Question

A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2 cm and 2 cm, find the diameter of the third ball.

Answer 1

Answer:

The diameter of third ball is $\dfrac{5}{2}$  cm

Step-by-step explanation:

Given as :

A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls.

The diameter of bigger spherical ball = D = 3 cm

Or, The radius of bigger spherical ball = R = $\dfrac{D}{2}$

i.e   Radius = R = $\dfrac{3}{2}$   cm

As bigger ball is recast into three small balls

So, The diameter of balls = $d_1$ = $\dfrac{3}{2}$   cm

Or,                                         $r_1$ = $\dfrac{3}{4}$   cm

                                             $d_2$ =  2  cm

                                             $r_2$ = 1 cm

Let The diameter of third ball = $d_3$   cm

And                                    radius = $r_3$   cm

Again

Volume of spherical ball = $\dfrac{4}{3}$ × π × Radius³

So, Volume of bigger ball = Sum of volume of three smaller balls

i.e  $\dfrac{4}{3}$ × π × R³  =  $\dfrac{4}{3}$ × π × $r_1$ ³ +  $\dfrac{4}{3}$ × π × $r_2$ ³  +  $\dfrac{4}{3}$ × π × $r_3$ ³

Or, R³ = $r_1$ ³ + $r_2$ ³ + $r_3$ ³

Or  (  $\dfrac{3}{2}$ )³  = (  $\dfrac{3}{4}$ )³  +  1³ + $r_3$ ³

Or, $r_3$ ³ = $\dfrac{27}{8}$  -  $\dfrac{27}{64}$ - 1  

Or, $r_3$ ³ = $\dfrac{125}{64}$

∴ , $r_3$   = ∛$\dfrac{125}{64}$

i.e  $r_3$  = $\dfrac{5}{4}$  cm

So, The radius of third ball = $r_3$  = $\dfrac{5}{4}$  cm

The diameter of third ball = $d_3$ = 2 ×  $\dfrac{5}{4}$  cm  = $\dfrac{5}{2}$  cm

Hence, The diameter of third ball is $\dfrac{5}{2}$  cm  . Answer

Answer 2

The diameter of the third ball is 2.5 cm

Step-by-step explanation:

Given data

Radius of main ball$\mathbf{=\frac{3}{2} \ cm}$

Radius of first ball$\mathbf{=\frac{3}{4} \ cm}$

Radius of second ball$\mathbf{=\frac{2}{2}=1 \ cm}$

Radius of third ball$\mathbf{=\frac{D}{2} \ cm}$

We know volume of spherical ball $\mathbf{=\frac{4}{3}\pi R^{3} }$

Volume of first ball+Volume of second ball+Volume of third ball =Volume of main ball

$\mathbf{\frac{4}{3}\pi \left ( \frac{3}{4} \right )^{3}+\frac{4}{3}\pi 1^{3}+\frac{4}{3}\pi \left ( \frac{D}{2} \right )^{3}=\frac{4}{3}\pi \left ( \frac{3}{2} \right )^{3} }$

On cancel common term on both side of above equation, we get

$\mathbf{\left ( \frac{3}{4} \right )^{3}+1+\left ( \frac{D}{2} \right )^{3}=\left ( \frac{3}{2} \right )^{3} }$

$\mathbf{\left ( \frac{D}{2} \right )^{3}=\left ( \frac{3}{2} \right )^{3}-\left ( \frac{3}{4} \right )^{3}-1}$

$\mathbf{\left ( \frac{D}{2} \right )^{3}=\frac{125}{64}}$

So

$\mathbf{\frac{D}{2}=\frac{5}{4}}$

$\mathbf{D=\frac{5}{2}=2.5 \ cm=}$ This is diameter of third ball.