Question

If $F = ax + b{t^2} + cxt$. Where F is force, x is distance and t is time. Then what is the dimension of $\dfrac{{a{x^2}c}}{{b{t^2}}}$?
\[\begin{array}{l}
A.\,[M{L^2}{T^{ - 1}}]\\
B.\,[ML{T^{ - 3}}]\\
C.\,[M{L^2}{T^{ - 2}}]\\
D.\,[M{L^3}{T^{ - 2}}]
\end{array}\]

Answer 1

Hint: In the above question we have to apply the concept of Principle of Homogeneity of Dimensions in an equation. By using the principle of homogeneity, we have to solve the dimension for a, b and c. And use that to calculate the dimensions of the given ratio.

Complete step by step answer:
The principle of homogeneity states that the Dimensions of the quantities on either sides of an equation must be equal, it also states that two quantities and only by subtracted or added if their dimensions are completely same. While taking ratio or taking product no such conditions has to be followed.
Therefore, if the relation $F = ax + b{t^2} + cxt$is true then the dimensions of the left hand side and the right hand side of the equation must be same, also in the right hand side all the quantities are added hence they must possess same dimensions and this makes their dimensions equal to the left hand side quantity.
Since F is force, its dimensional formula is $[ML{T^{ - 2}}]$.
And the quantities $ax$, $b{t^2}$and $cxt$ are added; their dimensions should be equal and should be equal to the dimensions of force.
Hence, the dimensions of $ax$, $b{t^2}$and $cxt$ are $[ML{T^{ - 2}}]$
We can calculate the dimensions of a, b and c and then find the dimension of the given ratio. But if we can break the given ratio to our advantage this question can be solved much easily.
If we try to break the question in the way given below:
$\dfrac{{ax \times cx}}{{b{t^2}}}$, we can easily see that the quantities $ax$ and $b{t^2}$ has the same dimensions and are in ratio. Therefore, their dimensions will get canceled.
Now we only have remaining, $cx$ and we have to find its dimensions.
Now we know the dimension of $cxt$, where t is time then the dimension of $cx$ will be:
$\begin{array}{l}
[cx]\,[T]\, = \,[ML{T^{ - 2}}]\\
[cx] = [ML{T^{ - 3}}]
\end{array}$
Hence the dimension of the quantity given in the question is $[ML{T^{ - 3}}]$.

So, the correct answer is “Option B”.

Note:
The above question can also be solved using a traditional method but it will require more time as the calculations will be way more comparative to the calculation done. In such approach after finding the dimensions of $ax$, $b{t^2}$and $cxt$ we have to put the dimensions of x and t and equate it with the calculated dimension, and find the dimensions of a, b and c. after that we have to put the dimension of a, b, c, x, and t in the ratio given in the question to solve for the answer.