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MHT CET 2026 (13 Apr, Shift 1) · Errors in Measurement

Errors in Measurement question from MHT CET 2026 (13 Apr, Shift 1)

Q28MHT CET 2026 · 13 APR · SHIFT 1Medium

Errors in Measurement

The period of oscillation of a simple pendulum is T=2π(L/g)1/2T = 2\pi(L/g)^{1/2}. Measured value of 'L' is 10 cm known to 1 mm accuracy and time for 100 oscillations is 50 second, using a watch of 1 second resolution. The percentage error in the measurement of 'g' is

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Answer: (c)

  1. g=4π2LT2g = \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 cm10 cm=1%\frac{\Delta L}{L} = \frac{0.1\text{ cm}}{10\text{ cm}} = 1\%; ΔTT=1 s50 s=2%\frac{\Delta T}{T} = \frac{1\text{ s}}{50\text{ s}} = 2\%
  3. Δgg=1%+2(2%)=5%\frac{\Delta g}{g} = 1\% + 2(2\%) = 5\%
  4. So the answer is (c) 5%5\%.
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