question_answer

small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the surface density of charge of each small drop with of the big drop is:

A)  64 : 1                                    

B)  1 : 64

C)  1 : 4                                      

D)  4 : 1

Correct Answer: C

Solution :

\[{{V}_{1}}=12V\,{{V}_{1}}=12\,V\] \[{{q}_{1}}={{C}_{1}}{{V}_{1}}=36\,\mu C\] \[{{q}_{2}}={{C}_{2}}{{V}_{1}}=72\,\mu C\] when the two are connected in series (\[+\,\,\upsilon e\]of each to \[-\,ve\]of each). \[{{C}_{1}}=3\,\mu F\] \[\therefore \]  \[{{C}_{2}}=9\,\mu F\] both are charged by 12V battery \[{{C}_{2}}\] \[\therefore \]charge on \[{{C}_{1}}=12\times 3=36\,\mu C\] charge on\[{{C}_{2}}=12\times 6=72\,\mu C\] q on \[{{A}_{1}}=+\,36\,\mu C\]on \[{{A}_{2}}=-36\,\mu C\] q on \[{{B}_{1}}=-72\,\mu C\]on \[{{B}_{2}}=+72\,\mu C\] Total charge on \[{{A}_{1}}\,\And {{B}_{1}}\,=(36-72)\mu C\] \[{{Q}_{1}}=-36\,\mu C\] Total charge on \[{{A}_{2}}\,\And \,{{B}_{2}}=(72-36)\mu C\] \[{{Q}_{2}}=36\,\mu C\] \[\therefore \]\[{{A}_{1}}\] and\[{{B}_{1}}\]have\[-\upsilon e\]charge and \[{{A}_{2}}{{B}_{2}}\]have \[+\,\upsilon e.\] So combination is in parallel \[{{C}_{eq}}=9\mu F\] V across each \[V=\frac{{{Q}_{1}}\times {{Q}_{2}}}{C}=\frac{36\,\mu C}{9\mu F}=4\,\text{volts}\]