A conducting wire of parabolic shape, initially y = x ^ 2 , is...

For calculating the motional emf across the length of the wire, let us project wire such that $\overset{\to}{B}, \overset{\to}{v} - , \overset{\to}{i}$ becomes mutually orthogonal. Thus, small emf 'dc' induced in elemental length $‘dy’$ of projected wire, will be -
$⇒ dε = B{v}_{0}dy = {B}_{0}$\left[1 + {\left(\dfrac{y}{L}\right)}^{\beta}\right]{V}_{0}dy$<br >$ε = \mathop{∫}_{0}^{L}{B}_{0}\left(1 + {\left(\dfrac{y}{L}\right)}^{\beta}\right){V}_{0}dy = \mathop{∫}_{0}^{L}{B}_{0} ⋅ {V}_{0} ⋅ dy + \mathop{∫}_{0}^{L}\dfrac{{B}_{0} ⋅ {V}_{0}}{{L}^{\beta}}{y}^{\beta} ⋅ dy
$= {B}_{0}{V}_{0}L\left[1 + \dfrac{1}{\beta + 1}\right]$
$∴$ emf in loop is proportional to $L$ for given value of $\beta$ .
$⇒$ Option $\left(A\right)$ is correct.
for
$\beta = 0; ε = 2{B}_{0}{V}_{0}L$
$\beta = 2; ε = {B}_{0}{V}_{0}L\left[1 + \dfrac{1}{3}\right] = \dfrac{4}{3}{B}_{0}{V}_{0}L ⇒$ Option $( B )$ is correct.
Now, the length of the projection of the wire $y = x$ and length $\sqrt{2}L$ on the y-axis is $L$ thus the answer remain unchanged $⇒$ Option $( C )$ is correct.