An experiment is performed to obtain value of acceleration due to gravity g by using a simple pendul

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1.Units, Dimensions and Measurement

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An experiment is performed to obtain the value of acceleration due to gravity $g$ by using a simple pendulum of length $L$. In this experiment time for $100\, oscillations$ is measured by using a watch of $1\, second$ least count and the value is $90.0\, seconds$. The length $L$ is measured by using a meter scale of least count $1\, mm$ and the value is $20.0\, cm$. The error in the determination of $g$ would be ........... $\%$

A$1.7$

B$2.7$

C$4.4$

D$2.27$

(JEE MAIN-2014)

Solution

Here, $T=2 \pi \sqrt{\frac{L}{g}}$ or $T^{2}=4 \pi^{2}(L / g)$
So, $g=\frac{4 \pi^{2} L}{T^{2}}$
Thus, $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2 \frac{\Delta T}{T}$
% error in $g=\frac{\Delta g}{g} \times 100$
$=\left(\frac{\Delta L}{L}+2 \frac{\Delta T}{T}\right) \times 100$
$=\left(\frac{(1 / 10)}{20}+2 \times \frac{1}{90}\right) \times 100=2.72 \%$

Standard 11

Physics