A) 2
B) 3
C) 4
D) 5
Correct Answer: A
Solution :
Given equation of the given line is \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OH\]and\[C{{H}_{3}}-C{{H}_{2}}-O-C{{H}_{3}}\]. The equation of the plane passing through the given line is \[C{{H}_{3}}-C{{H}_{2}}-O-C{{H}_{2}}-C{{H}_{3}}\] ?(i) \[\underset{n=5/3}{\mathop{3{{I}_{2}}}}\,\xrightarrow{\,}\underset{n=5}{\mathop{IO_{3}^{-}}}\,+\underset{n=1}{\mathop{5{{I}^{-}}}}\,\]\[\therefore \] \[Eq.\,\,wt\,\,of\,{{I}_{2}}=\frac{254}{5/3}=152.4\] If this plane is parallel to Z-axis whose DC's are (0, 0, 1), then normal to the plane will be perpendicular to Z-axis. \[-I\] \[-I\] \[-N{{O}_{2}}>-Cl>-OC{{H}_{3}}\] \[{{r}_{n}}={{r}_{0}}\times \frac{{{n}^{2}}}{z}\] From Eq. (i) \[z=1\] \[{{r}_{1}}={{r}_{0}}\] \[\therefore \] ?(ii) Now, the shortest distance is the distance of any point on the Z-axis for the plane (ii). Let us take (0, 0, 0) as a point on Z-axis. \[{{r}_{2}}=4{{r}_{0}}\] Shortest distance = Length of perpendicular from (0, 0, 0) on plane (ii) \[{{r}_{3}}=9{{r}_{0}}\]
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