A) \[\frac{6P}{7}\]
B) \[\frac{9P}{7}\]
C) \[\frac{36P}{7}\]
D) \[6P\]
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} \,\,\,a \\ a-x \end{smallmatrix}}{\mathop{X}}\,\underset{\begin{smallmatrix} a \\ x \end{smallmatrix}}{\mathop{Y}}\,+\underset{\begin{smallmatrix} a \\ x \end{smallmatrix}}{\mathop{Z}}\,\] \[\therefore \] \[P_{x}^{}=P\frac{(a-x)}{(a+x)}=\frac{P}{7}\] \[\therefore \] \[7a-7x=a+x\] \[6a=8x\] \[x=\frac{6a}{8}=\frac{3a}{4}\] \[{{K}_{p}}=\frac{{{x}^{2}}}{(a-x)}.\frac{[P]}{[a+x]}\] \[=\frac{{{\left( \frac{3a}{4} \right)}^{4}}[P]}{\left( a-\frac{3a}{4} \right)\left( a+\frac{3a}{4} \right)}\] \[=\frac{9{{a}^{2}}\times 4\times 4\times P}{16\times a\times 7a}\] \[{{K}_{p}}=\frac{9P}{7}\]
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