A pendulum clock nnormally shows correct time. On an extemely cold day, its length decreases by `0.2%`. Compute the error in time per day.

Explanation

The correct time period of pendulum clock is 2 seconds. Let L be its correct length.
`:. 2=2pisqrt((L)/(g))` …(i)
Decrease in length `=0.2%=(0.2)/(100)L`
Length after contraction,
`l=L-(0.2)/(100)L=l(1-(0.2)/(100))`
New time period t will be,
`t=2pisqrt((l)/(g))=wpisqrt((L)/(g)(1-(0.2)/(100)))` ...(ii)
Dividing (ii) by (i) , we get
`(t)/(2)=(1-(0.2)/(100))^(1//2)`
or `t=2(1-(0.2)/(100))^(1//2)=2(1-(1)/(2)xx(0.2)/(100)+...)`
`=(2-(0.2)/(100))s`
which is less than 2 seconds.
Time gained in 2 seconds`=(0.2)/(100)s`
Total time gained in 1 day `(=24xx60xx60s)`
`=(0.2)/(100)xx(24xx60xx60)/(2)=86.4s`