There are 27 drops of a conducting fluid- each drop has radius r and each of them is charged to same potential V-1 they are then combined to form a bigger drop- the potential of the bigger drop is V-2 the ratio V-2 -V-1 ignore the change in density of fluid on combining the drops-

Question

There are 27 drops of a conducting fluid- each drop has radius r and each of them is charged to same potential  V-1 they are then combined to form a bigger drop- the potential of the bigger drop is  V-2  the ratio  V-2 -V-1  ignore the change in density of fluid on combining the drops-

Toppr Admin · Accepted Answer

Let q be the charge on each drop - - V-1 - -frac -kq-r- -160-Let R be radius of single drop- then- -frac -4-3- -pi R-3 - 27 -frac -4-3- -pi r-3 -Rightarrow R - 3r - V-2 - -frac - k-27q- -R- - -frac - k27q-3r- - 9V-1 -Rightarrow -frac -V-2-V-1- - 9 -

let q be the charge on each drop : $$ V_1 = \frac {kq}{r} $$ Let R be radius of single drop. then$$ \frac {4}{3} \pi R^3 = 27 \frac {4}{3} \pi r^3 \Rightarrow R = 3r $$$$ V_2 = \frac { k(27q) }{R} = \frac { k27q}{3r} = 9V_1 \Rightarrow \frac {V_2}{V_1} = 9 $$

let q be the charge on each drop : $$ V_1 = \frac {kq}{r} $$
Let R be radius of single drop. then
$$ \frac {4}{3} \pi R^3 = 27 \frac {4}{3} \pi r^3 \Rightarrow R = 3r $$
$$ V_2 = \frac { k(27q) }{R} = \frac { k27q}{3r} = 9V_1 \Rightarrow \frac {V_2}{V_1} = 9 $$