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:

• A. 4
• B. 8
• C. 6
• D. 5

Answer 1

The Correct Option is C

The accuracy of the acceleration due to gravity, \( g \), is determined by calculating it from the provided measurements of the pendulum's length and oscillation time.

1. The formula for the time period \( T \) of a pendulum is \(T = 2\pi \sqrt{\frac{L}{g}}\), where \( L \) is the pendulum's length and \( g \) is the acceleration due to gravity.
2. Given that 50 oscillations take 40 seconds, the time period for one oscillation is \(T = \frac{40}{50} = 0.8\, \text{seconds}\).
3. Rearranging the formula to solve for \( g \) yields \(g = \frac{4\pi^2 L}{T^2}\).
4. The provided measurements and their associated errors are:
   • Length \( L = 20\, \text{cm} = 0.2\, \text{m} \) with an uncertainty of \( \pm 0.002\, \text{m} \).
   • The time for 50 oscillations is \( 40\, \text{seconds} \), with an error of \( \pm 1\, \text{second} \). This results in a time period error of \( \pm \frac{1}{50}\, \text{seconds} = \pm 0.02\, \text{seconds} \).
5. The relative error in the length measurement \( L \) is \( \frac{0.002}{0.2} = 0.01 \), or 1%.
6. The relative error in the time period \( T \) is \( \frac{0.02}{0.8} = 0.025 \), or 2.5%.
7. Applying the formula \( g = \frac{4\pi^2 L}{T^2} \) and considering error propagation:
   • The error in \( g \) contributed by \( L \) is 1%.
   • The error in \( g \) due to \( T^2 \) is twice the error in \( T \), which is \( 2 \times 2.5\% = 5\%\).
8. The total relative error in \( g \) is the sum of individual errors: \( 1\% + 5\% = 6\%\).

The accuracy in the measurement of the acceleration due to gravity is determined to be \(6\%\). The correct answer is 6%.