85×10−12 C2N−1m−2\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^2N^{-1}m^{-2}})

  1. A.3.3 W/m2^2
  2. B.6.6 W/m2^2
  3. C.1.7 W/m2^2
  4. D.13.3 W/m2^2

Correct answer: A

Explanation: I=12cε0E02=12×3×108×8.85×10−12×2500≈3.3I=\frac12c\varepsilon_0E_0^2=\frac12\times3\times10^8\times8.85\times10^{-12}\times2500\approx3.3 W/m2^2. (6.6 W/m2^2 omits the factor 12\frac12 of time-averaging.)

Question 11

Physics · 4 marks · −1

The electric field of a plane electromagnetic wave in vacuum is E⃗=60sin⁡(0.5×103x−ωt)j^\vec E=60\sin\left(0.5\times10^{3}x-\omega t\right)\hat j V/m. The corresponding magnetic field is

  1. A.B⃗=2×10−7sin⁡(0.5×103x−1.5×1011t)j^\vec B=2\times10^{-7}\sin\left(0.5\times10^{3}x-1.5\times10^{11}t\right)\hat j T
  2. B.B⃗=1.8×1010sin⁡(0.5×103x−1.5×1011t)k^\vec B=1.8\times10^{10}\sin\left(0.5\times10^{3}x-1.5\times10^{11}t\right)\hat k T
  3. C.B⃗=2×10−7sin⁡(0.5×103x−1.5×1011t)k^\vec B=2\times10^{-7}\sin\left(0.5\times10^{3}x-1.5\times10^{11}t\right)\hat k T
  4. D.B⃗=2×10−7sin⁡(0.5×103x−1.5×105t)k^\vec B=2\times10^{-7}\sin\left(0.5\times10^{3}x-1.5\times10^{5}t\right)\hat k T

Correct answer: C

Explanation: ω=ck=3×108×0.5×103=1.5×1011\omega=ck=3\times10^8\times0.5\times10^3=1.5\times10^{11} rad/s. B0=E0/c=60/(3×108)=2×10−7B_0=E_0/c=60/(3\times10^8)=2\times10^{-7} T. The wave travels along +x+x, so E^×B^=i^\hat E\times\hat B=\hat i; with E^=j^\hat E=\hat j, B^=k^\hat B=\hat k (since j^×k^=i^\hat j\times\hat k=\hat i). Option A has B⃗∥E⃗\vec B\parallel\vec E, B has B0=cE0B_0=cE_0, D has the wrong ω\omega.

Question 12

Physics · 4 marks · −1

A parallel-plate capacitor has circular plates of radius 5 cm. During charging, the electric field between the plates changes at the rate 101210^{12} V m−1^{-1} s−1^{-1}. The displacement current between the plates is approximately (ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12} SI units)

  1. A.7 mA
  2. B.22 mA
  3. C.0.7 A
  4. D.70 mA

Correct answer: D

Explanation: Id=ε0AdEdt=8.85×10−12×π(0.05)2×1012=8.85×7.85×10−3≈0.0695I_d=\varepsilon_0A\frac{dE}{dt}=8.85\times10^{-12}\times\pi(0.05)^2\times10^{12}=8.85\times7.85\times10^{-3}\approx0.0695 A ≈70\approx70 mA.

Question 13

Physics · 4 marks · −1

A laser beam of power 30 mW is incident on a plane, perfectly reflecting mirror at an angle of incidence of 60∘60^\circ; the whole beam strikes the mirror. The force exerted by the beam on the mirror is

  1. A.2×10−102\times10^{-10} N
  2. B.0.5×10−100.5\times10^{-10} N
  3. C.1×10−101\times10^{-10} N
  4. D.1.73×10−101.73\times10^{-10} N

Correct answer: C

Explanation: Momentum delivered per second by the beam =P/c=P/c along the beam. On reflection only the normal component reverses, so the force (normal to the mirror) is F=2Pccos⁡θ=2×0.03×0.53×108=1×10−10F=\frac{2P}{c}\cos\theta=\frac{2\times0.03\times0.5}{3\times10^8}=1\times10^{-10} N. (cos⁡2θ\cos^2\theta applies to pressure, where the illuminated area also grows as 1/cos⁡θ1/\cos\theta; the total force on the whole beam has a single cos⁡θ\cos\theta.)

