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Physics · Rotational Dynamics

JEE Main 2026 — 24 January, Morning Shift — Question 39

Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm . When released from rest the heavier mass is observed to fall 81 cm in 9 s . The rotational inertia of the pulley is ____\_\_\_\_ kg.m2\mathrm{kg} . \mathrm{m}^{2}. ( g=9.8 m/s2\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2} )

  1. Option A:

    9.5×10−39.5 \times 10^{-3}

    Correct
  2. Option B:

    4.75×10−34.75 \times 10^{-3}

  3. Option C:

    1.86×10−21.86 \times 10^{-2}

  4. Option D:

    8.3×10−38.3 \times 10^{-3}

Answer: A

Step-by-step solution

S=ut+12at2a=2 st2=2×0.8181=0.02 m/s2 m1 g−T1=m1a T2−m2 g=m2a( T1−T2)R=I⋅aR∴a=(m1−m2)g m1+m2+IR2\begin{aligned} & \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} & \mathrm{a}=\frac{2 \mathrm{~s}}{\mathrm{t}^{2}}=\frac{2 \times 0.81}{81}=0.02 \mathrm{~m} / \mathrm{s}^{2} & \mathrm{~m}_{1} \mathrm{~g}-\mathrm{T}_{1}=\mathrm{m}_{1} \mathrm{a} & \mathrm{~T}_{2}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{2} \mathrm{a} & \left(\mathrm{~T}_{1}-\mathrm{T}_{2}\right) \mathrm{R}=\mathrm{I} \cdot \frac{\mathrm{a}}{\mathrm{R}} & \therefore \mathrm{a}=\frac{\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\frac{\mathrm{I}}{\mathrm{R}^{2}}} \end{aligned}

0.02=(400−350)(10−3)g(400+350)(10−3)+IR20.02=\frac{(400-350)\left(10^{-3}\right) \mathrm{g}}{(400+350)\left(10^{-3}\right)+\frac{\mathrm{I}}{\mathrm{R}^{2}}} IR2=50×10−3 g0.02−750×10−3=23.75\frac{\mathrm{I}}{\mathrm{R}^{2}}=\frac{50 \times 10^{-3} \mathrm{~g}}{0.02}-750 \times 10^{-3}=23.75 I=23.75×4×10−4=9.5×10−3 kg−m2\mathrm{I}=23.75 \times 4 \times 10^{-4}=9.5 \times 10^{-3} \mathrm{~kg}-\mathrm{m}^{2}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Torque, Equation of Motion and Toppling