fluid doubts... from vmc modules... please help 3

∂ = \frac{\rho _{m}}{\rho _{w}}

mg - Upthrust = ma
\Rightarrow \pi r^{2}l\rho _{m}g-\pi r^{2}l\rho _{w}g= \pi r^{2}l\rho _{m}a
\Rightarrow \pi r^{2}lg(\rho _{m}-\rho _{w})=\pi r^{2}l\rho _{m}a
\Rightarrow g\left(\frac{\rho _{m}}{\rho _{m}} -\frac{\rho _{w}}{\rho _{m}}\right)=a
g(1 - 1∂) = a

BY using v² = u² + 2as

since initially rod at rest so u = 0
v^{2}=2aS \Rightarrow v= \sqrt{2al}
Substituting value of 'a' we get
\Rightarrow v= \sqrt{2lg\left(1-\frac{1}{\frac{\rho _{m}}{\rho _{w}}} \right)} here ∂ = \frac{\rho _{m}}{\rho _{w}}

THIS IS WAHT I GOT.....

Let at any isnt of time t, length of rod immersed in water = x.

So upthrust = Axdg (d= density of water, ρ = density of rod)

So writing force equation,
Alρg - Axdg = Alρa

=> (lρ - xd)g = lρa

∫adx = ∫vdv
=> ∫(1- xdlρ)gdx = v²/2 (limits 0 to l)

=> 2g(l-ld2ρ) = v²

=> v = \sqrt{2gl(1-\frac{1}{2\partial })}

seems like i made a mistake taking acceleration constant.
and using equation v² = u² + 2as..SRY for the mistake
[2][2] shuld have integrated as asish did.......