A pendulum clock loses 12 s a day if the temperature is ${40}^{0}C$  and goes fast by 4 s a day if the temperature is  ${20}^{0}C$. The temperature at which the clock will show correct time is given by  ${\left(20+k\right)}^{0}C$, then the value of $k$ is _____

Let T be the temperature at which the clock is correct. Time lost per day = ½ $\alpha$ (rise in temperature) $\times 86400$ 

$\Rightarrow 12=1/2\alpha \left(40-T\right)\times 86400$………….(i)
Time gained per day $=1/2\alpha$  (drop in temperature) $\times 86400$ 
$\Rightarrow 4=1/2\alpha \left(T-20\right)\times 86400$             …………..(ii)
$\frac{equation\left(1\right)}{equation\left(2\right)}=\frac{12}{4}=\frac{40-T}{T-20}$ 
$T={25}^{0}C$