Ncert Exercises & Solutions

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 caesium clock in measuring a time interval of 1 s?

$100 y e a r s = 100 \times 365 \times 24 \times 60 \times 60 = 3.154 \times 10^{9} s .$
$G i v e n, d i f f e r e n c e b e t w e e n t h e t w o c l o c k s a f t e r 100 y e a r s = 0.02 s$
$T i m e d i f f e r e n c e i n 1 s = \frac{0.02}{3.15 \times 10^{9}} = 6.35 \times 10^{- 12} s .$
$S o, a c c u r a c y i n m e a s u r i n g a t i m e i n t e r v a l o f 1 s = \frac{1}{6.35 \times 10^{- 12}}$
$= 1.57 \times 10^{11} = 10^{11} (A p p r o x .)$
$T h e r e f o r e, a c c u r a c y o f 1 p a r t i n 10^{11} t o 10^{12}$

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