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. `(2)/(5)MgR`
B. `(5)/(7)MgR`
C. `(7)/(8)MgR`
D. MgR

Answer

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

### Step-by-Step Solution:

1. **Understanding the Initial and Final States**:
- Initially, the mat is rolled into a cylinder of radius \( R \).
- When unrolled, it forms a cylinder of radius \( \frac{R}{2} \).

2. **Calculating the Mass Distribution**:
- The mass \( M \) of the mat is uniformly distributed along its length.
- The cross-sectional area of the cylinder formed by radius \( R \) is \( A_1 = \pi R^2 \).
- The cross-sectional area of the cylinder formed by radius \( \frac{R}{2} \) is \( A_2 = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \).
- Since the volume (and hence mass) is proportional to the cross-sectional area (assuming the length remains constant), the mass when the mat is rolled to radius \( \frac{R}{2} \) will be \( M_2 = \frac{M}{4} \).

3. **Calculating the Heights**:
- The center of mass of the cylinder when rolled to radius \( R \) is at height \( R \).
- The center of mass of the cylinder when rolled to radius \( \frac{R}{2} \) is at height \( \frac{R}{2} \).

4. **Calculating Potential Energy**:
- The initial potential energy \( PE_i \) when the mat is at radius \( R \) is given by:
\[ PE_i = M \cdot g \cdot R \]
- The final potential energy \( PE_f \) when the mat is at radius \( \frac{R}{2} \) is given by:
\[ PE_f = \frac{M}{4} \cdot g \cdot \frac{R}{2} = \frac{M \cdot g \cdot R}{8} \]

5. **Calculating the Decrease in Potential Energy**:
- The decrease in potential energy \( \Delta PE \) is given by:
\[ \Delta PE = PE_i - PE_f \]
- Substituting the values:
\[ \Delta PE = M \cdot g \cdot R - \frac{M \cdot g \cdot R}{8} \]
- Simplifying:
\[ \Delta PE = \left(1 - \frac{1}{8}\right) M \cdot g \cdot R = \frac{7}{8} M \cdot g \cdot R \]

### Final Answer:
The decrease in potential energy is \( \frac{7}{8} M g R \).

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