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JEE Advanced 2020 Paper 2 · Physics · Units and Measurements · medium

In an experiment on a simple pendulum, the string length from the suspension point to the top of the bob is measured as 90.0 cm with a metre scale of…

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In an experiment on a simple pendulum, the string length from the suspension point to the top of the bob is measured as 90.0 cm90.0\ \mathrm{cm} with a metre scale of least count 0.1 cm0.1\ \mathrm{cm}. The bob diameter, measured with a vernier calliper of least count 0.01 cm0.01\ \mathrm{cm}, is 2.00 cm2.00\ \mathrm{cm}. The time for 3030 oscillations is 57.2 s57.2\ \mathrm{s} with a stopwatch of least count 0.1 s0.1\ \mathrm{s}. Taking the maximum possible errors to be the least counts, the percentage error in the value of g=4π2L/T2g=4\pi^2L/T^2 (with LL the effective length) is. Give the answer correct to two decimal places.
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Answer: 0.47

The effective length is L=l+r=90.0+1.00=91.00 cmL=l+r=90.0+1.00=91.00\ \mathrm{cm} (radius is half the diameter). Errors add for a sum: the radius error is half the diameter least count, Δr=0.005 cm\Delta r=0.005\ \mathrm{cm}, so ΔL=0.1+0.005=0.105 cm\Delta L=0.1+0.005=0.105\ \mathrm{cm} and ΔL/L=0.105/91.00=0.1154%\Delta L/L=0.105/91.00=0.1154\%. For the period, T=t/30T=t/30 so ΔT/T=Δt/t=0.1/57.2=0.1748%\Delta T/T=\Delta t/t=0.1/57.2=0.1748\% (the number of oscillations is exact). Because g∝L T−2g\propto L\,T^{-2}, the maximum fractional error is Δgg=ΔLL+2ΔTT=0.1154%+0.3497%=0.4651%≈0.47%\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}=0.1154\%+0.3497\%=0.4651\%\approx0.47\%. A mistake is to use ΔL=0.1\Delta L=0.1 only, or to divide the time error by 3030 a second time. Percentage errors are simply added because the maximum-error rule is used, not the root-sum-square rule.
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