A block is released from the top of an inclined plane of inclination `theta`. Its velocity at the bottom of plane is `v`. If it slides down a rough inclined plane of the same inclination , its velocity at bottom is `v//2`. The coefficient of friction is

Options

A. `(3)/(4) cos theta`
B. `(1)/(4) cos theta`
C. `(1)/(4) tan theta`
D. `(3)/(4) tan theta`

Explanation

To solve the problem, we need to analyze the motion of a block sliding down two inclined planes: one smooth and one rough. We will derive the coefficient of friction based on the given conditions.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- A block is released from rest on a smooth inclined plane with inclination angle \( \theta \). Its velocity at the bottom is \( v \).
- When the block slides down a rough inclined plane of the same angle \( \theta \), its velocity at the bottom is \( \frac{v}{2} \).
- We need to find the coefficient of friction \( \mu \) on the rough inclined plane.

2. **Acceleration on the Smooth Inclined Plane:**
- For the smooth inclined plane, the only force acting down the incline is the component of gravitational force:
\[ a = g \sin \theta \]
- Using the equation of motion \( v^2 = u^2 + 2as \) (where \( u = 0 \)):
\[ v^2 = 0 + 2(g \sin \theta) s \implies v^2 = 2g \sin \theta \cdot s \quad (1) \]

3. **Acceleration on the Rough Inclined Plane:**
- For the rough inclined plane, the forces acting on the block include gravity and friction. The net force down the incline is:
\[ F_{\text{net}} = mg \sin \theta - F_{\text{friction}} \]
- The frictional force \( F_{\text{friction}} = \mu N = \mu mg \cos \theta \), where \( N = mg \cos \theta \).
- Therefore, the net force can be expressed as:
\[ mg \sin \theta - \mu mg \cos \theta = ma \]
- Dividing through by \( m \):
\[ g \sin \theta - \mu g \cos \theta = a \implies a = g \sin \theta - \mu g \cos \theta \quad (2) \]

4. **Using the Equation of Motion for the Rough Plane:**
- For the rough inclined plane, the final velocity is \( \frac{v}{2} \):
\[ \left(\frac{v}{2}\right)^2 = 0 + 2as \implies \frac{v^2}{4} = 2a s \]
- Substituting \( a \) from equation (2):
\[ \frac{v^2}{4} = 2(g \sin \theta - \mu g \cos \theta) s \]
- Rearranging gives:
\[ \frac{v^2}{4} = 2g \sin \theta s - 2\mu g \cos \theta s \quad (3) \]

5. **Substituting from Equation (1) into Equation (3):**
- From equation (1), we know \( v^2 = 2g \sin \theta s \). Substitute this into equation (3):
\[ \frac{1}{4}(2g \sin \theta s) = 2g \sin \theta s - 2\mu g \cos \theta s \]
- Simplifying:
\[ \frac{g \sin \theta s}{4} = 2g \sin \theta s - 2\mu g \cos \theta s \]
- Rearranging gives:
\[ 2\mu g \cos \theta s = 2g \sin \theta s - \frac{g \sin \theta s}{4} \]
- Factoring out \( g s \):
\[ 2\mu \cos \theta = 2\sin \theta - \frac{\sin \theta}{4} \]
- Combining terms:
\[ 2\mu \cos \theta = \frac{8\sin \theta}{4} - \frac{\sin \theta}{4} = \frac{7\sin \theta}{4} \]
- Thus:
\[ \mu = \frac{7\sin \theta}{8\cos \theta} = \frac{7}{8} \tan \theta \]

6. **Final Result:**
- The coefficient of friction \( \mu \) is:
\[ \mu = \frac{7}{8} \tan \theta \]