A block takes a twice as much time to slide down a `45^(@)` rough inclined plane as it takes to slide down a similar smooth , plane . The coefficient of friction is :

Options

A. `(3)/(4)`
B. `(sqrt3)/(2)`
C. `(1)/(4)`
D. `(1)/(3)`

Explanation

To solve the problem of finding the coefficient of friction for a block sliding down a rough inclined plane, we will follow these steps:

### Step 1: Analyze the Smooth Inclined Plane
For a smooth inclined plane, the only force acting on the block is its weight component along the incline. The weight of the block can be resolved into two components:
- Perpendicular to the incline: \( Mg \cos(45^\circ) \)
- Parallel to the incline: \( Mg \sin(45^\circ) \)

Since there is no friction on the smooth plane, the acceleration \( a_1 \) of the block can be calculated as:
\[ a_1 = g \sin(45^\circ) = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

### Step 2: Analyze the Rough Inclined Plane
For the rough inclined plane, we have to consider the frictional force acting against the motion of the block. The frictional force \( F_f \) can be expressed as:
\[ F_f = \mu N = \mu Mg \cos(45^\circ) = \mu Mg \cdot \frac{1}{\sqrt{2}} \]
The net force acting on the block down the incline is given by:
\[ F_{net} = Mg \sin(45^\circ) - F_f = Mg \sin(45^\circ) - \mu Mg \cos(45^\circ) \]
Substituting the values:
\[ F_{net} = Mg \cdot \frac{1}{\sqrt{2}} - \mu Mg \cdot \frac{1}{\sqrt{2}} = Mg \cdot \frac{1 - \mu}{\sqrt{2}} \]
Using Newton's second law, \( F_{net} = Ma \), we can write:
\[ Ma = Mg \cdot \frac{1 - \mu}{\sqrt{2}} \]
Thus, the acceleration \( a_2 \) for the rough inclined plane is:
\[ a_2 = g \cdot \frac{1 - \mu}{\sqrt{2}} \]

### Step 3: Relate the Times Taken on Both Planes
According to the problem, the time taken to slide down the rough incline is twice that of the smooth incline. Let \( T_1 \) be the time taken on the smooth incline and \( T_2 \) be the time taken on the rough incline:
\[ T_2 = 2T_1 \]
Using the kinematic equation \( S = ut + \frac{1}{2} a t^2 \) (where \( u = 0 \)):
\[ S = \frac{1}{2} a_1 T_1^2 \quad \text{(for smooth incline)} \]
\[ S = \frac{1}{2} a_2 T_2^2 \quad \text{(for rough incline)} \]
Substituting \( T_2 = 2T_1 \):
\[ S = \frac{1}{2} a_2 (2T_1)^2 = 2 a_2 T_1^2 \]

### Step 4: Set the Equations Equal
Since both expressions equal the same distance \( S \), we can set them equal:
\[ \frac{1}{2} a_1 T_1^2 = 2 a_2 T_1^2 \]
Dividing both sides by \( T_1^2 \) (assuming \( T_1 \neq 0 \)):
\[ \frac{1}{2} a_1 = 2 a_2 \]
Substituting the expressions for \( a_1 \) and \( a_2 \):
\[ \frac{1}{2} \cdot \frac{g}{\sqrt{2}} = 2 \cdot g \cdot \frac{1 - \mu}{\sqrt{2}} \]
Cancelling \( g \) and simplifying:
\[ \frac{1}{2\sqrt{2}} = \frac{2(1 - \mu)}{\sqrt{2}} \]
Multiplying through by \( \sqrt{2} \):
\[ \frac{1}{2} = 2(1 - \mu) \]
Solving for \( \mu \):
\[ \frac{1}{2} = 2 - 2\mu \implies 2\mu = 2 - \frac{1}{2} \implies 2\mu = \frac{4 - 1}{2} = \frac{3}{2} \]
\[ \mu = \frac{3}{4} \]

### Final Answer
The coefficient of friction \( \mu \) is \( \frac{3}{4} \).

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