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  4. In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of g is found to be x %. The value of x is

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Q. In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of $0.5\, s$ is measured from time of $100$ oscillation with a watch of $1 \, s$ resolution. If measured value of length is $10\, cm$ known to $1\, mm$ accuracy. The accuracy in the determination of $g$ is found to be $x \%$. The value of $x$ is

JEE MainJEE Main 2022Physical World, Units and Measurements

Solution:

$ T =2 \pi \sqrt{\frac{\ell}{ g }} $
$ g =\frac{1}{4 \pi^2} \frac{ T ^2}{\ell} $
$ \frac{\Delta g }{ g }=\frac{2 \Delta T }{ T }+\frac{\Delta \ell}{\ell}$
$ \frac{\Delta g }{ g }=2 \cdot \frac{1}{100 \times 0.5}+\frac{1 mm }{10 cm }$
$ \frac{\Delta g }{ g }=\frac{5}{100}$