If the length of a seconds pendulum is increased by 1%, then it loses number of oscillations in a day is 72p then p value is _____

$T=2\pi \sqrt{\frac{l}{g}}\Rightarrow T\alpha \sqrt{l}$  
$\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=0.5\%\Rightarrow$ the time period increases by 0.5%
The time loss perday = $\frac{0.5}{100}\times 86400=432\sec$
$\because$  Time period of seconds pendulum is 2 sec
The no. of oscillations lost perday 432/2..
$216=72x$ 
$x=3$