A particle starts from the origin at $t=0$ and moves in the $x-y$ plane with constant acceleration $b$ in the $x-$direction. The equation of motion is $x=2c{y}^2$. The $y-$component of its velocity $(\text{at}t=0)$ is

The correct option is A $\sqrt{\frac{b}{4c}}$
Given,
$a_x=b=\frac{d{v}_x}{dt}$

From equation of motion
$v_x=u_x+a_{x}t$
$v_x=u_x+bt$

Also given,
$a_y=0=\frac{d{v}_y}{dt}$

$v_y=u_y+a_{y}t$
$v_y=u_y$

Given equation,
$x=2c{y}^2$
differentiating w.r.t. time, we get

$\frac{dx}{dt}=2c\left(2y\right)(\frac{dy}{dt})=4cy{v}_y$

Again differentiating w.r.t. time, we get
$\frac{d{v}_x}{dt}=4c(\frac{dy}{dt}{v}_y+y\frac{d{v}_y}{dt})$

$a_x=4c(v_y^2+y{a}_y)$

Since, it is given at $t=0,x=0,y=0$
Therefore,

$a_x=4c{v}_y^2=b$

$\Rightarrow v_y=\sqrt{\frac{b}{4c}}$

Hence, option $\left(\text{B}\right)$ is the right choice.