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Iron exhibits `bcc` structure at roomj temperature. Above `9000^(@)C`, it transformers to `fcc` structure. The ratio of density of iron at room temperature to that at `900^(@)C` (assuming molar mass and atomic radius of iron remains constant with temperature) isA. `sqrt(3)/sqrt(2)`B. `(4sqrt(3))/(3sqrt(2))`C. `(3sqrt(3))/(4sqrt(2))`D. `(1)/(2)`

Answer» Correct Answer - C
BCC FCC
`4r=sqrt(3)a 4r=sqrt(2)a`
`a=(4r)/sqrt(3) a=(4r)/sqrt(2)`
`(d_(BC C))/(d_(FC C))=((Z_(BC C)xxM)/(N_(A)a^(3)))/((Z_(FC C)xxM)/(N_(A)a^(3)))=((2xxM)/(N_(A)((4r)/(sqrt(3)))^(3)))/((4xxM)/(N_(A)xx((4r)/(sqrt(2)))^(3)))=(3)/(4)sqrt((3)/(2))`


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