A) 5%
B) 6%
C) 1%
D) 8%
Correct Answer: D
Solution :
[d] As we know, time period of a simple pendulum \[T=2\pi \sqrt{\frac{L}{g}\Rightarrow g=\frac{4{{\pi }^{2}}L}{{{T}^{2}}}}\] The maximum percentage error in g \[\frac{\Delta g}{g}\times 100=\frac{\Delta L}{L}\times 100+2\left( \frac{\Delta T}{T}\times 100 \right)\] \[=2%+2(3%)=8%\]
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