A) 1.4\[=1.013\times {{10}^{5}}N/{{m}^{2}}\]
B) 2.5\[\overset{0}{\mathop{A}}\,\]
C) 3.5\[\overset{0}{\mathop{A}}\,\]
D) 1.2\[\overset{0}{\mathop{A}}\,\]
Correct Answer: A
Solution :
\[{{v}_{N}}>,{{v}_{o}}\,=\,{{v}_{H}}\] The capacitors \[=1.013\times {{10}^{5}}N/{{m}^{2}}\] and \[\overset{0}{\mathop{A}}\,\] are in parallel, their effective capacitance is \[\overset{0}{\mathop{A}}\,\] Now capacitors C and C are in series. So effective capacitance is \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\]
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