An object is placed on the surface of a smooth inclined plane of inclination . It takes time t to reach the bottom of the inclined plane. If the same object is allowed to slide down rough inclined plane of same inclination , it takes time nt to reach the bottom where n is a number greater than 1. The coefficient of friction  is given by -

Options

A. `mu=tantheta(1-1/n^(2))`
B. `mu=cottheta(1-1/n^(2))`
C. `mu=tantheta(1-1/n^(2))^(1//2)`
D. `mu=cottheta(1-1/n^(2))^(1//2)`

Answer

To solve the problem, we need to analyze the motion of an object on both a smooth inclined plane and a rough inclined plane. We will derive the coefficient of friction (μ) based on the given conditions.

### Step-by-Step Solution:

1. **Understanding the Motion on the Smooth Inclined Plane:**
- For the smooth inclined plane, the only force acting on the object along the incline is the component of gravitational force, which is \( mg \sin \theta \).
- Using Newton's second law, we have:
\[ ma = mg \sin \theta \]
This simplifies to:
\[ a = g \sin \theta \]

2. **Calculating the Distance Traveled:**
- The distance \( s \) traveled by the object in time \( t \) is given by the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ s = \frac{1}{2} g \sin \theta \cdot t^2 \]

3. **Understanding the Motion on the Rough Inclined Plane:**
- For the rough inclined plane, the forces acting on the object are \( mg \sin \theta \) down the incline and friction \( \mu mg \cos \theta \) up the incline.
- Applying Newton's second law:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
This simplifies to:
\[ a = g \sin \theta - \mu g \cos \theta \]

4. **Calculating the Distance Traveled on the Rough Plane:**
- The time taken to slide down the rough incline is \( nt \). The distance traveled is:
\[ s = \frac{1}{2} a (nt)^2 \]
Substituting for \( a \):
\[ s = \frac{1}{2} \left(g \sin \theta - \mu g \cos \theta\right) (nt)^2 \]

5. **Setting the Distances Equal:**
- Since the distance \( s \) is the same for both inclines, we can set the two equations for \( s \) equal to each other:
\[ \frac{1}{2} g \sin \theta \cdot t^2 = \frac{1}{2} \left(g \sin \theta - \mu g \cos \theta\right) (nt)^2 \]

6. **Simplifying the Equation:**
- Canceling \( \frac{1}{2} g \) from both sides:
\[ \sin \theta \cdot t^2 = \left(\sin \theta - \mu \cos \theta\right) n^2 t^2 \]
- Dividing both sides by \( t^2 \) (assuming \( t \neq 0 \)):
\[ \sin \theta = n^2 \left(\sin \theta - \mu \cos \theta\right) \]

7. **Rearranging to Solve for μ:**
- Expanding and rearranging gives:
\[ \sin \theta = n^2 \sin \theta - n^2 \mu \cos \theta \]
\[ n^2 \mu \cos \theta = n^2 \sin \theta - \sin \theta \]
\[ n^2 \mu \cos \theta = (n^2 - 1) \sin \theta \]
- Finally, solving for \( \mu \):
\[ \mu = \frac{(n^2 - 1) \sin \theta}{n^2 \cos \theta} \]

### Final Answer:
The coefficient of friction \( \mu \) is given by:
\[ \mu = \frac{(n^2 - 1) \sin \theta}{n^2 \cos \theta} \]