JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 18)
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 :-
$${1 \over {15\pi }}$$
$${1 \over {5\pi }}$$
$${1 \over {10\pi }}$$
$${2 \over \pi }$$
Explanation
_9th_April_Evening_Slot_en_18_2.png)
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 = $${\pi \over 4}{\left( {10} \right)^2}{{dh} \over {dt}}$$
$$ \Rightarrow $$ $${{dh} \over {dt}}$$ = $${1 \over {5\pi }}$$
Comments (0)
