A closed conical vessel is filled with water | vedclass

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(D) Let $R$ and $H$ be the radius and height of the full cone,respectively. The volume of the full cone is $V = \frac{1}{3} \pi R^2 H$.Since the water

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$3$

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(D) Let $R$ and $H$ be the radius and height of the full cone,respectively. The volume of the full cone is $V = \frac{1}{3} \pi R^2 H$.Since the water is let out at a constant speed,the rate of change of volume is constant,i.e.,$\frac{dV}{dt} = -k$ for some constant $k > 0$.At any height $h$,the radius $r$ of the water surface is given by $\frac{r}{h} = \frac{R}{H}$,so $r = \frac{R}{H} h$.The volume of water at height $h$ is $V(h) = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{R}{H} hight)^2 h = \frac{\pi R^2}{3 H^2} h^3$.Given that at $t = 0$,$h = H$,so $V(H) = \frac{1}{3} \pi R^2 H$.At $t = 21 \, min$,$h = \frac{H}{2}$,so the volume of water remaining is $V\left(\frac{H}{2}ight) = \frac{\pi R^2}{3 H^2} \left(\frac{H}{2}ight)^3 = \frac{1}{8} \left(\frac{1}{3} \pi R^2 Hight) = \frac{V}{8}$.The volume of water removed in $21 \, min$ is $V - \frac{V}{8} = \frac{7V}{8}$.Since the rate of outflow is constant,the time taken to remove the remaining volume $\frac{V}{8}$ is $t'$.Using the proportion: $\frac{	ext{Volume removed}}{	ext{Time taken}} = 	ext{constant}$.$\frac{7V/8}{21} = \frac{V/8}{t'}$.$t' = 21 	imes \frac{V/8}{7V/8} = 21 	imes \frac{1}{7} = 3 \, min$.Thus,it requires $3 \, min$ more to empty the vessel.

$A$ closed conical vessel is filled with water fully and is placed with its vertex down. The water is let out at a constant speed. After $21 \, min$,it was found that the height of the water column is half of the original height. How much more time in minutes does it require to empty the vessel?

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