A) 0.2%
B) 0.5%
C) 0.1%
D) 2%
Correct Answer: A
Solution :
Time period of a simple pendulum \[T=2\pi \sqrt{\frac{L}{g}}\] or \[g=\frac{4{{\pi }^{2}}L}{{{T}^{2}}}\] ?(i) Differentiating Eq. (i), we have \[\frac{\Delta g}{g}=\frac{\Delta L}{L}+\frac{2\Delta \Tau }{T}\] ?(ii) Given, \[L=100\,cm,\,T=2s,\] \[\Delta \Tau =\frac{0.1}{100}=0.001\,s,\] \[\Delta L=1mm=0.1\,cm\] Substituting the values in Eq. (ii), we have \[\therefore \] \[{{\left| \frac{\Delta g}{g} \right|}_{\max }}=\frac{\Delta L}{L}+\frac{2\Delta \Tau }{T}=\frac{0.1}{100}+2\times \frac{0.001}{2}\] Thus, maximum percentage error \[{{\left| \frac{\Delta g}{g} \right|}_{\max }}\times 100=\left( \frac{0.1}{100}\times 100 \right)\] \[+\,\left( \frac{2\times 0.001}{2}\times 100 \right)\] \[=0.1%+0.1%=0.2%\]
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