A) \[\sqrt{\frac{3}{2}}R\]
B) \[\frac{RT}{2V-b}-\frac{9}{4{{b}^{2}}},\]
C) \[{{M}_{1}}\]
D) \[{{L}_{1}}\]
Correct Answer: B
Solution :
The charge flowing through \[{{C}_{4}}=4C\] is \[{{C}_{2}}\] The series combination of \[{{C}_{4}}\] and \[\frac{22}{3}\] gives \[\frac{3}{22}\] \[\frac{7}{4}\] \[\frac{4}{7}\] \[\overset{0}{\mathop{A}}\,\] Now, \[2\mathbf{\hat{i}}+3\mathbf{\hat{j}}+8\mathbf{\hat{k}}\] and \[4\mathbf{\hat{j}}-4\mathbf{\hat{i}}+\alpha \mathbf{\hat{k}}\] form parallel combination giving \[-1\] \[\frac{1}{2}\] Net charge \[-\frac{1}{2}\] \[\frac{2}{5}\] Total charge flowing through \[\frac{3}{5}\] will be \[\frac{3}{7}\] \[\frac{3}{4}\] Since, \[\frac{R}{5}\] and \[\frac{4}{5}\] are in series combination hence, charge flowing through these will be same. Hence, \[\frac{5}{6}\] Thus, \[\frac{6}{7}\]
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