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The acceleration due to gravity is found from a simple pendulum using g = 4π^2 L/T^2. The length is measured as L = 100.0 cm with a least count of…

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The acceleration due to gravity is found from a simple pendulum using

g = 4\pi^2 L/T^2

. The length is measured as

L = 100.0\ \mathrm{cm}

with a least count of

0.1\ \mathrm{cm}

. The time for 20 oscillations is measured as

40.0\ \mathrm{s}

with a least count of

0.2\ \mathrm{s}

. The maximum percentage error in

g

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Answer: (B) $1.1\%$

Since

g\propto L\,T^{-2}

\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}

. The error in the measured total time

40.0\ \mathrm{s}

0.2\ \mathrm{s}

; the period is

T=40/20

but its fractional error equals that of the total time,

0.2/40=0.5\%

. Hence

\Delta g/g=0.1/100+2(0.5\%)=0.1\%+1.0\%=1.1\%

. Distractors come from forgetting the factor 2 on

T

(

0.6\%

) or dividing the error of

T

again by 20.

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