Question

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

• A. $(36\pi)^{1/3} \text{ cm}^2$
• B. $(24\pi)^{1/3} \text{ cm}^2$
• C. $(28\pi)^{1/3} \text{ cm}^2$
• D. $(12\pi)^{1/3} \text{ cm}^2$

Hint:

Whenever a shape of volume $V$ deforms into a perfect sphere, its surface area can be related directly via a consolidated geometric identity: $A = (36\pi V^2)^{1/3}$. Substituting $V = 1 \text{ cm}^3$ straight into this shortcut gives $(36\pi)^{1/3}$ instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An ice cube of edge $1 \text{ cm}$ melts completely into water. Since the container is in a gravity-free environment, the resulting liquid water minimizes its surface energy by pulling itself into a single, perfectly spherical droplet. We need to calculate the surface area of this sphere.

Step 2: Key Formula or Approach:
The volume of matter remains completely conserved during the melting process. Thus, the volume of the original cube must equal the volume of the final spherical water drop:
$$V_{\text{cube}} = V_{\text{sphere}}$$ The volume formulas are $V_{\text{cube}} = x^3$ and $V_{\text{sphere}} = \frac{4}{3}\pi r^3$.
The surface area of a sphere is given by:
$$A = 4\pi r^2$$

Step 3: Detailed Explanation:
First, calculate the volume of the ice cube with side length $x = 1 \text{ cm}$:
$$V_{\text{cube}} = (1 \text{ cm})^3 = 1 \text{ cm}^3$$ Equate this to the volume of the spherical drop to determine its radius $r$:
$$\frac{4}{3}\pi r^3 = 1 \implies r^3 = \frac{3}{4\pi} \implies r = \left(\frac{3}{4\pi}\right)^{1/3}$$ Now, square the radius to find $r^2$:
$$r^2 = \left(\frac{3}{4\pi}\right)^{2/3} = \left(\frac{9}{16\pi^2}\right)^{1/3}$$ Substitute this value into the surface area equation of the sphere:
$$A = 4\pi \cdot r^2 = 4\pi \cdot \left(\frac{9}{16\pi^2}\right)^{1/3}$$ To bring the outer coefficient $4\pi$ inside the cube root bracket, cubing it gives $(4\pi)^3 = 64\pi^3$:
$$A = \left(64\pi^3 \times \frac{9}{16\pi^2}\right)^{1/3}$$ Simplify the expression inside the radical:
$$A = \left(\frac{64}{16} \times 9 \times \frac{\pi^3}{\pi^2}\right)^{1/3} = (4 \times 9 \times \pi)^{1/3} = (36\pi)^{1/3} \text{ cm}^2$$

Step 4: Final Answer:
The approximate surface area of the water drop formed is $(36\pi)^{1/3} \text{ cm}^2$, which corresponds to option (A).