In an experiment to determine acceleration due to gravity by simple pendulum, a student committed 1% positive error in the measurement of length and 3% negative error in the measurement of time period. The maximum possible percentage error in the measured value of ‘g’ will be

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In an experiment to determine acceleration due to gravity by simple pendulum, a student committed 1% positive error in the measurement of length and 3% negative error in the measurement of time period. The maximum possible percentage error in the measured value of ‘g’ will be

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Detailed Solution

We know $T = 2 π \sqrt{\frac{L}{g}}$
$∴ T^{2} \propto \frac{L}{g} \Rightarrow g \propto \frac{L}{T^{2}}$
$\frac{Δ g}{g} \times 100 = \frac{Δ L}{L} \times 100 + 2 \frac{Δ T}{T} \times 100$
$= 1 % + (2 \times 3 %) = 7 %$