The period of oscillation of a simple pendulum is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}$. Measured value of ' L ' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of ' g ' will be :
Select the correct option:
A
1.13%
B
1.03%
C
1.33%
D
1.30%
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
T = 2π √(ℓ / g)
g = 4π² ℓ / T²
Δg / g = Δℓ / ℓ + 2ΔT / T
Δg / g = 1 × 10⁻³ / 1 + 2 × 0.01 / 1.95
Δg / g = 0.0113 or 1.13%
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