JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

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
$9.5 \times 1 0^{- 3}$
B
$4.75 \times 1 0^{- 3}$
C
$1.86 \times 1 0^{- 2}$
D
$8.3 \times 1 0^{- 3}$

Solution

From kinematics: $s = \frac{1}{2} a t^{2} \Rightarrow a = \frac{2 \times 0.81}{81} = 0.02$ m/s $^{2}$ .

For the Atwood system with a massive pulley (moment of inertia $I$ , radius $R$ , angular acceleration $α = a / R$ ):

$m_{1} g - T_{1} = m_{1} a, T_{2} - m_{2} g = m_{2} a, (T_{1} - T_{2}) R = I \frac{a}{R}$ .

Solving for $a$ :

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

Substituting $m_{1} = 0.400$ , $m_{2} = 0.350$ kg, $a = 0.02$ m/s $^{2}$ , $R = 0.02$ m:

$0.02 = \frac{0.05 \times 9.8}{0.75 + I / R ^{2}} \Rightarrow 0.75 + \frac{I}{R ^{2}} = \frac{0.49}{0.02} = 24.5$ .

$\frac{I}{R ^{2}} = 23.75 \Rightarrow I = 23.75 \times (0.02)^{2} = 23.75 \times 4 \times 1 0^{- 4} = 9.5 \times 1 0^{- 3}$ kg $\cdot$ m $^{2}$ .

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