Q. In the relation $p = \frac{\alpha}{\beta}\ e^{- \frac{\alpha z}{k\theta}},\ p$ is the pressure, $z$ the distance, $k$ is Boltzmann constant and $\theta$ is the temperature, the dimensional formula of $\beta$ will be
A) $\left\lbrack M^{0}L^{2}T^{0} \right\rbrack$
B) $\left\lbrack {ML}^{2}T \right\rbrack$
C) $\left\lbrack {ML}^{0}T^{- 1} \right\rbrack$
D) $\left\lbrack {ML}^{2}T^{- 1} \right\rbrack$
Explanation
In given equation, $\frac{az}{k\theta}$ should be dimensionless $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha = \frac{k\theta}{z}$ $\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \lbrack\alpha\rbrack = \frac{\left\lbrack ML^{2}T^{- 2}K^{- 1} \times K \right\rbrack}{\lbrack L\rbrack} = \left\lbrack MLT^{- 2} \right\rbrack$ And $\ \ \ \ \ \ \ \ \ \ p = \frac{\alpha}{\beta}$ $\Rightarrow \ \ \ \ \ \ \ \ \ \ \lbrack\beta\rbrack = \left\lbrack \frac{\alpha}{p} \right\rbrack = \frac{\left\lbrack MLT^{- 2} \right\rbrack}{\left\lbrack ML^{- 1}T^{- 2} \right\rbrack} = \left\lbrack M^{0}L^{2}T^{0} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.