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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2026 · 24 January · Morning Shift · Physics · Rotational Motion

Two masses 400 g and 350 g are suspended from the ends of a light...

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} \cdot \mathrm{m}^{2}. (g=9.8 m/s2)\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)

Options

  1. A4.75×10−34.75 \times 10^{-3}
  2. B9.5×10−39.5 \times 10^{-3}
  3. C1.86×10−21.86 \times 10^{-2}
  4. D8.3×10−38.3 \times 10^{-3}

Correct answer

B. 9.5×10−39.5 \times 10^{-3}

Step-by-step solution

From kinematics with s=0.81s = 0.81 m and t=9t = 9 s: a=2s/t2=0.02a = 2s/t^2 = 0.02 m/s².
For the pulley system, applying Newton's second law to both masses and torque equation: (m1−m2)g−(m1+m2)a=Ia/R2(m_1 - m_2)g - (m_1 + m_2)a = Ia/R^2.
Substituting values: 0.05×9.8−0.75×0.02=I×500.05 \times 9.8 - 0.75 \times 0.02 = I \times 50 gives 0.49−0.015=50I0.49 - 0.015 = 50I.
Therefore I=9.5×10−3I = 9.5 \times 10^{-3} kg·m²
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