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 $0.02~\text s.$What does this imply for the accuracy of the standard caesium clock in measuring a time interval of $1~\text s$?
1. $6\times10^{-12}~\text s$
2. $6\times10^{-10}~\text s$
3. $6\times10^{-8}~\text s$
4. $6\times10^{-6}~\text s$

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

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