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An ice block of $\frac{4 π}{3} {cm}^{3}$ is placed in an artificial satellite. When ice melts into water then its surface area will be, if the densities of ice and water are assumed to be same

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$\frac{4 π}{3} {cm}^{2}$

${(\frac{4 π}{3})}^{1 / 3} {cm}^{2}$

$4 π {cm}^{2}$

$(36 π)^{1 / 3} {cm}^{2}$

answer is B.

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Detailed Solution

Water shall assume spherical shape volume $V = \frac{4}{3} {πR}^{3}; R = Radius$
Equate volume $\frac{4}{3} {πR}^{3} = \frac{4 π}{3} \Rightarrow R = 1; Surface area = 4 {πR}^{2} = 4 π {cm}^{2}$

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