Discussion Forum

A large open tank containing water has two holes to its wall. A square hole of side a is made at a depth of y and a circular hole of radius r is made at a depth of 16 y from the surface of the water. If an equal amount of water comes out through both the holes per second, then  the relation between r and a will be

Answer:

Explanation:

 Using the principle  of continuity,

              A₁ V₁= A₂ V₂              .............(i)

 where, A₁  is the area of square hole, A₁= a²,

 A₂  is the area of circular  hole,  $A_{2}=\pi r^{2}$,

 v₁ is the velocity at depth y,  $v_{1}=\sqrt{2gy}$

 and v₂  is the velocity  at depth 16y,  $v_{2}=\sqrt{2g(16y)}$

 Substituting values in Eq.(i) , we get

$a^{2}(\sqrt{2gy})=\pi r^{2}(\sqrt{2g(16y)})$

 $a^{2}(\sqrt{2gy})=4\pi r^{2}(\sqrt{2gy})$

$r^{2}=\frac{a^{2}}{4\pi}$

$\therefore$        $ r=\frac{a^{}}{2\sqrt{\pi}}$