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 ___ $$kg.m^{2}$$.$$(g=9.8 m/s^{2})$$

We are asked to find the rotational inertia of a heavy pulley given that a 400 g mass falls 81 cm in 9 s when connected to a 350 g mass via a string over the pulley.

Here, $$m_1 = 400\text{ g} = 0.4\text{ kg}$$, $$m_2 = 350\text{ g} = 0.35\text{ kg}$$, the pulley radius is $$R = 2\text{ cm} = 0.02\text{ m}$$, the distance fallen is $$s = 81\text{ cm} = 0.81\text{ m}$$ in time $$t = 9\text{ s}$$, and $$g = 9.8\text{ m/s}^2$$.

Starting from rest, the acceleration can be found from the kinematic relation $$s = \tfrac12 a t^2$$, which gives

$$0.81 = \frac{1}{2} \times a \times 81$$ $$a = \frac{2 \times 0.81}{81} = 0.02\text{ m/s}^2.$$

Applying Newton’s second law to each mass and the torque equation for the pulley yields

For mass $$m_1$$ (heavier, descending): $$m_1 g - T_1 = m_1 a \quad (1)$$

For mass $$m_2$$ (lighter, ascending): $$T_2 - m_2 g = m_2 a \quad (2)$$

For the pulley (with angular acceleration $$\alpha = a/R$$): $$(T_1 - T_2)R = I \alpha = I \frac{a}{R} \quad (3)$$

From equation (1), $$T_1 = m_1(g - a) = 0.4(9.8 - 0.02) = 0.4 \times 9.78 = 3.912\text{ N}.$$

From equation (2), $$T_2 = m_2(g + a) = 0.35(9.8 + 0.02) = 0.35 \times 9.82 = 3.437\text{ N}.$$

Substituting these into equation (3) and solving for $$I$$ gives

$$I = \frac{(T_1 - T_2)R^2}{a} = \frac{(3.912 - 3.437)\times(0.02)^2}{0.02} = \frac{0.475 \times 0.0004}{0.02} = \frac{0.00019}{0.02} = 0.0095\text{ kg·m}^2 = 9.5 \times 10^{-3}\text{ kg·m}^2.$$

Therefore, the rotational inertia of the pulley is Option 1: $$9.5 \times 10^{-3}\text{ kg·m}^2.$$