Question:hard
A steel cube of side 10 cm is made by the sand-casting process. A cylindrical side-riser with diameter, \( d \), and height, \( h \), needs to be used.
Assume:
- No surface sharing between the riser and casting.
- \( d = h \).
- The connecting link between the riser and casting does not freeze before the casting.
- All the surfaces of the riser and casting are subjected to identical cooling conditions.
In this situation, which ONE or MORE among the following values of \( d \) (in cm) can theoretically fully compensate for shrinkage during casting?
A steel cube of side 10 cm is made by the sand-casting process. A cylindrical side-riser with diameter, \( d \), and height, \( h \), needs to be used.
Assume:
Assume:
- No surface sharing between the riser and casting.
- \( d = h \).
- The connecting link between the riser and casting does not freeze before the casting.
- All the surfaces of the riser and casting are subjected to identical cooling conditions.
Show Hint
Use Chvorinov's Rule: the riser's modulus \( d/6 \) must be at least the cube's modulus \( 10/6 \), so \( d \ge 10\ cm \).
Updated On: Aug 5, 2026
- 5
- 8
- 15
- 20
Show Solution
The Correct Option is C, D
Solution and Explanation
Step 1: State the riser design rule in modulus-ratio form:
A commonly used practical guideline (based on Chvorinov's Rule) is that the riser modulus should be at least equal to the casting modulus, often written as a ratio:
\[ \frac{M_{riser}}{M_{casting}} \ge 1 \]
If this ratio is 1 or greater, the riser solidifies at the same time or later than the casting, so it can keep feeding liquid metal right up until the casting is fully solid.
Step 2: Find the casting modulus numerically:
The steel cube has side $a = 10$ cm, giving volume $V_c = 1000\ cm^3$ and surface area $A_c = 600\ cm^2$ (all six faces exposed).
\[ M_c = \frac{1000}{600} = 1.667\ cm \]
Step 3: Test each candidate diameter directly instead of solving an inequality first:
For a cylinder with $h = d$ and no shared surface, general formulas give $V_r = \frac{\pi}{4}d^3$ and $A_r = \frac{3\pi}{2}d^2$, so $M_r = d/6$.
Now plug in each option:
For $d = 5$: $M_r = 5/6 = 0.833$ cm, ratio $= 0.833/1.667 = 0.5$, less than 1, riser freezes too soon. Fails.
For $d = 8$: $M_r = 8/6 = 1.333$ cm, ratio $= 1.333/1.667 = 0.8$, still less than 1. Fails.
For $d = 15$: $M_r = 15/6 = 2.5$ cm, ratio $= 2.5/1.667 = 1.5$, greater than 1. Works.
For $d = 20$: $M_r = 20/6 = 3.333$ cm, ratio $= 3.333/1.667 = 2.0$, greater than 1. Works.
Step 4: Collect the passing values:
Only $d = 15$ cm and $d = 20$ cm give a modulus ratio of 1 or more, so only these two riser sizes are large enough to remain liquid until after the casting solidifies.
Final Answer:
Testing each option against the modulus ratio confirms the same two valid diameters. \[ \boxed{d = 15\ cm\ \text{and}\ d = 20\ cm} \]
A commonly used practical guideline (based on Chvorinov's Rule) is that the riser modulus should be at least equal to the casting modulus, often written as a ratio:
\[ \frac{M_{riser}}{M_{casting}} \ge 1 \]
If this ratio is 1 or greater, the riser solidifies at the same time or later than the casting, so it can keep feeding liquid metal right up until the casting is fully solid.
Step 2: Find the casting modulus numerically:
The steel cube has side $a = 10$ cm, giving volume $V_c = 1000\ cm^3$ and surface area $A_c = 600\ cm^2$ (all six faces exposed).
\[ M_c = \frac{1000}{600} = 1.667\ cm \]
Step 3: Test each candidate diameter directly instead of solving an inequality first:
For a cylinder with $h = d$ and no shared surface, general formulas give $V_r = \frac{\pi}{4}d^3$ and $A_r = \frac{3\pi}{2}d^2$, so $M_r = d/6$.
Now plug in each option:
For $d = 5$: $M_r = 5/6 = 0.833$ cm, ratio $= 0.833/1.667 = 0.5$, less than 1, riser freezes too soon. Fails.
For $d = 8$: $M_r = 8/6 = 1.333$ cm, ratio $= 1.333/1.667 = 0.8$, still less than 1. Fails.
For $d = 15$: $M_r = 15/6 = 2.5$ cm, ratio $= 2.5/1.667 = 1.5$, greater than 1. Works.
For $d = 20$: $M_r = 20/6 = 3.333$ cm, ratio $= 3.333/1.667 = 2.0$, greater than 1. Works.
Step 4: Collect the passing values:
Only $d = 15$ cm and $d = 20$ cm give a modulus ratio of 1 or more, so only these two riser sizes are large enough to remain liquid until after the casting solidifies.
Final Answer:
Testing each option against the modulus ratio confirms the same two valid diameters. \[ \boxed{d = 15\ cm\ \text{and}\ d = 20\ cm} \]
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