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

A pendulum clock loses 12s a day if the temperature is `40^@C` and gains 4s 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:

A

`30^@C; alpha = 1.85 xx 10^-3//^@C`

B

`55^@C; alpha = 1.85 xx 10^-2//^@C`

C

`25^@C; alpha = 1.85 xx 10^-5//^@C`

D

`60^@C; alpha = 1.85 xx 10^-4//^@C`

Text Solution

Verified by Experts

The correct Answer is:
C
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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