A spherical ball of lead,3 , cm in diameter,i | vedclass

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(B) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Let the radius of the large ball be $R = \frac{3}{2} = 1.5 \, cm$.Let the radii of th

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$2.5$

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(B) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Let the radius of the large ball be $R = \frac{3}{2} = 1.5 \, cm$.Let the radii of the three smaller balls be $r_1, r_2,$ and $r_3$.Given $r_1 = \frac{1.5}{2} = 0.75 \, cm$ and $r_2 = \frac{2}{2} = 1 \, cm$.Since the volume remains constant during melting and recasting:$\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$$R^3 = r_1^3 + r_2^3 + r_3^3$$(1.5)^3 = (0.75)^3 + (1)^3 + r_3^3$$3.375 = 0.421875 + 1 + r_3^3$$3.375 = 1.421875 + r_3^3$$r_3^3 = 3.375 - 1.421875 = 1.953125$$r_3 = \sqrt[3]{1.953125} = 1.25 \, cm$.The diameter of the third ball is $2 	imes r_3 = 2 	imes 1.25 = 2.5 \, cm$.

$A$ spherical ball of lead,$3 \, cm$ in diameter,is melted and recast into three spherical balls. The diameters of $2$ of these are $1.5 \, cm$ and $2 \, cm$,respectively. The diameter of the third ball is $....... \, cm$.

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