Question & Answer
QUESTION

A spherical ball of lead 5 cm in diameter is melted and reset into three spherical balls. The diameter of these balls are $2$ cm and $2{(14.5)^{\dfrac{1}{3}}}$ cm. Find the diameter of the third ball.
$
  {\text{A}}{\text{. 8cm}} \\
  {\text{B}}{\text{. 5cm}} \\
  {\text{C}}{\text{. 4cm}} \\
  {\text{D}}{\text{. 1cm}} \\
 $

ANSWER Verified Verified
Hint:-In this question first we need to convert the diameter into the radius of given spheres. And then calculate their volumes separately. And Finally use the concept of constant volume of ball before or after the melting to get the radius of the third ball.
Complete step-by-step answer:
Let the radius of the spherical ball which is melted is R=5cm.
Volume of a sphere =$\dfrac{4}{3}\pi {(r)^3}$
where $r$ is the radius of circle
Volume of spherical ball which is melted=$\dfrac{4}{3}\pi {(2.5)^3}{\text{c}}{{\text{m}}^3}{\text{ \{ R = }}\dfrac{5}{2}{\text{\} eq}}{\text{.1}}$
Now, let the radius of small spherical balls be ${r_{1,}}{r_{2,}}{r_{3,}}$.
And
   $
   \Rightarrow {r_1} = \dfrac{2}{2} \\
   \Rightarrow {r_1} = 1{\text{cm}} \\
$
Similarly
$
   \Rightarrow {r_2} = \dfrac{{\;\;2{{(14.5)}^{\dfrac{1}{3}}}}}{2} \\
   \Rightarrow {r_2} = {(14.5)^{\dfrac{1}{3}}}{\text{cm}} \\
 $
Volume of spherical ball with radius ${r_1} = $$\dfrac{4}{3}\pi {(1)^3}$ eq.2
 Volume of spherical ball with radius ${r_2} = $
                                                                          $
   = \dfrac{4}{3}\pi {(14.4)^{3 \times \dfrac{1}{3}}} \\
   = \dfrac{4}{3}\pi (14.4){\text{ eq}}{\text{.3}} \\
$
And, Volume of spherical ball with radius ${r_3} = $ $\dfrac{4}{3}\pi {({r_3})^3}$ eq.4



Since, on melting a spherical ball into small spherical balls, the volume remains the same.
Then,
Volume of spherical ball with radius 2.5cm = Volume of spherical ball with radius 1cm
+ Volume of spherical ball with radius ${(14.5)^{\dfrac{1}{3}}}{\text{cm}}$+ Volume of spherical ball with radius ${r_3}{\text{cm}}$
Now put values of respective volumes from eq.1, eq.2, eq.3, eq.4 in above equation we get

 $ \Rightarrow {\text{ }}\dfrac{4}{3}\pi {(2.5)^3}{\text{ = }}\dfrac{4}{3}\pi {\text{ + }}\dfrac{4}{3}\pi (14.5){\text{ }} + \;{\text{ }}\dfrac{4}{3}\pi {({r_3})^3}$
On solving above equation, we get
$
   \Rightarrow {\text{ }}{(2.5)^3}{\text{ = 1 + }}14.5{\text{ }} + \;{\text{ }}{({r_3})^3} \\
   \Rightarrow {\text{ }}{({r_3})^3}{\text{ = 15}}{\text{.625 - 15}}{\text{.5}} \\
   \Rightarrow {\text{ }}{({r_3})^3}{\text{ = 0}}{\text{.125}} \\
   \Rightarrow {\text{ }}{r_3}{\text{ = 0}}{\text{.5cm}} \\
 $
Therefore, diameter of third ball =${r_3} \times 2$
                                                          =1cm
Hence, option D. is correct.
Note: -Whenever we get this type of problem the key concept of solving is to remember one thing that on transforming any shape to another shape then the volume remains the same even if its size may be changed. And never forget to convert the diameter into radius to use the volume formula of sphere($\dfrac{4}{3}\pi {(r)^3}$).