Question

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

• A. Error $\Delta T$ in measuring $T$, the time period, is $0.05$ seconds
• B. Error $\Delta T$ in measuring $T$, the time period, is $1$ second
• C. Percentage error in the determination of is $5\%$
• D. Both (a) and (c)

Answer 1

Answer: (D)

Both (a) and (c)

Answer 2

Relative error in measurement of time,
$\frac{\Delta t}{t}=\frac{1\,s}{40\,s}=\frac{1}{40}$
Time period, $T=\frac{40\,s}{20}=2\,s$
Error in measurement of time period,
$\Delta T=T\times\frac{\Delta t}{t}$
$=2\,s\times\frac{1}{40}=0.05\,s$
The time period of simple pendulum is
$Y=2\pi\sqrt{\frac{l}{g}}$ or $T^{2}=\frac{4\pi^{2}l}{g}$ or $g=\frac{4\pi^{2}l}{T^{2}}$
$\therefore \frac{\Delta g}{g}=\frac{2\Delta T}{T}=2\times\frac{1}{40}=\frac{1}{20}\quad\left(\because \frac{\Delta T}{T}=\frac{\Delta t}{t}\right)$
Percentage error in determination of $g$ is
$\frac{\Delta g}{g}\times100$
$=\frac{1}{20}\times100=5\%$