Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = 
where
R = resistance

f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:

where
x = separation between the parallel plates
Thus
C ∝ 
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,

Also,
Z ∝ I
Therefore,




Solving the above eqn:

When in the melting process particles start to move more freely when in the freezing process particles tend to slow and vibrate in place
The current is defined as the amount of charge Q that passes through a given point of a wire in a time

:

Since I=500 A and the time interval is

the charge is

One electron has a charge of

, therefore the number of electrons that pass a point in the wire during 4 minutes is

electrons