Question

An atomic clock has an accuracy of 1 part is \[{10^{10}}\]. If two such clocks are operated with precision, then after running for 2500 years they will record a difference of nearly:
A. 1 sec
B. 8 sec
C. 5 sec
D.\[{10^8}\,{\text{sec}}\]

Answer 1

Hint:From the given information, first determine rate of the difference occurring in the clock using the given value of accuracy of the clock. Convert the unit of time for which the two atomic clocks are run in the SI system of units. Then determine the time difference between the two clocks using the determined rate of occurring time difference.

Complete step by step answer:
We have given that the atomic clock has an accuracy of 1 part in \[{10^{10}}\]. This shows that there will be a difference of 1 second in \[{10^{10}}\] seconds when the time in atomic clock is compared with the accurate clock.
Hence, the rate of the difference occurring in the time of the clock is
\[R = \dfrac{1}{{{{10}^{10}}\,{\text{sec}}}}\]
Two atomic clocks have operated with precision for 2500 years.
\[t = 2500\,{\text{years}}\]
We have to determine the difference in the two clocks comparing the given accuracy of 1 part in \[{10^{10}}\].
Convert the time of 2500 years in seconds.
\[t = \left( {2500\,{\text{years}}} \right)\left( {\dfrac{{365.25\,{\text{days}}}}{{1{\text{ year}}}}} \right)\left( {\dfrac{{24\,{\text{h}}}}{{1\,{\text{day}}}}} \right)\left( {\dfrac{{60\,\min }}{{1\,{\text{h}}}}} \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)\]
\[ \Rightarrow t = 7.8894 \times {10^{10}}\,{\text{sec}}\]
So, we have to determine the time difference between the two atomic clocks after \[7.8894 \times {10^{10}}\,{\text{sec}}\].
The time difference \[\Delta t\] between the two atomic clocks after \[7.8894 \times {10^{10}}\,{\text{sec}}\] will be
\[\Delta t = Rt\]
Substitute \[\dfrac{1}{{{{10}^{10}}\,{\text{sec}}}}\] for \[R\] and \[7.8894 \times {10^{10}}\,{\text{sec}}\] for \[t\] in the above equation.
\[\Delta t = \left( {\dfrac{1}{{{{10}^{10}}\,{\text{sec}}}}} \right)\left( {7.8894 \times {{10}^{10}}\,{\text{sec}}} \right)\]
\[ \Rightarrow \Delta t = 7.8894\,{\text{sec}}\]
\[ \therefore \Delta t \approx 8\,{\text{sec}}\]
Therefore, the time difference recorded between the two atomic clocks after 2500 years will be \[8\,{\text{sec}}\].

Hence, the correct option is B.

Note: One should keep in mind that the given accuracy of the clock is always in seconds and not in any other units. The unit of this accuracy will be different only if it is given in the question. The students should not forget to convert the unit of the time in the SI system of units. Otherwise the rate of occurrence of time difference will be in seconds and the time will be in years.