Question

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 20 per square meter, find the inside surface area of the dome, the volume of the air inside the dome.

Answer 1

Hint:Area whitewashed is a curved surface area of the hemisphere because only walls are whitewashed, not the floors. We know that the curved surface area of the hemisphere is $$2\pi r^2$$ . The cost of white-washing is Rs. 20 per square meter and the total cost of whitewashing is Rs. 4989.60. Now, we can find the curved surface area by dividing the total cost of whitewashing by the cost of white-washing per square meter. Then, find the radius of the hemisphere by putting the value of curved surface area in the equation, curved surface area = $$2\pi r^2$$ .We know that the volume of the hemisphere is $${\displaystyle \frac{2}{3}}\pi r^3$$ . Put the value of radius in the formula $${\displaystyle \frac{2}{3}}\pi r^3$$ .

Complete step-by-step answer:

According to the question, we have the total cost of the whitewashing and cost of white washing per square meter.
Total cost of whitewashing = Rs. 4989.60 ……………….(1)
Cost of whitewashing per square meter = Rs. 20 …………….(2)
We know that,
$$\text{Area}\text{whitewashed}\times \text{cost}\text{of}\text{whitewash}\text{per}\text{square meter = Total cost}$$ ……………………(3)
From equation (1), equation (2), and equation (3), we get
$$\text{Area}\text{whitewashed}\times \text{cost}\text{of}\text{whitewash}\text{per}\text{square meter = Total cost}$$
$$\begin{array}{rl}& \Rightarrow \text{Area}\text{whitewashed}\times \text{Rs}\text{.20 per }{\text{m}}^2\text{= Rs}\text{.4989}\text{.60}\\ & \Rightarrow \text{Area}\text{whitewashed}={\displaystyle \frac{\text{Rs}\text{.4989}\text{.60}}{\text{Rs}\text{.20}}}{\text{m}}^2\end{array}$$
$$\Rightarrow \text{Area}\text{whitewashed}=249.48{\text{m}}^2$$
Area whitewashed = 249.48 $$m^2$$ .
Area whitewashed is the curved surface area of the hemisphere because only walls are whitewashed, not the floors.
Let the radius of the hemisphere be r.
We know that the curved surface area of the hemisphere = $$2\pi r^2$$ …………..(4)
Curved surface area = Area whitewashed = 249.48 $$m^2$$ ……………….(5)
From equation (4) and equation (5), we get
$$\begin{array}{rl}& 2\pi r^2=249.48\\ & \Rightarrow 2\times {\displaystyle \frac{22}{7}}\times r^2=249.48\\ & \Rightarrow r^2={\displaystyle \frac{249.48\times 7}{44}}\\ & \Rightarrow r^2=39.69\\ & \Rightarrow r=6.3\end{array}$$
Volume of the hemisphere is $${\displaystyle \frac{2}{3}}\pi r^3$$ .
Putting the value of radius in the formula $${\displaystyle \frac{2}{3}}\pi r^3$$ , we get
Volume =$${\displaystyle \frac{2}{3}}\pi r^3={\displaystyle \frac{2}{3}}\times {\displaystyle \frac{22}{7}}\times {\left(6.3\right)}^3=523.69m^3$$ .
Hence, the volume and surface area of the dome shaped hemisphere is 523.69 $$m^3$$ and 249.48 $$m^2$$ .

Note: In this question one can think that the total surface is getting whitewashed. In total surface area we include the area of the floor also. This is wrong because only the walls of the hemispherical dome are whitewashed, not the floor. Therefore, Curved surface area = Area whitewashed.