# A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $Rs. 4989.60$. If the cost of white-washing is $Rs. 20$ per square metre, find the(i) inside surface area of the dome,(ii) volume of the air inside the dome.

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Given:

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $Rs. 4989.60$.

The cost of white-washing is $Rs. 20$ per square metre.

To do:

We have to find the

(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

Solution:

Cost of white-washing the dome $=Rs.\ 4989.60$

Rate of white-washing $=Rs. 20$ per square metre.

(i) Cost of white washing the dome $=\text { Surface area of the dome } \times \text { Cost of white washing per square metre }$

This implies,

Surface area of the dome $=\frac{\text { Cost of white washing the dome }}{\text { Cost of white washing per square metre }}$

$=\frac{498.96}{2}$

$=249.48 \mathrm{sq} \mathrm{m}$

Inside surface area of the dome is $249.48 \mathrm{sq} \mathrm{m}$.

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

Surface area of the dome $=249.48 \mathrm{sq} \mathrm{m}$

This implies,

$2 \pi r^{2}=249.48$

$2 \times \frac{22}{7} \times r^{2}=249.48$

$r^{2}=\frac{249.48 \times 7}{44}$

$r^{2}=36.69$

$r=\sqrt{36.69}$

$r=6.3 \mathrm{~m}$

Therefore,

Volume of the air inside the dome $=$ Volume of the dome

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

$=\frac{2}{3} \times \frac{22}{7} \times(6.3)^{3}$

$=523.908\ m^3$

The volume of the air inside the dome is $523.908\ m^3$.

Updated on 10-Oct-2022 13:46:39