A tank is filled upto a height h with a liquid and is placed on a platform of height h from the ground. To get maximum range x_m a small hole is punched at a distance of y from the free surface of the liquid. Then

Velocity of liquid through orifice, $v=\sqrt{2gy}$
 and time taken by liquid to reach the ground 
 $t=\sqrt{\frac{2\left(h+h-y\right)}{g}}=\sqrt{\frac{2\left(2h-y\right)}{g}}$
Horizontal distance covered by liquid
$x=vt=\sqrt{2gy}\times \sqrt{\frac{2\left(2h-y\right)}{g}}=\sqrt{4y\left(2h-y\right)}$ 
or  $x^2=4y\left(2h-y\right)$
$\therefore \frac{d\left(x^2\right)}{dy}=8h-8y$
For x to be maximum $\frac{d}{dy}\left(x^2\right)=0$
$\therefore 8h-8y=0$  or  $h=y$
So,  $x_m=\sqrt{4h\left(2h-h\right)}=2h$
Hence the correct answer is option (a)