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A pendulum clock loses 12 s a day if the...

A pendulum clock loses 12 s a day if the temperature is `40^(@)C` and gains 4 s a day if the temperature is `20^(@)C`. The temperature at which the clock will show correct time, and the coefficient of linear expansion `(alpha)` of the metal of the pendulum shaft are respectively

A

`25^(@)C " and " 1.85 times 10^(-5)//^(@)C`

B

`60^(@)C " and " 1.85 times 10^(-4)//^(@)C`

C

`30^(@)C " and " 1.85 times 10^(-3)//^(@)C`

D

`55^(@)C " and " 1.85 times 10^(-2)//^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
A
Increase or decrease in time for change in temperature `DeltaT`,
`" "Deltat=1/2alphaDeltaTt`
`therefore " "1/2alpha(40-T_(0)) times 1 " ...(1)"`
`" ["T_(0)=` temperature at which clock gives accurate reading, t = 1 day]
and `4=1/2alpha(T_(0)-20) times 1 " ...(2)"`
(1) + (2) we get,
`" "3=(40-T_(0))/(T_(0)-20) " or, " 3T_(0)-60=T_(0)`
or, `" "T_(0)=100/4=25^(@)C`
Putting the value of `T_(0)` in equation (1) we get,
`" "15alpha=24/(24 times 60 times 60)`
or, `" "alpha=1/(15 times 60 times 60)=1.85 times 10^(-5)//^(@)C`
The option (A) is correct.
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