A pendulum clock, made of a material having coefficient of linear expansion `alpha=9xx10^(-7)//.^(@)C` has a period of 0.500 sec at `20^(@)C`. If the

Question

A pendulum clock, made of a material having coefficient of linear expansion `alpha=9xx10^(-7)//.^(@)C` has a period of 0.500 sec at `20^(@)C`. If the

A pendulum clock, made of a material having coefficient of linear expansion `alpha=9xx10^(-7)//.^(@)C` has a period of 0.500 sec at `20^(@)C`. If the clock is used in a climate where temperature averages `30^(@)C`, what correction is necessary at the end of 30 days to the time given by clock?
A. `2.5xx10^(-7)` s
B. `5xx10^(-7)s`
C. `1.125xx10^(-6)s`
D. `2.25xx10^(-6)` s

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
Given, the coefficient of linear expansion,
`alpha=9xx10^(-7)""^(@)C^(-1)`
Initial time period, `T_(0)=0.5 s`, initial temperature `T_(i) = 20°` C and final temperature, `T_(f) = 30°C`
Expansion in length, `Deltal = l alpha (T_(f) - T_(i))`
`!Deltal=lxx9xx10^(-7)(30-20)`
Now, the time period of pendulum,
`T=2pisqrt(l/g)`
Error in time period,
`(Deltat)/(T)=1/2(Deltal)/l+1/2(Deltag)/g`
Since, `Deltag=0`
`!(DeltaT)/T=1/2(Deltal)/l`
Now, substituting values in the above equation we get,
`!(Delta_(T))/(0.5)=1/2[(lxx9xx10^(-7))/l(30-20)]`
`!DeltaT=(0.5xx9xx10^(-7)xx10)/2`
`!DeltaT=2.25xx10^(-6)s`