Q 11-06-061JEE MainJEE Main 2026 (24 Jan, Shift 1)Medium
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 ______ $\text{kg}\cdot\text{m}^2$.
$(\text{g} = 9.8\ \text{m/s}^2)$
Answer: (D) $9.5\times10^{-3}$
From $s = \frac12at^2$: $a = \dfrac{2\times0.81}{81} = 0.02\ \text{m/s}^2$.
For an Atwood machine with a pulley of moment of inertia $I$:
$$a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/r^2}\Rightarrow 0.02 = \frac{0.05\times9.8}{0.75 + I/r^2}$$
$$0.75 + \frac{I}{r^2} = 24.5\Rightarrow\frac I{r^2} = 23.75\ \text{kg}$$
$$I = 23.75\times(0.02)^2 = 9.5\times10^{-3}\ \text{kg·m}^2$$
Solution by Sreeraj P, M.Sc Physics