Q63 / 90 · KVPY 2011 (SB/SX) · Physics
The surface of a planet is found to be uniformly charged. When a particle of mass $\mathrm{m}$ and no charge is thrown at an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontal range $\mathrm{L}$. A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$, with the same initial conditions has a range $\mathrm{L} / 2$. The range of particle of mass $\mathrm{m}$ and charge $2 \mathrm{q}$ with the same initial conditions is-
- A$\mathrm{L}$
- B$L / 2$
- Correct answerC$\mathrm{L} / 3$
- DL/4
Solution
For uncharged particle
$\mathrm{L}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$ ...(i)
Range for particle of mass $\mathrm{m}$ and charge $\mathrm{q}$.
$\frac{L}{2}=\frac{u^{2} \sin 2 \theta}{g+\frac{q E}{m}}$ ...(ii)
From (i) and (ii)
$\begin{array}{l}
\frac{u^{2} \sin 2 \theta}{2 g}=\frac{u^{2} \sin 2 \theta}{g+\frac{q E}{m}} \\
\Rightarrow m g=q E
\end{array}$
Range of particle of mass $m \&$ charge $2 \mathrm{q}$.
$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}+\frac{2 \mathrm{q} \mathrm{E}}{\mathrm{m}}}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}\left(1+\frac{2 \mathrm{qE}}{\mathrm{mg}}\right)}=\frac{\mathrm{L}}{3}$
Hence, the correct answer is option: 'C'.