MHT CET · Physics · Rotational Motion
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 above the rough floor. If the carpet is unrolled without sliding to a radius ' \(R / 2\) '. The change in potential energy is ( \(\mathrm{g}=\) acceleration due to gravity)
- A MgR
- B \(\frac{7}{8} \mathrm{MgR}\)
- C \(\frac{5}{7} \mathrm{MgR}\)
- D \(\frac{3}{4} \mathrm{MgR}\)
Answer & Solution
Correct Answer
(B) \(\frac{7}{8} \mathrm{MgR}\)
Step-by-step Solution
Detailed explanation
Density ( \(\rho=M / V\) ) of carpet remains the same after it is unrolled,
\(\rho_1=\rho_2\)
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,
\(\begin{aligned}
& \Delta U=U_1-U_2 \\
& \Delta U=M g R-\left(\frac{M}{4}\right) g\left(\frac{R}{2}\right) \\
& \Delta U=\frac{7}{8} M g R
\end{aligned}\)
\(\rho_1=\rho_2\)
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,
\(\begin{aligned}
& \Delta U=U_1-U_2 \\
& \Delta U=M g R-\left(\frac{M}{4}\right) g\left(\frac{R}{2}\right) \\
& \Delta U=\frac{7}{8} M g R
\end{aligned}\)
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