A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan-1 ((1/2)) . Water is poured into it at a constant rage 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 10 m; is :-

Question

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan-1 ((1/2)) . Water is poured into it at a constant rage 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 10 m; is :- (A) 2 / π  (B) 1 / 5 π  (C) 1/ 10 π  (D) 1/ 15 π

Tardigrade Admin · Accepted Answer

tanθ = (1/2) = (r/h) r = (h/2) V = (1/3) π r2h V = (1/3) π. (h3/4) (dV/dt) = (π/12) (3h)2 ((dh/dt)) 5 = (π/4) .(100) (dh/dt) ⇒ (dh/dt) = (1/5π)

$2/ π$

$1/5 π$

$1/10 π$

$1/15 π$

$tan θ = \frac{1}{2} = \frac{r}{h}$
$r = \frac{h}{2}$
$V = \frac{1}{3} π r^{2} h$
$V = \frac{1}{3} π . \frac{h ^{3}}{4}$
$\frac{d V}{d t} = \frac{π}{12} (3 h)^{2} (\frac{d h}{d t})$
$5 = \frac{π}{4} . (100) \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{1}{5 π}$

2/π

1/5π

1/10π

1/15π

tanθ=21​=hr​ r=2h​ V=31​πr2h V=31​π.4h3​ dtdV​=12π​(3h)2(dtdh​) 5=4π​.(100)dtdh​⇒dtdh​=5π1​

$2/ π$

$1/5 π$

$1/10 π$

$1/15 π$

$tan θ = \frac{1}{2} = \frac{r}{h}$
$r = \frac{h}{2}$
$V = \frac{1}{3} π r^{2} h$
$V = \frac{1}{3} π . \frac{h ^{3}}{4}$
$\frac{d V}{d t} = \frac{π}{12} (3 h)^{2} (\frac{d h}{d t})$
$5 = \frac{π}{4} . (100) \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{1}{5 π}$