Question

A water tank as the shape of an inverted right circular cone, whose semi-vertical angle is ${\rm{ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\dfrac{{\rm{1}}}{{\rm{2}}}} \right).$ Water is poured into it at a constant rage of $5$ cubic meter per minute. The rate (in m/min), at which the level of water is rising at the instant when the depth of water in the tank is ${\rm{10}}\,{\rm{m}}$ is
$\dfrac{{\rm{2}}}{{\rm{\pi }}}$
$\dfrac{1}{{{\rm{5\pi }}}}$
$\dfrac{1}{{{\rm{10\pi }}}}$
$\dfrac{1}{{{\rm{15\pi }}}}$

Answer 1

 Hint: In this solution, first, we have to find the radius from the semi-vertical angle of the inverted right circular cone which is ${\rm{ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\dfrac{{\rm{1}}}{{\rm{2}}}} \right).$ Then we have to put the value of the radius and height cone on the formula of volume of cone. After that we have to differentiate the volume of the cone. The volume of a cone is given by $\dfrac{1}{3}\pi {r^2}h$, where $r$ is the radius, and $h$ is the height of the cone.

Complete step-by-step answer:
Here,
It is given that,
Semi-vertical angle of the inverted right circular cone is ${\rm{ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\dfrac{{\rm{1}}}{{\rm{2}}}} \right).$
We can write,
${\rm{tan}}\theta = \dfrac{1}{2}$, where $\theta $ is the semi vertical angle.
Water is poured at a constant rage of $5$ cubic meter per minute.
Height of the water tank is ${\rm{10}}\,{\rm{m}}$.
Here,
We can write that, ${\rm{tan}}\theta = \dfrac{r}{h}$.
$\begin{array}{l}\therefore \dfrac{1}{2} = \dfrac{r}{h}\\ \Rightarrow r = \dfrac{h}{2}\end{array}$
Now,
We know that,
Volume of a cone $ = \dfrac{1}{3}\pi {r^2}h$ (1)
Now,
Putting the value of r and h in the equation (1), we get
$\begin{array}{c}{\rm{Volume}}\left( V \right) = \dfrac{1}{3}\pi {\left( {\dfrac{h}{2}} \right)^2}h\\{\rm{ }} = \dfrac{1}{3}\pi \dfrac{{{h^3}}}{4}\end{array}$ (2)
Now,
We need to differentiate the equation (2).
Therefore,
$\begin{array}{l}\dfrac{{dV}}{{dt}} = \dfrac{\pi }{{12}}{\left( {13h} \right)^2}\left( {\dfrac{{dh}}{{dt}}} \right)\\ \Rightarrow 5 = \dfrac{\pi }{4}\left( {100} \right)\dfrac{{dh}}{{dt}}\\ \Rightarrow \dfrac{{dh}}{{dt}} = \dfrac{1}{{5\pi }}\end{array}$
Thus, the rate at which the level of water is rising is $\dfrac{{\rm{1}}}{{{\rm{5\pi }}}}$.

Hence, the correct answer is B.

Note: A right circular cone has a circular base whose axis is perpendicular to the base at the centre point. Here we have to find the water level rising for the given situation. First, we have to find the value of ‘r’ from the semi-vertical angle of the inverted right circular cone which is ${\rm{ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\dfrac{{\rm{1}}}{{\rm{2}}}} \right).$Then we have to put the value of ‘r’ and the value of ‘h’ in the formula of volume of cone and then to find the rate at which the level of water is rising, we need to differentiate the answer from the volume of the cone.