Rotational Motion

JEE Main level, single correct, medium

100 of 280

Question 100 of 280

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)

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Answer: (B) 9.5×10−39.5 \times 10^{-3}

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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