All that matters is the constant velocity!

Take a small mass $\mathrm{d}m$ making an angle $\mathrm{d}\alpha$ at the centre.

So, to find Normal $\mathbb{N}$ on the pulley,

$N = 2T\sin\dfrac{\mathrm{d}\alpha}{2}\approx T\mathrm{d}\alpha$

Now for it to rotate with constant velocity $v$ , $N = \mathrm{d}ma_c = \dfrac{\mathrm{d}mv^2}{R}$

$\Rightarrow T\mathrm{d}\alpha = \rho R\mathrm{d}\alpha\dfrac{v^2}{R}= \rho v^2\mathrm{d}\alpha\\ \Rightarrow T = \rho v^2$

$\Huge{\therefore\boxed{T = \rho v^2}}\normalsize=3.25$

Note : The tension is independent of the radius of the pulley.

Cant we do it like this that...

$T=\dfrac{d(mv)}{d(t)}$

$T=\dfrac{dm}{dt} \times v$

Now we know $dm=\rho.dx$ . So $\dfrac{dm}{dt}=\rho \times \dfrac{dx}{dt} = \rho v$ .

So putting in the 1st eqn we get

$T=\rho v^2$

$T=3.25 N$ .

Is My solution logically ok?