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

Answer: Option A

Solution

Answer: Option A

Solution:

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

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