A right circular cone has height 9 text{ cm} | vedclass

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(B) For the conical vessel, height $H = 9 	ext{ cm}$ and base radius $R = 5 	ext{ cm}$.Full volume of the vessel is $V_{total} = \frac{1}{3} \pi R^2

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) For the conical vessel, height $H = 9 	ext{ cm}$ and base radius $R = 5 	ext{ cm}$.Full volume of the vessel is $V_{total} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (25)(9) = 75\pi 	ext{ cm}^3$.Let $h$ be the height of water and $r$ be the radius of the water surface at time $t$.By similar triangles, $\frac{r}{h} = \frac{R}{H} = \frac{5}{9}$, so $r = \frac{5h}{9}$.The area of the water surface is $A = \pi r^2 = \pi \left(\frac{5h}{9}ight)^2 = \frac{25\pi h^2}{81}$.Given the rate of change of water level is $\frac{dh}{dt} = \frac{\pi}{A} = \frac{\pi}{\frac{25\pi h^2}{81}} = \frac{81}{25h^2}$.Separating variables, $h^2 \, dh = \frac{81}{25} \, dt$.Integrating both sides, $\int h^2 \, dh = \int \frac{81}{25} \, dt \implies \frac{h^3}{3} = \frac{81}{25}t + C$.Since $h=0$ at $t=0$, we have $C=0$, so $h^3 = \frac{243}{25}t$.The volume of water at height $h$ is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{25h^2}{81}ight) h = \frac{25\pi h^3}{243}$.Substituting $h^3 = \frac{243}{25}t$, we get $V = \frac{25\pi}{243} \left(\frac{243}{25}tight) = \pi t$.For the cone to be completely filled, $V = V_{total} = 75\pi$.Therefore, $\pi t = 75\pi \implies t = 75 	ext{ seconds}$.

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