Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-
A.
$${M^0}{L^0}{T^0}$$
B.
$${M^{ - 1}}{L^{ - 1}}{T^{ - 1}}$$
C.
$${M^0}{L^2}{T^0}$$
D.
$${M^{ - 1}}{L^1}{T^2}$$
Answer :
$${M^0}{L^2}{T^0}$$
Solution :
Union of $$k$$ is joules per kelvin or dimensional formula of $$k$$ is $$\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]$$ Note: The power of an exponent is a number.
Therefore, dimensionally $$\frac{{\alpha z}}{{k\theta }} = {M^ \circ }{L^ \circ }{T^ \circ }$$
$$\eqalign{
& \therefore \alpha = \frac{{k\theta }}{z} \cr
& \therefore \alpha = \frac{{\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]\left[ \theta \right]}}{{\left[ L \right]}} = \left[ {ML{T^{ - 2}}} \right] \cr} $$
and dimensionally $$P = \frac{\alpha }{\beta } \Rightarrow \beta = \frac{\alpha }{P}$$
$$\therefore \left[ \beta \right] = \frac{{ML{T^{ - 2}}}}{{M{L^{ - 1}}{T^{ - 2}}}} = {M^0}{L^2}{T^0}$$
Releted MCQ Question on Basic Physics >> Unit and Measurement
Releted Question 1
The dimension of $$\left( {\frac{1}{2}} \right){\varepsilon _0}{E^2}$$ ($${\varepsilon _0}$$ : permittivity of free space, $$E$$ electric field)
A quantity $$X$$ is given by $${\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}}$$ where $${ \in _0}$$ is the permittivity of the free space, $$L$$ is a length, $$\Delta V$$ is a potential difference and $$\Delta t$$ is a time interval. The dimensional formula for $$X$$ is the same as that of-
Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-