
A spherical metal ball of radius $8cm$ is melted to make $8$ smaller identical balls. Find the radius of each new ball.
Answer
520.8k+ views
Hint:If an object is melted to make another object, the volume of the former and latter will be the same. Here since a larger ball is melted to make $8$ smaller identical balls, the volume of the larger ball will be equal to the sum of the volumes of the smaller balls.
Formula used:Volume of a sphere, $V = \dfrac{4}{3}\pi {r^3}$
where, $r$ is the radius of the sphere.
Complete step-by-step answer:
Given that a spherical ball of radius $8cm$ is melted to make $8$ smaller identical balls.
We have to find the radius of each new ball. Let it be $r$.
If an object is melted to make another object the volume of the former and latter ones will be the same.
Here, since a larger ball is melted to make $8$ smaller identical balls, the volume of the larger ball will be equal to the sum of the volumes of the smaller balls.
Volume of a sphere of radius $r$ is $\dfrac{4}{3}\pi {r^3}$
So, the volume of the larger sphere is $\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi \times {8^3}$
There are $8$ identical small balls with radius $r$
Equating volume of total number smaller spherical balls and volume of larger spherical ball we get,
$ \Rightarrow 8 \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {8^3}$
Cancelling $8 \times \dfrac{4}{3}\pi $ from both sides of the above equation we have,
$ \Rightarrow {r^3} = {8^2} = 64$
$ \Rightarrow r = \sqrt[3]{{64}} = 4$
Therefore the radius of each smaller sphere is $4cm$.
Note:Here the radius of the larger sphere is given directly. Sometimes instead of that, area or volume of the sphere will be mentioned. In those cases we have to find radius using this and then substitute it to find the answer.
Formula used:Volume of a sphere, $V = \dfrac{4}{3}\pi {r^3}$
where, $r$ is the radius of the sphere.
Complete step-by-step answer:
Given that a spherical ball of radius $8cm$ is melted to make $8$ smaller identical balls.
We have to find the radius of each new ball. Let it be $r$.
If an object is melted to make another object the volume of the former and latter ones will be the same.
Here, since a larger ball is melted to make $8$ smaller identical balls, the volume of the larger ball will be equal to the sum of the volumes of the smaller balls.
Volume of a sphere of radius $r$ is $\dfrac{4}{3}\pi {r^3}$
So, the volume of the larger sphere is $\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi \times {8^3}$
There are $8$ identical small balls with radius $r$
Equating volume of total number smaller spherical balls and volume of larger spherical ball we get,
$ \Rightarrow 8 \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {8^3}$
Cancelling $8 \times \dfrac{4}{3}\pi $ from both sides of the above equation we have,
$ \Rightarrow {r^3} = {8^2} = 64$
$ \Rightarrow r = \sqrt[3]{{64}} = 4$
Therefore the radius of each smaller sphere is $4cm$.
Note:Here the radius of the larger sphere is given directly. Sometimes instead of that, area or volume of the sphere will be mentioned. In those cases we have to find radius using this and then substitute it to find the answer.
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