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Question:

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 5cm. 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 2/3rd of the height of solid. What should be the radius (in cm) of the hole so that the volume of the hole is 1/3rd of the volume of the metallic solid after drilling?

Explanation:

Volume of original solid = volume of cylinder + volume of cone

Radius of original cylinder = radius of cone = 5 cm

Let height of the original cylinder be 3x cm. Hence, height of cone = 2x cm

Volume of solid = $π (5^{2}) (3 x) + \frac{π}{3} (5^{2}) (2 x)$

$= 25 π \times \frac{11 x}{3} = \frac{275 π x}{3}$

Let the radius of the cylindrical hole be r cm.

Also, height of cylindrical hole = (2/3)(5x) = (10x/3) cm

∴ Volume of hole = $π (r^{2}) \frac{10 x}{3}$

Hence, as per the given condition:

$\frac{π (r^{2}) \frac{10 x}{3}}{\frac{275 x}{3} - π (r^{2}) \frac{10 x}{3}} = \frac{1}{3}$

$∴ \frac{10 r^{2}}{275 - 10 r^{2}} = \frac{1}{3}$

∴ 275 – 10r² = 30r²
∴ 40r² = 245

$∴ r^{2} = \frac{245}{40} = \frac{55}{8}$

$∴ r = \sqrt{(\frac{55}{8})}$

Hence, option (d).

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