A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of 25.5 kVm$$^{-1}$$. The density of liquid is $$1.26 \times 10^3$$ kg m$$^{-3}$$. The radius of the drop is (neglect buoyancy).

A liquid drop with 6 excess electrons is held stationary under a uniform electric field of 25.5 kV/m. The density of the liquid is given as $$1.26 \times 10^3$$ kg/m³, and we neglect buoyancy. We need to find the radius of the drop.

Since the drop is stationary, the forces acting on it must balance. The gravitational force (weight) acts downward, and the electric force acts upward because the drop has excess electrons (negative charge) and the electric field exerts an upward force on negative charges.

The charge on one electron is $$-1.6 \times 10^{-19}$$ C. With 6 excess electrons, the total charge $$q$$ is:

$$ q = 6 \times (-1.6 \times 10^{-19}) = -9.6 \times 10^{-19} \text{ C} $$

The magnitude of the charge is $$|q| = 9.6 \times 10^{-19}$$ C. The electric field $$E$$ is given as 25.5 kV/m. Converting to V/m:

$$ E = 25.5 \times 1000 = 25500 \text{ V/m} = 2.55 \times 10^4 \text{ V/m} $$

The magnitude of the electric force $$F_e$$ acting upward is:

$$ F_e = |q| E = (9.6 \times 10^{-19}) \times (2.55 \times 10^4) $$

Calculating:

$$ 9.6 \times 2.55 = 24.48 $$

$$ 10^{-19} \times 10^4 = 10^{-15} $$

So,

$$ F_e = 24.48 \times 10^{-15} = 2.448 \times 10^{-14} \text{ N} $$

The gravitational force (weight) $$F_g$$ acting downward is $$F_g = m g$$, where $$m$$ is the mass of the drop and $$g$$ is the acceleration due to gravity (approximately 9.8 m/s²). The mass $$m$$ can be expressed in terms of density $$\rho$$ and volume $$V$$. Assuming the drop is spherical, its volume is:

$$ V = \frac{4}{3} \pi r^3 $$

where $$r$$ is the radius. Thus,

$$ m = \rho V = \rho \times \frac{4}{3} \pi r^3 $$

So,

$$ F_g = \rho \times \frac{4}{3} \pi r^3 \times g $$

For the drop to be stationary, the electric force upward must balance the gravitational force downward:

$$ F_e = F_g $$

Substituting the expressions:

$$ |q| E = \rho \frac{4}{3} \pi r^3 g $$

Solving for $$r^3$$:

$$ r^3 = \frac{|q| E \times 3}{4 \pi \rho g} $$

Plugging in the known values:

$$ |q| = 9.6 \times 10^{-19} \text{ C}, \quad E = 2.55 \times 10^4 \text{ V/m}, \quad \rho = 1.26 \times 10^3 \text{ kg/m}^3, \quad g = 9.8 \text{ m/s}^2, \quad \pi \approx 3.1416 $$

First, compute the numerator $$3 \times |q| \times E$$:

$$ 3 \times (9.6 \times 10^{-19}) \times (2.55 \times 10^4) = 3 \times 9.6 \times 2.55 \times 10^{-19+4} = 3 \times 9.6 \times 2.55 \times 10^{-15} $$

$$ 3 \times 9.6 = 28.8 $$

$$ 28.8 \times 2.55 = 73.44 $$

So, numerator $$= 73.44 \times 10^{-15} = 7.344 \times 10^{-14}$$

Now, compute the denominator $$4 \pi \rho g$$:

$$ \rho g = (1.26 \times 10^3) \times 9.8 = 1.26 \times 9.8 \times 10^3 = 12.348 \times 10^3 = 1.2348 \times 10^4 $$

$$ 4 \pi = 4 \times 3.1416 = 12.5664 $$

$$ 4 \pi \rho g = 12.5664 \times (1.2348 \times 10^4) = 12.5664 \times 1.2348 \times 10^4 $$

$$ 12.5664 \times 1.2348 = 15.51695472 \quad \text{(approximately)} $$

So, denominator $$= 15.51695472 \times 10^4 = 1.551695472 \times 10^5$$

Now,

$$ r^3 = \frac{7.344 \times 10^{-14}}{1.551695472 \times 10^5} = \frac{7.344}{1.551695472} \times 10^{-14-5} = 4.732 \times 10^{-19} $$

Taking the cube root to find $$r$$:

$$ r = \sqrt[3]{4.732 \times 10^{-19}} = \sqrt[3]{4.732} \times 10^{-19/3} = \sqrt[3]{4.732} \times 10^{-6.3333} $$

$$ 10^{-6.3333} = 10^{-6} \times 10^{-1/3} \approx 10^{-6} \times 0.4642 $$

$$ \sqrt[3]{4.732} \approx 1.679 \quad \text{(since } 1.679^3 \approx 4.732) $$

So,

$$ r \approx 1.679 \times 0.4642 \times 10^{-6} = 0.7793 \times 10^{-6} = 7.793 \times 10^{-7} \text{ m} $$

Approximating to two significant figures, $$r \approx 7.8 \times 10^{-7}$$ m.

Comparing with the options:

A. $$4.3 \times 10^{-7}$$ m

B. $$7.8 \times 10^{-7}$$ m

C. $$0.078 \times 10^{-7}$$ m (which is $$7.8 \times 10^{-9}$$ m)

D. $$3.4 \times 10^{-7}$$ m

The value $$7.8 \times 10^{-7}$$ m matches option B.

Hence, the correct answer is Option B.