A cubical cavity of 100 mm side undergoes volumetric solidification shrinkage and volumetric solid contraction of 5% each. No riser is used. Assume uniform cooling in all direction. What will be the side of the cube after solidification and contraction? [assume, (0.95)^1/3= 0.983, (0.95)^2/3= 0.9663]

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A cubical cavity of 100 mm side undergoes volumetric solidification shrinkage and volumetric solid contraction of 5% each. No riser is used. Assume uniform cooling in all direction. What will be the side of the cube after solidification and contraction? [assume, (0.95)^1/3= 0.983, (0.95)^2/3= 0.9663]

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9.83 cm
9.663 cm
96.63 cm
98.3 cm

Answer 1

Concept:

When a cubical cavity undergoes volumetric solidification shrinkage and solid contraction, its volume decreases in proportion to the given shrinkage percentages.

The final volume after shrinkage is given by:

\( V_{\text{final}} = V_{\text{initial}} \times (0.95)^2 \)

Since the cube shrinks uniformly in all directions, the final side length can be expressed as:

\( L_{\text{final}} = L_{\text{initial}} \times (0.95)^{2/3} \)

Given:

Initial side length of the cube: \(L_{\text{initial}} = 100 ~ mm = 10~ cm\)
Volumetric solidification shrinkage: 5%
Volumetric solid contraction: 5%
Given value: \((0.95)^{2/3} = 0.9663\)

Calculation:

Step 1: Compute the final side length

\( L_{\text{final}} = 10 \times 0.9663 \)

\( L_{\text{final}} = 9.663~cm\)

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