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In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using metre scale of least count 1 millimetre. Percentage error in the determination of g is close to (nearest integer):

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answer is 7.

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Detailed Solution

$l = \frac{g T^{2}}{4 π^{2}} = \frac{10 \times {(\frac{30}{20})}^{2}}{4 \times 10} = \frac{9}{16} m$

$g = 4 π^{2} \frac{l}{T^{2}} = 4 π^{2} \frac{l}{{(\frac{t}{N})}^{2}} = 4 π^{2} N^{2} \frac{l}{t^{2}}$

$\frac{d g}{g} = \frac{d l}{l} + \frac{2 d t}{t}$

$(\frac{d g}{g}) 100 = [\frac{10^{- 3}}{(\frac{9}{16})} + 2 \times \frac{1}{30}] 10^{2} = [\frac{1.6}{9} + \frac{20}{3}] = \frac{61.6}{9} \approx 7$

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