A floor-mat of mass M made up of extensible material, is rolled along its length so as to form a cylinder of radius R and kept on a rough horizontal surface. If the mat is now unrolled, without sliding, to a radius `(R )/(2)`, the decrease in potential energy is

Options

A. 1. `1/2MgR`
B. 2. `7/8MgR`
C. 3. `5/8MgR`
D. 4. `3/4 MgR`

Answer

To solve the problem, we need to calculate the decrease in potential energy when a floor mat of mass \( M \) is unrolled from a cylinder of radius \( R \) to a cylinder of radius \( \frac{R}{2} \).

### Step-by-Step Solution:

1. **Calculate Initial Potential Energy (PE_initial)**:
- When the mat is rolled into a cylinder of radius \( R \), the center of mass of the mat is at a height \( R \) from the ground.
- The potential energy of the entire mat can be calculated using the formula:
\[ PE_{\text{initial}} = M \cdot g \cdot R \]

2. **Determine the Mass Distribution**:
- When the mat is unrolled to a radius of \( \frac{R}{2} \), some part of the mat will be lying flat on the ground.
- The mass distribution is uniform, so we need to find the mass that remains in the cylinder of radius \( \frac{R}{2} \).
- The area of the original cylinder is \( \pi R^2 \) and the area of the new cylinder is \( \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \).
- The mass of the unrolled part (lying flat) can be calculated as:
\[ \text{Mass of unrolled part} = M - \frac{M}{4} = \frac{3M}{4} \]
- Therefore, the mass of the mat that remains in the cylinder is:
\[ M_{\text{remaining}} = \frac{M}{4} \]

3. **Calculate Final Potential Energy (PE_final)**:
- The center of mass of the remaining mass in the cylinder of radius \( \frac{R}{2} \) is at a height of \( \frac{R}{2} \).
- The potential energy of the remaining mass can be calculated as:
\[ PE_{\text{final}} = \left(\frac{M}{4}\right) \cdot g \cdot \left(\frac{R}{2}\right) = \frac{M \cdot g \cdot R}{8} \]

4. **Calculate the Decrease in Potential Energy (ΔPE)**:
- The decrease in potential energy is given by:
\[ \Delta PE = PE_{\text{initial}} - PE_{\text{final}} \]
- Substituting the values we calculated:
\[ \Delta PE = \left(M \cdot g \cdot R\right) - \left(\frac{M \cdot g \cdot R}{8}\right) \]
- Simplifying this:
\[ \Delta PE = M \cdot g \cdot R \left(1 - \frac{1}{8}\right) = M \cdot g \cdot R \cdot \frac{7}{8} \]

Thus, the decrease in potential energy is:
\[ \Delta PE = \frac{7M \cdot g \cdot R}{8} \]