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

To determine the percentage accuracy in the determination of 'g' using the given pendulum, we start by understanding the formula for the time period of a simple pendulum:

T = 2\pi \sqrt{\frac{l}{g}}

Where:

• T is the time period.
• l is the length of the pendulum.
• g is the acceleration due to gravity.

We are given that:

• The length of the pendulum l = 10 cm with an accuracy of 1 mm (0.1 cm).
• The total time for 200 oscillations is 100 seconds.
• The resolution of the clock is 1 second.

First, calculate the time period for one oscillation:

T = \frac{100}{200} = 0.5 seconds

The formula for percentage error in 'g' is derived from the formula:

\Delta T = \frac{1}{2} T \left( \frac{\Delta l}{l} + 2 \frac{\Delta g}{g} \right)

Rearranging for percentage error in 'g':

\frac{\Delta g}{g} \approx \frac{2 \Delta T}{T} + \frac{\Delta l}{l}

Given:

• \Delta l = 0.1 cm
• l = 10 cm
• \Delta T = 1/200 seconds
• T = 0.5 seconds

Calculate the individual percentage errors:

• \frac{\Delta l}{l} = \frac{0.1}{10} = 0.01
• \frac{2 \Delta T}{T} = \frac{2 \times (1/200)}{0.5} = 0.02

Now, substitute these into the formula:

\frac{\Delta g}{g} \approx 0.01 + 0.02 = 0.03

This means the percentage accuracy is approximately 3%.

Thus, the value of 'x' to the nearest integer is 3%.