$F=B l a$

where, $I=$ induced current in loop.

Also, $I=\frac{E}{R}$

where, $E=$ induced emf.

$\Rightarrow \quad I=\frac{-B a v}{R}$

So, opposing force is

$F=\frac{-B^{2} a^{2}}{R} \cdot v \Rightarrow m A=\frac{-B^{2} a^{2}}{R} \cdot v$

where, $A=$ acceleration of loop and $m=$ mass of loop.

$m \frac{d v}{d t}=\frac{-B^{2} a^{2} v}{R}$

$m \cdot \frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{-B^{2} a^{2} v}{R}$

$d v=\frac{-B^{2} a^{2}}{R m} \cdot d x \quad\left[\therefore v=\frac{d x}{d t}\right]$

Integrating above equation, we have

$\Rightarrow \quad v=\frac{-B^{2} a^{2} x}{R m}+C$

where, $C=$ constant of integration.

$\therefore v$ versus $x$ is a straight line with

negative slope. Also, emf is induced only when the loop is going in or emerging out of the region of magnetic field. So, correct graph is $(b)$.