(Q10).A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemi-sphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub. (Take π =
22
7
 )

The tub is in the shape of a cylinder, thus

Radius of the cylinder (r) = cm

Length of the cylinder (h) = cm

Volume of the cylinder (V) = πr2h

=
× × ×

Volume of the tub = 770 cm3.

Radius of the hemisphere (r) = 3.5 cm

Volume of the hemisphere =
2
3
 πr3
=
×
× × ×
=
×
=

Radius of the cone (r) = 3.5 cm

Height of the Cone (h) = 5 cm

Volume of the cone V =
1
3
 πr2h
=
×
× × ×
=

Volume of the solid = Volume of the hemisphere + Volume of the cone

=
+
=
= cm3

Now, when the solid is immersed in the tub, it replaces the water whose volume is equal to volume of the solid itself.

Thus the volume of the water replaced = cm3.

∴ Volume of the water left in the tub = Volume of the tub – Volume of the solid = – = cm3.