UPSEE Solved Model Paper 7

According to the question $\underset{\text{(bcc)}}{\mathrm{Fe}}\xrightarrow{> 916^{\circ}C}\underset{\text{(ccp)}}{\mathrm{Fe}}$ where, bcc = Body centred cube ccp = Cubic dose packed structure The density of unit cell is given as ρ = $\frac{Z\times M}{a^{3}\times N_{0}}$ ... (i) where, 2 = Number of atoms per unit cell M = Atomic mass a = Edge length No = Avogadro's constant = 6 023 × $10^{23}\mathrm{mol}^{-1}$ The density of iron (bcc structure) before heating For a bcc structure, Z = 2 Also, a = $\frac{4r}{\sqrt{3}}$ where, r = Radius of the sphere Substituting the values of Z and a in equation (i). we get ρ = $\frac{2M}{\left(\frac{4r}{\sqrt{3}}\right)^{3}\times N_{0}}$ ... (ii) The density of iron (ccp structure) after heating is For a ccp structure. Z = 4 Also, a = $2\sqrt{2}$ r (Since radius of atom remains the same) Substituting the values of T and a' in equation (i), we get ρ' = $\frac{4M}{(2\sqrt{2}r)^{3}\times N_{0}}$ ... (iii) Dividing equation (I) by equation (ii), we get The ratio of density $\frac{\rho}{\rho'}$ = 0.918 Therefore, the ratio of density of the crystal before heating and after heating is 0.918 Hence, option 'B' is correct