Question

A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations. For this observation, which of the following statements is true?

• A. Error $\Delta$T in measuring T, the time period, is 0.02 seconds
• B. Error $\Delta$T in measuring T, the time period, is 1 second
• C. Percentage error in the determination of g is 5$\%$
• D. Percentage error in the determination of g is 2.5

Answer 1

The correct option is C Percentage error in the determination of g is 5$\%$

Relative error in measurement of time,
$\frac{\Delta t}{t}=\frac{1s}{40s}=\frac{1}{40}$
Time period, $T=\frac{40s}{20}=2s$
Error in measurement of time period,
$\Delta T=T\times \frac{\Delta t}{t}=2s\times \frac{1}{40}=0.05s$
The time period of simple pendulum is
$T=2\pi \frac{l}{g}or{T}^2=\frac{4{\pi }^{2}l}{g}org=\frac{4{\pi }^{2}l}{T^2}\therefore \frac{\Delta g}{g}=\frac{2\Delta T}{T}=2\times \frac{1}{40}=\frac{1}{20}(\because \frac{\Delta T}{T}=\frac{\Delta T}{t})$
Percentage error in determination of g is
$\frac{\Delta g}{g}\times 100=\frac{1}{20}\times 100=5\%$