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?
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?
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
A scientist performs an experiment in order to measure a certain physical quantity and takes $100$ observations. He repeats the same experiment and takes $400$ observations. By doing so,
A scientist performs an experiment in order to measure a certain physical quantity and takes $100$ observations. He repeats the same experiment and takes $400$ observations. By doing so,
The period of oscillation of a simple pendulum is $T=2\pi \sqrt {\frac{l}{g}} $. Measured value of $L$ is $20.0\; cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90\ s$ using a wrist watch of $1\; s$ resolution. The accuracy in the determination of $g$ is ........ $\%$
The period of oscillation of a simple pendulum is $T=2\pi \sqrt {\frac{l}{g}} $. Measured value of $L$ is $20.0\; cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90\ s$ using a wrist watch of $1\; s$ resolution. The accuracy in the determination of $g$ is ........ $\%$
- [JEE MAIN 2015]
The temperatures of two bodies measured by a thermometer are $t_{1}=20^{\circ} C \pm 0.5^{\circ} C$ and $t_{2}=50^{\circ} C \pm 0.5^{\circ} C$ Calculate the temperature difference and the error theirin.
The temperatures of two bodies measured by a thermometer are $t_{1}=20^{\circ} C \pm 0.5^{\circ} C$ and $t_{2}=50^{\circ} C \pm 0.5^{\circ} C$ Calculate the temperature difference and the error theirin.
A wire has a mass $0.3 \pm 0.003\,g$, radius $0.5 \pm 0.005\,mm$ and length $6 \pm 0.06\,cm$. The maximum percentage error in the measurement of its density is .......... $\%$
A wire has a mass $0.3 \pm 0.003\,g$, radius $0.5 \pm 0.005\,mm$ and length $6 \pm 0.06\,cm$. The maximum percentage error in the measurement of its density is .......... $\%$
- [IIT 2004]
The following observations were taken for determining surface tension $T$ of water by capillary method:
diameter of capillary, $D= 1.25 \times 10^{-2}\; m$
rise of water, $h=1.45 \times 10^{-2}\; m $
Using $g= 9.80 \;m/s^2$ and the simplified relation $T = \frac{{rhg}}{2}\times 10^3 N/m$ , the possible error in surface tension is ........... $\%$
The following observations were taken for determining surface tension $T$ of water by capillary method:
diameter of capillary, $D= 1.25 \times 10^{-2}\; m$
rise of water, $h=1.45 \times 10^{-2}\; m $
Using $g= 9.80 \;m/s^2$ and the simplified relation $T = \frac{{rhg}}{2}\times 10^3 N/m$ , the possible error in surface tension is ........... $\%$
- [JEE MAIN 2017]