While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be

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answer is B.

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Detailed Solution

$\begin{aligned} \begin{array}{l} \begin{array}{l} T = 2 π \sqrt{\frac{L}{g}} or T^{2} = 4 π^{2} \frac{L}{g} \\ g = 4 π^{2} \frac{L}{T^{2}}; \frac{Δ g}{g} = \frac{Δ L}{L} - 2 \frac{Δ T}{T} \\ \frac{Δ g}{g} \times 100 = \frac{Δ L}{L} \times 100 - 2 \frac{Δ T}{T} \times 100 \end{array} \end{array} \\ Actual % error in g = \frac{Δ L}{L} \times 100 - 2 \frac{Δ T}{T} \times 100 \\ = + 2 % - 2 \times 1 % = 0 % \end{aligned}$