Question

Pressure depends on distance as, $p=\frac{\alpha }{\beta }\exp (-\frac{\alpha z}{k\theta })$, where $\alpha ,\beta$ are constants, $z$ is distance, $k$ is Boltzman's constant and $\theta$ is temperature. The dimension of $\beta$ are :

• A. $M^o{L}^o{T}^o$
• B. $M^{-1}{L}^{-1}{T}^{-1}$
• C. $M^o{L}^2{T}^o$
• D. $M^{-1}{L}^1{T}^2$

Answer 1

The correct option is C $M^o{L}^2{T}^o$

$p=\frac{\alpha }{\beta }{e}^{-\left(\alpha z/k\theta \right)}$, By dimension analysis

$\left(a\right)\therefore dim\left[\alpha z\right]=dim\left[k\theta \right]$

$\therefore dim\left[\alpha \right]+dim\left[z\right]=dim\left[k\right]+dim\left[\theta \right]$

$\therefore dim\left[\alpha \right]+\left[M^0{L}^1{T}^0\right]=\left[M^1{L}^2{T}^{-2}{\theta }^{-1}\right]+\left[M^0{L}^0{T}^0{\theta }^{+1}\right]$

$\therefore dim\left[\alpha \right]=\left[M^1{L}^1{T}^{-2}\right]$

$\left(b\right)dim\left[\frac{\alpha }{\beta }\right]=dim\left[p\right]$

$dim\left[\alpha \right]-dim\left[\beta \right]=dim\left[p\right]$

$\therefore dim\left[\beta \right]=dim\left[\alpha \right]-dim\left[p\right]$

$\therefore dim\left[\beta \right]=\left[M^1{L}^1{T}^{-2}\right]-\left[M^1{L}^{-1}{T}^{-2}\right]$

$\therefore dim\left[\beta \right]=\left[M^0{L}^2{T}^0\right]$