JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Rotational Mechanics

WhatsApp Email

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}

)

(A) $9.5 \times 1 0^{- 3}$ ✓ Correct
(B) $4.75 \times 1 0^{- 3}$
(C) $1.86 \times 1 0^{- 2}$
(D) $8.3 \times 1 0^{- 3}$

Answer: (A)

Solution

From

s = \frac{1}{2} a t^{2}

with

s = 0.81

m and

t = 9

a = \frac{2 \times 0.81}{81} = 0.02 m/s^{2}

. For two masses over a pulley of moment of inertia

I

and radius

R

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2} + I / R ^{2}}

. Substituting

m_{1} = 0.400

kg,

m_{2} = 0.350

kg and

g = 9.8 m/s^{2}

0.02 = \frac{0.05 \times 9.8}{0.75 + I / R ^{2}}

, so

0.75 + \frac{I}{R ^{2}} = 24.5

and

\frac{I}{R ^{2}} = 23.75

. With

R = 0.02

I = 23.75 \times (0.02)^{2} = 9.5 \times 1 0^{- 3} kg\cdotm^{2}

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Rotational Mechanics →

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics