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A metallic solid is made up of a solid cylindrical base with a solid cone on its top. The radius of the base of the cone is 5 cm. and the ratio of the height of the cylinder and the cone is 3:2. A cylindrical hole is drilled through the solid with height equal to 2/$$3^{rd}$$ of the height of solid. What should be the radius (in cm) of the hole so that the volume of the hole is 1/$$3^{rd}$$ of the volume of the metallic solid after drilling?
H - volume of hole
R - volume of remaining solid
T - Total volume
It is given, $$H\ =\ \frac{1}{3}R$$
H + R = T
T = 4H
$$\frac{1}{3}\pi\ \left(25\right)\left(2x\right)+\pi\left(25\right)\left(3x\right)=4\times\pi\ r^2\left(\frac{10}{3}x\right)$$
$$r^2=\frac{275}{40}$$
$$r\ =\sqrt{\frac{\ 55}{8}}$$
The answer is option D.
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