MHT CET · Physics · Mechanical Properties of Fluids
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)
- A \((36 \pi)^{1 / 3} \mathrm{~cm}^2\)
- B \((24 \pi)^{1 / 3} \mathrm{~cm}^2\)
- C \((28 \pi)^{1 / 3} \mathrm{~cm}^2\)
- D \((12 \pi)^{1 / 3} \mathrm{~cm}^2\)
Answer & Solution
Correct Answer
(A) \((36 \pi)^{1 / 3} \mathrm{~cm}^2\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{x}=1 \mathrm{~cm}
\)
\(\therefore\) 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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