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. 1.33 %
• B. 1.03 %
• C. 1.13 %
• D. 1.30 %

Hint:

When a quantity depends on multiple variables raised to powers, the percentage error is the sum of the products of the power and the percentage error of each variable: $\frac{\Delta y}{y} = \sum |n_i| \frac{\Delta x_i}{x_i}$.

Answer 1

The Correct Option is C

Step 1: From $T = 2\pi \sqrt{\frac{L}{g}}$, we get $g = \frac{4\pi^2 L}{T^2}$.
Step 2: The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.
Step 3: Substitute the given values: $L = 1.0 \text{ m}$, $\Delta L = 1 \text{ mm} = 0.001 \text{ m}$, $T = 1.95 \text{ s}$, and $\Delta T = 0.01 \text{ s}$. \[ \frac{\Delta g}{g} \times 100 = \left( \frac{0.001}{1.0} + 2 \times \frac{0.01}{1.95} \right) \times 100 \] \[ % \text{ error} = (0.001 + 0.010256) \times 100 \approx 1.1256% \approx 1.13% \]