Explanation

Time period of a simple pendulum is given by:

$$ T = 2\pi \sqrt{\frac{l}{g}} $$

Number of oscillations $n$ in a given time $t$ is:

$$ n = \frac{t}{T} $$

For the first pendulum,

$$ n_1 = 20,\quad t = 10\ \text{s} $$

So,

$$ T_1 = \frac{10}{20} = 0.5\ \text{s} $$

For the second pendulum,

$$ n_2 = 40,\quad t = 10\ \text{s} $$

So,

$$ T_2 = \frac{10}{40} = 0.25\ \text{s} $$

Since $T \propto \sqrt{l}$, we write:

$$ \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} $$

Substitute values:

$$ \frac{0.5}{0.25} = \sqrt{\frac{30}{l_2}} $$

$$ 2 = \sqrt{\frac{30}{l_2}} $$

Squaring both sides:

$$ 4 = \frac{30}{l_2} $$

$$ l_2 = \frac{30}{4} = 7.5\ \text{cm} $$

Hence, the required length of the string is 7.5 cm.

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