Question

The period of oscillation of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$ , where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has $2000$ divisions. If the measured value L is $50\, cm,$ the accuracy in the determination of g is $1.1\%$ and the time taken for 100 oscillations is 100 seconds, what should be the resolution of the clock (in milliseconds) .

• A. 1
• B. 2
• C. 5
• D. 0.25
• E. 0.1

Answer 1

Answer: (C)

Given,
$T=2 \pi \sqrt{\frac{l}{g}}$
or $T^{2}=4 \pi^{2} \frac{l}{g}$
$\therefore 2 \frac{\Delta T}{T}=\frac{\Delta l}{l}+\frac{\Delta g}{g}\,...(i)$
Now, $l=50\, cm, \Delta l=2\, mm =0.2\, cm$
$\frac{\Delta g}{g}=1.1 \%=\frac{1.1}{100}$
Put these values in Eq. (i), then we get
$\frac{\Delta T}{T} =\frac{1}{2}\left[\frac{0.2}{50}+\frac{1.1}{100}\right]$
$=7.5 \times 10^{-3} s$ or $7.5\, ms$
$\because$ In $100\, s$, resolution of clock is $7.5\, ms$
$\therefore $ In $60\, s$ resolution of clock is
$\frac{7.5 \times 60}{100} \approx 5\, ms$