Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

Question

Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is (A) 8 times (B) 4 times (C) 2 times (D) 32 times

Tardigrade Admin · Accepted Answer

Volume of 8 small drops = Volume of big drop 8 × (4/3) π r3=(4/3) π R3 ⇒ R=2 r As capacity is proportional to r, hence capacity becomes 2 times.

Q. Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

8 times

4 times

2 times

32 times

Volume of 8 small drops $=$ Volume of big drop
$8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$\Rightarrow R = 2 r$
As capacity is proportional to $r$ , hence capacity becomes 2 times.

Q. Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

8 times

4 times

2 times

32 times

Volume of 8 small drops = Volume of big drop 8×34​πr3=34​πR3 ⇒R=2r As capacity is proportional to r, hence capacity becomes 2 times.

Volume of 8 small drops $=$ Volume of big drop
$8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$\Rightarrow R = 2 r$
As capacity is proportional to $r$ , hence capacity becomes 2 times.