A) 8%
B) 3%
C) 4%
D) 6%
E) 5%
Correct Answer: E
Solution :
Time period of simple pendulum is \[T=2\pi \sqrt{\frac{l}{g}}\] Or \[\frac{\Delta T}{T}=\frac{1}{2}\left( \frac{\Delta l}{l}-\frac{\Delta g}{g} \right)\] Or \[\frac{\Delta g}{g}=\frac{\Delta l}{l}-\frac{2\Delta T}{T}\] \[\therefore \]Maximum percentage error in g, \[\frac{\Delta g}{g}\times 100=\frac{\Delta l}{l}\times 100+\frac{2\Delta T}{T}\times 100\] \[=1+2\times 2\] \[=5%\]
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