Mathematics in Physics

JEE Main level, numerical answer, medium

77 of 121

Question 77 of 121

In an experiment to find acceleration due to gravity (g)\left(g\right) using simple pendulum, time period of 0.5s0.5s is measured from time of 100100 oscillation with a watch of 1s1s resolution. If measured value of length is 10cm10cm known to 1mm1mm accuracy. The accuracy in the determination of gg is found to be xx%. The value of xx is

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Answer: 5

Time period of oscillation is given by T=2πlgT=2\pi \sqrt{\dfrac{l}{g}} , here, ll is length and gg is acceleration due to gravity.
From above relation, we have g=14π2T2lg=\dfrac{1}{4{\pi }^{2}}\dfrac{{T}^{2}}{l}.
Fractional error in gg is stated as
Δgg=2ΔTT+Δll\dfrac{\Delta g}{g}=\dfrac{2\Delta T}{T}+\dfrac{\Delta l}{l}
⇒Δgg×100=2×1100×0.5×100+1mm10cm×100\Rightarrow \dfrac{\Delta g}{g}\times 100=2\times \dfrac{1}{100\times 0.5}\times 100+\dfrac{1mm}{10cm}\times 100
⇒Δgg×100=5100×100=5\Rightarrow \dfrac{\Delta g}{g}\times 100=\dfrac{5}{100}\times 100=5%
Hence, value of x=5x=5.

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