27 small drops each having charge q and radius r coalesce to form big drop. How many times charge and capacitance will become ?

Question

27 small drops each having charge q and radius r coalesce to form big drop. How many times charge and capacitance will become ? (A)  3, 27  (B)  27, 3  (C)  27, 27  (D)  3, 3

Tardigrade Admin · Accepted Answer

In coalescing into a single drop charge remains conserved. Also volume before and after coalescing remains same.
Let

R

and

r

be the radii of bigger and each smaller drop. Charge remains conserved.
Hence, charge on bigger drop

= 27 \times

charge on smaller drop
i.e.,

q^{'} = 27 q

Now, before and after coalescing, volume remains same. That is,

\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}

∴ R = 3 r

Hence,capacitance of bigger drop

C^{'} = 4 π ε_{0} R

= 4 π ε_{0} (3 r)

= 3 (4 π ε_{0} r)

= 3 C

Q. $27$ small drops each having charge $q$ and radius $r$ coalesce to form big drop. How many times charge and capacitance will become ?

$3, 27$

$27, 3$

$27, 27$

$3, 3$

Q. 27 small drops each having charge q and radius r coalesce to form big drop. How many times charge and capacitance will become ?

3,27

27,3

27,27

3,3

Q. $27$ small drops each having charge $q$ and radius $r$ coalesce to form big drop. How many times charge and capacitance will become ?

$3, 27$

$27, 3$

$27, 27$

$3, 3$