An object is placed on the surface of a smooth inclined plane of inclination `theta` . 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 `theta`, it takes time nt to reach the bottom where n is a number greater than 1. The coefficient of friction `mu` is given by -

Options

A. `mu=tan theta(1-1//n^(2))`
B. `mu=cot theta(1-1//n^(2))`
C. `mu=tan theta(1-1//n^(2))^(1//2)`
D. `mu=cot theta(1-1//n^(2))^(1//2)`

Answer

To solve the problem, we need to analyze the motion of an object sliding down two inclined planes: one smooth and one rough. Let's break it down step by step.

### Step 1: Analyze the smooth inclined plane
1. **Identify the forces**: On a smooth inclined plane, the only force causing the object to slide down is the component of gravitational force parallel to the incline, which is \( mg \sin \theta \).
2. **Calculate acceleration**: Using Newton's second law, we have:
\[ mg \sin \theta = m a \]
This simplifies to:
\[ a = g \sin \theta \]
3. **Use the equation of motion**: The distance \( s \) covered by the object is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ s = \frac{1}{2} (g \sin \theta) t^2 \]
Let's label this as Equation (1).

### Step 2: Analyze the rough inclined plane
1. **Identify the forces**: On a rough inclined plane, the forces acting on the object include the gravitational force and the frictional force. The frictional force \( F_f \) is given by \( \mu N \), where \( N = mg \cos \theta \).
2. **Calculate net force and acceleration**: The net force acting on the object is:
\[ mg \sin \theta - F_f = ma \]
Substituting for \( F_f \):
\[ mg \sin \theta - \mu mg \cos \theta = ma \]
This simplifies to:
\[ g \sin \theta - \mu g \cos \theta = a \]
Thus, the acceleration \( a \) is:
\[ a = g (\sin \theta - \mu \cos \theta) \]
Let's label this as Equation (2).

3. **Use the equation of motion**: The time taken to slide down the rough incline is \( nt \). Thus, the distance \( s \) can also be expressed as:
\[ s = \frac{1}{2} a (nt)^2 \]
Substituting for \( a \):
\[ s = \frac{1}{2} g (\sin \theta - \mu \cos \theta) (nt)^2 \]
Let's label this as Equation (3).

### Step 3: Equate the distances
Since the distance \( s \) is the same for both cases, we can set Equation (1) equal to Equation (3):
\[ \frac{1}{2} g \sin \theta t^2 = \frac{1}{2} g (\sin \theta - \mu \cos \theta) (nt)^2 \]
Cancelling \( \frac{1}{2} g \) from both sides (assuming \( g \neq 0 \)):
\[ \sin \theta t^2 = (\sin \theta - \mu \cos \theta) n^2 t^2 \]
Dividing both sides by \( t^2 \) (assuming \( t \neq 0 \)):
\[ \sin \theta = (\sin \theta - \mu \cos \theta) n^2 \]

### Step 4: Solve for the coefficient of friction \( \mu \)
Rearranging the equation:
\[ \sin \theta = n^2 \sin \theta - n^2 \mu \cos \theta \]
Bringing all terms involving \( \mu \) to one side:
\[ n^2 \mu \cos \theta = n^2 \sin \theta - \sin \theta \]
Factoring out \( \sin \theta \):
\[ n^2 \mu \cos \theta = \sin \theta (n^2 - 1) \]
Now, solving for \( \mu \):
\[ \mu = \frac{\sin \theta (n^2 - 1)}{n^2 \cos \theta} \]
This can be simplified to:
\[ \mu = \tan \theta \left(1 - \frac{1}{n^2}\right) \]

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