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JEE Main 2024 (31 Jan, Shift 2) · Errors in Measurement

Errors in Measurement question from JEE Main 2024 (31 Jan, Shift 2)

Q64JEE Main 2024 · 31 JAN · SHIFT 2Medium

Errors in Measurement

The measured value of the length of a simple pendulum is 20 cm20\text{ cm} with 2 mm2\text{ 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%N\%. The value of NN is:

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Answer: (a)

  1. g=4π2lT2⇒Δgg=Δll+2ΔTTg = \frac{4\pi^2 l}{T^2} \Rightarrow \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}
  2. Δll=0.2 cm20 cm=1%\frac{\Delta l}{l} = \frac{0.2\text{ cm}}{20\text{ cm}} = 1\%
  3. ΔTT=Δtt=140=2.5%\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{40} = 2.5\%
  4. Δgg=1%+5%=6%\frac{\Delta g}{g} = 1\% + 5\% = 6\%
  5. So the answer is (a) 6.
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