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 $$2\frac{1}{3}$$ cm is made. What will be the diameter of the base of the cylinder ?

Solution (By Examveda Team)

Let the radius of the cylinder be R
Then,
$$\eqalign{ & \pi \times {R^2} \times \frac{7}{3} = \frac{4}{3}\pi \times 7 \times 7 \times 7 \cr & \Rightarrow {R^2} = \left( {\frac{{4 \times 7 \times 7 \times 7}}{3} \times \frac{3}{7}} \right) \cr & \Rightarrow {R^2} = 196 \cr & \Rightarrow {R^2} = {\left( {14} \right)^2} \cr & \Rightarrow {R^2} = 14{\text{ cm}} \cr} $$
∴ Diameter = 2R = 28