Two metallic spheres of radii ${{r}_{1}}$ and ${{r}_{2}}$are melted and cast into a single solid sphere of radius ${{r}_{3}}$. Prove that ${{r}_{3}}={{\left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)}^{\dfrac{1}{2}}}$ .
Answer
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Hint: When we cast one solid into another solid, the net Volume of the material does not change. So, compare the Volume of the new sphere with the sum of the volumes of the initial spheres and solve for ${{r}_{3}}$ .
Complete step-by-step answer:
We know that Volume of a sphere = $\dfrac{4}{3}\pi {{r}^{3}}$
Hence the Volume of the first sphere = $\dfrac{4}{3}\pi {{r}_{1}}^{3}$
And the Volume of the second sphere = $\dfrac{4}{3}\pi {{r}_{2}}^{3}$
Net Volume of the material = $\dfrac{4}{3}\pi {{r}_{1}}^{3}+\dfrac{4}{3}\pi {{r}_{2}}^{3}$
Also, net Volume of material = Volume of the new sphere
$\Rightarrow \dfrac{4}{3}\pi {{r}_{3}}^{3}=\dfrac{4}{3}\pi {{r}_{1}}^{3}+\dfrac{4}{3}\pi {{r}_{2}}^{3}$
Taking \[\dfrac{4}{3}\pi \] common from RHS, we get
$\Rightarrow \dfrac{4}{3}\pi {{r}_{3}}^{3}=\dfrac{4}{3}\pi \left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)$
Multiplying both sides by $\dfrac{3}{4\pi }$ we get
$\begin{align}
& \Rightarrow \dfrac{4}{3}\pi {{r}_{3}}^{3}\times \dfrac{3}{4\pi }=\dfrac{4}{3}\pi \left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)\times \dfrac{3}{4\pi } \\
& \Rightarrow {{r}_{3}}^{3}={{r}_{1}}^{3}+{{r}_{2}}^{3} \\
\end{align}$
Raising power $\dfrac{1}{3}$ on both sides, we get
$\Rightarrow {{\left( {{r}_{3}}^{3} \right)}^{\dfrac{1}{3}}}={{\left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)}^{\dfrac{1}{3}}}$
i.e. ${{r}_{3}}={{\left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)}^{\dfrac{1}{3}}}$
Hence proved.
Note:
[a] Casting one or more two solids to other solids is like pouring water from one or more utensils into other one or more utensils. Although the shape will change, the net quantity of water will remain constant. Therefore, we compared the value of the Volume of new solid to the sum of volumes of the individual solids.
[b] Some important formulae to remember:
[1] Volume of a Cube = ${{a}^{3}}$
[2] Volume of a Cuboid = $lbh$
[3] Volume of a right circular cylinder = $\pi {{r}^{2}}h$
[4] Volume of a right circular cone = $\dfrac{1}{3}\pi {{r}^{2}}h$
[5] Volume of a sphere = $\dfrac{4}{3}\pi {{r}^{3}}$
These formulae often come in handy while solving such questions based on the volume of solid objects and hence, should be memorized.
Complete step-by-step answer:
We know that Volume of a sphere = $\dfrac{4}{3}\pi {{r}^{3}}$
Hence the Volume of the first sphere = $\dfrac{4}{3}\pi {{r}_{1}}^{3}$
And the Volume of the second sphere = $\dfrac{4}{3}\pi {{r}_{2}}^{3}$
Net Volume of the material = $\dfrac{4}{3}\pi {{r}_{1}}^{3}+\dfrac{4}{3}\pi {{r}_{2}}^{3}$
Also, net Volume of material = Volume of the new sphere
$\Rightarrow \dfrac{4}{3}\pi {{r}_{3}}^{3}=\dfrac{4}{3}\pi {{r}_{1}}^{3}+\dfrac{4}{3}\pi {{r}_{2}}^{3}$
Taking \[\dfrac{4}{3}\pi \] common from RHS, we get
$\Rightarrow \dfrac{4}{3}\pi {{r}_{3}}^{3}=\dfrac{4}{3}\pi \left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)$
Multiplying both sides by $\dfrac{3}{4\pi }$ we get
$\begin{align}
& \Rightarrow \dfrac{4}{3}\pi {{r}_{3}}^{3}\times \dfrac{3}{4\pi }=\dfrac{4}{3}\pi \left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)\times \dfrac{3}{4\pi } \\
& \Rightarrow {{r}_{3}}^{3}={{r}_{1}}^{3}+{{r}_{2}}^{3} \\
\end{align}$
Raising power $\dfrac{1}{3}$ on both sides, we get
$\Rightarrow {{\left( {{r}_{3}}^{3} \right)}^{\dfrac{1}{3}}}={{\left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)}^{\dfrac{1}{3}}}$
i.e. ${{r}_{3}}={{\left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)}^{\dfrac{1}{3}}}$
Hence proved.
Note:
[a] Casting one or more two solids to other solids is like pouring water from one or more utensils into other one or more utensils. Although the shape will change, the net quantity of water will remain constant. Therefore, we compared the value of the Volume of new solid to the sum of volumes of the individual solids.
[b] Some important formulae to remember:
[1] Volume of a Cube = ${{a}^{3}}$
[2] Volume of a Cuboid = $lbh$
[3] Volume of a right circular cylinder = $\pi {{r}^{2}}h$
[4] Volume of a right circular cone = $\dfrac{1}{3}\pi {{r}^{2}}h$
[5] Volume of a sphere = $\dfrac{4}{3}\pi {{r}^{3}}$
These formulae often come in handy while solving such questions based on the volume of solid objects and hence, should be memorized.
Last updated date: 25th Sep 2023
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