A metallic rod of length 'l' is tied to a string of length '2l' and...

Let the fixed end of the string be the origin. The rod is situated between the distances $r_1 = 2l$ and $r_2 = 2l + l = 3l$ from the fixed end.

Consider a small element of the rod of length $dx$ at a distance $x$ from the fixed end. The velocity of this element is $v = x(2\omega)$.

The motional emf induced in this small element is $de = Bv\,dx = B(x \cdot 2\omega)\,dx$.

The total induced emf across the ends of the rod is obtained by integrating $de$ from $x = 2l$ to $x = 3l$:

$e = \int_{2l}^{3l} 2B\omega x\,dx$

$e = 2B\omega \left[ \dfrac{x^2}{2} \right]_{2l}^{3l}$

$e = B\omega ((3l)^2 - (2l)^2)$

$e = B\omega (9l^2 - 4l^2)$

$e = 5B\omega l^2$

Answer: $5\,B\omega\,l^2$