Question

The period of oscillation of a simple pendulum is $T =2 \pi \sqrt{\frac{ L }{ g }}$. Measured value of 'L' is $1.0 m$ from meter scale having a minimum division of $1 mm$ and time of one complete oscillation is 1.95 s measured from stopwatch of $0.01 s$ resolution. The percentage error in the determination of 'g' will be :

• A. $1.13 \%$
• B. $1.03 \%$
• C. $1.33 \%$
• D. $1.30 \%$

Answer 1

The Correct Option is A

1. We begin by analyzing the period of oscillation formula for a simple pendulum: T = 2 \pi \sqrt{\frac{L}{g}}, where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
2. To find the percentage error in g, we need the formula for g in terms of the measured quantities: g = \frac{4 \pi^2 L}{T^2}.
3. We must consider percentage error propagation for both L and T:
   • The percentage error in L is computed as: \frac{\Delta L}{L} \times 100\%. Given L = 1.0 \text{ m}, and the minimum division of the meter scale is 1 \text{ mm} = 0.001 \text{ m}, the percentage error in L is \frac{0.001}{1.0} \times 100\% = 0.1\%.
   • The percentage error in T is: \frac{\Delta T}{T} \times 100\%. Given T = 1.95 \text{ s} and resolution of the stopwatch is 0.01 \text{ s}, the percentage error in T is \frac{0.01}{1.95} \times 100\% \approx 0.513\%.
4. According to the error propagation rules in multiplication and division, the percentage error in g is calculated as: \text{Percentage error in } g = 2 \times \text{Percentage error in } T + \text{Percentage error in } L. Substituting the known values, it becomes 2 \times 0.513\% + 0.1\% = 1.126\%.
5. Rounding this value to the appropriate number of significant figures gives us approximately 1.13\%.
6. Thus, the percentage error in the determination of g is 1.13\%, which matches the given correct answer. This indicates that the computations and assumptions align with the problem's requirements.