Question

L A cracker rocket is in the shape of a right circular cone standing on a right circular cylinder.If the height of the conical portion is half the height of the cylinder, radius of the cylinder is$3.5 cm$, whereas the radius of the conical portion is two times the radius of the cylinder. I
the total height of the solid is $21 cm$, find the volume of the rocket.($\pi=\frac{22}{7}$)

Answer 1

If the total height of the solid is  , then the volume of the rocket is 898.34 cm³.

Step-by-step explanation:

We have,

A cracker rocket is in the shape of a right circular cone standing on a right circular cylinder.

If we take the height of the conical portion to be “h₁” and the cylindrical portion to be “h₂” and also, the radius of the cone to be “r₁” and of the cylinder to be “r₂”.

Then, h₁ = ½ * h₂ (given)

The radius of the cylinder, r₂ = 3.5 cm

Then, the radius of the cone, r₁ = 2 * r₂ = 2 * 3.5 = 7 cm  

The total height of the rocket is given as, h = 21 cm

i.e., h₁ + h₂ = 21

⇒ h₁ + 2h₁ = 21

⇒ h₁ = 21/3

⇒ h₁ = 7 cm

∴ h₂ = 2 * h₁ = 2 * 7 = 14 cm

Now,

The volume of the conical portion is given by,

= 1/3 * π * r₁² * h₁  

= 1/3 * (22/7) * (7²) *7

= 359.34 cm³

And

The volume of the cylindrical portion is given by,

= π * r₂² * h₂

= (22/7) * (3.5)² * (14)

= 539 cm³

Thus,  

The total volume of the cracker rocket is,

= [volume of the conical portion] + [volume of the cylindrical portion]

= 359.34 + 539

= 898.34 cm³

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