Experimental Physics

JEE Main level, single correct, medium

19 of 70

Question 19 of 70

The measured value of the length of a simple pendulum is 20cm20cm with 2mm2mm accuracy. The time for 5050 oscillations was measured to be 4040 seconds with 11 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is NN%. The value of NN is:

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Answer: (C) 66

Time period of a simple pendulum is given by, T=2π(lg)T=2\pi \left(\dfrac{l}{g}\right).
⇒g=4π2lT2\Rightarrow g=\dfrac{4{\pi }^{2}l}{{T}^{2}}
Therefore, fractional error will be
Δgg=Δll+2ΔTT\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+\dfrac{2\Delta T}{T}
=0.220+2(140)=\dfrac{0.2}{20}+2\left(\dfrac{1}{40}\right)
=1.220=\dfrac{1.2}{20}
Percentage change =1.220×100=6=\dfrac{1.2}{20}\times 100=6%

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