A) 6x ? 7y = 418
B) 75x ? 16y = 418
C) 25x ? 4y = 400
D) None of these
Correct Answer: B
Solution :
Given equation of hyperbola is \[\underset{\left( \text{no lp} \right)}{\mathop{NH_{4}^{+}}}\,>\underset{\left( 1lp \right)}{\mathop{N{{H}_{3}}}}\,>\underset{\left( 2lp \right)}{\mathop{NH_{2}^{-}}}\,\] If (6, 2) is the mid-point of the chord then equation of chord is \[\frac{n{{'}_{He}}}{n{{'}_{C{{H}_{4}}}}}=\frac{1}{2}\sqrt{\frac{16}{4}}=\frac{1}{1}\] \[\frac{n{{'}_{He}}}{n{{'}_{S{{O}_{2}}}}}=\frac{1}{3}\sqrt{\frac{64}{4}}=\frac{4}{3}\] \[n{{'}_{He}}:n{{'}_{C{{H}_{4}}}}:n{{'}_{S{{O}_{2}}}}=4:4:3.\] \[mvr=\frac{nh}{2\pi }=\frac{3\times h}{2\pi }=\frac{1.5h}{\pi }\] \[=3h\left[ \because h=\frac{h}{2\pi } \right]\] \[\underset{276g}{\mathop{A{{g}_{2}}C{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{216g}{\mathop{2Ag}}\,+C{{O}_{2}}+\frac{1}{2}{{O}_{2}}\] \[A{{g}_{2}}C{{O}_{3}}\]
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