A large open tank containing water has two holes to its wall. A square hole of side 'a' is made at a depth…
A large open tank containing water has two holes to its wall. A square hole of side 'a' is made at a depth 'y' and a circular hole of radius 'r' is made at a depth '16y'from the surface of water. If equal amount of water comes out through both the holes per second, then the relation between 'r' and 'a' will be
$r=\frac{2 a}{\pi}$
$r=\frac{a}{2 \sqrt{\pi}}$
$r=\frac{a}{2 \pi}$
$r=\frac{2 a}{\sqrt{\pi}}$
Solution
$v_{1} A_{1}=v_{2} A_{2}$
$\sqrt{2 g h_{1}} A_{1}=\sqrt{2 g h_{2}} A_{2}$
$\sqrt{y} a^{2}=\sqrt{16} y \pi r^{2}$
$a^{2}=4 \pi r^{2}$
$r=\frac{a}{2 \sqrt{\pi}}$