Question

The diameter of the iron ball used for the shot -put game is 14 cm. It is melted and then a solid cylinder of height $2\dfrac{1}{3}cm$ is made . what will be the diameter of the base of the cylinder?
A. 14cm
B. $\dfrac{{14}}{3}cm$
C. 28cm
D. $\dfrac{{28}}{3}cm$

Answer 1

Hint: In this particular type of question we first need to calculate the volume of the spherical ball and then equate the derived volume with the volume of the solid cylinder to get the desired answer .
Volume of sphere = $\dfrac{4}{3}\pi {r^3}$

Complete step-by-step answer:
Let the radius of ball be r = $\dfrac{{diameter}}{2} = \dfrac{{14}}{2} = 7cm$
Volume = $\dfrac{4}{3}\pi \times {7^3} = \dfrac{{1372\pi }}{3}c{m^3}$
Now volume of cylinder = $\pi {R^2}h$
$ = \pi {R^2} \times \dfrac{7}{3}$ ( since h = $2\dfrac{1}{3} = \dfrac{7}{3}cm$ )
where R and h are the radius and height of the cylinder respectively .
since we know that the volumes of both solids are same
$\therefore \dfrac{{1372\pi }}{3} = \pi {R^2} \times \dfrac{7}{3}$
$
   \Rightarrow {R^2} = \dfrac{{1372}}{7} = 196 \\
   \Rightarrow R = 14cm \\
$
Therefore diameter of cylinder =28cm

Note: We should remember to recall the basic formulas of volume of spherical balls and cylinders. Also note that in this question the diameter is provided rather than the radius, therefore we have to convert diameter into radius by multiplying it with $\dfrac{1}{2}$. Always keep in mind that the volume doesn't change if any solid's shape is recast.