JEE Main Practice — Physics Set 7

JEE Main · 2026 · with answers and explanations

questions
18
Total marks
72
Duration
44 minutes
Negative marking
−1

Subjects: Physics

Question 1

Physics · 4 marks · −1

Unpolarised light of intensity 64 W/m2^2 passes through two ideal polarisers whose transmission axes are at 60∘60^\circ to each other. The intensity of the emergent light (in W/m2^2) is ______.

Correct answer: 8

Explanation: After the first polariser: 642=32\frac{64}{2}=32 W/m2^2. Malus's law: 32cos⁡260∘=32×14=832\cos^260^\circ=32\times\frac14=8 W/m2^2.

Question 2

Physics · 4 marks · −1

A Young's double-slit apparatus with slit separation 0.6 mm and screen distance 1 m is illuminated simultaneously by wavelengths 450 nm and 600 nm. The minimum distance from the central maximum (in mm) at which bright fringes of the two wavelengths coincide is ______.

Correct answer: 3

Explanation: Need n1(450)=n2(600)n_1(450)=n_2(600) with smallest integers: n1=4n_1=4, n2=3n_2=3 (path difference 1800 nm). y=n1λ1Dd=1800×10−9×10.6×10−3=3×10−3y=\frac{n_1\lambda_1D}{d}=\frac{1800\times10^{-9}\times1}{0.6\times10^{-3}}=3\times10^{-3} m =3=3 mm.

Question 3

Physics · 4 marks · −1

A transverse wave on a string is described by y=0.05sin⁡[2π(10t−0.5x)]y=0.05\sin\left[2\pi(10t-0.5x)\right] with xx, yy in metre and tt in second. The speed of the wave is

  1. A.5 m/s
  2. B.20 m/s
  3. C.0.05 m/s
  4. D.10 m/s

Correct answer: B

Explanation: ω=20π\omega=20\pi rad/s and k=πk=\pi rad/m. v=ωk=20v=\frac\omega k=20 m/s (equivalently fλ=10×2f\lambda=10\times2).

Question 4

Physics · 4 marks · −1

The speed of sound in air at 27∘27^\circC is 340 m/s. Its speed at 127∘127^\circC is approximately

  1. A.453 m/s
  2. B.340 m/s
  3. C.393 m/s
  4. D.294 m/s

Correct answer: C

Explanation: v∝Tv\propto\sqrt T: v=340400300=340×1.155≈393v=340\sqrt{\frac{400}{300}}=340\times1.155\approx393 m/s.

Question 5

Physics · 4 marks · −1

A string 1 m long, fixed at both ends, has linear mass density 0.01 kg/m and tension 100 N. The frequency of its third harmonic is

  1. A.150 Hz
  2. B.50 Hz
  3. C.100 Hz
  4. D.300 Hz

Correct answer: A

Explanation: v=T/μ=104=100v=\sqrt{T/\mu}=\sqrt{10^4}=100 m/s. fn=nv2Lf_n=\frac{nv}{2L}: f3=3×1002=150f_3=\frac{3\times100}{2}=150 Hz.

Question 6

Physics · 4 marks · −1

The fundamental frequency of an open organ pipe is 300 Hz. A pipe of the same length closed at one end will have its first overtone at

  1. A.600 Hz
  2. B.150 Hz
  3. C.300 Hz
  4. D.450 Hz

Correct answer: D

Explanation: Open pipe: v2L=300⇒v4L=150\frac{v}{2L}=300\Rightarrow\frac{v}{4L}=150 Hz (closed pipe fundamental). The closed pipe has only odd harmonics; first overtone =3×150=450=3\times150=450 Hz.

Question 7

Physics · 4 marks · −1

A tuning fork A of frequency 256 Hz produces 4 beats per second with fork B. When B is loaded with a little wax, the beat frequency becomes 2 per second. The original frequency of B was

  1. A.252 Hz
  2. B.258 Hz
  3. C.260 Hz
  4. D.254 Hz

Correct answer: C

Explanation: fB=256±4f_B=256\pm4, i.e. 252 or 260 Hz. Loading with wax lowers fBf_B. If fB=260f_B=260, lowering it towards 256 reduces beats (e.g. to 2 at 258 Hz) — consistent. If fB=252f_B=252, lowering would increase beats. So fB=260f_B=260 Hz.

Question 8

Physics · 4 marks · −1

A car moves at 34 m/s towards a large vertical wall, sounding a horn of frequency 500 Hz. The speed of sound is 340 m/s. The frequency of the echo heard by the driver is approximately

  1. A.611 Hz
  2. B.556 Hz
  3. C.550 Hz
  4. D.409 Hz

Correct answer: A

Explanation: Frequency received by the wall (source approaching): f1=500340340−34f_1=500\frac{340}{340-34}. The wall re-emits f1f_1 as a stationary source; the driver approaches it: f2=f1340+34340=500×374306≈611f_2=f_1\frac{340+34}{340}=500\times\frac{374}{306}\approx611 Hz.

Question 9

Physics · 4 marks · −1

A string 60 cm long, fixed at both ends, vibrates in three loops at a frequency of 150 Hz. The speed of transverse waves on the string (in m/s) is ______.

Correct answer: 60

Explanation: Three loops: 3⋅λ2=0.63\cdot\frac\lambda2=0.6 m ⇒λ=0.4\Rightarrow\lambda=0.4 m. v=fλ=150×0.4=60v=f\lambda=150\times0.4=60 m/s.

