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MHT CET 2025 (22 Apr, Shift 1) · Errors in Measurement

Errors in Measurement question from MHT CET 2025 (22 Apr, Shift 1)

Q49MHT CET 2025 · 22 APR · SHIFT 1Medium

Errors in Measurement

The period of oscillating simple pendulum is T=2πlgT = 2\pi\sqrt{\frac{l}{g}} where length ' ll ' is 100 cm with error 1 mm. Period is 2 second. The time of 100 oscillations is measured by a stopwatch of least count 0.1s. The percentage error in gravitational acceleration ' g ' is

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Answer: (a)

  1. T=2πl/g⇒g=4π2lT2T = 2\pi\sqrt{l/g} \Rightarrow g = \frac{4\pi^2 l}{T^2}, so Δgg=Δll+2ΔTT\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}.
  2. Δll=0.1 cm100 cm=0.1%\frac{\Delta l}{l} = \frac{0.1\text{ cm}}{100\text{ cm}} = 0.1\%
  3. Time for 100 oscillations t=200 st = 200\text{ s} with Δt=0.1 s\Delta t = 0.1\text{ s}: ΔTT=0.1200=0.05%\frac{\Delta T}{T} = \frac{0.1}{200} = 0.05\%
  4. Δgg=0.1%+2(0.05%)=0.2%\frac{\Delta g}{g} = 0.1\% + 2(0.05\%) = 0.2\%
  5. So the answer is (a) 0.2%0.2\%.
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