The lengths of smooth & rough inclined planes of inclination `45^(@)` is same Times of sliding of a body on two surface `t_(1)t_(2)` and `mu =0.75` then `t_(1):t_(2)=` .

Options

A. `2:1`
B. `2:3`
C. `1:2`
D. `3:2`

Explanation

To solve the problem, we need to find the ratio of the times taken by a body to slide down a smooth inclined plane (t1) and a rough inclined plane (t2), given that the angle of inclination is 45 degrees and the coefficient of friction (μ) is 0.75.

### Step-by-Step Solution:

1. **Identify the accelerations on the inclined planes**:
- For the smooth inclined plane, the only force acting along the incline is the component of gravitational force. Thus, the acceleration (a1) is given by:
\[ a_1 = g \sin \theta \]
- For the rough inclined plane, the frictional force opposes the motion. The acceleration (a2) is given by:
\[ a_2 = g \sin \theta - \mu g \cos \theta \]

2. **Substituting the angle**:
- Since the angle of inclination θ is 45 degrees, we have:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \quad \text{and} \quad \cos 45^\circ = \frac{1}{\sqrt{2}} \]
- Therefore, we can rewrite the accelerations:
\[ a_1 = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]
\[ a_2 = g \cdot \frac{1}{\sqrt{2}} - 0.75 \cdot g \cdot \frac{1}{\sqrt{2}} = g \cdot \frac{1}{\sqrt{2}} \cdot (1 - 0.75) = g \cdot \frac{1}{\sqrt{2}} \cdot 0.25 = \frac{g}{4\sqrt{2}} \]

3. **Using the equation of motion to find time**:
- The time taken to slide down an incline can be found using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
- Since the initial velocity (u) is 0, we can simplify this to:
\[ L = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2L}{a}} \]
- For the smooth incline:
\[ t_1 = \sqrt{\frac{2L}{a_1}} = \sqrt{\frac{2L}{\frac{g}{\sqrt{2}}}} = \sqrt{\frac{2L \sqrt{2}}{g}} = \frac{\sqrt{4L}}{\sqrt{g}} = \frac{2\sqrt{L}}{\sqrt{g}} \]
- For the rough incline:
\[ t_2 = \sqrt{\frac{2L}{a_2}} = \sqrt{\frac{2L}{\frac{g}{4\sqrt{2}}}} = \sqrt{\frac{8L}{g}} = \frac{2\sqrt{2L}}{\sqrt{g}} \]

4. **Finding the ratio of times**:
- Now, we can find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{\frac{2\sqrt{L}}{\sqrt{g}}}{\frac{2\sqrt{2L}}{\sqrt{g}}} = \frac{\sqrt{L}}{\sqrt{2L}} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \]

5. **Final Answer**:
- Thus, the ratio \( t_1 : t_2 \) is:
\[ t_1 : t_2 = 1 : 2 \]