JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Rotational Mechanics

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The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M . Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is

____

(assume no slipping of string on pulley.)

(A) $\frac{( M - m ) g}{[ ( \frac{13}{6} ) M + m ]}$
(B) $\frac{( M - m ) g}{M + m}$
(C) $\frac{( M - m ) g}{[ ( \frac{8}{3} ) M + m ]}$ ✓ Correct
(D) $\frac{( M - m ) g}{2 M + m}$

Answer: (C)

Solution

Apply Newton's second law to both blocks and the pulley. For block M:

M g - T_{2} = M a

. For block m:

T_{1} - m g = ma

. For pulley:

(T_{2} - T_{1}) r = I α = I \frac{a}{r}

. The moment of inertia of the pulley (rim + two rods):

I = M r^{2} + 2 \times \frac{M ( 2 r ) ^{2}}{12} = M r^{2} + \frac{2 M r ^{2}}{3} = \frac{5 M r ^{2}}{3}

. Adding the three equations:

(M - m) g = (M + m + \frac{I}{r ^{2}}) a = (M + m + \frac{5 M}{3}) a = (\frac{8 M}{3} + m) a

. Therefore

a = \frac{( M - m ) g}{\frac{8 M}{3} + m}

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Rotational Mechanics

Solution

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