Question

# In $F = Av + Bt + \dfrac{{Cx}}{{D + At}}$ , where F = force, v = speed, t = time, x = position, the dimensions of C will be:-(A) $\left[ {{M^2}{L^{ - 2}}{T^0}} \right]$(B) $\left[ {{M^1}{L^{ - 1}}{T^0}} \right]$(C) $\left[ {{M^2}{L^0}{T^{ - 2}}} \right]$(D) $\left[ {{M^1}{L^0}{T^{ - 2}}} \right]$

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Hint: - For the correctness of an equation the dimension on either side of the equation must be the same. This is well-known as the principle of homogeneity of dimensions. In equations having more than two terms, the dimension of each term needs to be the same.
The dimension of force is $\left[ {{M^1}{L^1}{T^{ - 2}}} \right]$ .

Complete Step-by-step solution:
We can check the relation between various physical quantities by finding the dimension of physical quantities. By dimensional analysis, we can check the rightness of an equation, we can derive the correct relationship between different physical quantities and it can also be used to convert one system of the unit into another.
For the rightness of an equation, the dimension on either side of the equation must be the same. In equations having more than two terms, the dimension of each term must be the same.
In the given equation $Av$ , $Bt$ and $\dfrac{{Cx}}{{D + At}}$ must have a force dimension.
Force is mass times acceleration. We know that the unit of force is $kg - m{s^{ - 2}}$ .
By evaluating the unit, we can write the dimension of force as $\left[ {ML{T^{ - 2}}} \right]$ .
And, velocity is the rate of change of displacement. We know that unit of velocity is $m{s^{ - 1}}$
By evaluating the unit, we can write the dimension of velocity as $\left[ {L{T^{ - 1}}} \right]$ .
Using the principle of homogeneity of dimensions, we can write;
$\Rightarrow F = Av$ ........... $\left( 1 \right)$
On substituting all the dimensions in the equation $\left( 1 \right)$, we get ;
$\Rightarrow \left[ {ML{T^{ - 2}}} \right] = A\left[ {L{T^{ - 1}}} \right]$
$\Rightarrow A = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {L{T^{ - 1}}} \right]}}$
On further simplifying the above equation, we get;
$\Rightarrow A = \left[ {M{T^{ - 1}}} \right]$
In the given term $\dfrac{{Cx}}{{D + At}}$ , $D$ and $At$ must have the same dimension. Therefore,
$\Rightarrow D = At$ ........... $\left( 2 \right)$
Where $t$ stands for time and its SI unit is second.
On evaluating the unit, we can write the dimension of time as $\left[ T \right]$ .
Substitute all the dimensions in the equation $\left( 2 \right)$, we get;
$\Rightarrow D = \left[ {M{T^{ - 1}}} \right] \times \left[ T \right]$
$\Rightarrow D = \left[ M \right]$
Using the principle of homogeneity of dimensions, we can also write;
$\Rightarrow F = \dfrac{{Cx}}{{D + At}}$ ........... $\left( 3 \right)$
Where $x$ stands for length and its SI unit is meter.
On evaluating the unit, we can write the dimension of length as $\left[ L \right]$.
Substitute all the dimensions in the equation $\left( 3 \right)$ , we get;
$\Rightarrow \left[ {ML{T^{ - 2}}} \right] = \dfrac{{C \times \left[ L \right]}}{{\left[ M \right]}}$
$\Rightarrow C = \dfrac{{\left[ {ML{T^{ - 2}}} \right] \times \left[ M \right]}}{{\left[ L \right]}}$
$\Rightarrow C = \dfrac{{\left[ {{M^2}L{T^{ - 2}}} \right]}}{{\left[ L \right]}}$
$\Rightarrow C = \left[ {{M^2}{L^0}{T^{ - 2}}} \right]$
$\therefore$ The dimension of $C$ is $\left[ {{M^2}{L^0}{T^{ - 2}}} \right]$ .

The correct answer is (C) $\left[ {{M^2}{L^0}{T^{ - 2}}} \right]$ .

Note: Firstly, Mass, length, and time are most commonly encountered fundamental quantities so they must be specified in all dimensional formulas. The square bracket symbolization is used only for dimensional formulas. Secondly, to simplify dimensional formulas we can apply multiplication and division properties of the exponent.