- $\text{h}=\frac{2\text{S}\cos\theta}{\rho\text{rg}}$
$\text{LHS}=[\text{L}]$
Surface tension $=\text{S}=\frac{\text{F}}{\text{I}}=\frac{\text{MLT}^{-2}}{\text{L}}=[\text{MT}^{-2}]$
Density $=\rho=\frac{\text{M}}{\text{V}}=[\text{ML}^{-3}\text{T}^{0}]$
Radius $=\text{r}=[\text{L],g = [LT}^{-2}]$
$\text{RHS}=\frac{2\text{S}\cos\theta}{\rho\text{rg}}=\frac{[\text{MT}^{-2}]}{[\text{ML}^{-3}\text{T}^0][\text{L}][\text{LT}^{-2}]}=[\text{M}^0\text{L}^1\text{T}^0]=[\text{L}]$
$\text{LHS = RHS}$
So, the relation is correct
- $\text{v}=\sqrt{\frac{\text{P}}{\rho}}$ where v = velocity
$\text{LHS = Dimension of v = [LT}^{-1}]$
Dimension of $\text{p}=\frac{\text{F}}{\text{A}}=[\text{ML}^{-1}\text{T}^{-2}]$
Dimension of $\rho=\frac{\text{m}}{\text{v}} = \text{[ML}^{-2}]$
$\text{RHS}=\sqrt{\frac{\text{p}}{\rho}}=\sqrt{\frac{[\text{ML}^{-1}\text{T}^{-2}]}{[\text{ML}^{-3}]}}=[\text{L}^2\text{T}^{-2}]^{\frac{1}{2}}=[\text{LT}^{-1}]$
So, the relation is correct.
- $\text{V}=\frac{(\pi\text{pr}^4\text{t})}{(8\eta\text{I})}$
$\text{LHS = Dimension of V = [L}^3]$
Dimension of $\text{p}=[\text{ML}^{-1}\text{T}^{-2}],\text{r}^4=[\text{L}^4],\text{t}=[\text{T}]$
Coefficient of viscosity $=[\text{ML}^{-1}\text{T}^{-1}]$
$\text{RHS}=\frac{\pi\text{pr}^4\text{t}}{8\eta\text{I}}=\frac{[\text{ML}^{-1}\text{T}^{-2}][\text{L}^4][\text{T}]}{[\text{ML}^{-1}\text{T}^{-1}][\text{L}]}$
So, the relation is correct.
- $\text{v}=\frac{1}{2\pi}\sqrt{(\text{mgI}/\text{I})}$
$\text{LHS = dimension of v = [T}^{-1}]$
$\text{RHS}=\sqrt{(\text{mgI}/\text{I})}=\sqrt{\frac{[\text{M}][\text{LT}^{-2}][\text{L}]}{[\text{ML}^2]}}=[\text{T}^{-1}]$
$\text{LHS = RHS}$
So, the relation is correct.