# 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 ${{r}_{3}}={{\left( {{r}_{1}}^{3}+{{r}_{2}}^{3} \right)}^{\dfrac{1}{2}}}$ .

Last updated date: 16th Mar 2023

•

Total views: 303.3k

•

Views today: 3.83k

Answer

Verified

303.3k+ views

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE