A dome of a building is in the form of a hemisphere. From inside,

Question.

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs $498.96$. If the cost of white-washing is Rs $2.00$ per square meter, find the

(i) inside surface area of the dome,

(ii) volume of the air inside the dome. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$


Solution:

(i) Cost of white-washing the dome from inside $=$ Rs $498.96$

Cost of white-washing $1 \mathrm{~m}^{2}$ area $=$ Rs 2

Therefore, CSA of the inner side of dome $=\left(\frac{498.96}{2}\right) \mathrm{m}^{2}$

$=249.48 \mathrm{~m}^{2}$

(ii) Let the inner radius of the hemispherical dome be $r$.

CSA of inner side of dome $=249.48 \mathrm{~m}^{2}$

$2 \pi r^{2}=249.48 \mathrm{~m}^{2}$

$\Rightarrow 2 \times \frac{22}{7} \times r^{2}=249.48 \mathrm{~m}^{2}$

$\Rightarrow r^{2}=\left(\frac{249.48 \times 7}{2 \times 22}\right) \mathrm{m}^{2}=39.69 \mathrm{~m}^{2}$

$\Rightarrow r=6.3 \mathrm{~m}$

Volume of air inside the dome = Volume of hemispherical dome

$=\frac{2}{3} \pi r^{3}$

$=\left[\frac{2}{3} \times \frac{22}{7} \times(6.3)^{3}\right] \mathrm{m}^{3}$

$=523.908 \mathrm{~m}^{3}$

$=523.9 \mathrm{~m}^{3}$ (approximately)

Therefore, the volume of air inside the dome is $523.9 \mathrm{~m}^{3}$.

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