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(B) The density $\rho = M/V$ of the carpet remains constant.Let $M_1 = M$ and $R_1 = R$. The volume $V = \pi R^2 l$,where $l$ is the length of the car

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$\frac{7}{8} MgR$

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(B) The density $ho = M/V$ of the carpet remains constant.Let $M_1 = M$ and $R_1 = R$. The volume $V = \pi R^2 l$,where $l$ is the length of the carpet.When unrolled to radius $R_2 = R/2$,the mass $M_2$ of the carpet remaining in the rolled form is proportional to the volume of the cylinder.$M_2 = \frac{M}{\pi R^2 l} 	imes \pi (R/2)^2 l = \frac{M}{4}$.The potential energy of the initial rolled carpet (assuming center of mass at height $R$) is $U_1 = MgR$.The potential energy of the final rolled carpet (with radius $R_2 = R/2$ and mass $M_2 = M/4$) is $U_2 = M_2 g R_2 = (M/4) g (R/2) = \frac{1}{8} MgR$.The change in potential energy is $\Delta U = U_1 - U_2 = MgR - \frac{1}{8} MgR = \frac{7}{8} MgR$.

$A$ carpet of mass $M$ made of a material is rolled along its length in the form of a cylinder of radius $R$ and kept on a rough floor. If the carpet is unrolled without sliding to a radius $R/2$,the change in potential energy is ($g=$ acceleration due to gravity).

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