Question

A dome of a building is in the form of a hemisphere. From inside, it was whitewashed, at the cost of Rs. 498.96. If the cost of whitewashing is Rs. 2.00 per square meter, find
(i) the inside surface area of the dome
(ii) the volume of the air inside the dome.

Answer 1

Hint: First of all take the inside surface area of the hemispherical dome as S. Then divide the total cost of whitewashing – \[\text{cost/}{{\text{m}}^{2}}\] to find the value of S. Use \[S=2\pi {{r}^{2}}\] to find the value of the radius of the dome and then use the formula of the volume \[V=\dfrac{2}{3}\pi {{r}^{3}}\] to get the value of the volume.

Complete step-by-step answer:
We are given a dome of the building is in the form of the hemisphere. From inside it is whitewashed at the cost of Rs. 498.96. If the cost of whitewashing is Rs. 2 per square meter, we have to find the inside surface area and volume of the air inside the dome. The dome would look like the figure below,

(i) Let us assume that the inside surface area of the hemispherical dome is ‘S’ \[{{m}^{2}}\].
Now, we are given that this inside surface of the dome is whitewashed at \[Rs.\text{ }2/{{m}^{2}}\]. That means, cost of whitewashing per square meter area = Rs. 2
So, we get the total cost of whitewashing per ‘S’ square meter area \[=\left( Rs.\text{ }2/{{m}^{2}} \right).\left( S\text{ }{{m}^{2}} \right)=Rs.\text{ }2S\]
Now, we are given that the total cost of whitewashing the dome from inside = Rs. 498.96. So, we get,
Rs. 2S = Rs. 498.96
Or, 2S = 498.96
By dividing 2 on both the sides of the above equation, we get,
\[S=249.48\text{ }{{m}^{2}}\]
So, we get the inside surface area of the hemispherical dome as \[249.48\text{ }{{m}^{2}}\].
(ii) Now we have to find the volume of the air inside the dome or volume of the hemispherical dome.
We know that the inside surface area of the dome, \[S=2\pi {{r}^{2}}\]
Also, we have found that the inside surface area of the dome = \[249.48\text{ }{{m}^{2}}\]
By equating above two values, we get,
 \[2\pi {{r}^{2}}=249.48\]
By substituting \[\pi =\dfrac{22}{7}\] and dividing both the sides by 2, we get,
\[\left( \dfrac{22}{7} \right){{r}^{2}}=124.74\]
Now by multiplying \[\dfrac{7}{22}\] on both the sides of the above equation, we get,
\[{{r}^{2}}=\dfrac{7}{22}\times 124.74\]
So, we get \[{{r}^{2}}=39.69\]
By taking the square root on both the sides of the above equation, we get,
\[r=\sqrt{39.69}\]
\[\Rightarrow r=6.3\text{ }m\]
Now, we know that the volume of a hemisphere \[=\dfrac{2}{3}\pi \left( {{r}^{3}} \right)\].
So, we get the volume of the air inside the dome \[=\dfrac{2}{3}\pi \left( {{r}^{3}} \right)\].
By substituting r = 6.3 m, we get,
The volume of the air inside the dome \[=\dfrac{2}{3}\times \dfrac{22}{7}\times \left( 6.3 \right)\times \left( 6.3 \right)\times \left( 6.3 \right)\]\[=523.908\text{ }{{m}^{3}}\]
So, we get the volume of the air inside the dome as \[523.908\text{ }{{m}^{3}}\].

Note: Students often get confused about whether they should take the surface area of the hemisphere as \[2\pi {{r}^{2}}\] or \[3\pi {{r}^{2}}\]. But they must know that the surface area of the hemisphere without the circular base is \[2\pi {{r}^{2}}\] and with circular base is \[3\pi {{r}^{2}}\]. \[2\pi {{r}^{2}}\] must not be taken into use unless it is given that the base area of the hemisphere is also considered. When nothing is given, take the surface as \[2\pi {{r}^{2}}\]. Also, in the hemisphere, if the curved surface area is given then also take \[2\pi {{r}^{2}}\] only.