Question

A spherical drop of liquid metal 'r' mm radius takes 12 seconds to solidify. What would be the time taken to solidify if the diameter of the drop is doubled?

• A. 24 seconds
• B. 36 seconds
• C. 48 seconds
• D. 96 seconds

Hint:

Chvorinov's Rule is crucial for calculating solidification time. Remember that for spheres, solidification time is proportional to the square of the radius (\(T \propto r^2\)). If the diameter (and thus radius) doubles, the solidification time becomes four times longer.

Answer 1

The Correct Option is C

Step 1: Understand Chvorinov's Rule.
Chvorinov's Rule is used to estimate the solidification time of a casting. It states that the solidification time (\(T\)) is directly proportional to the square of the ratio of the volume (\(V\)) to the surface area (\(A\)) of the casting. \[ T = K \left( \frac{V}{A} \right)^2 \] where \(K\) is the mold constant.
Step 2: Apply Chvorinov's Rule to a spherical drop.
For a sphere:
Volume \(V = \frac{4}{3} \pi r^3\)
Surface Area \(A = 4 \pi r^2\)
The volume to surface area ratio is: \[ \frac{V}{A} = \frac{\frac{4}{3} \pi r^3}{4 \pi r^2} = \frac{r}{3} \] So, the solidification time \(T\) for a spherical drop is proportional to \( \left( \frac{r}{3} \right)^2 \), which means \( T \propto r^2 \).
Step 3: Calculate the new solidification time when the diameter is doubled.
Let the initial radius be \(r_1 = r\). The initial solidification time is \(T_1 = 12\) seconds.
When the diameter of the drop is doubled, the radius also doubles. So, the new radius is \(r_2 = 2r\).
Since \( T \propto r^2 \), we can write the ratio of solidification times: \[ \frac{T_2}{T_1} = \frac{K r_2^2}{K r_1^2} = \left( \frac{r_2}{r_1} \right)^2 \] Substitute the values: \[ \frac{T_2}{12} = \left( \frac{2r}{r} \right)^2 \] \[ \frac{T_2}{12} = (2)^2 \] \[ \frac{T_2}{12} = 4 \] \[ T_2 = 4 \times 12 \] \[ T_2 = 48 \text{ seconds} \] Therefore, the time taken to solidify if the diameter of the drop is doubled would be 48 seconds. The final answer is \( \boxed{\text{3}} \).