Question:
The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:
The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:
Updated On: Apr 19, 2026
- 4
- 8
- 6
- 5
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The Correct Option is C
Approach Solution - 1
Let's solve this problem by calculating the accuracy of the acceleration due to gravity, \( g \), based on the measurements provided for the pendulum's length and the time of oscillations.
- First, understand the formula for the time period \( T \) of a pendulum: \(T = 2\pi \sqrt{\frac{L}{g}}\), where \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
- The time period for one oscillation, given 50 oscillations take 40 seconds, is: \(T = \frac{40}{50} = 0.8\, \text{seconds}\).
- Let's rearrange the equation to find \( g \): \(g = \frac{4\pi^2 L}{T^2}\).
- Given the measurements:
- Length \( L = 20\, \text{cm} = 0.2\, \text{m}\) with an accuracy of \( \pm 0.002\, \text{m} \).
- Time for 50 oscillations \( = 40\, \text{seconds} \), leading to an error of \( \pm 1\, \text{second} \) in measurement, so the time period error is \( \pm \frac{1}{50}\, \text{seconds} = \pm 0.02\, \text{seconds} \).
- The relative error in \( L \) is \(\frac{0.002}{0.2} = 0.01\) or 1%.\)
- The relative error in \( T \) is \(\frac{0.02}{0.8} = 0.025\) or 2.5%.\)
- Using the formula for \( g \): \(g = \frac{4\pi^2 L}{T^2}\), the errors combine as follows:
- The error in \( g \) from \( L \) is 1%.
- The error in \( g \) from \( T^2 \) is twice the error in \( T \) (applying error propagation for powers) - therefore, it is \( 2 \times 2.5\% = 5\%\).
- The total relative error in \( g \) is \( 1\% + 5\% = 6\%\).
Therefore, the accuracy in the measurement of acceleration due to gravity is \(6\%\). Thus, the correct answer is 6%.
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Approach Solution -2
The period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{\ell}{g}}. \]Rearrange to solve for \( g \):
\[ g = \frac{4\pi^2 \ell}{T^2}. \]The percentage error in \( g \) is given by:
\[ \frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}. \]Substitute the values:
\[ \Delta \ell = 0.2 \, \text{cm} = 0.002 \, \text{m}, \quad \ell = 0.2 \, \text{m}, \] \[ \Delta T = 1 \, \text{s}, \quad T = \frac{40}{50} = 0.8 \, \text{s}. \]Calculate the percentage errors:
\[ \frac{\Delta \ell}{\ell} = \frac{0.002}{0.2} = 0.01 = 1\%. \] \[ 2 \frac{\Delta T}{T} = 2 \times \frac{1}{40} = 0.05 = 5\%. \]Therefore, the total percentage error in \( g \) is:
\[ 1\% + 5\% = 6\%. \]Thus, \( N = 6 \).
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