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A right circular iron cone is cut from a right circular cylinder in such a way that the radius and the height of both are same. Radius is half of height and the diameter of cone’s base is 12 cm. If the weight of iron is 8 gm per cm3, then what is the weight of the wasted iron after cutting the cone from the cylinder?

A right circular iron cone is cut from a right circular cylinder in such a way that the radius and the height of both are same. Radius is half of height and the diameter of cone’s base is 12 cm. If the weight of iron is 8 gm per cm3, then what is the weight of the wasted iron after cutting the cone from the cylinder? Correct Answer 7241

Given:

The radius and the height of both cylinder and cone are same.

Radius is half of height and the diameter of cone’s base is 12 cm.

The weight of iron is 8 gm per cm3.

Formula Used:

Volume of Cylinder = 22/7 × radius2 × height

Volume of Cube = 1/3 × 22/7 × radius2 × height

Calculation:

We have diameter = 12 cm 

So, Radius = 6 cm 

So, Height = 2 × 6 cm = 12 cm 

Now, 

Volume of Cylinder = (22/7 × 36 × 12) cm3

Volume of Cone = (1/3 × 22/7 × 36 × 12) cm3

Volume of the wasted part = {22/7 × 36 × 12 × (1 – 1/3)} cm3 = (22/7 × 36 × 12 × 2/3) cm3

The weight of the wasted iron = (22/7 × 36 × 12 × 2/3 × 8) gm/cm3

≈ 7241 gm/cm3

∴ The wasted iron’s weight is 7241 gm/cm3.

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