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 \[g=2{{\pi }^{2}}\frac{L}{{{T}^{2}}}\] ...(i) Taking log on both sides, we have \[\log g=4{{\pi }^{2}}[\log L-2\log T]\] Differentiating, we get \[\frac{\Delta g}{g}=\frac{\Delta L}{L}-\frac{2\Delta T}{T}\] \[{{\left| \frac{\Delta g}{g} \right|}_{\max }}=\frac{\Delta L}{L}+\frac{2\Delta T}{T}\] ??. (ii) Given, \[L=100\text{ }cm,\text{ }T=2s,\text{ }\Delta T=\frac{0.1}{100}=0.001s\] \[\Delta L=1\text{ }mm=0.1\text{ }cm\] Substituting the given values in Eq. (ii), we get \[\frac{\Delta g}{g}=\frac{0.1}{100}+2\times \frac{0.001}{2}\] Thus, maximum percentage error \[{{\left| \frac{\Delta g}{g} \right|}_{\max }}\times 100=\left( \frac{0.1}{100}\times 100 \right)+\left( 2\times \frac{0.001}{2}\times 100 \right)\] \[=\text{ }0.1%\text{ }+\text{ }0.1%\] \[=\text{ }0.2%\]
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