Question 14

Physics · 4 marks · −1

An electromagnetic wave of frequency 5 MHz passes from vacuum into a non-magnetic medium of relative permittivity 4. Its wavelength in the medium (in metre) is ______.

Correct answer: 30

Explanation: v=cμrεr=3×1082=1.5×108v=\frac{c}{\sqrt{\mu_r\varepsilon_r}}=\frac{3\times10^8}{2}=1.5\times10^8 m/s. Frequency is unchanged, so λ=vf=1.5×1085×106=30\lambda=\frac{v}{f}=\frac{1.5\times10^8}{5\times10^6}=30 m.

Question 15

Physics · 4 marks · −1

A plane electromagnetic wave of intensity 90 W/m2^2 travels in vacuum. The average total energy density of the wave, in nJ/m3^3, is ______. (c=3×108c=3\times10^8 m/s)

Correct answer: 300

Explanation: I=uavc⇒uav=903×108=3×10−7I=u_{av}c\Rightarrow u_{av}=\frac{90}{3\times10^8}=3\times10^{-7} J/m3^3 =300=300 nJ/m3^3.

Question 16

Physics · 4 marks · −1

Two point charges repel each other with a force FF. If each charge is doubled and the distance between them is halved, the new force is

  1. A.4F4F
  2. B.8F8F
  3. C.16F16F
  4. D.FF

Correct answer: C

Explanation: F∝q1q2r2F\propto\frac{q_1q_2}{r^2}: F′=F×2×2(1/2)2=16FF'=F\times\frac{2\times2}{(1/2)^2}=16F.

Question 17

Physics · 4 marks · −1

The electrostatic potential energy of a system of two point charges +2 μ+2\ \muC and −3 μ-3\ \muC separated by 30 cm in air is (14πε0=9×109\frac{1}{4\pi\varepsilon_0}=9\times10^9 SI units)

  1. A.+0.18+0.18 J
  2. B.−0.18-0.18 J
  3. C.−0.54-0.54 J
  4. D.−1.8-1.8 J

Correct answer: B

Explanation: U=kq1q2r=9×109×(2×10−6)(−3×10−6)0.3=−0.18U=\frac{kq_1q_2}{r}=\frac{9\times10^9\times(2\times10^{-6})(-3\times10^{-6})}{0.3}=-0.18 J.

Question 18

Physics · 4 marks · −1

A parallel-plate air capacitor has capacitance C0C_0. A slab of dielectric constant 4 and thickness equal to half the plate separation is inserted parallel to the plates. The new capacitance is

  1. A.8C05\frac{8C_0}{5}
  2. B.5C02\frac{5C_0}{2}
  3. C.4C04C_0
  4. D.5C08\frac{5C_0}{8}

Correct answer: A

Explanation: The air layer (d/2d/2) and slab (d/2d/2, K=4K=4) are in series: C=ε0Ad2+d2K=ε0Ad⋅112+18=85C0C=\frac{\varepsilon_0A}{\frac d2+\frac{d}{2K}}=\frac{\varepsilon_0A}{d}\cdot\frac{1}{\frac12+\frac18}=\frac{8}{5}C_0.

Question 19

Physics · 4 marks · −1

A point charge qq is placed at one corner of a cube. The total electric flux through all six faces of the cube is

  1. A.q24ε0\frac{q}{24\varepsilon_0}
  2. B.qε0\frac{q}{\varepsilon_0}
  3. C.q6ε0\frac{q}{6\varepsilon_0}
  4. D.q8ε0\frac{q}{8\varepsilon_0}

Correct answer: D

Explanation: Eight identical cubes meeting at the corner would enclose the charge completely, so this cube receives 18\frac18 of the total flux qε0\frac{q}{\varepsilon_0}, i.e. q8ε0\frac{q}{8\varepsilon_0}. (The three faces touching the corner get zero flux; each of the other three gets q24ε0\frac{q}{24\varepsilon_0}.)

Question 20

Physics · 4 marks · −1

A 2 μ2\ \muF capacitor charged to 100 V is connected across an uncharged 3 μ3\ \muF capacitor. The energy lost in the process is

  1. A.10 mJ
  2. B.4 mJ
  3. C.6 mJ
  4. D.2.4 mJ

Correct answer: C

Explanation: Initial energy =12(2×10−6)(100)2=10=\frac12(2\times10^{-6})(100)^2=10 mJ. Common potential =2005=40=\frac{200}{5}=40 V; final energy =12(5×10−6)(40)2=4=\frac12(5\times10^{-6})(40)^2=4 mJ. Loss =6=6 mJ =12C1C2C1+C2V2=\frac12\frac{C_1C_2}{C_1+C_2}V^2.

