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An ice cube of edge $1 \mathrm{~cm}$ melts in a gravity free container. The approximate surface area of water formed is (water is in the form of spherical drop)
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Verified Answer
The correct answer is:
$(36 \pi)^{1 / 3} \mathrm{~cm}^2$
$$
\mathrm{x}=1 \mathrm{~cm}
$$
$\therefore$ Volume of the cube $\mathrm{v}=\mathrm{x}^3=1 \mathrm{~cm}^3$, volume of drop $=$ volume of cube.
$$
\begin{aligned}
& \frac{4}{3} \pi r^3=x^3=1 \mathrm{~cm}^3 \\
& \therefore r^3=\frac{3}{4 \pi} \text { or } r=\left(\frac{3}{4 \pi}\right)^{\frac{1}{3}} \\
& \therefore r^2=\left(\frac{9}{16 \pi^2}\right)^{\frac{1}{3}}
\end{aligned}
$$
Surface area of drop $=4 \pi r^2=4 \pi\left(\frac{9}{16 \pi^2}\right)^{\frac{1}{3}}$
$$
=\left(64 \pi^3 \times \frac{9}{16 \pi^2}\right)^{\frac{1}{3}}=(36 \pi)^{\frac{1}{3}}
$$
\mathrm{x}=1 \mathrm{~cm}
$$
$\therefore$ Volume of the cube $\mathrm{v}=\mathrm{x}^3=1 \mathrm{~cm}^3$, volume of drop $=$ volume of cube.
$$
\begin{aligned}
& \frac{4}{3} \pi r^3=x^3=1 \mathrm{~cm}^3 \\
& \therefore r^3=\frac{3}{4 \pi} \text { or } r=\left(\frac{3}{4 \pi}\right)^{\frac{1}{3}} \\
& \therefore r^2=\left(\frac{9}{16 \pi^2}\right)^{\frac{1}{3}}
\end{aligned}
$$
Surface area of drop $=4 \pi r^2=4 \pi\left(\frac{9}{16 \pi^2}\right)^{\frac{1}{3}}$
$$
=\left(64 \pi^3 \times \frac{9}{16 \pi^2}\right)^{\frac{1}{3}}=(36 \pi)^{\frac{1}{3}}
$$
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