In an experiment of simple pendulum time period measured was 50,sec for 25 vibrations when length of

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1.Units, Dimensions and Measurement

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In an experiment of simple pendulum time period measured was $50\,sec$ for $25$ vibrations when the length of the simple pendulum was taken $100\,cm$ . If the least count of stop watch is $0.1\,sec$ . and that of meter scale is $0.1\,cm$ then maximum possible error in value of $g$ is .......... $\%$

A

$0.5$

B

$1$

C

$0.4$

D

$0.1$

Solution

${\rm{T}} = 2\pi \sqrt {\frac{l}{{\rm{g}}}} $

$ \Rightarrow {\rm{g}} = \frac{{4{\pi ^2}l}}{{{{\rm{T}}^2}}}$

$\frac{{\Delta g}}{g} = \frac{{\Delta l}}{l} + \frac{{2\Delta T}}{T}$

$=\left[\frac{0.1}{100}+2\left(\frac{0.1}{50}\right)\right] \times 100$

$=(0.1+0.4) \%=0.5 \%$

Standard 11

Physics

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