Mathematics in Physics

JEE Main level, single correct, medium

99 of 121

Question 99 of 121

A simple pendulum is being used to determine the value of gravitational acceleration gg at a certain place. The length of the pendulum is 25.0cm25.0cm and a stopwatch with 1s1s resolution measures the time taken for 4040 oscillations to be 50s50s. The accuracy in gg is:

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Answer: (C) 4.404.40%

Time period of the simple pendulum, T=2πlg⇒g=4π2lT2T=2\pi \sqrt{\dfrac{l}{g}}\Rightarrow g=\dfrac{4{\pi }^{2}l}{{T}^{2}}
∆gg=2∆TT+∆LL⇒∆gg=2(150)+0.125.0\dfrac{∆g}{g}=\dfrac{2∆T}{T}+\dfrac{∆L}{L} \Rightarrow \dfrac{∆g}{g}=2\left(\dfrac{1}{50}\right)+\dfrac{0.1}{25.0}
⇒∆gg=4.4\Rightarrow \dfrac{∆g}{g}=4.4%

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