Question

A force is given by $F=at+b{{t}^{2}}$ , where t is time. The dimension of a and b are:
$\begin{aligned} & \text{A}\text{. }\left[ ML{{T}^{-3}} \right]\text{and}\left[ ML{{T}^{-4}} \right] \\\ & \text{B}\text{. }\left[ ML{{T}^{-4}} \right]\text{and}\left[ ML{{T}^{-3}} \right] \\\ & \text{C}\text{. }\left[ ML{{T}^{-1}} \right]\text{and}\left[ ML{{T}^{-2}} \right] \\\ & \text{D}\text{. }\left[ ML{{T}^{-2}} \right]\text{and}\left[ ML{{T}^{0}} \right] \\\ \end{aligned}$

Answer 1

When we write an equation, the equation should be always dimensionally correct. We can check the dimensional correctness of the equation by finding the dimension of each quantity in the equation related with addition and subtraction. Here, find the dimension of force and then equate it with the dimension of the other quantities.

Complete step by step answer:
For an equation to be dimensionally correct, the dimension of every physical quantity on both sides of the equation related with addition and subtraction should be the same.
For example, if we consider a physical equation $x=y+z$ , the dimension of each quantity x, y and z should be the same.

For the equation $F=at+b{{t}^{2}}$to be dimensionally correct, the dimension of F, at and $b{{t}^{2}}$ should be the same.

So, the dimension of F will be equal to the dimension of at.

$\left[ F \right]=\left[ at \right]$

Dimension of F can be obtained by expressing it as the product of mass and acceleration of an object.
The dimension of mass is $\left[ M \right]$
The dimension of acceleration is $\left[ L{{T}^{-2}} \right]$
The dimension of force will be, $=\left[ M \right]\left[ L{{T}^{-2}} \right]=\left[ ML{{T}^{-2}} \right]$
Now dimension of a can be found as,

$\left[ a \right]=\dfrac{\left[ F \right]}{\left[ t \right]}=\dfrac{\left[ ML{{T}^{-2}} \right]}{\left[ T \right]}=\left[ ML{{T}^{-3}} \right]$
Again, the dimension of $b{{t}^{2}}$ will be equal to the dimension of F.
So, we can write,

$\begin{aligned} & \left[ b{{t}^{2}} \right]=\left[ F \right] \\\ & \left[ b \right]=\dfrac{\left[ F \right]}{\left[ {{t}^{2}} \right]} \\\ & \left[ b \right]=\dfrac{\left[ ML{{T}^{-2}} \right]}{{{\left[ T \right]}^{2}}}=\left[ ML{{T}^{-4}} \right] \\\ \end{aligned}$

So, the dimension of a and b is $\left[ ML{{T}^{-3}} \right]$and $\left[ ML{{T}^{-4}} \right]$respectively. Hence, the correct answer is option A.

Note:
When we try to find the dimensional formula of a physical quantity, always try to break the physical quantity in terms of the fundamental physical quantity. Since, we know the dimensions of all fundamental quantities it will be easier to find the answer this way.