Question:
A simple pendulum is being used to determine the value of gravitational acceleration \(g\) at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in \(g\) is :
A simple pendulum is being used to determine the value of gravitational acceleration \(g\) at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in \(g\) is :
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\(g=4\pi^2L/T^2\), so \(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}\). Use least count 0.1 cm for L and 1 s on the total time of 50 s.
Updated On: Nov 5, 2026
- 4.40%
- 3.40%
- 2.40%
- 5.40%
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The Correct Option is A
Solution and Explanation
\(T=2\pi\sqrt{L/g}\Rightarrow g=\frac{4\pi^2L}{T^2}\).
\(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}\).
\(L=25.0\) cm has least count 0.1 cm: \(\frac{\Delta L}{L}=\frac{0.1}{25.0}=0.004\).
Total time for 40 oscillations \(=50\) s with resolution 1 s: \(\frac{\Delta T}{T}=\frac{1}{50}=0.02\) (same ratio for one period).
\(\frac{\Delta g}{g}=0.004+2\times0.02=0.044=4.40\%\).
Answer: (1) 4.40%
\(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}\).
\(L=25.0\) cm has least count 0.1 cm: \(\frac{\Delta L}{L}=\frac{0.1}{25.0}=0.004\).
Total time for 40 oscillations \(=50\) s with resolution 1 s: \(\frac{\Delta T}{T}=\frac{1}{50}=0.02\) (same ratio for one period).
\(\frac{\Delta g}{g}=0.004+2\times0.02=0.044=4.40\%\).
Answer: (1) 4.40%
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