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A spherical drop of molten iron of radius 2 mm solidifies in 10 s. A similar drop of radius 4 mm will solidify in:


1. 10 s
2. 30 s
3. 20 s
4. 40 s

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Correct Answer - Option 4 : 40 s

Concept:

i) Chvorinov’s Principle

\({t_s} \propto \frac{V}{A} \Rightarrow {t_s} = K{\left( {\frac{V}{A}} \right)^2}\)

ts = Solidification time, V = Volume, A = surface area, K = solidification factor

ii) For sphere, \(\frac{V}{A} = \frac{r}{3} = \frac{d}{6}\)

Calculation:

Given:

Chvorinov’s Principle:

\({t_s} \propto {\left( {\frac{V}{A}} \right)^2}\)

For sphere:

\(\frac{V}{A} = \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = \frac{r}{3}\)

\(\therefore \frac{{{t_{s1}}}}{{{t_{s2}}}} = \frac{{{{\left( {\frac{{{V_1}}}{{{A_1}}}} \right)}^2}}}{{{{\left( {\frac{{{V_2}}}{{{A_2}}}} \right)}^2}}} = \frac{{{{\left( {\frac{{{r_1}}}{3}} \right)}^2}}}{{{{\left( {\frac{{{r_2}}}{3}} \right)}^2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\)

r1 = 2 mm, r2 = 2r1 = 4 mm, ts1 = 10 sec

\(\Rightarrow \frac{{10}}{{{t_{s2}}}} = {\left( {\frac{2}{4}} \right)^2}\)

ts2 = 40 sec

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