Question

A spherical solid of iron having radius of 12cm is melted and recasted into three small solid spherical spheres of different sizes.if the radii of two spheres are 6cm and 8cm .Find radii of the third side

Answer 1

4/3 π r^3 = 4/3 π r1^3 + 4/3 π r2^3 + 4/3 π r3^3
since all the 4/3 π get canceled
r^3 = r1^3 + r2^3 +r3^3
12×12×12=6×6×6 + 8×8×8 + r3^3
1728=728+ r3^3
r3^3=1728-728
r3^3=1000
r3=10cm

Answer 2

Formula to be used:
(i) Volume of a sphere= $\frac{4}{3} \pi r^3$, where r= radius of the sphere.
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The radius of the original solid sphere is 12cm
Plug in r=12 in the formula:
Volume of the original solid = $\frac{4}{3} \pi r^3=\frac{4}{3} \pi (12)^3=\frac{4}{3} \pi (1728)cm^3$

The volume of the sphere with radius 6cm is given by,
Volume =  $\frac{4}{3} \pi r^3=\frac{4}{3} \pi (6)^3=\frac{4}{3} \pi (216)cm^3$

The volume of the sphere with radius 8cm is given by,
Volume =  $\frac{4}{3} \pi r^3=\frac{4}{3} \pi (8)^3=\frac{4}{3} \pi (512)cm^3$

Let the radius of the third sphere  be Rcm.
Then, the volume of the sphere with radius Rcm is given by,
Volume =  $\frac{4}{3} \pi r^3=\frac{4}{3} \pi (R)^3=\frac{4}{3} \pi R^3cm^3$


Volume of the original sphere must be equal to the sum of the volumes of the three small spheres.

Volume of original sphere = Sum of the volumes  of the three sphere.
[tex]\text{ Factor out } \frac{4}{3} \pi \text{ from the right side: } \\ \frac{4}{3} \pi (1728)= \frac{4}{3} \pi (216+ 512 + R^3) \\ \text { Divide both sides by }\frac{4}{3} \pi \text { : } \\ 1728=216+512+R^3 \\ 1728=728+R^3\\\text{ Subtract }728 \text{ from both sides:}\\ 1000=R^3 \\\text{ Cube root both sides :} \\ 10=R\\ R=10 [/tex] [/tex]

Therefore, required radius of the third sphere is 10cm

Answer : 10cm