Surface Areas and Volumes — Question 10

(i) Given,

Inside of dome was white-washed at the cost of ₹ 4989.60.

Cost of white washing = ₹ 20 per square metre.

Inside surface area of the dome

= $\dfrac{\text{Total cost for white-washing the dome inside}}{\text{Rate of white-washing}}$

⇒ $\dfrac{4989.60}{20}$ = 249.48 m²

Hence, inner surface area of the dome = 249.48 m².

(ii) Let 'r' be the radius of a hemispherical dome.

Inner surface area of the hemispherical dome = 2πr²

Substituting values we get :

$\Rightarrow 2πr^2 = 249.48 \\[1em] \Rightarrow r^2 = \dfrac{249.48}{2π} \\[1em] \Rightarrow r^2 = \dfrac{249.48}{2 \times \dfrac{22}{7}} \\[1em] \Rightarrow r^2 = \dfrac{249.48 \times 7}{44} \\[1em] \Rightarrow r^2 = 5.67 \times 7\\[1em] \Rightarrow r^2 = 39.69 \\[1em] \Rightarrow r = \sqrt{39.69} \\[1em] \Rightarrow r = 6.3 \text{ m}$

The volume of the air inside the dome will be the same as the volume of the hemisphere.

Now the volume of the air inside the dome (v) = $\dfrac{2}{3}πr^3$

$v = \dfrac{2}{3} \times \dfrac{22}{7} \times 6.3 \times 6.3 \times 6.3 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 6.3 \times 39.69 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 250.047 \\[1em] = \dfrac{44}{21} \times 250.047 \\[1em] = \dfrac{11002.068}{21} \\[1em] = 523.9 \text{ m}^3$

Hence, the volume of the air inside the dome is 523.9 m³.