A second's pendulum is oscillating at a place where g = 9.8 `m//s^(2)`. If the pendulum is shifted to a place where g = 9.6 `m//s^(2)`, then to keep this time period unchanged, how much change of length should be introduced ?

Explanation

To solve the problem, we need to find out how much the length of the pendulum should be changed in order to keep the time period unchanged when moving from a place with a gravitational acceleration of \( g = 9.8 \, \text{m/s}^2 \) to a place with \( g = 9.6 \, \text{m/s}^2 \).

### Step 1: Write the formula for the time period of a simple pendulum.
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

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