Question

$\dfrac{\alpha }{{{t^2}}} = Fv + \dfrac{\beta }{{{x^2}}}$ Find the dimensional formula for $\left[ \alpha \right]$ and $\left[ \beta \right]$ (here t = time, F = force, v = velocity, x = distance).
(A) $\left[ \beta \right] = {M^1}{L^2}{T^{ - 3}};\left[ \alpha \right] = {M^1}{L^2}{T^{ - 1}}$
(B) $\left[ \beta \right] = {M^1}{L^3}{T^{ - 3}};\left[ \alpha \right] = {M^1}{L^3}{T^{ - 1}}$
(C) $\left[ \beta \right] = {M^1}{L^4}{T^{ - 3}};\left[ \alpha \right] = {M^1}{L^2}{T^{ - 1}}$
(D) $\left[ \beta \right] = {M^1}{L^5}{T^{ - 3}};\left[ \alpha \right] = {M^1}{L^2}{T^{ - 1}}$

Answer 1

Hint
The basic rule to solve this problem is the physical quantities can be added or equated only if the quantities have the same dimensions. So first we equate the dimensions of quantities on the R.H.S of the equation and then we compare the dimensions with L.H.S of the equation
- Dimensional formula of force (F) = $\left[ {{M^1}{L^1}{T^{ - 2}}} \right]$
- Dimensional formula of velocity (v)= $\left[ {{L^1}{T^{ - 1}}} \right]$
- Dimensional formula of time (t) = $\left[ {{T^1}} \right]$
- Dimensional formula of distance (x) = $\left[ {{L^1}} \right]$

Complete step by step answer
From the given equation $\dfrac{\alpha }{{{t^2}}} = Fv + \dfrac{\beta }{{{x^2}}}$ , we can see that the terms $Fv$ and $\dfrac{\beta }{{{x^2}}}$ are added. Two physical quantities can be added only if they have the same dimensions. So by equating the dimensions,
$ \Rightarrow \left[ {Fv} \right] = \left[ {\dfrac{\beta }{{{x^2}}}} \right]$ …..$(1)$
Dimension of Fv = $\left[ {Fv} \right] = \left[ {{M^1}{L^1}{T^{ - 2}}} \right]\left[ {{L^1}{T^{ - 1}}} \right] = \left[ {{M^1}{L^2}{T^{ - 3}}} \right]$ …..$(2)$
From equations $(1)$ and $(2)$ , we get
$ \Rightarrow \left[ {\dfrac{\beta }{{{x^2}}}} \right] = \left[ {{M^1}{L^2}{T^{ - 3}}} \right]$
By substituting the dimensional formula of x in the above equation, we get
$ \Rightarrow \left[ \beta \right] = \left[ {{M^1}{L^2}{T^{ - 3}}} \right]\left[ {{L^2}} \right]$
$ \Rightarrow \left[ \beta \right] = \left[ {{M^1}{L^4}{T^{ - 3}}} \right]$
$\therefore $ The dimensional formula of $\beta $ is $\left[ {{M^1}{L^4}{T^{ - 3}}} \right]$
In the given equation, the terms $\dfrac{\alpha }{{{t^2}}}$ and $Fv + \dfrac{\beta }{{{x^2}}}$ are equated. Two physical quantities can be equated only if they have the same dimensions. So by equating the dimensions,
$ \Rightarrow \left[ {\dfrac{\alpha }{{{t^2}}}} \right] = \left[ {Fv + \dfrac{\beta }{{{x^2}}}} \right]$
$ \Rightarrow \left[ {\dfrac{\alpha }{{{t^2}}}} \right] = \left[ {{M^1}{L^2}{T^{ - 3}}} \right]$
By substituting the dimensional formula of t in the above equation, we get
$ \Rightarrow \left[ \alpha \right] = \left[ {{M^1}{L^2}{T^{ - 3}}} \right]\left[ {{T^2}} \right]$
$ \Rightarrow \left[ \alpha \right] = \left[ {{M^1}{L^2}{T^{ - 1}}} \right]$
$\therefore $ The dimensional formula of $\alpha $ is $\left[ {{M^1}{L^2}{T^{ - 1}}} \right]$
The correct option is (C) $\left[ \beta \right] = {M^1}{L^4}{T^{ - 3}};\left[ \alpha \right] = {M^1}{L^2}{T^{ - 1}}$.

Additional Information
- If two physical quantities are multiplied, the dimensional formula of the resulting physical quantity will be obtained by the multiplication of dimensional formulas of physical quantities.
- If two physical quantities are divided, the dimensional formula of the resulting physical quantity will be obtained by the division of dimensional formulas of physical quantities.

Note
In the second case, while equating the dimensions on both sides of the equation, there is a chance of equating the dimension of $\dfrac{\alpha }{{{t^2}}}$ with $\left[ \beta \right]$ and it should not be done. The dimension of $\dfrac{\alpha }{{{t^2}}}$ should be only compared with either $F_v$ or $\dfrac{\beta }{{{x^2}}}$ as both will have the same dimensions.