Question

A model of a rocket consists of a cylinder surmounted by a cone at one end. The dimensions of the model areas follows: common radius = 3 cm, height of the cone = 4 cm and total height = 14 cm.
If the model is made to a scale of 1 : 500, then find the total surface area of the rocket (in $m^2$).

• A. $6800m^2$
  
• B. $6000m^2$
• C. $6600m^2$
  
• D. None of These

Answer 1

The correct option is C $6600m^2$


Common radius, r = 3 cm
Height of the cone, $h_1=4cm$
Total height of the model = 14 cm
Height of the cylinder, $h_2=(14–4)cm=10cm$
Slant height of cone, $l=\sqrt{r^2+{\left(h_1\right)}^2}$
$=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}$
$=\sqrt{9+16}$
$=\sqrt{25}$
= 5 cm
If the model is made to made to scale of 1 : 500 the actual dimensions of the rocket are:
Common radius = r = 3 × 500 = 1500 cm
= 15 m
Height of the cone = h1 = 4 × 500 = 2000 cm
= 20 m
Height of the cylinder = h2 = 10 × 500 = 5000 cm
= 50 m
Slant height of the cone = l = 5 × 500 = 2500 cm
= 25 m
Total surface area of the rocket =
Curved Surface Area of cone + Curved Surface Area of the Cylinder + Area of Circular
Base of Cylinder
$=\pi rl+2\pi r{h}_2+\pi r^2$
$=\pi \times 15\times 25+2\times \pi \times 15\times 50+\pi \times 15\times 15$
$=\pi \times 375+\pi \times 1500+\pi \times 225$
$=\pi (375+1500+225)=2100\pi$
$=2100\times \frac{22}{7}=6600m^2$

So,option c is correct.