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Two clocks are being tested against a standard clock located in a national laboratory. At $12: 00: 00$ noon by the standard clock, the readings of the two clocks are
$\begin{array}{ccc} & \text {Clock} 1 & \text {Clock} 2 \\ \text { Monday } & 12: 00: 05 & 10: 15: 06 \\ \text { Tuesday } & 12: 01: 15 & 10: 14: 59 \\ \text { Wednesday } & 11: 59: 08 & 10: 15: 18 \\ \text { Thursday } & 12: 01: 50 & 10: 15: 07 \\ \text { Friday } & 11: 59: 15 & 10: 14: 53 \\ \text { Saturday } & 12: 01: 30 & 10: 15: 24 \\ \text { Sunday } & 12: 01: 19 & 10: 15: 11\end{array}$
If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer?
Solution
The range of variation over the seven days of observations is $162 \;s$ for clock $1$ , and $31 \,s$ for clock $2 .$ The average reading of clock $1$ is much closer to the standard time than the average reading of clock $2 .$ The important point is that a clock's zero error is not as significant for precision work as its vartation, because a zero-error can always be easily corrected. Hence clock $2$ is to be preferred to clock $1$
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,