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Q. Time period $T$ of a simple pendulum of length $l$ is given by $T=2\pi\sqrt{\frac{l}{g}}.$ If the length is increased by $2\%$, then an approximate change in the time period is

WBJEEWBJEE 2016Application of Derivatives

Solution:

$\frac{dT}{d\ell}=\frac{2\pi}{\sqrt{g}}. \frac{1}{2\sqrt{\ell}}$
$\therefore \Delta T=\frac{dT}{d\ell}. \Delta\ell=\frac{\pi}{\sqrt{g\ell}}.\left(\frac{2\ell}{100}\right)$
$=2\pi\sqrt{\frac{\ell}{g}}. \frac{1}{100}=\frac{T}{100}$
$\therefore \frac{\Delta T}{T}=\frac{1}{100}$
$\therefore 1\%$