Question

8. A metallic ball of radius 9 cm is melted and recast into three metallic balls. If the radii of two balls are 1 cm and 6 cm, find the radius of third ball.​

Answer 1

Answer:

The radius of third ball is 8 cm.

Step-by-step-explanation:

We have given that,

• Radius of metallic ball ( R ) = 9 cm

The metallic spherical ball is melted and recast into three small metallic spherical balls.

• Radius of first ball ( r₁ ) = 1 cm
• Radius of second ball ( r₂ ) = 6 cm

We have to find the radius of third ball.

Let the radius of third ball be r₃.

As the metallic ball is melted and recast into small 3 spheres, the volume of original metallic ball will be equal to sum of volumes of small 3 metallic balls.

$\displaystyle{\therefore\:\sf\:Volume\:of\:origianl\:ball\:=\:Volume\:of\:first\:ball\:+\:Volume\:of\:second\:ball\:+\:Volume\:of\:third\:ball}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:Volume\:of\:sphere\:=\:\dfrac{4}{3}\:\pi\:r^3\:}}}$

$\displaystyle{\implies\sf\:\dfrac{4}{3}\:\pi\:R^3\:=\:\dfrac{4}{3}\:\pi\:r_1^3\:+\:\dfrac{4}{3}\:\pi\:r_2^3\:+\:\dfrac{4}{3}\:\pi\:r_3^3}$

$\displaystyle{\implies\sf\:\cancel{\dfrac{4}{3}\:\pi}\:(\:R^3\:)\:=\:\cancel{\dfrac{4}{3}\:\pi}\:(\:r_1^3\:+\:r_2^3\:+\:r_3^3\:)}$

$\displaystyle{\implies\sf\:R^3\:=\:r_1^3\:+\:r_2^3\:+\:r_3^3}$

$\displaystyle{\implies\sf\:9^3\:=\:1^3\:+\:6^3\:+\:r_3^3}$

$\displaystyle{\implies\sf\:r_3^3\:=\:9^3\:-\:1^3\:-\:6^3}$

$\displaystyle{\implies\sf\:r_3^3\:=\:(\:3^2\:)^3\:-\:(\:3\:\times\:2\:)^3\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:3^2\:\times\:3^2\:\times\:3^2\:-\:3\:\times\:2\:\times\:3\:\times\:2\:\times\:3\:\times\:2\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:3^2\:\times\:3^2\:\times\:3^2\:-\:3^2\:\times\:2^2\:\times\:3\:\times\:2\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:3^2\:(\:3^2\:\times\:3^2\:-\:2^2\:\times\:6\:)\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:9\:(\:9\:\times\:9\:-\:4\:\times\:6\:)\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:9\:(\:81\:-\:24\:)\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:9\:\times\:57\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:513\:-\:1}$

$\displaystyle{\implies\sf\:r_3^3\:=\:512}$

By taking cube root on both sides, we get,

$\displaystyle{\implies\sf\:\sqrt[\sf\:3]{\sf\:r_{3}^3}\:=\:\sqrt[\sf\:3]{\sf\:512}}$

$\displaystyle{\implies\sf\:r_3\:=\:\sqrt[\sf3]{\sf2\:\times\:256}}$

$\displaystyle{\implies\sf\:r_3\:=\:\sqrt[\sf3]{\sf2\:\times\:2\:\times\:128}}$

$\displaystyle{\implies\sf\:r_3\:=\:\sqrt[\sf3]{\sf2\:\times\:2\:\times\:2\:\times\:64}}$

$\displaystyle{\implies\sf\:r_3\:=\:\sqrt[\sf3]{\underline{\sf2\:\times\:2\:\times\:2}\:\times\:\underline{\sf4\:\times\:4\:\times\:4}}}$

$\displaystyle{\implies\sf\:r_3\:=\:2\:\times\:4}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:r_3\:=\:8\:cm\:}}}}$

∴ The radius of third ball is 8 cm.

Answer 2

Step-by-step explanation:

We have given that,

Radius of metallic ball ( R ) = 9 cm

The metallic spherical ball is melted and recast into three small metallic spherical balls.

Radius of first ball ( r₁ ) = 1 cm

Radius of second ball ( r₂ ) = 6 cm

We have to find the radius of third ball.

Let the radius of third ball be r₃.

As the metallic ball is melted and recast into small 3 spheres, the volume of original metallic ball will be equal to sum of volumes of small 3 metallic balls.

\displaystyle{\therefore\:\sf\:Volume\:of\:origianl\:ball\:=\:Volume\:of\:first\:ball\:+\:Volume\:of\:second\:ball\:+\:Volume\:of\:third\:ball}∴Volumeoforigianlball=Volumeoffirstball+Volumeofsecondball+Volumeofthirdball

We know that,

\displaystyle{\boxed{\pink{\sf\:Volume\:of\:sphere\:=\:\dfrac{4}{3}\:\pi\:r^3\:}}}

Volumeofsphere=

3

4

πr

3

\displaystyle{\implies\sf\:\dfrac{4}{3}\:\pi\:R^3\:=\:\dfrac{4}{3}\:\pi\:r_1^3\:+\:\dfrac{4}{3}\:\pi\:r_2^3\:+\:\dfrac{4}{3}\:\pi\:r_3^3}