
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Answer
513.3k+ views
Hint: We will find the volume of the inner side of the tank by taking the radius as 1m. Then we will find the volume of the outer side of the tank by taking the radius as (1m + 1 cm). Finally, we will subtract the inner volume from the outer volume to get the volume of iron used to make the tank.
Formulas used: \[V = \dfrac{2}{3}\pi {r^3}\] where \[V\] is the volume of the hemisphere and \[r\] is the radius of the hemisphere.
Complete step-by-step answer:
Let us find the inner volume of the hemisphere. Let us convert the radius into cm.
\[1{\rm{m}} = 100{\rm{cm}}\]
Substitute 100 for \[r\] in the formula.
\[\begin{array}{l}{V_i} = \dfrac{2}{3}\pi {\left( {100} \right)^3}\\ \Rightarrow {V_i} = \dfrac{{2000000\pi }}{3}{\rm{c}}{{\rm{m}}^3}\end{array}\]
Let us find the outer volume of the hemisphere. Let’s find its radius. Its radius is the sum of the inner radius of the hemisphere and the thickness of the iron sheet.
\[\begin{array}{l}r = 100{\rm{cm}} + 1{\rm{cm}}\\ \Rightarrow r = 101{\rm{cm}}\end{array}\]
\[\begin{array}{l} \Rightarrow {V_o} = \dfrac{2}{3}\pi {\left( {101} \right)^3}\\ \Rightarrow {V_o} = \dfrac{{2\left( {1030301} \right)\pi }}{3}\\ \Rightarrow {V_o} = \dfrac{{2060602\pi }}{3}{\rm{c}}{{\rm{m}}^3}\end{array}\]
Let us subtract the volume of inner hemisphere from volume of outer hemisphere and find the volume of iron used to make the hemisphere:-
\[\begin{array}{l}V = \dfrac{{\left( {2060602 - 2000000} \right)\pi }}{3}\\ \Rightarrow V = \dfrac{{60602}}{3} \times \dfrac{{22}}{7}\\ \Rightarrow V = 63487.8{\rm{c}}{{\rm{m}}^3}\end{array}\]
Note: Alternatively, we can convert the given length of radius into metre from centimetre and then find the volume of iron used in \[{{\rm{m}}^3}\]. Radius of inner hemisphere will be taken as 1m and radius of outer hemisphere will be taken as 1.01 m ( \[1 + \dfrac{1}{{100}}\] ) m because 1 m contains 100cm. The volume will come out to be \[0.06348{{\rm{m}}^3}\].
Formulas used: \[V = \dfrac{2}{3}\pi {r^3}\] where \[V\] is the volume of the hemisphere and \[r\] is the radius of the hemisphere.
Complete step-by-step answer:

Let us find the inner volume of the hemisphere. Let us convert the radius into cm.
\[1{\rm{m}} = 100{\rm{cm}}\]
Substitute 100 for \[r\] in the formula.
\[\begin{array}{l}{V_i} = \dfrac{2}{3}\pi {\left( {100} \right)^3}\\ \Rightarrow {V_i} = \dfrac{{2000000\pi }}{3}{\rm{c}}{{\rm{m}}^3}\end{array}\]
Let us find the outer volume of the hemisphere. Let’s find its radius. Its radius is the sum of the inner radius of the hemisphere and the thickness of the iron sheet.
\[\begin{array}{l}r = 100{\rm{cm}} + 1{\rm{cm}}\\ \Rightarrow r = 101{\rm{cm}}\end{array}\]
\[\begin{array}{l} \Rightarrow {V_o} = \dfrac{2}{3}\pi {\left( {101} \right)^3}\\ \Rightarrow {V_o} = \dfrac{{2\left( {1030301} \right)\pi }}{3}\\ \Rightarrow {V_o} = \dfrac{{2060602\pi }}{3}{\rm{c}}{{\rm{m}}^3}\end{array}\]
Let us subtract the volume of inner hemisphere from volume of outer hemisphere and find the volume of iron used to make the hemisphere:-
\[\begin{array}{l}V = \dfrac{{\left( {2060602 - 2000000} \right)\pi }}{3}\\ \Rightarrow V = \dfrac{{60602}}{3} \times \dfrac{{22}}{7}\\ \Rightarrow V = 63487.8{\rm{c}}{{\rm{m}}^3}\end{array}\]
Note: Alternatively, we can convert the given length of radius into metre from centimetre and then find the volume of iron used in \[{{\rm{m}}^3}\]. Radius of inner hemisphere will be taken as 1m and radius of outer hemisphere will be taken as 1.01 m ( \[1 + \dfrac{1}{{100}}\] ) m because 1 m contains 100cm. The volume will come out to be \[0.06348{{\rm{m}}^3}\].
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