A water tank has the shape of an inverted rig | vedclass

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Question

(A) Let the semi-vertical angle be $\alpha = 	an^{-1}\left(\frac{1}{2}ight)$.Then,$	an \alpha = \frac{r}{h} = \frac{1}{2}$,which implies $r = \fra

Vedclass · Accepted Answer

$\frac{1}{5 \pi}$

Vedclass · Answer

(A) Let the semi-vertical angle be $\alpha = 	an^{-1}\left(\frac{1}{2}ight)$.Then,$	an \alpha = \frac{r}{h} = \frac{1}{2}$,which implies $r = \frac{h}{2}$.The volume of the cone is given by $V = \frac{1}{3} \pi r^2 h$.Substituting $r = \frac{h}{2}$,we get $V = \frac{1}{3} \pi \left(\frac{h}{2}ight)^2 h = \frac{1}{12} \pi h^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = \frac{1}{12} \pi (3h^2) \frac{dh}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt}$.Given $\frac{dV}{dt} = 5 	ext{ m}^3/	ext{min}$,we have $5 = \frac{1}{4} \pi h^2 \frac{dh}{dt}$,so $\frac{dh}{dt} = \frac{20}{\pi h^2}$.At $h = 10 	ext{ m}$,the rate of change of the water level is $\frac{dh}{dt} = \frac{20}{\pi (10)^2} = \frac{20}{100 \pi} = \frac{1}{5 \pi} 	ext{ m/min}$.

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