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MHT CET 2026 (11 Apr, Shift 1) · Errors in Measurement

Errors in Measurement question from MHT CET 2026 (11 Apr, Shift 1)

Q29MHT CET 2026 · 11 APR · SHIFT 1Medium

Errors in Measurement

The period of oscillation of a simple pendulum is given by T=2πlgT = 2\pi\sqrt{\frac{l}{g}} where ll is about 8080 cm and is known to have 0.10.1 cm accuracy. The period is about 1.51.5 s. The time of 5050 oscillations is measured by a stop watch of least count 0.10.1 s. The percentage error in gg is nearly

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Answer: (b)

  1. Δgg=Δll+2ΔTT\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}
  2. Δll=0.180=0.125%\frac{\Delta l}{l} = \frac{0.1}{80} = 0.125\%
  3. Time of 50 oscillations =50×1.5=75 s= 50 \times 1.5 = 75\text{ s}, so ΔTT=0.175=0.133%\frac{\Delta T}{T} = \frac{0.1}{75} = 0.133\%
  4. Δgg=0.125%+0.267%≈0.4%\frac{\Delta g}{g} = 0.125\% + 0.267\% \approx 0.4\%
  5. So the answer is (b) 0.40.4.
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