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# The dimensions of $\dfrac{\alpha }{\beta }$ in the equation $F = \dfrac{{\alpha - {t^2}}}{{\beta {v^2}}}$ where $F$ is the force, $v$ is velocity and $t$ is time, is (1) $[ML{T^{ - 1}}]$(2) $[M{L^{ - 1}}{T^{ - 2}}]$(3) $[M{L^3}{T^{ - 4}}]$(4) $[M{L^2}{T^{ - 4}}]$

Last updated date: 17th Apr 2024
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Hint The dimensions of a physical quantity and the dimensional formula of the physical quantity unit are the same. All physical quantities can be articulated in terms of seven fundamental (base) quantities which are mass, length, time, temperature, electric current, luminous intensity, and amount of substance. These seven quantities are called the seven dimensions of the physical world.

It is given in the question that $F = \dfrac{{\alpha - {t^2}}}{{\beta {v^2}}}$ where $F$ is the force, $v$ is velocity and $t$ is time.
We know the dimensions of force, velocity, and time which are,
$F = [ML{T^{ - 2}}]$
$t = [{M^0}{L^0}T]$
$v = [{M^0}L{T^{ - 1}}]$
$M$, $L$ and $T$ are used to represent the dimensions of the three mechanical quantities mass, length, and time respectively and $K$ is used for temperature, $I$ is used for electric current, $cd$ is used for luminous intensity and $mol$ is used for the amount of substance.
Only the quantities with the same dimensional formula can be added or subtracted from each other. Hence from this, we can say that the dimensions of $\alpha = [{T^2}]$.
Inserting this dimensional formula into the given formula of force and then doing dimensional analysis
$[ML{T^{ - 2}}] = \dfrac{{[{T^2}]}}{{\beta {{[L{T^{ - 1}}]}^2}}}$
$\Rightarrow [ML{T^{ - 2}}] = \dfrac{{[{T^2}]}}{{\beta [{L^2}{T^{ - 2}}]}}$
$\Rightarrow\beta = [{M^{ - 1}}{L^{ - 3}}{T^6}]$
Hence dimension of $\beta$is $[{M^{ - 1}}{L^{ - 3}}{T^6}]$
Using dimensions of $\beta$ and $\alpha$ we can find the dimensions of $\dfrac{\alpha }{\beta }$.
$\dfrac{\alpha }{\beta } = \dfrac{{[{T^2}]}}{{[{M^{ - 1}}{L^{ - 3}}{T^6}]}}$
$\Rightarrow \dfrac{\alpha }{\beta } = [{M^1}{L^3}{T^{ - 4}}]$

Therefore the correct answer is (3) $[M{L^3}{T^{ - 4}}]$

Note
There are certain limitations to the use of dimensional analysis which are that it does not tell anything about the dimensional constant present in an equation, it fails where there are more than three quantities whose dimensions are not known to us, there are certain quantities whose dimensional formula is same so we cannot differentiate them with this.