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Mathematics in Physics

JEE Main level, single correct, medium

102 of 121

Question 102 of 121

In a simple pendulum experiment for determination of acceleration due to gravity $(g)$ , time taken for $20$ oscillations is measures by using a watch of $1$ second least count. The mean value of time taken comes out to be $30s$ . The length of the 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

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Answer: (B) $\text{6.8}%$

g=\dfrac{4{\pi }^{2}l}{{T}^{2}}

\dfrac{∆g}{g}=\dfrac{∆l}{l}+\dfrac{2∆T}{T}

=\dfrac{0.1}{55}+2\times \left(\dfrac{1}{30}\right)

=0.06848

Percentage error

\approx 6.8%

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