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Physics · Rotational Dynamics

JEE Main 2026 — 21 January, Evening Shift — Question 31

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.)

Question figure
  1. Option A:

    (M−m)g[(136)M+m]\frac{(M-m) g}{\left[\left(\frac{13}{6}\right) M+m\right]}

  2. Option B:

    (M−m)gM+m\frac{(M-m) g}{M+m}

  3. Option C:

    (M−m)g[(83)M+m]\frac{(M-m) g}{\left[\left(\frac{8}{3}\right) M+m\right]}

    Correct
  4. Option D:

    (M−m)g2M+m\frac{(M-m) g}{2 M+m}

Answer: C

Step-by-step solution

\mathrm{Mg} - T_2 &= Ma \\ T_1 - \mathrm{mg} &= ma \\ (T_2 - T_1) r &= I \frac{a}{r} \\ (1) + (2) + (3) &: \quad (\mathrm{M} - \mathrm{m}) \mathrm{g} = \left( \mathrm{M} + \mathrm{m} + \frac{\mathrm{I}}{r^2} \right) a \end{aligned}$$ Here $I=M r^{2}+\frac{M \times(2 r)^{2}}{12} \times 2$ $=\left(1+\frac{2}{3}\right) \mathrm{Mr}^{2}$ $=\frac{5}{3} \mathrm{Mr}^{2}$ $(M-m) g=\left[M+m+\frac{5 M}{3}\right] a$ $a=\frac{(M-m) g}{\left[M+m+\frac{5 M}{3}\right]}$
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Evening Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Moment of Inertia

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