Question

A spherical ball of lead $3$ cm in diameter is melted and recast into three spherical balls. The diameter of two of these are $1.5$ cm and $2$ cm respectively. The diameter of the third ball is ___________.

• A. $2.66$ cm
• B. $2.5$ cm
• C. $3$ cm
• D. $3.5$ cm

Answer 1

Answer: (B)

$2.5$ cm

In this case the volume of the initial spherical ball will be the sum of the volumes of the new three balls.

Volume of a sphere $=\frac{4}{3}\pi r^3$

As the diameter of the given sphere is $3cm$ then its radius $=\frac{3}{2}=1.5cm$ ......... as $[radius=\frac{diameter}{2}$

The radius of other two balls will be $.75cm$ and $1cm$.

Let the radius of the third new ball be $Rcm$.

Volume of the three new sphere $=$ Volume of old sphere

$\Rightarrow$ $\frac{4}{3}\pi {.75}^3$ + $\frac{4}{3}\pi {.1}^3$ + $\frac{4}{3}\pi .R^3$ $=$ $\frac{4}{3}\pi {1.5}^3$

$\Rightarrow$ ${.75}^3$ + $1^3$ + $R^3$ $=$ ${1.5}^3$

$\Rightarrow$ $0.421875+1+R^3=3.375$

$\Rightarrow$ $1.421875+R^3=3.375$

$\Rightarrow$ $R^3=1.953125$

$\Rightarrow$ $R={1.953125}^{\frac{1}{3}}$

$\Rightarrow$ $R=1.25$

So, the radius of the third new ball is $1.25cm$ .

Therefore, the diameter $=2.5$ cm