A) \[1927\text{ }sec.\]
B) \[2727\text{ }sec.\]
C) \[2427\text{ }sec.\]
D) \[864\text{ }sec.\]
Correct Answer: D
Solution :
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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