A metallic ball of radius 9 ,cm is melted and | vedclass

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(B) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.Let the radius of the original ball be $R = 9 \,cm$.Let the radii of the

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$8$

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(B) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.Let the radius of the original ball be $R = 9 \,cm$.Let the radii of the three smaller balls be $r_1 = 1 \,cm$,$r_2 = 6 \,cm$,and $r_3$ (the unknown radius).Since the volume remains conserved during melting and recasting,the volume of the original ball equals the sum of the volumes of the three smaller balls:$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3$Dividing both sides by $\frac{4}{3} \pi$,we get:$R^3 = r_1^3 + r_2^3 + r_3^3$Substituting the given values:$9^3 = 1^3 + 6^3 + r_3^3$$729 = 1 + 216 + r_3^3$$729 = 217 + r_3^3$$r_3^3 = 729 - 217$$r_3^3 = 512$Taking the cube root on both sides:$r_3 = \sqrt[3]{512} = 8 \,cm$.Thus,the radius of the third ball is $8 \,cm$.

$A$ metallic ball of radius $9 \,cm$ is melted and recast into three metallic balls. If the radii of two balls are $1 \,cm$ and $6 \,cm$,find the radius of the third ball (in $cm$).

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