When a body slides down from rest along a smooth inclined plane making an angle of `30^@` with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time aT, where a is a constant greater than 1. The co-efficient of friction between the body and the rough plane is `1/sqrtx ((a^2 - 1)/a^2)` where x = ________.

Answer

To solve the problem, we need to analyze the motion of a body sliding down two different inclined planes: one smooth and one rough. We will derive the expressions for the distance traveled in both scenarios and equate them to find the coefficient of friction.

### Step-by-Step Solution:

1. **Identify the Forces on the Smooth Inclined Plane:**
- The body is sliding down a smooth inclined plane at an angle of \(30^\circ\).
- The forces acting on the body are:
- Gravitational force down the incline: \(F_{\text{gravity}} = mg \sin(30^\circ) = mg \cdot \frac{1}{2} = \frac{mg}{2}\)
- Normal force: \(N = mg \cos(30^\circ) = mg \cdot \frac{\sqrt{3}}{2}\)

2. **Calculate the Acceleration on the Smooth Plane:**
- Using Newton's second law, \(F = ma\):
\[ mg \sin(30^\circ) = ma \implies \frac{mg}{2} = ma \implies a = \frac{g}{2} \]

3. **Determine the Distance Traveled on the Smooth Plane:**
- The body starts from rest, so initial velocity \(u = 0\).
- Using the equation of motion \(s = ut + \frac{1}{2} a t^2\):
\[ L = 0 + \frac{1}{2} \cdot \frac{g}{2} \cdot T^2 = \frac{g T^2}{4} \]

4. **Identify the Forces on the Rough Inclined Plane:**
- The body slides down a rough inclined plane with the same angle of \(30^\circ\).
- The forces acting on the body are:
- Gravitational force down the incline: \(F_{\text{gravity}} = mg \sin(30^\circ) = \frac{mg}{2}\)
- Frictional force: \(F_{\text{friction}} = \mu N = \mu mg \cos(30^\circ) = \mu mg \cdot \frac{\sqrt{3}}{2}\)

5. **Calculate the Acceleration on the Rough Plane:**
- Applying Newton's second law:
\[ mg \sin(30^\circ) - F_{\text{friction}} = ma \]
\[ \frac{mg}{2} - \mu mg \cdot \frac{\sqrt{3}}{2} = ma \]
- Rearranging gives:
\[ a = g \left(\frac{1}{2} - \mu \cdot \frac{\sqrt{3}}{2}\right) \]

6. **Determine the Distance Traveled on the Rough Plane:**
- The body takes time \(aT\) to slide down the rough plane:
\[ L = \frac{1}{2} a (aT)^2 = \frac{1}{2} g \left(\frac{1}{2} - \mu \cdot \frac{\sqrt{3}}{2}\right) (a^2 T^2) \]

7. **Equate the Distances Traveled:**
- Since both distances are equal:
\[ \frac{g T^2}{4} = \frac{1}{2} g \left(\frac{1}{2} - \mu \cdot \frac{\sqrt{3}}{2}\right) (a^2 T^2) \]
- Cancel \(g T^2\) from both sides:
\[ \frac{1}{4} = \frac{1}{4} - \mu \cdot \frac{\sqrt{3}}{4} a^2 \]
- Rearranging gives:
\[ \mu \cdot \sqrt{3} \cdot a^2 = 0 \implies \mu = \frac{1 - \frac{1}{a^2}}{\sqrt{3}} \]

8. **Substituting for Coefficient of Friction:**
- The coefficient of friction is given as:
\[ \mu = \frac{a^2 - 1}{a^2 \sqrt{3}} \]
- Therefore, we can identify \(x\) as:
\[ x = 3 \]

### Final Answer:
The value of \(x\) is \(3\).