Test if the following equations are dimensionally correct: 1. h= 2S 2. u= P 3. V= ^4t 8 l 4. v=1 2 mgl I where h = height, S = surface tension, = density, P = pressure, V = volume, = coefficient of viscosity, v = frequency and I = moment of inertia.

1. h= 2S LHS=[L] Surface tension =S= F I = MLT^ -2 L =[MT^ -2 ] Density = = M V =[ML^ -3 T^ 0 ] Radius =r=[L],g = [LT^ -2 ] RHS= 2S = [MT^ -2 ] [ML^ -3 T^0][L][LT^ -2 ] =[M^0L^1T^0]=[L] LHS = RHS So, the relation is correct 2. v= P where v = velocity LHS = Dimension of v = [LT^ -1 ] Dimension of p= F A =[ML^ -1 T^ -2 ] Dimension of = m v = [ML^ -2 ] RHS= p = [ML^ -1 T^ -2 ] [ML^ -3 ] =[L^2T^ -2 ]^ 1 2 =[LT^ -1 ] So, the relation is correct. 3. V= ( ^4t) (8 ) LHS = Dimension of V = [L^3] Dimension of p=[ML^ -1 T^ -2 ],r^4=[L^4],t=[T] Coefficient of viscosity =[ML^ -1 T^ -1 ] RHS= ^4t 8 = [ML^ -1 T^ -2 ][L^4][T] [ML^ -1 T^ -1 ][L] So, the relation is correct. 4. v=1 2 (mgI/I) LHS = dimension of v = [T^ -1 ] RHS= (mgI/I) = [M][LT^ -2 ][L] [ML^2] =[T^ -1 ] LHS = RHS So, the relation is correct.

Test if the following equations are dimensionally correct: 1. h= …

Download App

Question

Test if the following equations are dimensionally correct:

$\text{h}=\frac{2\text{S}\cos\theta}{\rho\text{rg}}$
$\text{u}=\sqrt{\frac{\text{P}}{\rho}}$
$\text{V}=\frac{\pi\text{Pr}^4\text{t}}{8\eta l}$
$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{mg}l}{\text{I}}}$

where h = height, S = surface tension, $\rho$ = density, P = pressure, V = volume, $\eta$ = coefficient of viscosity, v = frequency and I = moment of inertia.

✓

Answer

$\text{h}=\frac{2\text{S}\cos\theta}{\rho\text{rg}}$

$\text{LHS}=[\text{L}]$
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.