Question

The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:

• A. 4
• B. 8
• C. 6
• D. 5

Answer 1

The Correct Option is C

The period \( T \) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{\ell}{g}}. \]

Rearrange to solve for \( g \):

\[ g = \frac{4\pi^2 \ell}{T^2}. \]

The percentage error in \( g \) is given by:

\[ \frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}. \]

Substitute the values:

\[ \Delta \ell = 0.2 \, \text{cm} = 0.002 \, \text{m}, \quad \ell = 0.2 \, \text{m}, \] \[ \Delta T = 1 \, \text{s}, \quad T = \frac{40}{50} = 0.8 \, \text{s}. \]

Calculate the percentage errors:

\[ \frac{\Delta \ell}{\ell} = \frac{0.002}{0.2} = 0.01 = 1\%. \] \[ 2 \frac{\Delta T}{T} = 2 \times \frac{1}{40} = 0.05 = 5\%. \]

Therefore, the total percentage error in \( g \) is:

\[ 1\% + 5\% = 6\%. \]

Thus, \( N = 6 \).