Question 21

Physics · 4 marks · −1

A thin fixed ring of radius RR carries a uniformly distributed charge Q>0Q>0. A particle of mass mm and charge −q-q is placed on the axis of the ring at a small distance from its centre and released. The time period of its small oscillations is

  1. A.2π4πε0mR2Qq2\pi\sqrt{\frac{4\pi\varepsilon_0 mR^2}{Qq}}
  2. B.2π4πε0mR3Qq2\pi\sqrt{\frac{4\pi\varepsilon_0 mR^3}{Qq}}
  3. C.2π2πε0mR3Qq2\pi\sqrt{\frac{2\pi\varepsilon_0 mR^3}{Qq}}
  4. D.2πε0mR3Qq2\pi\sqrt{\frac{\varepsilon_0 mR^3}{Qq}}

Correct answer: B

Explanation: On the axis E=14πε0Qx(R2+x2)3/2≈Qx4πε0R3E=\frac{1}{4\pi\varepsilon_0}\frac{Qx}{(R^2+x^2)^{3/2}}\approx\frac{Qx}{4\pi\varepsilon_0R^3} for x≪Rx\ll R. Force on −q-q: F=−Qq4πε0R3xF=-\frac{Qq}{4\pi\varepsilon_0R^3}x (restoring). ω2=Qq4πε0mR3\omega^2=\frac{Qq}{4\pi\varepsilon_0mR^3}, so T=2π4πε0mR3QqT=2\pi\sqrt{\frac{4\pi\varepsilon_0mR^3}{Qq}}.

Question 22

Physics · 4 marks · −1

An electric dipole of moment 4×10−94\times10^{-9} C m is placed in a uniform electric field of 5×1045\times10^{4} N/C. The work required to rotate it from the position of stable equilibrium to that of unstable equilibrium, in μ\muJ, is ______.

Correct answer: 400

Explanation: W=pE(cos⁡0∘−cos⁡180∘)=2pE=2×4×10−9×5×104=4×10−4W=pE(\cos0^\circ-\cos180^\circ)=2pE=2\times4\times10^{-9}\times5\times10^4=4\times10^{-4} J =400 μ=400\ \muJ.

Question 23

Physics · 4 marks · −1

Two isolated conducting spheres of radii 3 cm and 6 cm carry charges 12 nC and 6 nC respectively and are far apart. They are now connected by a thin conducting wire. The common potential of the spheres (in volt) is ______. (14πε0=9×109\frac{1}{4\pi\varepsilon_0}=9\times10^9 SI units)

Correct answer: 1800

Explanation: Total charge 18 nC is shared so that potentials are equal; for well-separated spheres V=kQtotr1+r2=9×109×18×10−90.09=1800V=\frac{kQ_{tot}}{r_1+r_2}=\frac{9\times10^9\times18\times10^{-9}}{0.09}=1800 V. (Charges become 6 nC and 12 nC, each giving V=kq/r=1800V=kq/r=1800 V.)

Question 24

Physics · 4 marks · −1

If gg is the acceleration due to gravity on the Earth's surface, its value at a height equal to half the Earth's radius above the surface is

  1. A.g2\frac{g}{2}
  2. B.4g9\frac{4g}{9}
  3. C.2g3\frac{2g}{3}
  4. D.g4\frac{g}{4}

Correct answer: B

Explanation: gh=g(1+h/R)2=g(3/2)2=4g9g_h=\frac{g}{(1+h/R)^2}=\frac{g}{(3/2)^2}=\frac{4g}{9}.

Question 25

Physics · 4 marks · −1

A planet has 8 times the mass and twice the radius of the Earth. If the escape speed from the Earth is 11.2 km/s, the escape speed from the planet is

  1. A.11.2 km/s
  2. B.44.8 km/s
  3. C.22.4 km/s
  4. D.5.6 km/s

Correct answer: C

Explanation: ve=2GMR∝MR=82=2v_e=\sqrt{\frac{2GM}{R}}\propto\sqrt{\frac MR}=\sqrt{\frac82}=2. So v=22.4v=22.4 km/s.

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