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JEE Main 2026 (24 Jan, Shift 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2026 (24 Jan, Shift 1)

Q9JEE Main 2026 · 24 JAN · SHIFT 1Difficult

Rotational Kinematics & Dynamics

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⋅(g=9.8 m/s2)\underline{\qquad}\ \text{kg}\cdot\text{m}^{2}\cdot\left(g = 9.8\text{ m/s}^{2}\right)

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Answer: (d)

  1. From rest: s=12at2⇒0.81=12a(81)⇒a=0.02s = \tfrac{1}{2}at^2 \Rightarrow 0.81 = \tfrac{1}{2}a(81) \Rightarrow a = 0.02 m/s2^2.
  2. a=(m1−m2)gm1+m2+I/r2⇒0.75+Ir2=0.05×9.80.02=24.5a = \dfrac{(m_1 - m_2)g}{m_1 + m_2 + I/r^2} \Rightarrow 0.75 + \dfrac{I}{r^2} = \dfrac{0.05 \times 9.8}{0.02} = 24.5
  3. Ir2=23.75⇒I=23.75×(0.02)2=9.5×10−3\dfrac{I}{r^2} = 23.75 \Rightarrow I = 23.75 \times (0.02)^2 = 9.5 \times 10^{-3} kg m2^2
  4. So the answer is (d) 9.5×10−39.5 \times 10^{-3}.
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