A) \[{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\]
B) \[-{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\]
C) \[{{Q}^{2}}/(2\pi {{\varepsilon }_{0}}{{a}^{2}})\]
D) \[2\mu F\]
Correct Answer: B
Solution :
Key Idea: Volume of drops after coalescing, remains same. Let radius of one drop of mercury be r since drop is spherical is shape, its volume is \[3q/{{\varepsilon }_{0}}\]Total volume of two drops \[2q/{{\varepsilon }_{0}}\] Radius of large drop formed be R. Volume of 2 drops = Volume of large drop \[q/{{\varepsilon }_{0}}\] \[\beta \] \[x\] The surface area of the two drops is \[y\] Surface area of resultant; drop is \[x\]. Surface energy of two drops before coalescing is \[{{i}_{1}},\] and surface energy after coalescing is \[{{i}_{2}},\] \[{{h}^{1/2}}\] \[h\]
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