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(A) Given,edge of the cube $x = 1 \ cm$. Volume of the cube $V = x^3 = (1 \ cm)^3 = 1 \ cm^3$. Since the ice melts in a gravity-free container,the wat

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$(36 \pi)^{1/3} \ cm^2$

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(A) Given,edge of the cube $x = 1 \ cm$. Volume of the cube $V = x^3 = (1 \ cm)^3 = 1 \ cm^3$. Since the ice melts in a gravity-free container,the water forms a spherical drop. Volume of the spherical drop = Volume of the cube = $1 \ cm^3$. Let $r$ be the radius of the spherical drop. $\frac{4}{3} \pi r^3 = 1 \implies r^3 = \frac{3}{4 \pi} \implies r = \left(\frac{3}{4 \pi}ight)^{1/3}$. Surface area of the spherical drop $A = 4 \pi r^2$. $A = 4 \pi \left(\frac{3}{4 \pi}ight)^{2/3} = 4 \pi \frac{3^{2/3}}{(4 \pi)^{2/3}}$. $A = (4 \pi)^{1 - 2/3} 	imes 3^{2/3} 	imes 4^{2/3} / 4^{2/3} 	ext{ (simplifying powers)} = (4 \pi)^{1/3} 	imes (3^2)^{1/3} = (4 \pi 	imes 9)^{1/3} = (36 \pi)^{1/3} \ cm^2$.

An ice cube of edge $1 \ cm$ melts in a gravity-free container. The approximate surface area of the water formed is (water is in the form of a spherical drop)

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