Three spherical balls of radius 2 , cm,4 , cm | vedclass

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(A) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Total volume of the three spheres $= \frac{4}{3} \pi (2^3 + 4^3 + 6^3) = \frac{4}{3}

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Total volume of the three spheres $= \frac{4}{3} \pi (2^3 + 4^3 + 6^3) = \frac{4}{3} \pi (8 + 64 + 216) = \frac{4}{3} \pi (288) = 384 \pi \, cm^3$.Since $25 \%$ of the material is lost,the volume of the new sphere is $75 \%$ of the total initial volume.Volume of the new sphere $= 0.75 	imes 384 \pi = 288 \pi \, cm^3$.Let the radius of the new sphere be $R$. Then,$\frac{4}{3} \pi R^3 = 288 \pi$.$R^3 = 288 	imes \frac{3}{4} = 72 	imes 3 = 216$.$R = \sqrt[3]{216} = 6 \, cm$.

Three spherical balls of radius $2 \, cm$,$4 \, cm$,and $6 \, cm$ are melted to form a new spherical ball. In this process,there is a loss of $25 \%$ of the material. What is the radius (in $cm$) of the new ball?

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