Question

# A water tank has the slope of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan−1(0.5). Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m. (π=227)

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Solution

## Let r,h and α be the radius, height and the semi-vertical angle of the given conical water tank. ⇒tanα=rh⇒α=tan−1(rh) As α=tan−1(0.5) (given) Therefore, rh=0.5 ⇒r=h2 ⋯(1) Let the volume of the cone be V Therefore, V=13πr2h Using (1) in the above equation, we get - V=πh312 Differentiating the above equation with respect to time, we get - dVdt=ddh(πh312)⋅dhdt (Using chain rule) =14πh2⋅dhdt Its is given that dVdt=5 m3/h and h=4 m ∴5=π4(4)2⋅dhdt ⇒dhdt=54π m/h =5×74×22 m/h =3588 m/h Therefore, the rate of change of water level is 3588 m/h

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