A rocket is in the form of a right circular cylinder closed at the

Leave a Comment / surface areas and volumes / By Sapling Academy

Q) A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of cylinder. The diameter and height of cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, then find the total surface area and volume of rocket. (Use π = 3.14)

Ans:

We are given that, diameter of cylinder, D = 6 cm ∴ radius of cylinder, R = 3 cm

Since radius of cylinder = radius of cone,

∴ diameter of cone, d = 6 cm ∴ radius of cone, r = 3 cm

Height of the cylinder, H = 12 cm

Slant height of the cone, l = 5 cm

Step 1: Height of the cone, h = $\sqrt$ (slant ~ height^2 – radius ~ of ~ cone^2)

= $\sqrt (5^2 - 3^2) = \sqrt 16$ = 4 cm

Step 2:

Curved surface area of the cone = $\pi$ r l = $\pi$ (3) (5) = 15 $\pi$

Curved surface area of the Cylinder = 2 $\pi$ R H = 2 $\pi$ (3)(12) = 72 $\pi$

Base area of the Cylinder = $\pi$ R^2 = $\pi$ (3)^2 = 9 $\pi$

∴ Total Surface Area of the rocket = Surface Area of Cone + Surface Area of Cylinder + Base Area of Cylinder

= 15 $\pi$ + 72 $\pi$ + 9 $\pi$ = 96 $\pi$ = 96 x 3.14 = 301.44 cm²

Therefore, the Surface Area of the Rocket is 301.44 cm²

Note: Here, base area of cylinder is taken because it is given that the cylinder is closed.

Step 3:

Volume of the Cone = $\frac{1}{3} \pi$ r² h = $\frac{1}{3} \pi$ (3)² (4) = 12 $\pi$

Volume of the Cylinder = $\pi$ R² H = $\pi$ (3)²(12) = 108 $\pi$

Volume of the rocket = Volume of Cone + Volume of Cylinder

= 12 $\pi$ + 108 $\pi$ = 120 $\pi$ = 120 x 3.14 = 376.8 cm³

Therefore, the Volume of the Rocket is 376.8 cm³

Leave a Comment Cancel Reply