Correct solution is option A ) \(2a\alpha B\)

Explaination:

"a" at any time "t" , the side of the square a= \((a_0-\alpha_t)\) , where \(\alpha_0\) side at t=0 .

at this instant , flux through the square

⇒\(\phi = BA\cos0^o = B(a_o-\alpha_t)^2 \)

⇒\(\therefore emf \ induced \ E= -\frac{d\phi}{dt}\)

⇒\(E=-B\times 2 (a_0-\alpha_t)\)

⇒E= \(2a\alpha B\)