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Units and Measurements question

2024 · 31 Jan · Shift 2 · Q73
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Units and Measurements question

2024 · 31 Jan · Shift 2 · Q73

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The measured value of the length of a simple pendulum is 20 cm20 \mathrm{~cm}20 cm with 2 mm2 \mathrm{~mm}2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%\mathrm{N} \%N%. The value of N\mathrm{N}N is:
  1. A
    6
  2. B
    5
  3. C
    4
  4. D
    8
View written solutionFree

Correct answer: A

  1. For a simple pendulum, T=2πlgT=2\pi\sqrt{\frac{l}{g}}T=2πgl​​ so g=4π2lT2.g=\frac{4\pi^2 l}{T^2}.g=T24π2l​.

  2. Hence 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:

    • Measured length: l=20 cm=200 mml=20\text{ cm}=200\text{ mm}l=20 cm=200 mm
    • Accuracy: Δl=2 mm\Delta l=2\text{ mm}Δl=2 mm Δll=2200=0.01=1%.\frac{\Delta l}{l}=\frac{2}{200}=0.01=1\%.lΔl​=2002​=0.01=1%.

  4. Error in time period:

    • Time for 50 oscillations = 40 s40\text{ s}40 s with resolution 1 s1\text{ s}1 s
    • So error in total time Δt=1 s\Delta t=1\text{ s}Δt=1 s
    • Since T=t/50T=t/50T=t/50, fractional error in TTT is same as fractional error in ttt: ΔTT=Δtt=140=0.025=2.5%.\frac{\Delta T}{T}=\frac{\Delta t}{t}=\frac{1}{40}=0.025=2.5\%.TΔT​=tΔt​=401​=0.025=2.5%.

  5. Therefore, Δgg=1%+2(2.5%)=1%+5%=6%.\frac{\Delta g}{g}=1\%+2(2.5\%)=1\%+5\%=6\%.gΔg​=1%+2(2.5%)=1%+5%=6%.

  6. So, N=6.N=6.N=6.

  7. Checking options:

    • A: 666 ✅
    • B: 555 ❌
    • C: 444 ❌
    • D: 888 ❌

Therefore, the correct answer is A.

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