A pendulum clock loses 24 sec a day if the temperature is `60^(@)C` and gains 8 sec a day if the temperature is `40^(@)C` .The temperature at which the clock will show correct time.

Answer

18.9

Explanation

Let it keep correct time at `theta^@C`
So, `2pi sqrt((l_(theta))/(g))=2` sec
At `30^@C`, time for one oscillation = `T_(30)`
`=2 pi sqrt((l_(30))/(g))=2pi sqrt((l_1{1+alpha(30-theta)})/(g))=2pi sqrt((l_1)/(g)){1+alpha(30-theta)}^(1//2)`
`~~2 pi sqrt ((l_(theta))/(g)){1+((30-theta)alpha)/(2)}=2{1+((30-theta)alpha)/(2)}=2+(30-theta)alpha`
So, loss of time for 2 secs is `(30-theta) alpha` sec.
So, loss for 1 day, 86400`(30-theta)alpha` sec. .........(i)
Similarly, gain for 1 day at `10^@C=86400(theta-10)alpha=8` ..... (ii)
Dividing equation (i) by (ii), `therefore (30-theta)/(theta-10)=(10)/(8)=5/4`
or 120-4`theta`=5`theta`-50
or `9 theta=170` or `theta=18.9^@C`.