
qE = mg ….(i)
q = ne
V = Ed ⇒ E = \(\frac{V}{d}\)
from equation (i)
ne (v/d) = mg
n = mgd/eV
= \(900 \times \frac{4\pi}{3} \times \frac{8 \times 8 \times 8 \times 10^{-21} \times 10 \times 0.01}{1.6 \times 10^{-19} \times 200}\)
n = 6 (approx)
$XY$ is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration $\mathrm{c}_{1}$ and $\mathrm{c}_{2}\left(\mathrm{c}_{1}>\mathrm{c}_{2}\right) \mathrm{mol} \mathrm{L}^{-1}$. For the reverse osmosis to take place identify the correct condition} (Here $\mathrm{p}_{1}$ and $\mathrm{p}_{2}$ are pressures applied on chamber 1 and 2 ) 
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.