DIMENSIONAL ANALYSIS AND ITS APPLICATIONS

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

a = dimensionless of b

a = MLT^-2

Using F = MA = $m\frac{V}{T}$ $\frac{}{}$

m = FTV¹

F = ηAdv / dx
η = F (dx/dv) (1/A) = m * a * time * (1/area) = MV/A

Since, $\frac{x^2}{\alpha kT}$ should be dimensionless.

So, dimension of  $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$

Dimension of  $\alpha {\beta }^2$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

The dimensional formula for energy E is [ML²T? ²].
The dimensional formula for the gravitational constant G is [M? ¹L³T? ²].
The ratio E/G has dimensions: [ML²T? ²] / [M? ¹L³T? ²] = [M²L? ¹T? ].

x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]

 K = (Q?)Δx / AΔT
⇒ (ML²T?²)(L) / (L²)(θ)(T)
⇒ M¹L¹T?³θ?¹

Bonus $\mathrm{}V=K\left(h{)}^a\right(I{)}^b\left(G{)}^c\right(C{)}^d\mathrm{}(V$

is voltage   $)$ $$

We know,   $\begin{array}{rr}& \left[\mathrm{h}\right]={\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-1}\\ & \left[\mathrm{l}\right]=\mathrm{A}\\ & \left[\mathrm{G}\right]={\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\\ & \left[\mathrm{C}\right]=\mathrm{L}{\mathrm{T}}^{-1}\\ & \left[\mathrm{V}\right]=\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\end{array}$

$M{L}^2{T}^{-3}{A}^{-1}={\left(M{L}^2{T}^{-1}\right)}^a(A{)}^b{\left(M^{-1}{L}^3{T}^{-2}\right)}^c{\left(L{T}^{-1}\right)}^d$

$M{L}^2{T}^{-3}{A}^{-1}=M^{a-c}{L}^{2a+3c+d}{T}^{-a-2c-d}{A}^b$

$\mathrm{a}-\mathrm{c}=1$

$2\mathrm{a}+3\mathrm{c}+\mathrm{d}=2$

$-a-2c-d=-3$

$b=-1$

On solving,

$c=-1$ $a=0$

$\mathrm{d}=5,\mathrm{b}=-1$

$V=K\left(h{)}^{?}\right(I{)}^{-1}\left(G{)}^{-1}\right(C{)}^5$

Since, $\frac{x^2}{\alpha kT}$  should be dimensionless.

So, dimension of   $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$

Dimension of  $\alpha {\beta }^2$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$  

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$