A pendulum clock having a metal rod keeps correct time at  $20°$C. When temperature falls to $0°C$ , it gains 12 seconds per day.  The coefficient of linear expansion of the metal is $x\times {10}^{-6}/°C$  then  $x$  ( rounded to nearest integer) is___________.

 $T=2\pi \sqrt{\frac{L}{g}}\Rightarrow L=\frac{T^{2}g}{4{\pi }^2}\frac{\Delta L}{L}=\frac{2\Delta T}{T}or\alpha \Delta \theta =2\times \left(\frac{T_{20}-T_0}{T_0}\right)$
Or   $\frac{T_{20}-T_0}{T_0}=\frac{1}{2}\alpha \Delta \theta =\frac{1}{2}\times \alpha \times 20°C=10°C\times \alpha$
Here   is the correct time period .  Since the length L of the rod decreases when cooled, the time period decreases.  So the clock runs fast at  . The time gained in a day  $(=24hrs=24\times 60\times 60s)$ is $10\alpha \times 24hrs=10\alpha \times (24\times 60\times 60s)$
 Thus $12s=10°C\alpha \times 24\times 60s$
 $\Rightarrow \alpha =\frac{12s}{10°C\times (24\times 60\times 60s)}=14\times {10}^{-6}per°C$