Question

Two long, fixed, parallel, thick rails are lying in a region having uniform magnetic field, perpendicular to the plane of the rails. Two identical sliders of mass m and resistance R each (initially at rest) can slide over the rails. If one of the slider is given velocity $V_o$, the velocity of the other slider after considerably long time will be [Neglect friction] we have given $v_A=\frac{5v_o}{8}$

• A. $v_0$
• B. zero
• C. $2v_0$
• D. $\frac{v_0}{4}$

Answer 1

The correct option is A $v_0$

$\begin{array}{c}P=-\left(\frac{p_o}{v_o}\right)V+p_o\\ Now\\ w={\int }_{v_A}^{v}pdv\Delta u=\frac{3}{2}R\left(T-T_A\right)\\ =-\frac{p_o}{v_o}\frac{\left(v^2-v_A^2\right)}{2}+p_{{}_o}\left(v-v_A\right)+\frac{3}{2}R\left(T-T_A\right)\\ w+\Delta u=0\\ \Rightarrow -\frac{p_o}{v_o}\frac{\left(v^2-v_A^2\right)}{2}+p_{{}_o}\left(v-v_A\right)+\frac{3}{2}R\left(\frac{pv}{R}-\frac{P_A{v}_A}{R}\right)=0\\ \Rightarrow p_o\left(v-v_A\right)+\frac{3}{2}\left[\left[-\frac{p_o}{v_o}\left(v\right)+p_o\right]v-\left(-\frac{p_o}{v_o}{v}_A+p_o\right)\right]=\frac{p_o}{v_o}\left(\frac{v^2-v_1^2}{2}\right)\\ \Rightarrow p_o\left(v_1-v_A\right)+\frac{3}{2}\left[-\frac{p_o}{v_o}{v}^2+p_{o}v+\frac{p_o}{v_o}{v}_A^2-p_o{v}_A\right]=\frac{p_o}{v_o}\left(\frac{v^2-v_A^2}{2}\right)\\ \Rightarrow v-v_A\times \frac{3v^2}{2.v_o}+\frac{3v}{2}+\frac{3v_A^2}{2v_o}-\frac{3v_A}{2}=\frac{1}{v_o}\left(\frac{v^2-v_A^2}{2}\right)\\ \Rightarrow \frac{5v}{2}-\frac{5v}{2}a-\frac{3}{2v_o}\left(v^2-v_A^2\right)=\frac{1}{2v_o}\left(v^2-v_A^2\right)\\ \Rightarrow \frac{v}{2}-\frac{5v}{2}A=\frac{2\left(v^2-v_A^2\right)}{v_o}\\ \Rightarrow \frac{5}{2}\left(v-v_A\right)=\frac{2v+v_1\left(v-v_A\right)}{v_o}\\ \Rightarrow \frac{5}{4}{v}_o=v+v_A\\ take{v}_A=\frac{5v_o}{8}\\ \therefore v=\frac{5v_o}{8}\end{array}$

Hence, the option $A$ is the correct answer.