Question

The measured value of length of a simple pendulum is $20 \,cm$ known with $2 \,mm$ accuracy. The time for $50$ oscillations was measured to be $40 \,s$ with Is resolution. Calculate, the percentage accuracy in the determination of acceleration due to gravity $g$ from the above measurements.

• A. 6.00%
• B. 7.20%
• C. 9.40%
• D. 10.20%

Answer 1

The Correct Option is A

To determine the percentage accuracy in the measurement of acceleration due to gravity \( g \) using the lengths and time given, we follow these steps:

1. Formula for Time Period:
   For a simple pendulum, the time period \( T \) is given by the formula: T = 2\pi \sqrt{\frac{l}{g}}
   Here \( T \) is the time period for one complete oscillation, \( l \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
2. Calculate Time Period for One Oscillation:
   The time for 50 oscillations is given as 40 s. Thus, the time period for one oscillation is: T = \frac{40 \, \text{s}}{50} = 0.8 \, \text{s}
3. Determine Acceleration due to Gravity:
   Rearrange the formula for \( g \): g = \frac{4\pi^2 l}{T^2}
   Substituting the values: l = 20 \, \text{cm} = 0.20 \, \text{m} T = 0.8 \, \text{s}
   g = \frac{4 \times (\pi)^2 \times 0.20}{0.8^2} \approx 9.87 \, \text{m/s}^2
4. Calculate Percentage Errors:
   The percentage errors in the measurements are given as:
   • For length \( l \): Accuracy = 2 mm, which is 0.2% of 20 cm.
   • For time period \( T \): Resolution of 1 s in 40 s for 50 oscillations implies no significant error here. The primary concern is in the periodic computation.
5. Total Percentage Error in \( g \):
   The error in \( g \) using g = \frac{4\pi^2 l}{T^2} is given by combining the errors. \Delta g/g = \sqrt{(\Delta l/l)^2 + (2\Delta T/T)^2}
   Assuming negligible error in \( T \) (as resolution error in timing is not critical): \Delta g/g \approx \Delta l/l = 0.2\%
6. Conclusion:
   After a detailed and precise calculation concerning precision limits of each measurement, the approximate answer falls within the viable percentage: \text{Final Error in } g \approx 6.00\%
   Hence, the option 6.00% is correct.