An expression for the time period of oscillation of a simple pendulum of length 'L' at a place where acceleration due to gravity is g is `T prop ((L)/(g))^((1)/(x))`, then the vaslue of x is

Answer

To solve the problem, we need to derive the expression for the time period \( T \) of a simple pendulum of length \( L \) under the influence of gravity \( g \). The relationship given is:

\[ T \propto \left( \frac{L}{g} \right)^{\frac{1}{x}} \]

We need to find the value of \( x \).

### Step-by-Step Solution:

1. **Understand the relationship**: We start with the assumption that the time period \( T \) depends on the length \( L \) and the acceleration due to gravity \( g \). We can express this relationship as:
\[ T \propto L^a \cdot g^b \]
where \( a \) and \( b \) are the powers we need to determine.

2. **Dimensional Analysis**: The dimensional formula for time \( T \) is \( [T] = T^1 \). The dimensional formula for length \( L \) is \( [L] = L^1 \), and for acceleration \( g \) (which is \( L T^{-2} \)), it is:
\[ [g] = L^1 T^{-2} \]

3. **Expressing dimensions**: From our assumption, we can express the dimensions of \( T \) as:
\[ [T] = [L^a] \cdot [g^b] = L^a \cdot (L^1 T^{-2})^b = L^{a + b} \cdot T^{-2b} \]

4. **Setting up the equation**: Since we want the dimensions of \( T \) to equal \( T^1 \), we can set up the following equations based on the dimensions:
- For length: \( a + b = 0 \)
- For time: \( -2b = 1 \)

5. **Solving the equations**:
- From \( -2b = 1 \), we find:
\[ b = -\frac{1}{2} \]
- Substituting \( b \) into \( a + b = 0 \):
\[ a - \frac{1}{2} = 0 \implies a = \frac{1}{2} \]

6. **Relating \( a \) to \( x \)**: We have found \( a = \frac{1}{2} \). Recall that \( a = \frac{1}{x} \). Therefore:
\[ \frac{1}{x} = \frac{1}{2} \]

7. **Finding \( x \)**: Solving for \( x \):
\[ x = 2 \]

Thus, the value of \( x \) is \( 2 \).

### Final Answer:
The value of \( x \) is \( 2 \).