**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 = (slant ~ height^2 – radius ~ of ~ cone^2)

= = 4 cm

**Step 2:**

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

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

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

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

= 15 + 72 + 9 = 96 = 96 x 3.14 = 301.44 cm^{2}

**Therefore, the Surface Area of the Rocket is 301.44 cm ^{2}**

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

**Step 3:**

Volume of the Cone = r^{2} h = (3)^{2} (4) = 12

Volume of the Cylinder = R^{2} H = (3)^{2 }(12) = 108

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

= 12 + 108 = 120 = 120 x 3.14 = 376.8 cm^{3}

**Therefore, the Volume of the Rocket is 376.8 cm ^{3}**