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

Text Solution

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Let T be the temperature at whoch the clock is correct. Time lost per day`=1//2alpha ("rise in temperature")xx86400`
`implies12=(1)/(2alpha)(40-T)xx86400` ..(i)
time gained per day `=1//2alpha("drop in temperature")xx86400`
`implies4=(1)/(2alpha)(T-20)xx86400` ..(ii)
Adding Eqs. (i) and (ii) we get
`32=86400alpha(40-20)impliesalpha=1.85xx10^(-5)//^@C`
Dividing Eq. (i) by Eq. (ii) we get
`12(T-20)=4(40-T)impliesT=25^@C`
`implies` clock shows correct time at `25^@C`
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