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A student uses a simple pendulum of exactly $1 \mathrm{~m}$ length to determine $g$, the acceleration due to gravity. He uses a stop watch with the least count of $1 \mathrm{~s}$ for this and records $40 \mathrm{~s}$ for 20 oscillations. For this observation, which of the following statement(s) is/are true?
Options:
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Verified Answer
The correct answers are:
Error $\Delta T$ in measuring $T$, the time period, is $0.05 \mathrm{~s}$
,
Percentage error in the determination of $g$ is $5 \%$
Error $\Delta T$ in measuring $T$, the time period, is $0.05 \mathrm{~s}$
,
Percentage error in the determination of $g$ is $5 \%$
$T=\frac{40 \mathrm{~s}}{20}=2 \mathrm{~s}$.
Further, $t=n T=20 T$ or $\Delta t=20 \Delta T$
$\therefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T}$
or $\Delta T=\frac{T}{t} \cdot \Delta t=\left(\frac{2}{40}\right)(1)=0.05 \mathrm{~s}$
Further, $T=2 \pi \sqrt{\frac{l}{g}}$ or $\quad T \propto g^{-1 / 2}$
$$
\therefore \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100
$$
or $\%$ error in determination of $g$ is
$$
\begin{aligned}
\frac{\Delta g}{g} \times 100 & =-200 \times \frac{\Delta T}{T} \\
& =-\frac{200 \times 0.05}{2}=-5 \%
\end{aligned}
$$
$\therefore$ correct options are (a) and (c).
Further, $t=n T=20 T$ or $\Delta t=20 \Delta T$
$\therefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T}$
or $\Delta T=\frac{T}{t} \cdot \Delta t=\left(\frac{2}{40}\right)(1)=0.05 \mathrm{~s}$
Further, $T=2 \pi \sqrt{\frac{l}{g}}$ or $\quad T \propto g^{-1 / 2}$
$$
\therefore \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100
$$
or $\%$ error in determination of $g$ is
$$
\begin{aligned}
\frac{\Delta g}{g} \times 100 & =-200 \times \frac{\Delta T}{T} \\
& =-\frac{200 \times 0.05}{2}=-5 \%
\end{aligned}
$$
$\therefore$ correct options are (a) and (c).
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