Question

A force F is given by \[F=at+b{{t}^{2}}\], where t is time. The dimensions of a and b are
\[A.\,[\text{ML}{{\text{T}}^{-3}}]\,\text{and}\,\,[\text{ML}{{\text{T}}^{-4}}]\]
\[B.\,[\text{ML}{{\text{T}}^{-4}}]\,\text{and}\,\,[\text{ML}{{\text{T}}^{-3}}]\]
\[C.\,[\text{ML}{{\text{T}}^{-1}}]\,\text{and}\,\,[\text{ML}{{\text{T}}^{-2}}]\]
\[D.\,[\text{ML}{{\text{T}}^{-2}}]\,\text{and}\,\,[\text{ML}{{\text{T}}^{0}}]\]

Answer 1

Hint: This question is based on the formation of the dimensional formula using the dimensional formula of the known quantity. Firstly we will derive the dimensional formula for the force using the basic formula that relates the force, mass and acceleration. Then, we will separately consider the equations of the terms ‘a’ and ‘b’ and compare these with the dimensional formula of the force obtained to find the dimensions of these terms.

Formula used: \[F=ma\]

Complete step by step answer:
The dimensional formula of the force is derived from the equation that relates the force with mass and acceleration. So, we have,
\[F=ma\]
Here, the unit of the mass is kg and the unit of the acceleration is meter per second-square. Substitute the units of the mass and acceleration in terms of the dimensional formulae in the above equation.
\[\begin{align}
  & F=[M][L{{T}^{-2}}] \\
 & F=[{{M}^{1}}{{L}^{1}}{{T}^{-2}}] \\
\end{align}\]….. (1)
Thus, we have obtained the dimensional formula of the force.
Now, we will consider the given equation of the force to find the dimensions of a and b.
\[F=at+b{{t}^{2}}\]…… (2)
Consider and compare the equations (1) and (2) for the following calculations.
So, firstly consider the term “a”.
The dimensional formula of the force must be equated with the equation (2) considering onlt the equation related to the term ‘a’.
\[at=a[T]=[F]\]
Now substitute the dimensional formula of the force in the above equation.
\[a[T]=[ML{{T}^{-2}}]\]
Take all the dimensional terms on one side of the equation.
\[\begin{align}
  & a=[ML{{T}^{-2}}]\times [{{T}^{-1}}] \\
 & \Rightarrow a=[ML{{T}^{-3}}] \\
\end{align}\]
Therefore, the dimensions of the term ‘a’ is \[a=[ML{{T}^{-3}}]\].
So, now consider the term “b”.
The dimensional formula of the force must be equated with the equation (2) considering only the equation related to the term ‘b’.
\[b{{t}^{2}}=b[{{T}^{2}}]=[F]\]
Now substitute the dimensional formula of the force in the above equation.
\[b[{{T}^{2}}]=[ML{{T}^{-2}}]\]
Take all the dimensional terms on one side of the equation.
\[\begin{align}
  & b=[ML{{T}^{-2}}]\times [{{T}^{-2}}] \\
 & \Rightarrow b=[ML{{T}^{-4}}] \\
\end{align}\]
Therefore, the dimensions of the term ‘b’ is \[b=[ML{{T}^{-4}}]\].
As, the dimensions of the terms ‘a’ and ‘b’ of the equation\[F=at+b{{t}^{2}}\]are obtained to be equal to\[A.\,[\text{ML}{{\text{T}}^{-3}}]\,\text{and}\,\,[\text{ML}{{\text{T}}^{-4}}]\].

So, the correct answer is “Option A”.

Note: The units of the parameters should be taken care of. As the dimensional formulae are derived from the units and the units are derived from the formulae, so, the formula for calculating the parameters should be known to solve this type of question.