Question

27 small drops of mercury having the same radius collage to form one big drop .Find the ratio of the capacitance of big drop to small drop.​

Answer 1

Answer:

Explanation:

Direct formula

Answer 2

The ratio of the capacitance of big drop to small drop is 9 : 1

Explanation:

Number of small drops = 27

Find the ratio of the capacitance of big drop to small drop

Here 27 small drops combined to form a big drop. So, both of their volume will be equal.

Volume of bigger drop = Volume of smaller drop

$\text{Volume of the sphere} = \frac{4}{3} \times \pi r^3$

Let us consider 'R' be the radius of bigger drop and 'r' be the radius of the smaller drop.

$\text{Volume of the bigger drop} = \frac{4}{3} \times \pi R^3$

$\text{Volume of the smaller drop} = 27 \times \frac{4}{3} \times \pi r^3$

$\text{Potential energy of bigger drop V'} = \frac{\text{Q}}{4\pi\varepsilon_0 \text{R}}$

Substitute Q = 27q and R = 3r and 'q' is the charge

$\text{Potential energy of bigger drop V'} = \frac{27\text{q}}{4\pi\varepsilon_0 (3\text{r})}$

$\text{Potential energy of bigger drop V'} = \frac{9\text{q}}{4\pi\varepsilon_0 \text{r}} ------>(1)$

$\text{Potential energy of smaller drop V} = \frac{\text{q}}{4\pi\varepsilon_0 \text{r}} ------>(2)$

Equate (1) and (2) ,

     V' = V

$\frac{9\text{q}}{4\pi\varepsilon_0 \text{r}} = \frac{\text{q}}{4\pi\varepsilon_0 \text{r}}$

$9[ \frac{\text{q}}{4\pi\varepsilon_0 \text{r}}] = \frac{\text{q}}{4\pi\varepsilon_0 \text{r}}$

From the above relation, V' = 9 V

Therefore the ratio of the capacitance of big drop to small drop is 9 : 1, when 27 drops of mercury with same radius collage to form one big drop.

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