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Q.

The time period of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}} .$ The measured value of the length of pendulum is 10cm known to a 1 mm accurancy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of ‘g’ using this pendulum is ‘x’. The value of ‘x’ to the nearest integer is

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a

$2 %$

b

$3 %$

c

$5 %$

d

$4 %$

answer is B.

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Detailed Solution

$\begin{array}{l} g = \frac{4 π^{2} ℓ}{T^{2}} \\ \frac{Δg}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{ΔT}{T} \\ = \frac{0.1}{10} + 2 (\frac{\frac{1}{200}}{0.5}) \\ \frac{Δg}{g} = \frac{1}{100} + \frac{1}{50} \\ \frac{Δg}{g} \times 100 = 3 % \end{array}$

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The time period of a simple pendulum is given by T=2πlg. The measured value of the length of pendulum is 10cm known to a 1 mm accurancy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of ‘g’ using this pendulum is ‘x’. The value of ‘x’ to the nearest integer is