
A carpet of mass $M$, made of an extensible 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 $\dfrac{R}{2}$, the decrease in potential energy is
\[\begin{align}
& A.\dfrac{1}{2}MgR \\
& B.\dfrac{7}{8}MgR \\
& C.\dfrac{5}{8}MgR \\
& D.\dfrac{3}{4}MgR \\
\end{align}\]
Answer
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Hint: We know that potential energy is energy possessed by an object due to its position. To find the change in the potential energy, we need to calculate the individual energies of the cylindrical carpet and the carpet on the floor. Then calculating the difference between the both gives the change in the potential energy.
Formula used:
$P.E=mgh$, where \[P.E\] is the potential energy of a body with mass $m$ which is at a height $h$ from the ground and $g$ is acceleration due to gravity.
Complete step-by-step answer:
Given that the cylinder of carpet of mass $M$ and radius $R$ is rolled on the floor to give a radius $\dfrac{R}{2}$.
Let us assume that the density of the carpet is given as $\rho$ then, since the same cylinder is rolled to a radius$\dfrac{R}{2}$, the density of the carpet on the floor is also $\rho$ .
Then, equating the densities, we get $\dfrac{M}{\pi R^{2}}=\dfrac{m}{\pi \left(\dfrac{R}{2}\right)^{2}}$
Then, we see that $m=\dfrac{M}{4}$
Now potential energy is given as $P.E=mgh$, here the radius of the carpet will give the height of the carpet.
Then, the potential energy when the carpet is kept as a cylinder is given as $P.E_{1}=MgR$
Similarly, the potential energy when the carpet when unrolled on the floor is given as $P.E_{2}=mg\dfrac{R}{2}=g\cdot\dfrac{M}{4}\cdot\dfrac{R}{2}=\dfrac{MgR}{8}$
Then the difference in potential energy is given as $\Delta P.E=P.E_{2}-P.E_{1}$
Substituting, we get $\Delta P.E=\dfrac{MgR}{8}-MgR=\dfrac{7}{8}MgR$
Hence the answer is option \[B.\dfrac{7}{8}MgR\]
So, the correct answer is “Option B”.
Note: The initial potential energy is straight forward. Clearly, here the radius or the height of the carpet varies when it is rolled on the floor. But the density of the carpet is a constant. Since density depends on mass, then we can find the new mass and then the new potential energy.
Formula used:
$P.E=mgh$, where \[P.E\] is the potential energy of a body with mass $m$ which is at a height $h$ from the ground and $g$ is acceleration due to gravity.
Complete step-by-step answer:
Given that the cylinder of carpet of mass $M$ and radius $R$ is rolled on the floor to give a radius $\dfrac{R}{2}$.
Let us assume that the density of the carpet is given as $\rho$ then, since the same cylinder is rolled to a radius$\dfrac{R}{2}$, the density of the carpet on the floor is also $\rho$ .
Then, equating the densities, we get $\dfrac{M}{\pi R^{2}}=\dfrac{m}{\pi \left(\dfrac{R}{2}\right)^{2}}$
Then, we see that $m=\dfrac{M}{4}$
Now potential energy is given as $P.E=mgh$, here the radius of the carpet will give the height of the carpet.
Then, the potential energy when the carpet is kept as a cylinder is given as $P.E_{1}=MgR$
Similarly, the potential energy when the carpet when unrolled on the floor is given as $P.E_{2}=mg\dfrac{R}{2}=g\cdot\dfrac{M}{4}\cdot\dfrac{R}{2}=\dfrac{MgR}{8}$
Then the difference in potential energy is given as $\Delta P.E=P.E_{2}-P.E_{1}$
Substituting, we get $\Delta P.E=\dfrac{MgR}{8}-MgR=\dfrac{7}{8}MgR$
Hence the answer is option \[B.\dfrac{7}{8}MgR\]
So, the correct answer is “Option B”.
Note: The initial potential energy is straight forward. Clearly, here the radius or the height of the carpet varies when it is rolled on the floor. But the density of the carpet is a constant. Since density depends on mass, then we can find the new mass and then the new potential energy.
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