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  4. While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 1 % in the length of the pendulum and a negative error of 3 % in the value of the time period. His percentage error in the measurement of the value of g will be

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Q. While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of $1\%$ in the length of the pendulum and a negative error of $3\%$ in the value of the time period. His percentage error in the measurement of the value of $g$ will be

Physical World, Units and Measurements

Solution:

$T=2\pi\sqrt{\frac{l}{g}}\,\Rightarrow T^{2}=\frac{4\pi^{2}l}{g}\,\Rightarrow g=\frac{4\pi^{2}l}{T^{2}}$
$\frac{\Delta g}{g}\times100=\frac{\Delta l}{l}\times 100+2 \frac{\Delta T}{T}\times100$
$=1\%+2\times3\%=7\%$