Iron exhibits bcc structure at room temperature. Above $900^{o} C$ , it transforms to fcc structure. The ratio of density of iron at room temperature to that at $900^{o} C$ (assuming molar mass and atomic radii of iron remains constant with temperature) is:

\frac{\sqrt{3}}{\sqrt{2}}

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\frac{4 \sqrt{3}}{3 \sqrt{2}}

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\frac{3 \sqrt{3}}{4 \sqrt{2}}

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\frac{1}{2}

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Solution

The correct option is C $\frac{3 \sqrt{3}}{4 \sqrt{2}}$
For bcc lattice: $Z = 2, a = \frac{4 r}{\sqrt{3}}$
For FCC lattice, $Z = 4, a = 2 \sqrt{2} r$
$\frac{d_{R . T}}{d_{900^{o} C}}$ = $\frac{(\frac{Z M}{N_{A} a^{3}})_{b c c}}{(\frac{Z M}{N_{A} a^{3}})_{f c c}}$
Given, molar mass and atomic radii are constant.
$= \frac{2}{4} (\frac{2 \sqrt{2} r}{\frac{4 r}{\sqrt{3}}})^{3} = \frac{3 \sqrt{3}}{4 \sqrt{2}}$

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