A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed. What fraction of water overflows?
38
Radius of the conical vessel, R = AC = 6 cm.
Height of the conical vessel, H = OC = 8 cm.
Let the radius of the sphere be r.
Then, PC = PD = r.
Now, AC = AD = 6 cm.
[Since lengths of two tangents from an external point to a circle are equal]
OA=√OC2+AC2=√82+62=√100=10cm
OD= (OA - AD) = (10 - 6) cm = 4 cm.
OP = (OC - PC) = (8 - r).
In right angled ΔODP, we have:
OP2=OD2+PD2
(8−r)2=42+r2
64−16r+r2=16+r2
16r=48⇒r=3.
Volume of water overflown = Volume of sphere
=43πr3=[43π×(3)3]cm3=(36π)cm3 .
Volume of water in the cone before immersing the sphere = volume of cone
=13πr2h=(13π×(6)2×8)cm3=(96π)cm3
∴ Fraction of water overflown = Volume of water overflownOriginal volume of water
⇒ Fraction of water overflown =36π96π=38