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Question

An inverted cone has a depth of 40cm and a base of radius 5cm. Water is poured into it at a rate of 1.5 cubic centimetres per minute. Find the rate at which the level of water in the cone is rising when the depth is 4cm.

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Solution

Let V be the volume in the inverted cone.
Volume (V)=13πr2h
By similar triangles,
540=rh
r=h8
V=13π(h8)2h
V=1192πh3
As the water is getting filled in funnel at the rate of 1.5cm3/sec, i.e.,
dVdt=1.5
ddt(1192πh3)=32
164πh2dhdt=32
dhdt=96πh2
The rate of change of water level, i.e.,
dhdt=96πh2
When ater is at 4cm, i.e., when h=4,
dhdt=96π42
dhdt=6(227)=6×722=2111cm/s
Hence the correct answer is 2111cm/s.

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