A spherical ball of radius 1 cm is dropped into a conical vessel of radius 3 cm and slant height 6 cm. The volume of water (in cm³), that can just immerse the ball, is

Solution (By Examveda Team)

$$\eqalign{ & \Delta ABC = {\text{equilateral }}\Delta \cr & \therefore \angle ACB = {60^ \circ } \cr & \& \angle BCP = {30^ \circ } \cr & \Delta CDO,\,\angle CDO = {90^ \circ } \cr & \left( {{\text{Angle between radius and tangent is }}{{90}^ \circ }} \right) \cr} $$

$$\eqalign{ & OD = 1P = 1{\text{ cm}} \cr & OC = 2P = 2\left( 1 \right) = 2{\text{ cm}} \cr & {\text{Then, }}CZ = OC + OZ = 2 + 1 = 3{\text{ cm}} \cr & \Delta CZY,\,\angle CZY = {90^ \circ } \cr & CZ = \sqrt 3 P = 3{\text{ cm}} \cr & YZ = 1P = \sqrt 3 {\text{ cm}} \cr & {\text{Now, in cone }}XYC \cr & r = ZY = \sqrt 3 {\text{ cm}} \cr & h = CZ = 3{\text{ cm}} \cr & {\text{Volume of cone}} = \frac{1}{3}\pi {r^2}h \cr & = \frac{1}{3}\pi {\left( {\sqrt 3 } \right)^2}\left( 3 \right) \cr & = 3\pi {\text{ c}}{{\text{m}}^2} \cr & {\text{Volume of sphere}} = \frac{4}{3}\pi r_s^3 \cr & \left( {\therefore {r_s} = 1{\text{ cm}}} \right) \cr & = \frac{4}{3}\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{Volume of water that can immerse the ball}} \cr & = \left( {3\pi - \frac{{4\pi }}{3}} \right){\text{c}}{{\text{m}}^3} \cr & = \frac{{5\pi }}{3}{\text{ c}}{{\text{m}}^3} \cr} $$