A carpet of mass ' $M$ ' made of a material is rolled along its length in the form of a cylinder of radius '…
- MgR
- $\frac{7}{8} \mathrm{MgR}$
- $\frac{5}{7} \mathrm{MgR}$
- $\frac{3}{4} \mathrm{MgR}$
Solution
If $\mathrm{M}_2$ and $\mathrm{V}_2$ are mass and volume respectively of unrolled carpet, $\begin{aligned} & \frac{\mathrm{M}}{\mathrm{~V}}=\frac{\mathrm{M}_2}{\mathrm{~V}_2} \\ & \mathrm{M}_2=\frac{\mathrm{M}}{\pi \mathrm{R}^2 l} \times \pi \mathrm{R}_2 l=\frac{\mathrm{M}}{\pi \mathrm{R}^2 l} \times \pi\left(\frac{\mathrm{R}^2}{4}\right) l \end{aligned}$ $\mathrm{M}_2=\frac{\mathrm{M}}{4}...(i)$
Potential energy of rolled carpet, $\mathrm{U}_1=\mathrm{MgR}$
Potential energy of unrolled carpet, $\mathrm{U}_2=\mathrm{M}_2 \mathrm{gR}_2=\left(\frac{\mathrm{M}}{4}\right) \mathrm{g}\left(\frac{\mathrm{R}}{2}\right)...[From(i)]$
Change in potential energy, $\begin{aligned} & \Delta U=U_1-U_2 \\ & \Delta U=M g R-\left(\frac{M}{4}\right) g\left(\frac{R}{2}\right) \\ & \Delta U=\frac{7}{8} M g R \end{aligned}$ ~
Asked in: MHT CET 2024 (04 May Shift 2)