A) 400 V
B) 500 V
C) 1000 V
D) 1300 V
Correct Answer: B
Solution :
Equivalent capacitance between B and C is \[\text{5}.\text{54}\times \text{1}{{0}^{-\text{7}}}\text{rad}\] Now equivalent capacitance between A and C is \[\text{2}.\text{54}\times \text{1}{{0}^{-\text{4}}}\text{ rad}\] \[\text{6}.\text{54}\times \text{1}{{0}^{-\text{7}}}\text{rad}\] Charge on condenser of capacity \[\alpha =\text{4},\beta =\text{6}~~~~\] is \[\alpha =\text{5},\beta =\text{5}\] \[\alpha =\text{6},\beta =\text{4}\] (since, potential at the point C will be zero) Now potential difference across condenser of \[\alpha =\text{6},\beta =\text{6}\] is \[[M{{L}^{2}}{{T}^{-2}}{{\theta }^{-1}}]\] \[[{{M}^{2}}L{{T}^{-2}}\theta ]\] \[[M{{L}^{3}}{{T}^{-1}}{{\theta }^{-1}}]\] As \[\frac{6}{\sqrt{2}}\] Hence, \[6\sqrt{2}\] Charge on nucleus \[\text{2}.\text{34 se}{{\text{c}}^{\text{-1}}}\] \[\text{S}{{\text{O}}_{\text{2}}}\] Potential at the surface of the nucleus \[\text{N}{{\text{H}}_{3}}\] \[{{\text{H}}_{\text{2}}}\] \[\text{C}{{\text{O}}_{\text{2}}}\]
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