In a simple pendulum experiment to determine the acceleration due to gravity gg, the length LL of the pendulum is measured as 100100 cm with a meter scale of least count 11 mm. The time for 100100 oscillations is 200200 s measured with a stopwatch of least count 11 s. What is the maximum percentage error in the value of gg?

Model Answer & Options

Source: Extra Practice

0.1%

2.1%

1.1%

0.6%

Explanation

The formula for the time period of a simple pendulum is T=2πL/gT = 2\pi\sqrt{L/g}, which gives g=4π2LT2g = 4\pi^2 \frac{L}{T^2}. The relative error in gg is Δgg=ΔLL+2ΔTT\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}. Given: L=100L = 100 cm, ΔL=1\Delta L = 1 mm = 0.10.1 cm. Total time t=200t = 200 s for n=100n=100 oscillations, so T=t/nT = t/n and ΔT/T=Δt/t\Delta T/T = \Delta t/t. Here Δt=1\Delta t = 1 s. Substituting: Δgg=0.1100+2(1200)=0.001+0.01=0.011\frac{\Delta g}{g} = \frac{0.1}{100} + 2\left(\frac{1}{200}\right) = 0.001 + 0.01 = 0.011. Expressed as a percentage: 0.011×100=1.1%0.011 \times 100 = 1.1\%. Option 0.1% only considers length, and 2.1% or 0.6% result from incorrect application of the power rule for TT or arithmetic errors.

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