In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured for time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10cm known to 1mm accuracy, the accuracy in the determination of g is found to be x %. The value of x is ……………

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answer is 5.

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Detailed Solution

Time period of a simple pendulum, $T = 2 π \sqrt{\frac{l}{g}}$

Squaring on both sides

$\begin{array}{l} T^{2} = 4 π^{2} \frac{l}{g} \Rightarrow g = 4 π^{2} (\frac{l}{T^{2}}) \\ ∴ \frac{Δ g}{g} = \frac{Δ I}{I} + 2 \frac{Δ T}{T} \end{array}$

Percentage error in g is,
$\begin{array}{l} \frac{Δ g}{g} \times 100 = \frac{Δ l}{l} \times 100 + 2 (\frac{Δ T}{T} \times 100) \\ = (\frac{0.1}{10} \times 100) + (2 \times \frac{1}{50} \times 100) \\ = 1 % + 4 % = 5 % \end{array}$