Mathematics in Physics

JEE Main level, single correct, medium

91 of 121

Question 91 of 121

The time period of a simple pendulum is given by T=2πlg.T=2\pi \sqrt{\dfrac{l}{g}}. The measured value of the length of the pendulum is 10cm10cm known to a 1mm1mm accuracy. The time for 200200 oscillations of the pendulum is found to be 100100 second using a clock of 1s1s resolution. The percentage accuracy in the determination of gg using this pendulum is x.x. The value of xx to the nearest integer is:-

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Answer: (B) 33%

g=4π2lT2g=\dfrac{4{\pi }^{2}l}{{T}^{2}}
Δgg=Δll+2ΔTT=0.110+2(12000.5)\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T}=\dfrac{0.1}{10}+2\left(\dfrac{\dfrac{1}{200}}{0.5}\right)
Δgg=1100+150\dfrac{\Delta g}{g}=\dfrac{1}{100}+\dfrac{1}{50}
Δgg×100=3\dfrac{\Delta g}{g}\times 100=3%

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