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Units and Measurements question

2022 · 28 Jul · Shift 2 · Q66
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Units and Measurements question

2022 · 28 Jul · Shift 2 · Q66

JEE MainPhysicsUnits and MeasurementsNumerical+4 / −1
In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s0.5 \mathrm{~s}0.5 s is measured from time of 100 oscillation with a watch of 1 s1 \mathrm{~s}1 s resolution. If measured value of length is 10 cm10 \mathrm{~cm}10 cm known to 1 mm1 \mathrm{~mm}1 mm accuracy, The accuracy in the determination of g\mathrm{g}g is found to be x%x \%x%. The value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 5

  1. For a simple pendulum, T=2πlgT = 2\pi\sqrt{\frac{l}{g}}T=2πgl​​ so g=4π2lT2g = \frac{4\pi^2 l}{T^2}g=T24π2l​

  2. The fractional error in ggg is Δgg=Δll+2ΔTT\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}gΔg​=lΔl​+2TΔT​

  3. Error in length measurement:

    • Measured length: l=10 cml = 10\,\text{cm}l=10cm
    • Accuracy: 1 mm=0.1 cm1\,\text{mm} = 0.1\,\text{cm}1mm=0.1cm

    Hence, Δll=0.110=0.01=1%\frac{\Delta l}{l} = \frac{0.1}{10} = 0.01 = 1\%lΔl​=100.1​=0.01=1%

  4. Error in time period measurement:

    • Time period T=0.5 sT = 0.5\,\text{s}T=0.5s
    • It is measured from time of 100 oscillations.

    Total time for 100 oscillations: t=100T=100(0.5)=50 st = 100T = 100(0.5) = 50\,\text{s}t=100T=100(0.5)=50s

    Watch resolution = 1 s1\,\text{s}1s, so error in measuring ttt is taken as Δt=1 s\Delta t = 1\,\text{s}Δt=1s

    Therefore, Δtt=150=0.02=2%\frac{\Delta t}{t} = \frac{1}{50} = 0.02 = 2\%tΔt​=501​=0.02=2%

    Since T=t100,T = \frac{t}{100},T=100t​, we have ΔTT=Δtt=2%\frac{\Delta T}{T} = \frac{\Delta t}{t} = 2\%TΔT​=tΔt​=2%

  5. Now, Δgg=1%+2(2%)=1%+4%=5%\frac{\Delta g}{g} = 1\% + 2(2\%) = 1\% + 4\% = 5\%gΔg​=1%+2(2%)=1%+4%=5%

  6. Therefore, x=5x = 5x=5

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