Question

In a simple pendulum experiment, the maximum percentage error in the measurement of length is 2% and that in the observation of the time-period is 3%. Then the maximum percentage error in determination of the acceleration due to gravity \( g \) is

• A. 5%
• B. 6%
• C. 7%
• D. 8%
• E. 10%

Hint:

Always multiply the percentage error of a quantity by the power it is raised to in the formula. Since \( T \) is squared in the denominator for \( g \), its error contribution is doubled.

Answer 1

The Correct Option is D

Concept: This problem involves the propagation of errors in physical measurements. When a physical quantity depends on other measured quantities through a formula, the errors in those quantities combine to give the total error.
• Time Period of a Simple Pendulum: \( T = 2\pi \sqrt{\frac{l}{g}} \).
• Formula for \( g \): Rearranging for \( g \), we get \( g = \frac{4\pi^2 l}{T^2} \).
• Relative Error Rule: For a quantity \( X = \frac{A^a}{B^b} \), the maximum relative error is \( \frac{\Delta X}{X} = a \frac{\Delta A}{A} + b \frac{\Delta B}{B} \).

Step 1: Derive the error relationship for \( g \).
From the formula \( g = \frac{4\pi^2 l}{T^2} \), since \( 4\pi^2 \) is a constant, it does not contribute to the error. The relative error in \( g \) is: \[ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T} \]

Step 2: Calculate the maximum percentage error.
Percentage error is simply the relative error multiplied by 100. \[ \left( \frac{\Delta g}{g} \times 100 \right) = \left( \frac{\Delta l}{l} \times 100 \right) + 2 \left( \frac{\Delta T}{T} \times 100 \right) \] Substituting the given percentage errors: \[ % \text{ error in } g = 2% + 2(3%) \] \[ % \text{ error in } g = 2% + 6% = 8% \]