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

Answer: (C)

6

Answer 2

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, $N = 6$.