Q127 / 180 · KCET 2012 · Physics
In the uniform electric field of $\mathrm{E}=1 \times 10^{4} \mathrm{NC}^{-1}$,
an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $2 \times 10^{-2} \mathrm{~m}$ is nearly
$$
\left(\frac{\mathrm{e}}{\mathrm{m}} \text { of electron } \approx 1.8 \times 10^{-11} \mathrm{C} \mathrm{} \mathrm{kg}^{-1}\right)
$$
- Correct answerA$8.5 \times 10^{6} \mathrm{~ms}^{-1}$
- B$1.6 \times 10^{6} \mathrm{~ms}^{-1}$
- C$0.85 \times 10^{6} \mathrm{~ms}^{-1}$
- D$0.425 \times 10^{6} \mathrm{~ms}^{-1}$
Solution
By work-energy theorem,
$\begin{aligned} \text { Work done by electric field }=& \begin{array}{l}\text { Change in KE of } \\ \text { electron }\end{array} \end{aligned}$
or $\mathrm{qE} x=\frac{1}{2} \mathrm{mv}^{2}$
or $\quad v=\sqrt{\frac{2 \mathrm{q} E x}{m}}$
$$
\begin{aligned}
&=\sqrt{2 \times 1.8 \times 10^{11} \times 1 \times 10^{4} \times 2 \times 10^{-2}} \\
&=8.5 \times 10^{6} \mathrm{~ms}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.