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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 rr and change qq coalesce to form a big drop. The ratio of the surface charge density of each smell drop with that of big drop is
(A) 4:14:1
(B) 1:41:4
(C) 1:641:64
(D) 64:164:1

Explanation

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  =  qA \sigma \; = \;\dfrac{q}{A}{\text{ }}
Where,
σ  \sigma \;is surface charge density
qq is charge
AA is surface area

Complete step by step answer
Given by,
Let the radius of big drop =R = R
Volume of the big drop =43πR3 = \dfrac{4}{3}\pi {R^3}
Volume of a small drop =43πr3 = \dfrac{4}{3}\pi {r^3}
Volume of 6464 small drops =64×43πr3 = 64 \times \dfrac{4}{3}\pi {r^3}
Here the above equation is equal,
Volume of the big drop is equal to Volume of 6464 small drops
We get,
43πr3=64×43πr3\dfrac{4}{3}\pi {r^3} = 64 \times \dfrac{4}{3}\pi {r^3}
Common factor is canceled,
Here,
R3=64r3{R^3} = 64{r^3}
On simplifying,
R=4rR = 4r
Charge on a small drop =q = q
Charge on a big drop=Q = Q
The value of Q=64qQ = 64q
Small drop σ1=q4πr2{\sigma _1} = \dfrac{q}{{4\pi {r^2}}}
Big drop σ2=Q4πR2{\sigma _2} = \dfrac{Q}{{4\pi {R^2}}}
Substituting the given value in above equation,
We get,
σ2=64q4π(4r)2{\sigma _2} = \dfrac{{64q}}{{4\pi {{(4r)}^2}}}
On simplifying,
σ2=64q64πr2{\sigma _2} = \dfrac{{64q}}{{64\pi {r^2}}}
Common factor canceled,
σ2=qπr2{\sigma _2} = \dfrac{q}{{\pi {r^2}}}
According to the surface density,
Here,
σ1σ2=q4πr2×qπr2\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{q}{{4\pi {r^2}}} \times \dfrac{q}{{\pi {r^2}}}
Again, the common factors are canceled,
We get,
σ1σ2=14\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{1}{4}
Hence,
σ1:σ2=1:4{\sigma _1}:{\sigma _2} = 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.