Question 10

Physics · 4 marks · −1

In a resonance-tube experiment with a tuning fork of 500 Hz, the first and second resonances occur at air-column lengths of 16 cm and 50 cm. The speed of sound in air (in m/s) is ______.

Correct answer: 340

Explanation: Successive resonances differ by λ2\frac\lambda2: λ=2(50−16)=68\lambda=2(50-16)=68 cm. v=500×0.68=340v=500\times0.68=340 m/s. (End correction =50−3×162=1=\frac{50-3\times16}{2}=1 cm, which cancels in the difference.)

Question 11

Physics · 4 marks · −1

A force F=3x2F=3x^2 N acts on a particle along the xx-axis. The work done by the force as the particle moves from x=0x=0 to x=2x=2 m is

  1. A.12 J
  2. B.6 J
  3. C.4 J
  4. D.8 J

Correct answer: D

Explanation: W=∫023x2 dx=[x3]02=8W=\int_0^2 3x^2\,dx=\left[x^3\right]_0^2=8 J.

Question 12

Physics · 4 marks · −1

A pump lifts 600 kg of water per minute to a height of 20 m. The minimum power of the pump is (g=10g=10 m/s2^2)

  1. A.2 kW
  2. B.120 kW
  3. C.1 kW
  4. D.20 kW

Correct answer: A

Explanation: P=mght=600×10×2060=2000P=\frac{mgh}{t}=\frac{600\times10\times20}{60}=2000 W =2=2 kW.

Question 13

Physics · 4 marks · −1

A 2 kg block moving at 4 m/s on a smooth horizontal surface hits a light spring of force constant 800 N/m fixed at its other end. The maximum compression of the spring is

  1. A.0.1 m
  2. B.0.4 m
  3. C.0.2 m
  4. D.0.02 m

Correct answer: C

Explanation: 12mv2=12kx2⇒x=vmk=42800=4×0.05=0.2\frac12mv^2=\frac12kx^2\Rightarrow x=v\sqrt{\frac mk}=4\sqrt{\frac{2}{800}}=4\times0.05=0.2 m.

Question 14

Physics · 4 marks · −1

A 0.5 kg stone tied to a string of length 0.4 m is whirled in a vertical circle with just enough speed at the lowest point to complete the circle. The tension in the string at the lowest point is (g=10g=10 m/s2^2)

  1. A.5 N
  2. B.30 N
  3. C.25 N
  4. D.15 N

Correct answer: B

Explanation: Minimum speed at the bottom: v2=5gRv^2=5gR. T=mg+mv2R=mg+5mg=6mg=6×0.5×10=30T=mg+\frac{mv^2}{R}=mg+5mg=6mg=6\times0.5\times10=30 N.

Question 15

Physics · 4 marks · −1

A 2 kg ball moving at 6 m/s collides head-on elastically with a stationary 4 kg ball. The fraction of the initial kinetic energy transferred to the 4 kg ball is

  1. A.89\frac89
  2. B.49\frac49
  3. C.13\frac13
  4. D.23\frac23

Correct answer: A

Explanation: v2=2m1m1+m2u=46×6=4v_2=\frac{2m_1}{m_1+m_2}u=\frac{4}{6}\times6=4 m/s. KE transferred =12(4)(16)=32=\frac12(4)(16)=32 J out of 12(2)(36)=36\frac12(2)(36)=36 J: fraction 89\frac{8}{9} (=4m1m2(m1+m2)2=\frac{4m_1m_2}{(m_1+m_2)^2}).

Question 16

Physics · 4 marks · −1

A bullet of mass 20 g moving horizontally at 300 m/s embeds itself in a 1.98 kg block attached to a horizontal spring of force constant 900 N/m on a smooth floor. The maximum compression of the spring is approximately

  1. A.10 cm
  2. B.14.1 cm
  3. C.20 cm
  4. D.7.1 cm

Correct answer: B

Explanation: Collision (momentum only): V=0.02×3002=3V=\frac{0.02\times300}{2}=3 m/s. Then energy: 12(2)(3)2=12(900)x2⇒x=32900=210\frac12(2)(3)^2=\frac12(900)x^2\Rightarrow x=3\sqrt{\frac{2}{900}}=\frac{\sqrt2}{10} m ≈14.1\approx14.1 cm. (Using the bullet's full KE of 900 J would be wrong — 891 J is lost in the inelastic collision.)

Question 17

Physics · 4 marks · −1

A 2 kg particle moves along the xx-axis in a potential U(x)=x2−4x+6U(x)=x^2-4x+6 (J, with xx in metre). If its total mechanical energy is 11 J, its maximum speed (in m/s) is ______.

Correct answer: 3

Explanation: UU is minimum at x=2x=2: Umin=4−8+6=2U_{min}=4-8+6=2 J. Kmax=11−2=9K_{max}=11-2=9 J =12(2)v2⇒v=3=\frac12(2)v^2\Rightarrow v=3 m/s.

Question 18

Physics · 4 marks · −1

A 5 kg block starts from rest and is pushed 10 m along a rough horizontal floor (μk=0.2\mu_k=0.2) by a constant horizontal force of 30 N. Its kinetic energy (in joule) at the end of the 10 m is ______. (g=10g=10 m/s2^2)

Correct answer: 200

Explanation: Friction =0.2×5×10=10=0.2\times5\times10=10 N. Work–energy theorem: K=(30−10)×10=200K=(30-10)\times10=200 J.

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