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(C) The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$, which implies $T \propto \sqrt{\ell}$.Given that the time d

Vedclass · Accepted Answer

$7.5$

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(C) The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$, which implies $T \propto \sqrt{\ell}$.Given that the time duration $t$ is constant, the number of oscillations $n$ is inversely proportional to the time period $T$ $(n = \frac{t}{T})$.Therefore, $n \propto \frac{1}{\sqrt{\ell}}$, which means $n^2 \propto \frac{1}{\ell}$ or $\ell \propto \frac{1}{n^2}$.Initially, $n_1 = 20$ and $\ell_1 = 30 \ cm$.Finally, $n_2 = 40$.Using the ratio: $\frac{\ell_2}{\ell_1} = \left( \frac{n_1}{n_2} ight)^2$.$\ell_2 = 30 	imes \left( \frac{20}{40} ight)^2 = 30 	imes \left( \frac{1}{2} ight)^2 = 30 	imes \frac{1}{4} = 7.5 \ cm$.

$A$ simple pendulum of string length $30 \ cm$ performs $20$ oscillations in $10 \ s$. The length of the string required for the pendulum to perform $40$ oscillations in the same time duration is . . . . . . $cm$. [Assume that the mass of the pendulum remains same.]

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