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A student uses a simple pendulum of exactly $1 \mathrm{~m}$ length to determine $\mathrm{g}$, the acceleration due to gravity. He uses a stop watch with the least count of $1 \mathrm{sec}$ for this and records $40$ seconds for $20$ oscillations. For this observation, which of the following statement$(s)$ is (are) true?
$(A)$ Error $\Delta T$ in measuring $T$, the time period, is $0.05$ seconds
$(B)$ Error $\Delta \mathrm{T}$ in measuring $\mathrm{T}$, 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 $2.5 \%$
$(B,D)$
$(A,D)$
$(C,D)$
$(A,C)$
Solution
$ \frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{\Delta \mathrm{t}}{\mathrm{t}}=\frac{1}{40} $
$ \Delta \mathrm{T}=0.05 \mathrm{sec} $
$ \mathrm{g}=\frac{4 \pi^2 \mathrm{Ln}^2}{\mathrm{t}^2} $
$ \frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{2 \Delta \mathrm{t}}{\mathrm{t}} $
$ \Rightarrow \% \text { error }=\frac{2 \Delta \mathrm{t}}{\mathrm{t}} \times 100=5 \%$