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A cube shaped casting solidifies In 10 minutes. Time taken by the same material cube 27 times heavier than the original one in minutes will be:
1. 90
2. 75
3. 10
4. 60

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Correct Answer - Option 1 : 90

Concepts:

From Chvorinov's rule solidification time,

\({t_s} = K{\left( {\frac{V}{{SA}}} \right)^2}\)

For cube 

\(\frac{V}{{SA}} = \;\frac{{{a^3}}}{{6\; × \ {a^2}}} = \;\frac{a}{6}\)

Where K is constant, V is the volume of the solid, SA is the surface area of the solid.

Calculation:

Given:

First cube solidification time (t1) = 10 min

Mass = density × volume

Mass = ρ × V

 Let the mass of the first cube is M1 and the mass of the second cube is M2.

∴ M2 = 27 × M1

ρ2 × V2 = 27 × ρ1 × V1

The same material means the density is constant.

∴ ρ1 = ρ2

∴ V2 = 27 × V1

\(a_2^3 = 27 × a_1^3\)

a2 = 3 a1

Now,

\(\frac{t_2}{\left ( a_2/6 \right )^2} = \frac{t_1}{\left ( a_1/6 \right )^2}\)

\(t_2 = t_1\left ( \frac{a_2}{a_1} \right )^2\)

\(t_2 = 10\left ( \frac{3 a_1}{a_1} \right )^2\)

t2 = 10 × 9

t2 = 90 min

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