A particle of $\frac{\mathrm{e}}{\mathrm{m}}$ ratio equal to $4\times {10}^{5}coul/kg$ is accelerated from rest through a potential difference of 20 volts. The speed of the particle is:

Given,

$\mathrm{e}/\mathrm{m} =4 \mathrm{x} 10 5 \mathrm{coul}/\mathrm{kg}$

$\mathrm{Potential} \mathrm{difference},\mathrm{V}= 20 \mathrm{volt}$
The speed of particle is found as,

$KE\mathit{ }= qV\mathit{ }= \frac{1}{2}m{V}^2$

$\frac{q}{m}=\frac{1}{2}\frac{{\mathrm{V}}^2}{\mathrm{V}}$

$V^2=2\times \frac{q}{m}\times V=2\times 4\times {10}^5\times 20$

$= 40 x {410}^1=16 x {10}^6$

$V\mathit{ }= 4\times {10}^3\mathrm{m}/s$
Hence, option D is correct.