Question

The measured value of length of a simple pendulum is $20 \,cm$ known with $2 \,mm$ accuracy. The time for $50$ oscillations was measured to be $40 \,s$ with Is resolution. Calculate, the percentage accuracy in the determination of acceleration due to gravity $g$ from the above measurements.

• A. 6.0%
• B. 7.2%
• C. 9.4%
• D. 10.2%

Answer 1

Answer: (A)

The errors in both $I$ and $T$ are least count errors.
$T=2 \pi \sqrt{l / g} $
$\Rightarrow T^{2}=4 \pi^{2} . l / g $
$\therefore g=4 \pi^{2} l / T^{2}$
The errors in both $I \,\&\, T$ are least count errors
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T}{T}$
Length of simple pendulum $l=20\, cm$
acuracy $\Delta l=2\, mm =0.2\, cm$
Time for $50$ oscillations $T=40 \, s$
resolution $\Delta T=1 \, s$
$\therefore \frac{\Delta g}{g}=\left(\frac{0.2}{20}\right)+2\left(\frac{1}{40}\right)$
$=\frac{0.2}{20}+\frac{2}{40}=\frac{1.2}{20}$
Percentage error
$\frac{\Delta g}{g} \times 100=\frac{1.2}{20} \times 100=\pm 6 \%$
$\%$ accuracy in $g=6 \%$