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)

$T=\frac{40 s }{20}=2 s$
Further, $t=n T=20 T$
or $\Delta t=20 \Delta T$
$\therefore \frac{\Delta t}{t}=\frac{\Delta T}{T}$
or $\Delta T=\frac{T}{t} \cdot \Delta t=\left(\frac{2}{40}\right)(1)=0.05 s$
Further, $T=2 \pi \sqrt{\frac{l}{g}}$
or $T \propto g^{-1 / 2}$
$\therefore \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100$
or $\%$ error in determination of $g$ is
$ \frac{\Delta g}{g} \times 100 =-200 \times \frac{\Delta T}{T} $
$=-\frac{200 \times 0.05}{2}=-5 \%$