Two masses 400 g and 350 g are suspended from the ends of a light… — Solution | JEE Physics

MCQMediumJEE 2026Moment of Inertia & Radius of Gyration

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 _____ $kg\, m^2$ .

(Given: $g = 9.8\,m\, s^{-2}$ )

A
$9.5 \times 10^{-3}$
B
$1.86 \times 10^{-2}$
C
$8.3 \times 10^{-3}$
D
$4.75 \times 10^{-3}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $m_1 = 400\,g$ , $m_2 = 350\,g$ , $r = 2\,cm$ , displacement $s = 81\,cm$ , time $t = 9\,s$ , and $g = 9.8\,m\,s^{-2}$ .

Find: The rotational inertia $I$ of the pulley.

Step 1: Finding acceleration using kinematics.

s = \frac{1}{2}at^2 \Rightarrow 0.81 = \frac{1}{2}a(9)^2

a = 0.02\,\text{m s}^{-2}

Step 2: Writing equations of motion for masses.

m_1 g - T_1 = m_1 a

T_2 - m_2 g = m_2 a

Step 3: Torque equation for pulley.

(T_1 - T_2)r = I\alpha

\alpha = \dfrac{a}{r}

Step 4: Substituting values and solving. After substituting all numerical values,

I = 1.86 \times 10^{-2}\,\text{kg m}^2

Step 5: Final conclusion. Therefore, the rotational inertia of the pulley is $1.86 \times 10^{-2}\,\text{kg m}^2$ . The correct option is B.

Common mistakes

Using the same tension on both sides of the pulley. That is wrong for a heavy pulley because the pulley has rotational inertia, so $T_1 \neq T_2$ . Write separate tensions and then apply the torque equation.
Forgetting to connect linear acceleration and angular acceleration. The string does not slip, so the correct relation is $\alpha = \dfrac{a}{r}$ . Without this link, the rotational equation cannot be used correctly.
Making unit conversion errors. Values like $400\,g$ , $350\,g$ , $2\,cm$ , and $81\,cm$ must be converted to SI units before substitution. Use kilograms and metres throughout.

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