Question

Water is filled into a right inverted conical tank at a constant rate of $3{\text{m}}^3\text{/sec},$ whose semi vertical angle is ${\cos }^{-1}\frac{4}{5}.$ The rate $\left(\text{in}\text{m}\text{/sec}\right),$ at which the level of water is rising at the instant when the depth of water in the tank is $4\text{m},$ is

• A. $\frac{1}{\pi }$
• B. $\frac{1}{2\pi }$
• C. $\frac{1}{3\pi }$
• D. $\frac{1}{4\pi }$

Answer 1

The correct option is C $\frac{1}{3\pi }$

As, $\theta ={\cos }^{-1}\frac{4}{5},\Rightarrow \cos \theta =\frac{4}{5}$
From figure :

$r=h\tan \theta =\frac{3h}{4}$
$\Rightarrow V=\frac{\pi r^{2}h}{3}=\frac{3\pi h^3}{16}\Rightarrow \frac{dV}{dt}=\frac{9\pi h^2}{16}\cdot \frac{dh}{dt}=3$
At $h=4:\frac{dh}{dt}=\frac{1}{3\pi }$