2. Solved Papers
  3. NEET

NEET NEET SOLVED PAPER 2018

  • question_answer
    Iron exhibits bcc structure at room temperature. Above\[\text{900 }\!\!{}^\circ\!\!\text{ C}\], it transforms to fcc structure. The ratio of density of iron at room temperature to that at \[\text{900 }\!\!{}^\circ\!\!\text{ C}\](assuming molar mass and atomic radii of iron remains constant with temperature) is [NEET - 2018]

    A)  \[\frac{3\sqrt{3}}{4\sqrt{2}}\]                     

    B)  \[\frac{4\sqrt{3}}{3\sqrt{2}}\]

    C)  \[\frac{\sqrt{3}}{\sqrt{2}}\]                         

    D)  \[\frac{1}{2}\]

    Correct Answer: A

    Solution :

    For BCC lattice: \[Z=2,a=\frac{4r}{\sqrt{3}}\] For FCC lattice: \[\text{Z=4, a=2}\sqrt{\text{2}}\text{r}\] \[\therefore \frac{{{d}_{25{}^\circ C}}}{{{d}_{900{}^\circ C}}}=\frac{{{\left( \frac{ZM}{{{N}_{A}}{{a}^{3}}} \right)}_{BCC}}}{{{\left( \frac{ZM}{{{N}_{A}}{{a}^{3}}} \right)}_{FCC}}}\]                 \[=\frac{2}{4}{{\left( \frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}} \right)}^{3}}\]                 \[=\left( \frac{3\sqrt{3}}{4\sqrt{2}} \right)\]


You need to login to perform this action.
You will be redirected in 3 sec spinner