A rocket is in the form of a right circular cylinder closed at the lower end with a cone

Solution;

Given diameter of the cylinder = 6 cm

Radius of the cylinder, r = 6/2 = 3 cm

Height of the cylinder, H = 12 cm

Slant height of the cone, l = 5 cm

Radius of the cone, r = 3 cm

Height of the cone, h = √(l2-r2)

h = √(52-32)

h = √(25-9)

h = √16

h = 4 cm

Total surface area of the rocket = curved surface area of cylinder +base area of cylinder+ curved surface area of cone

= 2rH+r2+rl

= r(2H+r+l)

= 3.14×3×(2×12+3+5)

= 3.14×3×(24+3+5)

= 3.14×3×32

= 301.44 cm2

Hence the Total surface area of the rocket is 301.44 cm2.

Volume of the rocket = Volume of the cone + volume of the cylinder

= (1/3)r2h +r2H

= r2((h/3)+H)

= 3.14×32×((4/3)+12)

= 3.14×9×((4+36)/3)

= 3.14×9×(40/3)

= 3.14×3×40

= 376.8 cm3

Hence the volume of the rocket is 376.8 cm3.

Text Solution

Solution :  Since, rocket is the combination of a right circular cylinder and a cone. <br> Given, diameter of the cylinder = 6 cm <br> `therefore` Radius of the cylinder = 12 cm <br> and height of the cylinder `= pir^(2)h= 3.14 xx (3)^(3) xx 12`<br> `" "= 339.12 cm^(3)` <br> and height of the cylinder = 12 cm <br> `therefore` Volume of the cylinder `= pir^(2)h = 3.14 xx (3)^(3) xx 12` <br> `= 339.12 cm^(3)` <br> and curved surface area `= 2pirh` <br> `= 2 xx 3.14 xx 3 xx 12 = 226.08` <br> Now, in right angled `DeltaAOc` <br> `" " h = sqrt(5^(2) -3^(2)) = sqrt(25-9) = sqrt(16) = 4` <br> ` therefore` Height of the cone h = 4 cm <br> Radius of the cone ,r= 3cm <br> Now, volume of the cone <br> `" " =(1)/(3) pir^(2)h = (1)/(3) xx 3.14 xx (3)^(2) xx 4` <br> `" " = (113.04)/(3) = 37.68 cm^(3)` <br> and curved surface area `= pirl = 3.14 xx 3xx5 = 47.1` <br> Hence, total volume of the rocket`=339.12 +37.68 = 376. 8 cm ^(3)` <br> and total surface area of the rocket = CSA of cone + CSA of cylinder + Area of base of cylinder <br> `" " = 47.1 + 226.08 + 28.26` <br> `" " = 301.44 cm^(2)` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_MATH_X_C12_S01_056_S01.png" width="80%">

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter

In fig. cone ABC is cut out by a plane parallel to the base FG. DEFG is the frustum so obtained. Let O be the centre of the base of the cone and O’ the centre of the base of the frustum.

And, C =  

 volume of the frustum = 

         Volume of wire =