• # question_answer If the length of second pendulum is decreased by 2% how many seconds it will lose per day? A) $1927\text{ }sec.$                 B) $2727\text{ }sec.$C) $2427\text{ }sec.$                 D) $864\text{ }sec.$

We know that, the correct time period of the second pendulum in 2 seconds. Suppose I in the correct length). $2=2\pi \,\sqrt{\frac{1}{g}}$                        ?...(i) Given, Decrease in length $=2%=\frac{2h}{100}l$ $\therefore$ length after contraction $=L=\frac{2l}{100}$ $\Rightarrow$            $i=L\,\left( 1-\frac{2}{100} \right)$ New time period is given by $t=2\pi \sqrt{\frac{l}{g}\left( 1-\frac{2}{100} \right)}$            ??(ii) $\left( \because \,\,\,t=2\pi \sqrt{\frac{l}{g}} \right)$ From equations (i) and (ii), we have, $\frac{t}{2}=\sqrt{1-\frac{2}{100}}$                     ??(iii) $={{\left( 1-\frac{2}{100} \right)}^{1/2}}$ Using binomial theorem, we have $t=2\left( 1-\frac{1}{2}\times \frac{2}{100} \right)$ $\Rightarrow$ $t=2\left( 2-\frac{2}{100} \right)\sec$ It is clear that it less than 2. $\therefore$  The clock gains time and time gained in 2 seconds, $=\frac{2}{100}\,\,\sec .$ $\therefore$ Total time gained per day be clock or time loss by day. $=\frac{2}{100}\times \frac{24\times 60\times 60}{2}=864\sec .$
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