A 1.0 m long metallic rod is rotated with an angular frequency of

Length of the rod, l = 1 m

Angular frequency,ω = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.

Average linear velocity of the rod, $v=\frac{l \omega+0}{2}=\frac{l \omega}{2}$

Emf developed between the centre and the ring,

$e=B l v=B l\left(\frac{l \omega}{2}\right)=\frac{B l^{2} \omega}{2}$

$=\frac{0.5 \times(1)^{2} \times 400}{2}=100 \mathrm{~V}$

Hence, the emf developed between the centre and the ring is 100 V.