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JEE Main 2021 · Physics · Units and Measurements · easy

In an experiment to find g with a simple pendulum, the length is measured as L = 98.0 cm with a least count of 0.1 cm, and the total time for 50…

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In an experiment to find gg with a simple pendulum, the length is measured as L=98.0 cmL = 98.0\ \mathrm{cm} with a least count of 0.1 cm0.1\ \mathrm{cm}, and the total time for 50 oscillations is 100.0 s100.0\ \mathrm{s} with a least count of 0.1 s0.1\ \mathrm{s}. The maximum percentage error in gg obtained from g=4π2L/T2g = 4\pi^2 L/T^2 is closest to
  1. A. 0.1 %0.1\ \%
  2. B. 0.2 %0.2\ \%
  3. C. 0.3 %0.3\ \%
  4. D. 0.5 %0.5\ \%
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Answer: (C) 0.3 %0.3\ \%

Since g∝LT−2g\propto L T^{-2}, Δgg=ΔLL+2ΔTT\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}. The period is T=100.0/50T=100.0/50, and the fractional error in TT equals that in the 50-oscillation time: 0.1/100.0=0.1%0.1/100.0=0.1\%. Also ΔL/L=0.1/98.0≈0.102%\Delta L/L = 0.1/98.0\approx 0.102\%. Hence Δg/g≈0.102+2(0.1)=0.302%≈0.3%\Delta g/g\approx 0.102+2(0.1)=0.302\%\approx 0.3\%. Forgetting the square on TT gives 0.2%0.2\%.
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