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Friday, 9 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 3 August 2021, Shift 2, Physics Q5

Question 5 of 90 in this shift, Physics question 5 of 30, Section A.

One end of a metal wire is fixed at the centre of a uniform disc of radius 4.0 cm and mass 100 g. The other end of the wire is fixed with a clamp. The hanging disc is rotated about the wire through a small angle and is released. If the disc makes torsional oscillations with time period 0.20 s, the torsional constant of wire is (Given π2=10\pi^2 = 10)

The figure, in words

Disc hanging horizontally from a ceiling by a metal wire attached at its centre; a curved double arrow shows rotation about the wire.
  1. (1)4×10−2 kg m2s−24\times10^{-2}\ \mathrm{kg\,m^2 s^{-2}}
  2. (2)8×10−2 kg m2s−28\times10^{-2}\ \mathrm{kg\,m^2 s^{-2}}Official answer
  3. (3)1.2×10−2 kg m2s−21.2\times10^{-2}\ \mathrm{kg\,m^2 s^{-2}}
  4. (4)8×10−1 kg m2s−28\times10^{-1}\ \mathrm{kg\,m^2 s^{-2}}

Official answer

Option 2

NTA final key.

Chapter
Oscillations
Topic
SHM Equations
Idea tested
Torsional Pendulum Time Period

Same topic in other shifts

All SHM Equations questions
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  6. 2 Apr 2025, Shift 1 · Q33A particle is subjected to two simple harmonic motions as : x1=7sin⁡5tx_1=\sqrt{7}\sin5t cm and x2=27sin⁡(5t+π3)x_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.