The pulley shown in figure is made using a thin rim and... | null (Physics)

JEE Mains 2026 21st January Shift 2

Physics

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.)

\\

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

\frac{(M-m) g}{M+m}

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

\frac{(M-m) g}{2 M+m}

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Correct Answer:

Explanation

\begin{align*} & \mathrm{Mg}-\mathrm{T}_{2}=\mathrm{Ma} \tag....{1}\\ & \mathrm{~T}_{1}-\mathrm{mg}=\mathrm{ma} \tag....{2}\\ & \left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right) \mathrm{r}=\mathrm{I} \frac{\mathrm{a}}{\mathrm{r}} \tag{3}\\ & (1)+(2)+(3) \\ & (\mathrm{M}-\mathrm{m}) \mathrm{g}=\left(\mathrm{M}+\mathrm{m}+\frac{\mathrm{I}}{\mathrm{r}^{2}}\right) \mathrm{a} \end{align*}\\

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]}

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