Question

A spherical drop of molten metal of radius 5 mm was found to solidify in 12 s. In how much time will a similar drop of radius 10 mm solidify?

• A. 12 s
• B. 24 s
• C. 48 s
• D. 96 s

Hint:

Chvorinov’s rule states that solidification time \( t \propto \left( \frac{V}{A} \right)^2 \). For spheres, \( t \propto r^2 \), so doubling the radius increases the time by a factor of 4.

Answer 1

The Correct Option is C

Step 1: Understand the solidification process and Chvorinov’s rule.
The solidification time of a casting is governed by Chvorinov’s rule, which states that the solidification time \( t \) is proportional to the square of the volume-to-surface-area ratio (\( \frac{V}{A} \)) of the casting: \[ t \propto \left( \frac{V}{A} \right)^2. \] For spherical droplets, we can calculate \( \frac{V}{A} \) and compare the solidification times. Step 2: Calculate the volume-to-surface-area ratio for a sphere.
For a sphere of radius \( r \):
Volume \( V = \frac{4}{3} \pi r^3 \),
Surface area \( A = 4 \pi r^2 \),
Ratio \( \frac{V}{A} = \frac{\frac{4}{3} \pi r^3}{4 \pi r^2} = \frac{r}{3} \).
So, \( \frac{V}{A} \propto r \), and the solidification time: \[ t \propto \left( \frac{V}{A} \right)^2 \propto r^2. \] Step 3: Set up the proportionality for the two droplets.
First droplet: Radius \( r_1 = 5 \, \text{mm} \), solidification time \( t_1 = 12 \, \text{s} \).
Second droplet: Radius \( r_2 = 10 \, \text{mm} \), solidification time \( t_2 = ? \).
Using Chvorinov’s rule: \[ \frac{t_2}{t_1} = \left( \frac{r_2}{r_1} \right)^2, \] \[ \frac{t_2}{12} = \left( \frac{10}{5} \right)^2, \] \[ \frac{t_2}{12} = (2)^2 = 4, \] \[ t_2 = 12 \times 4 = 48 \, \text{s}. \] Step 4: Evaluate the options.
(1) 12 s: Incorrect, as the time should increase with the square of the radius ratio. Incorrect.
(2) 24 s: Incorrect, as \( 12 \times 2 \neq 48 \). Incorrect.
(3) 48 s: Matches the calculated solidification time. Correct.
(4) 96 s: Incorrect, as \( 12 \times 8 \neq 48 \). Incorrect.
Step 5: Select the correct answer.
The solidification time for the 10 mm radius droplet is 48 s, matching option (3).