
The hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is Rs 5 per 100 ${\text{c}}{{\text{m}}^2}$.
Answer
609.6k+ views
Hint- Here, we will proceed by using the formula for the circumference of any circle of radius r which is given by $2\pi r$ and from here we will find the value of radius of the base of the hemispherical dome and then we will use the formula i.e., curved surface area of a hemisphere with base radius r = $2\pi {r^2}$.
Complete step-by-step answer:
Given, Circumference of the base of dome = 17.6 m
Cost of painting 100 ${\text{c}}{{\text{m}}^2}$ = Rs 5
As we know that 1 ${{\text{m}}^2}$ = 10000 ${\text{c}}{{\text{m}}^2}$
$
\Rightarrow {\text{1 c}}{{\text{m}}^2} = \dfrac{1}{{10000}}{\text{ }}{{\text{m}}^2} \\
\Rightarrow {\text{100 c}}{{\text{m}}^2} = \left( {100} \right)\left( {\dfrac{1}{{10000}}} \right){\text{ }}{{\text{m}}^2} \\
\Rightarrow {\text{100 c}}{{\text{m}}^2} = \dfrac{1}{{100}}{\text{ }}{{\text{m}}^2} \\
$
So, Cost of painting $\dfrac{1}{{100}}{\text{ }}{{\text{m}}^2}$ = Rs 5
$ \Rightarrow $Cost of painting $1{\text{ }}{{\text{m}}^2}$ = Rs $\left( {5 \times 100} \right)$ = Rs 500
Let r be the radius of the base of the hemispherical dome as shown in the figure
Also, circumference of any circular surface having radius r = $2\pi r$ where $\pi = \dfrac{{22}}{7}$
Circumference of the base of dome = $2\pi r$
$
\Rightarrow 17.6 = 2\pi r \\
\Rightarrow r = \dfrac{{17.6}}{{2\pi }} = \dfrac{{17.6}}{{2\left( {\dfrac{{22}}{7}} \right)}} = \dfrac{{17.6 \times 7}}{{44}} \\
\Rightarrow r = 2.8{\text{ m}} \\
$
So, the radius of the base of the hemispherical dome is 2.8 m
It is given that only the hemispherical dome in the building coloured red in the figure needs to be painted which means only the curved surface area of the hemispherical dome will be painted.
Since, Curved surface area of a hemisphere with base radius r = $2\pi {r^2}$
Curved surface area of the hemispherical dome painted (red colour in figure) = $2\left( {\dfrac{{22}}{7}} \right){\left( {2.8} \right)^2}$
$ \Rightarrow $ Curved surface area of the hemispherical dome painted (red colour in figure) = 49.28 ${{\text{m}}^2}$
Also, Total cost of painting the hemispherical dome = (Cost of painting $1{\text{ }}{{\text{m}}^2}$)$ \times $( Curved surface area of the hemispherical dome painted in ${{\text{m}}^2}$)
$ \Rightarrow $Total cost of painting the hemispherical dome = $500 \times 49.28 = {\text{Rs }}24640$
Therefore, the total cost of painting the hemispherical dome of a building is Rs 24640.
Note- In this particular problem, the base of the hemispherical dome is open and is not painted that’s why only the curved surface of the hemisphere is painted. Also, the base of the hemisphere is always a circular surface. Here, it is important to make sure that the units are identical. Here, we have converted every unit into ${{\text{m}}^2}$.
Complete step-by-step answer:
Given, Circumference of the base of dome = 17.6 m
Cost of painting 100 ${\text{c}}{{\text{m}}^2}$ = Rs 5
As we know that 1 ${{\text{m}}^2}$ = 10000 ${\text{c}}{{\text{m}}^2}$
$
\Rightarrow {\text{1 c}}{{\text{m}}^2} = \dfrac{1}{{10000}}{\text{ }}{{\text{m}}^2} \\
\Rightarrow {\text{100 c}}{{\text{m}}^2} = \left( {100} \right)\left( {\dfrac{1}{{10000}}} \right){\text{ }}{{\text{m}}^2} \\
\Rightarrow {\text{100 c}}{{\text{m}}^2} = \dfrac{1}{{100}}{\text{ }}{{\text{m}}^2} \\
$
So, Cost of painting $\dfrac{1}{{100}}{\text{ }}{{\text{m}}^2}$ = Rs 5
$ \Rightarrow $Cost of painting $1{\text{ }}{{\text{m}}^2}$ = Rs $\left( {5 \times 100} \right)$ = Rs 500
Let r be the radius of the base of the hemispherical dome as shown in the figure
Also, circumference of any circular surface having radius r = $2\pi r$ where $\pi = \dfrac{{22}}{7}$
Circumference of the base of dome = $2\pi r$
$
\Rightarrow 17.6 = 2\pi r \\
\Rightarrow r = \dfrac{{17.6}}{{2\pi }} = \dfrac{{17.6}}{{2\left( {\dfrac{{22}}{7}} \right)}} = \dfrac{{17.6 \times 7}}{{44}} \\
\Rightarrow r = 2.8{\text{ m}} \\
$
So, the radius of the base of the hemispherical dome is 2.8 m
It is given that only the hemispherical dome in the building coloured red in the figure needs to be painted which means only the curved surface area of the hemispherical dome will be painted.
Since, Curved surface area of a hemisphere with base radius r = $2\pi {r^2}$
Curved surface area of the hemispherical dome painted (red colour in figure) = $2\left( {\dfrac{{22}}{7}} \right){\left( {2.8} \right)^2}$
$ \Rightarrow $ Curved surface area of the hemispherical dome painted (red colour in figure) = 49.28 ${{\text{m}}^2}$
Also, Total cost of painting the hemispherical dome = (Cost of painting $1{\text{ }}{{\text{m}}^2}$)$ \times $( Curved surface area of the hemispherical dome painted in ${{\text{m}}^2}$)
$ \Rightarrow $Total cost of painting the hemispherical dome = $500 \times 49.28 = {\text{Rs }}24640$
Therefore, the total cost of painting the hemispherical dome of a building is Rs 24640.
Note- In this particular problem, the base of the hemispherical dome is open and is not painted that’s why only the curved surface of the hemisphere is painted. Also, the base of the hemisphere is always a circular surface. Here, it is important to make sure that the units are identical. Here, we have converted every unit into ${{\text{m}}^2}$.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Questions & Answers - Ask your doubts

Master Class 9 Social Science: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
The average rainfall in India is A 105cm B 90cm C 120cm class 10 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

Who Won 36 Oscar Awards? Record Holder Revealed

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

