Answer

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**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,

\[\sigma \; = \;\dfrac{q}{A}{\text{ }}\]

Where,

\[\sigma \;\]is 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 $ = \dfrac{4}{3}\pi {R^3}$

Volume of a small drop $ = \dfrac{4}{3}\pi {r^3}$

Volume of $64$ small drops $ = 64 \times \dfrac{4}{3}\pi {r^3}$

Here the above equation is equal,

Volume of the big drop is equal to Volume of $64$ small drops

We get,

\[\dfrac{4}{3}\pi {r^3} = 64 \times \dfrac{4}{3}\pi {r^3}\]

Common factor is canceled,

Here,

${R^3} = 64{r^3}$

On simplifying,

$R = 4r$

Charge on a small drop $ = q$

Charge on a big drop$ = Q$

The value of $Q = 64q$

Small drop ${\sigma _1} = \dfrac{q}{{4\pi {r^2}}}$

Big drop \[{\sigma _2} = \dfrac{Q}{{4\pi {R^2}}}\]

Substituting the given value in above equation,

We get,

\[{\sigma _2} = \dfrac{{64q}}{{4\pi {{(4r)}^2}}}\]

On simplifying,

\[{\sigma _2} = \dfrac{{64q}}{{64\pi {r^2}}}\]

Common factor canceled,

\[{\sigma _2} = \dfrac{q}{{\pi {r^2}}}\]

According to the surface density,

Here,

\[\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{q}{{4\pi {r^2}}} \times \dfrac{q}{{\pi {r^2}}}\]

Again, the common factors are canceled,

We get,

\[\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{1}{4}\]

Hence,

\[{\sigma _1}:{\sigma _2} = 1:4\]

**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.

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