Question

A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5m and height 21m and the cone has a slant height 8m. Calculate total surface area and volume of the rocket?

Answer 1

Given radius of cylinder (a) = 2.5m

Height of cylinder (h) = 21m

Slant height of cylinder (l) = 8m

Curved surface area of cone(S₁) = πrl

S₁ = π(2.5)(8)cm²            ...........(1)

Curbed surface area of a cone${\displaystyle =2\pi rh+\pi r^2}$

${\displaystyle S_2=2\pi \left(2.5\right)\left(21\right)+\pi {\left(2.5\right)}^{2}c{m}^2}$         ...........(2)

∴Total curved surface area = (1) + (2)

S = S₁ + S₂

S = π(2.5)(8) + 2π(2.5)(21) + π(2.5)²

S = 62.831 + 329.86 + 19.63

S = 412.3m²

∴Total curved surface area = 412.3m²

Volume of a cone ${\displaystyle =\frac{1}{3}\pi r^{2}h}$

${\displaystyle V_1=\frac{1}{3}\times \pi {\left(2.5\right)}^{2}hc{m}^3}$      .........(3)

Let ‘h’ be height of cone

${\displaystyle l^2=r^2+h^2}$

⇒ ${\displaystyle l^2-r^2=h^2}$

⇒ ${\displaystyle h=\sqrt{l^2-r^2}}$

⇒ ${\displaystyle h=\sqrt{8^2-{25}^2}}$

⇒ h =23.685m

Subtracting ‘h’ value in(3)

Volume of a cone ${\displaystyle \left(V_1\right)=\frac{1}{3}\times \pi {\left(2.5\right)}^2\left(23.685\right) c{m}^2}$      ........(4)

Volume of a cylinder ${\displaystyle \left(V_2\right)=\pi r^{2}h}$

${\displaystyle =\pi {\left(2.5\right)}^{2}21m^3}$           ...........(5)

Total volume = (4) + (5)

V = V₁ + V₂

⇒ ${\displaystyle V=\frac{1}{3}\times \pi {\left(2.5\right)}^2\left(23.685\right)+\pi {\left(2.5\right)}^2=1}$

⇒ V = 461.84m²

Total volume (V) = 461.84m²