3 spherical balls of radii 1, cm,2, cm,and 3, | vedclass

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(C) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Total volume of the three spherical balls $= \frac{4}{3} \pi (1^3 + 2^3 + 3^3) = \fra

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$3$

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(C) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Total volume of the three spherical balls $= \frac{4}{3} \pi (1^3 + 2^3 + 3^3) = \frac{4}{3} \pi (1 + 8 + 27) = \frac{4}{3} \pi (36) \, cm^3$.Given that $25\%$ of the material is lost during the melting process.Remaining volume $= \frac{4}{3} \pi (36) 	imes (1 - 0.25) = \frac{4}{3} \pi (36) 	imes 0.75 = \frac{4}{3} \pi (36) 	imes \frac{3}{4} = \frac{4}{3} \pi (27) \, cm^3$.Let the radius of the new spherical ball be $R$.Then,$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (27)$.$R^3 = 27$.$R = 3 \, cm$.

$3$ spherical balls of radii $1\, cm$,$2\, cm$,and $3\, cm$ are melted to form a single spherical ball. In the process,the loss of material is $25\%$. The radius of the new ball is.........$cm$.

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