Answer
$ v\propto \sqrt{\frac{T}{R^2\rho}}$
Work Step by Step
We can show the required relation as follows:
$v=R^aT^b\rho^c$.....eq(1) where $v, R,T~and~\rho$ represent speed, radius, tension and density respectively
Now we use dimensional analysis
$m/s=m^a(Kgm/s^2)^b(Kg/m^3)^c$
This simplifies to:
$mKg^0s^{-1}=m^{a+b-3c}Kg^{b+c}s^{-2b}$
We compare both sides to obtain:
$1=a+b-3c$...eq(2)
$-1=-2b$...eq(3)
$0=b+c$.....eq(4)
Solving eq(2), (3) and (4), we obtain:
$b=\frac{1}{2}$
$c=-\frac{1}{2}$
$a=-1$
We plug in the values of a, b and c in eq(1) to obtain:
$v=R^{-1}T^{\frac{1}{2}}\rho^{-\frac{1}{2}}$
$\implies v=\frac{1}{R}\sqrt{\frac{T}{\rho}}$
$\implies v\propto \sqrt{\frac{T}{R^2\rho}}$
