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Question:

In a rocket shape firecracker, explosive powder is to be filled up inside the metallic enclosure. The metallic enclosure is made up of a cylindrical base and conical top with the base of radius 8 centimeter. The ratio of height of cylinder and cone is 5:3. A cylindrical hole is drilled through the metal solid with height one third the height of the metal solid. What should be the radius of the hole, so that volume of the hole (in which gun powder is to be filled up) is half of the volume of metal solid after drilling?

Explanation:

Let the height of the cylinder be 5h than that of the conical part be 3h.
∴ Height of hole = $\frac{1}{3}$ (3h + 5h)
= $\frac{8 h}{3}$

Radius of cone = radius of cylinder = 8

Let radius of hole = r

Now, 1/2(Total volume – Volume of hole) = Volume of hole

∴ Total volume = 3Volume of hole

∴ π × 8² × 5h + $\frac{1}{3}$ × π × 8² × 3h = 3πr² $(\frac{8 h}{3})$

∴ 8² × 6 = 8r²

∴ r² = 8 × 6

∴ r = 4 $\sqrt{3}$ cm

Hence, option (a).

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