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A sphere of diameter 42 cm is melted and bullets are formed from it. The bullet has circular cylindrical shape from one side and conical from other side. The radius of the base of the cylindrical part of the bullet is 6 cm and the cylindrical part has thrice as much length as the conical part. Find how many such bullets can be formed if the curved surface area of the cylindrical part of a bullet is 79.2 cm2.

A sphere of diameter 42 cm is melted and bullets are formed from it. The bullet has circular cylindrical shape from one side and conical from other side. The radius of the base of the cylindrical part of the bullet is 6 cm and the cylindrical part has thrice as much length as the conical part. Find how many such bullets can be formed if the curved surface area of the cylindrical part of a bullet is 79.2 cm2. Correct Answer 147

Suppose ‘r’ and ‘H’ are the radius and height of the cylindrical part and ‘h’ is the height of conical part

Given r = 6 cm and curved surface area of the cylindrical part of a bullet is 79.2 cm2

⇒ 2πrH = 79.2

⇒ 2 × 22/7 × 6 × H = 79.2

⇒ H = 2.1 cm

⇒ Height of the cylindrical part of the bullet = H = 2.1 cm

⇒ Height of the conical part of the bullet = h = 2.1/3 = 0.7 cm

⇒ Volume of a bullet = Volume of Cylindrical part + volume of conical part = πr2(h/3 + H)

⇒ 22/7 × 36 × (0.7/3 + 2.1)

⇒ 22/7 × 36 × 7/3

⇒ 264 cm3

Now

Volume of the sphere with diameter 42 cm which is melted = 4/3 × πr3

⇒ 4/3 × 22/7 × 21 × 21 × 21

⇒ 38808 cm3

∴ Number of bullets = 38808/264 = 147

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