Question
Question: Small drops of mercury each of radius \(r\) and change \(q\) coalesce to form a big drop. The ratio ...
Small drops of mercury each of radius r and change q coalesce to form a big drop. The ratio of the surface charge density of each smell drop with that of big drop is
(A) 4:1
(B) 1:4
(C) 1:64
(D) 64:1
Solution
Hint Here we know the radius value of mercury and change so we find the large drop and small drop by using the formula ratio of surface density change according to some change in the formula for calculating the ratio than the volume of the small and large drops in the problem.
Useful formula:
The surface charge density formula is given by,
σ=Aq
Where,
σis surface charge density
q is charge
A is surface area
Complete step by step answer
Given by,
Let the radius of big drop =R
Volume of the big drop =34πR3
Volume of a small drop =34πr3
Volume of 64 small drops =64×34πr3
Here the above equation is equal,
Volume of the big drop is equal to Volume of 64 small drops
We get,
34πr3=64×34πr3
Common factor is canceled,
Here,
R3=64r3
On simplifying,
R=4r
Charge on a small drop =q
Charge on a big drop=Q
The value of Q=64q
Small drop σ1=4πr2q
Big drop σ2=4πR2Q
Substituting the given value in above equation,
We get,
σ2=4π(4r)264q
On simplifying,
σ2=64πr264q
Common factor canceled,
σ2=πr2q
According to the surface density,
Here,
σ2σ1=4πr2q×πr2q
Again, the common factors are canceled,
We get,
σ2σ1=41
Hence,
σ1:σ2=1:4
Thus, The option B is correct answer
Note Charging density is known, according to electromagnetism, as a measure of the electrical charge per unit volume of space in one, two or three dimensions. The linear surface or volume charge density is the quantity of electric charge per surface area or volume, to be specific.