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 \[\frac{\Delta g}{g}+\frac{\Delta L}{L}+\frac{2\Delta T}{T}\] ?. (ii) Given, \[L=100\,\,cm,\,\,T=2\,s\], \[\Delta T=\frac{0.1}{100}=0.001\,\,s\], \[\Delta L=1\,mm=0.1\,\,cm\] Substituting the values in Eq. (ii), we have \[\therefore \] \[{{\left| \frac{\Delta g}{g} \right|}_{\max }}=\frac{\Delta L}{L}+\frac{2\Delta T}{T}=\frac{0.1}{100}+2\times \frac{0.001}{2}\] Thus, maximum percentage error \[{{\left| \frac{\Delta g}{g} \right|}_{\max }}\times 100=\left( \frac{0.1}{100}\times 100 \right)\] \[+\left( \frac{2\times 0.001}{2}\times 100 \right)\]
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