A) increases by 2%
B) decreases by 2%
C) increases by 1%
D) decreases by 1%
Correct Answer: C
Solution :
Given: \[{{l}_{2}}=1.02\,{{l}_{1}}.\] We know that time period \[(T)=2\pi \times \frac{l}{g}\propto \sqrt{l}.\] Therefore\[=\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}}=\sqrt{\frac{1:02\,{{l}_{1}}}{{{l}_{1}}}}=1.01\] Thus time period increases by 1%.
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