JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

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
$\frac{( M - m ) g}{[ ( \frac{8}{3} ) M + m ]}$
D
$\frac{( M - m ) g}{2 M + m}$

Solution

The pulley's moment of inertia about its centre is the rim's plus the two rods'. Let $r$ be the rim radius, so each rod has length $2 r$ .

Rim (thin ring): $I_{r im} = M r^{2}$ . Each rod about its centre (perpendicular axis): $I_{r o d} = \frac{1}{12} M (2 r)^{2} = \frac{M r ^{2}}{3}$ . Two rods: $\frac{2 M r ^{2}}{3}$ .

Total: $I = M r^{2} + \frac{2 M r ^{2}}{3} = \frac{5 M r ^{2}}{3}$ .

Newton's second law on the blocks and torque equation on the pulley (with $a = r α$ ):

$M g - T_{2} = M a$ , $T_{1} - m g = ma$ , $(T_{2} - T_{1}) r = I \frac{a}{r}$ .

Adding the three: $(M - m) g = (M + m + \frac{I}{r ^{2}}) a = (M + m + \frac{5 M}{3}) a = (\frac{8 M}{3} + m) a$ .

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

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