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The least count of stop watch is $\frac{1}{5}\,second$. The time of $20$ oscillations of pendulum is measured to be $25\,seconds$. Then percentage error in the measurement of time will be.......... $\%$
$8$
$1$
$0.8$
$16$
Solution
$\frac{\Delta \mathrm{t}}{\mathrm{t}} \times 100=\frac{0.2}{25} \times 100=0.8 \%$
Similar Questions
Students $I$, $II$ and $III$ perform an experiment for measuring the acceleration due to gravity $(g)$ using a simple pendulum.
They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.
Least count for length $=0.1 \mathrm{~cm}$
Least count for time $=0.1 \mathrm{~s}$
Student | Length of the pendulum $(cm)$ | Number of oscillations $(n)$ | Total time for $(n)$ oscillations $(s)$ | Time period $(s)$ |
$I.$ | $64.0$ | $8$ | $128.0$ | $16.0$ |
$II.$ | $64.0$ | $4$ | $64.0$ | $16.0$ |
$III.$ | $20.0$ | $4$ | $36.0$ | $9.0$ |
If $\mathrm{E}_{\mathrm{I}}, \mathrm{E}_{\text {II }}$ and $\mathrm{E}_{\text {III }}$ are the percentage errors in g, i.e., $\left(\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100\right)$ for students $\mathrm{I}, \mathrm{II}$ and III, respectively,