Question

An inverted conical tank of w m radius and 4m height is initially full of water, has an outlet at bottom. The outlet is opened at some instant. The rate of flow through the outlet at ny time t is 6h^3/2, where h is height of water level above the outlet at time t. Then the time it takes to empty the tank, is –

• A. $\frac{2\pi}{9}$ units
• B. $\frac{\pi}{9}$ units
• C. $\frac{2\pi}{8}$units
• D. None

Answer 1

Answer: (A)

$\frac{2\pi}{9}$ units

Answer 2

Given that $\frac{dv}{dt}$ = 6h^3/2 ... (1)

At time 't', the volume of water in the tank is v = $\frac{1}{3}$pr²h

D OAB and D OA'B' are similar

Ž $\frac{OA'}{A'B'}$ = $\frac{OA}{AB}$ Ž $\frac{h}{r}$ = $\frac{4}{2}$ Ž r = $\frac{h}{2}$

Ž v = $\frac{1}{3}$ p $\frac{h^{3}}{4}$ Ž h³ = $\frac{12v}{\pi}$

Ž $\frac{dv}{dt}$ = $\frac{–6\sqrt{12}}{\sqrt{\pi}}$ $\sqrt{v}$

Ž $\int_{v_{0}}^{0}\frac{dv}{\sqrt{v}}$ = $\frac { - 6 \sqrt { 12 } } { \sqrt { \pi } }$ $\int_{0}^{t}{dt}$

Ž $\left( 2\sqrt{v} \right)_{v_{0}}^{0}$ = $\frac{- 6\sqrt{12}}{\sqrt{\pi}}$t

Ž $\frac{2\sqrt{v_{0}}\sqrt{\pi}}{6\sqrt{12}}$ = t = $\frac { 2 \sqrt { \pi } } { 6 \sqrt { 12 } }$ $\sqrt{\frac{16}{3}\pi}$

Ž t = $\frac{2\pi}{9}$units.

Hence (1) is the correct answer.