Question

Liquid is filled in a container upto a height of H. A small hle is made at the bottom of the tank. Time taken to empty from H to (H)/(3) is t_(0). Find the time taken to empty tank from (H)/(3) to zero.

Answer 1

The time taken to empty tank from H/3 to zero is:

• Liquid is filled in a container up to a height of H. A small hole is made at the bottom of the tank. Time taken to empty from H to H/3 is t₀.

• Time taken for tank, $T = \frac{A}{a} * \sqrt{ \frac{2 *H}{g} }$

               where A = area of tank

                           a = area of orifice

                           H = height emptied in time T

                           g = acceleration due to gravity

•    Time from H to 0 = Time from H to H/3 + Time from H/3 to 0

        ⇒ Time from H/3 to 0 = $\frac{A}{a} * \sqrt{ \frac{2 *H}{g} }$ - t₀ = $\frac{A}{a} * \sqrt{ \frac{2 *H}{3 * g} }$

        ⇒  t₀ = $\frac{A}{a} * \sqrt{ \frac{2 *H}{g} } * \frac{\sqrt{3} -1 }{\sqrt_{3}}$

• Time from H/3 to 0 = $\frac{A}{a} * \sqrt{ \frac{2 *H}{3 * g} }$ = t₀* $\frac{1} {\sqrt {3}-1}$