Solution:
Here, the given formula is \(h=\frac{2T\cos\left(\theta\right)}{r\rho g}\)
Here,
the dimension of h = [L]
the dimension of r = [L]
the dimension of ρ = [ML-3]
the dimension of g = [LT-2]
the dimension of T = [MT-2] and 2 cos(θ) is dimensionless.
Now the dimension of the quantity of the left-hand side is [L]
And, the dimension of the quantity of the right-hand side is
\(\frac{2T\cos\left(\theta\right)}{r\rho g}=\frac{\left[ML^{-2}\right]}{\left[L\right]\left[ML^{-3}\right]\left[LT^{-2}\right]}=\left[M^{1-1}L^{-1+3-1}T^{-2+2}\right]=\left[L\right]\)
The, dimension on left hand side are equal to the dimensions to the right side of the formula. So, the formula is dimensionally correct.