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.,