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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.
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 $523.908\ m^3$.
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