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