Starting from rest, the time taken by a body sliding down on a rough inclined plane at `45^(@)` with the horizontal is twice the time taken to travel on a smooth plane of same inclination and same distance. Then, the coefficient of kinetic friction is

Options

A. 0.25
B. 0.33
C. 0.5
D. 0.75

Explanation

To solve the problem step by step, we will use the information provided and apply the relevant physics concepts.

### Step 1: Understand the problem
We have two inclined planes: one is smooth and the other is rough. The angle of inclination for both planes is \(45^\circ\). The body starts from rest and slides down the rough plane in twice the time it takes to slide down the smooth plane.

### Step 2: Define the variables
Let:
- \( t_s \) = time taken on the smooth plane
- \( t_r \) = time taken on the rough plane
- Given that \( t_r = 2t_s \)
- The angle of inclination \( \theta = 45^\circ \)

### Step 3: Use the relationship between time and acceleration
For a smooth inclined plane, the acceleration \( a_s \) can be calculated using the formula:
\[ a_s = g \sin(\theta) \]
For \( \theta = 45^\circ \), \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), hence:
\[ a_s = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

The distance \( d \) traveled down the plane can be expressed in terms of time and acceleration:
\[ d = \frac{1}{2} a_s t_s^2 \]
Substituting for \( a_s \):
\[ d = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} \cdot t_s^2 \]

### Step 4: Calculate the acceleration on the rough plane
For the rough inclined plane, the acceleration \( a_r \) is given by:
\[ a_r = g \sin(\theta) - \mu g \cos(\theta) \]
Using \( \theta = 45^\circ \):
\[ \sin(45^\circ) = \cos(45^\circ) = \frac{1}{\sqrt{2}} \]
Thus:
\[ a_r = g \cdot \frac{1}{\sqrt{2}} - \mu g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1 - \mu) \]

### Step 5: Write the equation for distance on the rough plane
The distance traveled on the rough plane can also be expressed as:
\[ d = \frac{1}{2} a_r t_r^2 \]
Substituting for \( a_r \):
\[ d = \frac{1}{2} \cdot \frac{g}{\sqrt{2}}(1 - \mu) \cdot (2t_s)^2 \]
This simplifies to:
\[ d = \frac{1}{2} \cdot \frac{g}{\sqrt{2}}(1 - \mu) \cdot 4t_s^2 = 2 \cdot \frac{g}{\sqrt{2}}(1 - \mu) t_s^2 \]

### Step 6: Set the distances equal
Since both distances are equal, we can set the two equations for distance equal to each other:
\[ \frac{1}{2} \cdot \frac{g}{\sqrt{2}} t_s^2 = 2 \cdot \frac{g}{\sqrt{2}}(1 - \mu) t_s^2 \]

### Step 7: Simplify the equation
Dividing both sides by \( \frac{g}{\sqrt{2}} t_s^2 \) (assuming \( t_s \neq 0 \)):
\[ \frac{1}{2} = 2(1 - \mu) \]
This leads to:
\[ \frac{1}{2} = 2 - 2\mu \]
Rearranging gives:
\[ 2\mu = 2 - \frac{1}{2} \]
\[ 2\mu = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]
\[ \mu = \frac{3}{4} \]

### Step 8: Final answer
Thus, the coefficient of kinetic friction \( \mu \) is:
\[ \mu = 0.75 \]