**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.