Question

A spherical ball of radius $3$ cm is melted and recast into three spherical balls. The radii of two of the balls are $1.5$ cm and $2$ cm. Find the diameter of the third ball.

• A. $1cm$
• B. $3cm$
• C. $4cm$
• D. $5cm$

Answer 1

The correct option is D $5cm$

Volume of the large ball of radius 3 cm

$=\frac{4}{3}\times \frac{22}{7}\times 3\times 3\times 3=\frac{792}{7}c{m}^3$

Total volume of the two smaller balls of radii 1.5 cm and 2 cm

$\frac{4}{3}\times \frac{22}{7}\times \frac{3}{2}\times \frac{3}{2}\times \frac{3}{2}+\frac{4}{3}\times \frac{22}{7}\times 2\times 2\times 2$

$[\therefore V=\frac{4}{3}\pi r^3]$

$\frac{4}{3}\times \frac{22}{7}(\frac{27}{8}+8)=\frac{143}{3}c{m}^3$

$\therefore$ Volume of the third ball

$=\frac{792}{7}-\frac{143}{3}=\frac{2376-1001}{21}$

$\frac{1375}{21}c{m}^3$

Let r be the radius of the 3rd ball

$\therefore \frac{4}{3}\times \frac{22}{7}\times r^3=\frac{1375}{21}$

$r^3\frac{1375}{21}\times \frac{3}{4}\times \frac{7}{22}=\frac{125}{8}$

$\therefore r=\sqrt[3]{3\frac{125}{8}}=\frac{5}{2}cm$

Hence diameter $=2r=2\times \frac{5}{2}=5cm$