Question

The time period of a simple pendulum is given by T = 2π √(l/g). The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is :

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

When calculating percentage error, remember that powers in the formula (like $T^2$) become multipliers for the relative error.

Answer 1

The Correct Option is B

Step 1: $g = 4\pi^2 \frac{l}{T^2}$.
Step 2: $\frac{\Delta g}{g} \times 100 = (\frac{\Delta l}{l} + 2\frac{\Delta T}{T}) \times 100$. Note: $\frac{\Delta T}{T} = \frac{\Delta t}{t}$ where $t$ is total time.
Step 3: $%$ error $= (\frac{0.1}{10} + 2\frac{1}{100}) \times 100 = (0.01 + 0.02) \times 100 = 3%$.