A single solid sphere is melted into three solid spheres of lead whose radii are 1 cm, 6 cm, 8 cm respectively. Find the radius of the original sphere
Solution
Correct Answer: Option A
Let the radius of the original sphere be r
∴ Volume of the solid sphere is given by \(= \frac{4}{3}\pi {r^3}\)
∴ Volume of the first solid sphere \(= \frac{4}{3}\pi {\left( 1 \right)^3}\)
∴ Volume of the second solid sphere \(= \frac{4}{3}\pi {\left( 6 \right)^3}\)
∴ Volume of the third solid sphere \(= \frac{4}{3}\pi {\left( 8 \right)^3}\)
Since a single sphere is melted into three solid sphere
∴ Volume of the original solid sphere = Volume of the three smaller solid spheres
\(\therefore {\rm{\;}}\frac{4}{3}\pi {r^3} = {\rm{\;}}\frac{4}{3}\pi {\left( 1 \right)^3} + {\rm{\;}}\frac{4}{3}\pi {\left( 6 \right)^3} + {\rm{\;}}\frac{4}{3}\pi {\left( 8 \right)^3}\)
⟹ r3= 13 + 63 + 83
⟹ r3= 729
⟹ r = 9 cm
∴ Radius of the original sphere (r) = 9 cm
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