Write and explain principle of homogeneity. Check dimensional consistency of given equation.
Magnitude of physical quantities having same dimension can be added together or subtracted from one another.
This principle is called principle of homogeneity of dimension.
This rule is very useful to check dimensional consistency of given equation.
To check dimensional consistency of given equation all terms on both side of the equation should have same dimension.
Dimensional consistency do not guarantee correctness of equation.
It is uncertain to the extent of dimensionless quantities or function.
For example,
$x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}$
Here, $x$ is distance covered by object in time $t$.
$x_{0}=\text { initial position of object during motion }$
$x=\text { final position }$
$v_{0}=\text { initial velocity }$
$a=\text { acceleration }$
$\text { LHS }=x=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
In $RHS$ there are three terms,
$x_{0}=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
$v_{0} t=\left[\mathrm{L}^{1} \mathrm{~T}^{-1}\right]\left[\mathrm{T}^{1}\right]=\mathrm{M}$$^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
In $\frac{1}{2} a t^{2}$ is constant term which is dimensionless.
$\therefore a t^{2} =\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]\left[\mathrm{T}^{2}\right]$
$=\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$
Here all terms in given equation have same dimension. Hence, given equation,
The potential energy of a particle varies with distance $x$ from a fixed origin as $U=\frac{A \sqrt{x}}{x^2+B}$, where $A$ and $B$ are dimensional constants then dimensional formula for $A B$ is
Force $(F)$ and density $(d)$ are related as $F\, = \,\frac{\alpha }{{\beta \, + \,\sqrt d }}$ then dimension of $\alpha $ are
A famous relation in physics relates 'moving mass' $m$ to the 'rest mass' $m_{0}$ of a particle in terms of its speed $v$ and the speed of light, $c .$ (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant $c$. He writes:
$m=\frac{m_{0}}{\left(1-v^{2}\right)^{1 / 2}}$
Guess where to put the missing $c$
The potential energy of a point particle is given by the expression $V(x)=-\alpha x+\beta \sin (x / \gamma)$. A dimensionless combination of the constants $\alpha, \beta$ and $\gamma$ is