A pendulum clock gives correct time at `20^@C`at a place where `g=9.800 m s^(-2)`.The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where `g=9.788 m s^(-2)`. At what temperature will it give correct time ? coefficient of linear expansion of steel `= 12 xx 10^(-6)C^(-1)`.

Explanation

To solve the problem, we need to determine the temperature at which the pendulum clock will give the correct time at a new location with a different value of gravitational acceleration.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Pendulum**:
The time period \( T \) of a simple pendulum is given by the formula:
\[
T = 2\pi \sqrt{\frac{L}{g}}
...