A water tank has the shape of an inverted right circular cone, whose semi-vertical angle i…

Question

A water tank has the shape of an inverted right
circular cone, whose semi-vertical angle is
$${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$. Water is poured into it at a constant
rate of 5 cubic meter per minute. The 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 10m; is :-

Accepted Answer

Given $${{dv} \over {dt}}$$ = 5 cm3/min

and $$\theta $$ = $${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$

$$ \Rightarrow $$ tan $$\theta $$ = $${1 \over 2}$$ = $${r \over h}$$

$$ \Rightarrow $$ h = 2r

Volume of the cone,

v = $${1 \over 3}\pi {r^2}h$$ = $${1 \over 3}\pi {\left( {{h \over 2}} \right)^2}h$$ = $${{\pi {h^3}} \over {12}}$$

$$ \therefore $$ $${{dv} \over {dt}} = {\pi  \over {12}}\left( {3{h^2}} \right){{dh} \over {dt}}$$

$$ \Rightarrow $$ 5 = $${\pi  \over 4}{h^2}{{dh} \over {dt}}$$

$$ \Rightarrow $$ 5 = $${\pi  \over 4}{\left( {10} \right)^2}{{dh} \over {dt}}$$

$$ \Rightarrow $$ $${{dh} \over {dt}}$$ = $${1 \over {5\pi }}$$

Solutions · 1