A) 6.0%
B) 7.2%
C) 9.4%
D) 10.2%
Correct Answer: A
Solution :
We know that, \[g=\frac{4{{\pi }^{2}}l}{{{T}^{2}}}=4{{\pi }^{2}}l{{T}^{-2}}\] The fractional error in g is given by \[\frac{{{\delta }_{g}}}{g}=\frac{{{\delta }_{l}}}{l}-2\frac{{{\delta }_{T}}}{T}\](ignoring constant terms \[4{{\pi }^{2}}\]) The maximum fractional error in g is \[{{\left| \frac{{{\delta }_{g}}}{g} \right|}_{\max }}=\frac{{{\delta }_{l}}}{l}+2\left( \frac{{{\delta }_{T}}}{T} \right)\] The percentage error in g is \[{{\left| \frac{{{\delta }_{g}}}{g} \right|}_{\max }}\times 100=\frac{{{\delta }_{l}}}{l}\times 100+2\left( \frac{{{\delta }_{T}}}{T}\times 200 \right)\] \[=\frac{2}{20}\times 100+2\left( \frac{0.001}{0.50} \right)\times 100\] \[=2%+2\times (2%)\] \[=2%+4%=6.0%\]
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