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Q.

A conical vessel of height 10 ft and semivertical angle $30^{\circ}$ is full of water. If empties in such away that the height at the water in the vessel is decreasing at a constant rate of 1 ft/min. Then the rate at which the volume of water in the vessel is decreasing when its height is 6 ft. is

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a

$π$

b

$12 π$

c

$\frac{π}{2}$

d

$3 π$

answer is B.

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Detailed Solution

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$h = 10 θ = 30, \frac{dh}{dt} = 1 \tan 30^{0} = \frac{r}{h} \Rightarrow r = \frac{h}{\sqrt{3}} h = 6 V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \times \frac{h^{2}}{3} \times h V = \frac{1}{9} π h^{3} \frac{dv}{dt} = \frac{1}{9} π (3 h^{2}) \frac{dh}{dt} = \frac{1}{3} π \times 36 \times 1 = 12 π$

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