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 $30s$. The length of pendulum is measured by using a meter scale of least count $1mm$ and the value obtained is $55.0cm$. The percentage error in the determination of g is close to :-

A

$0.7%$

B

$0.2%$

C

$3.5%$

D

$6.8%$

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Solution

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Correct option is D)

$T=2030sec $$ΔT=201 sec$ $L=55cm$$ΔL=1mm=0.1cm$ $g=T_{2}4π_{2}L $ percentage error in g is $gΔg ×100%=(LΔL +T2ΔT )100%$ $=⎝⎜⎜⎜⎛ 550.1 +2030 2(201 ) ⎠⎟⎟⎟⎞ 100%=6.8%$