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 (t_{1}) = 10 min

Mass = density × volume

Mass = ρ × V

Let the mass of the first cube is M_{1} and the mass of the second cube is M_{2}.

∴ M_{2} = 27 × M_{1}

ρ_{2} × V_{2} = 27 × ρ_{1 }× V_{1}

The same material means the density is constant.

∴ ρ_{1} = ρ_{2}

∴ V_{2} = 27 × V_{1}

\(a_2^3 = 27 × a_1^3\)

a_{2} = 3 a_{1}

Now,

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

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

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

t_{2} = 10 × 9

**t**_{2} = 90 min