Question 2.26
NCERT Class XI Physics
It is claimed that two caesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
Solution
$ \begin{aligned} 1 \text{ y} &\equiv 365 \times 24 \times 3600\\ & = 31536000 \text{ s}\\\\ \therefore 100\text{ y} &\equiv 100 \times 31536000 \text{ s}\\ &=3153600000 \text{ s}\\ &\approx 3 \times 10^{9} \text{ s} \end{aligned} $
error given as $\displaystyle 0.02 \text{ s in } 100 \text{ y } [\equiv 3 \times 10^9 \text{ s}]$
$\displaystyle \therefore \text{in }1 \text{ s } \equiv \bigg( \frac{3 \times 10^9}{0.02} \bigg) = 1.5 \times 10^{11} \text{ s}$
Significant figures given in 100 years are 1, and in 0.02 are 1. The result is obtained by multipication/division. So SF will be 1.
$\displaystyle \therefore \text{error per }1 \text{ s } \approx 2 \times 10^{11} \text{ s}$
Order of error is 1 part in [order] $\displaystyle 10^{11} \text{ seconds } \: \underline {Ans}$
Video Explanation
Please watch this youtube video for a quick explanation of the solution:
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