Question

The least count of a stop watch is 0.2 s. The time of 20 oscillations of a pendulum is measured to be 25 s. The percentage error in the measurement of time will be:

• A. 16 %
• B. 1.8 %
• C. 0.8 %
• D. 0.1 %

Answer 1

Answer: (C)

0.8 %

Number of oscillations $n=20$

Time taken $=25$ $s$

$\therefore$ Time period of oscillation $T=\frac{25}{20}=1.25$ $s$

Given :
Error in time for $20$ oscillations, $\Delta T=0.2$ $s$
Hence error in time for 1 oscilations, $\Delta T^{\prime }=\frac{0.2}{n}=$ $s$

$\therefore$ Percentage error $=\frac{\Delta T}{T}\times 100=\frac{0.2}{20\times 1.25}\times 100=0.8$ $\%$