Question

In the equation $F = A\sin B{x^2} + \dfrac{C}{t}{e^{Dt}}$, $F$, $x$ and $t$ are force, position and time respectively, then give the dimensions of $\dfrac{A}{{CB}}$.

Answer 1

Hint: In order to solve the question of dimensional analysis we need to know some rules of it.
> Two physical quantities can only be equated if they have the same dimensions.
> Two physical quantities can only be added if they have the same dimensions.
> The dimensions of the multiplication of two quantities are given by the multiplication of the dimensions of the two quantities.

Complete step by step answer:
The given equation is $F = A\sin B{x^2} + \dfrac{C}{t}{e^{Dt}}$.
Using $2nd$law of dimensional analysis we have:
$A\sin B{x^2}$ and $\dfrac{C}{t}{e^{Dt}}$ both have dimensions of force.
We know that the dimensional formula of Force $ = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}}$
Thus, the dimensions of the $A\sin B{x^2}$ will be of force.
Now from $1st$ law we have to equate $A\sin B{x^2}$ to $\left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}}$ as the dimensions of both are same. But we know the argument of sine function is dimensionless.
$\left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}} = A\sin B{x^2}$
$\left[ A \right] = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}} --- (i)$
And,
 $ B{x^2} = {\left[ M \right]^0}{\left[ L \right]^0}{\left[ T \right]^0} \\
  \left[ B \right]{\left[ L \right]^2} = {\left[ M \right]^0}{\left[ L \right]^0}{\left[ T \right]^0} \\
  \left[ B \right] = {\left[ L \right]^{ - 2}} - - (ii) $
Similarly, for $\dfrac{C}{t}{e^{Dt}}$ we know that its dimensions are the same as force.
So, Dimensions of $\dfrac{C}{t}$ are the same as force because ${e^{Dt}}$ is a dimensionless quantity.
 Now,
$\dfrac{C}{t} = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}} \\
 \Rightarrow \dfrac{{\left[ C \right]}}{{\left[ T \right]}} = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}} \\
 \Rightarrow \left[ C \right] = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 1}} - - (iii) $
Now, we have all the values required i.e., dimensions of $\left[ A \right]$,$\left[ B \right]$and $\left[ C \right]$
From equation $\left( i \right)$,$(ii)$ and $(iii)$ we get,
$\Rightarrow \dfrac{A}{{CB}} = \dfrac{{\left[ M \right]\left[ L \right]{{\left[ T \right]}^{ - 2}}}}{{\left[ {\left[ {\left[ M \right]\left[ L \right]{{\left[ T \right]}^{ - 1}}{{\left[ L \right]}^{ - 2}}} \right]} \right]}} \\
 \Rightarrow \dfrac{A}{{CB}} = {\left[ L \right]^2}{\left[ T \right]^{ - 1}} $
Final answer: The dimensions of $\dfrac{A}{{CB}} = {\left[ L \right]^2}{\left[ T \right]^{ - 1}}$.

Note: We need to keep some points in our mind solving these types of problems:
>Dimensional formulas of important quantities should be remembered.
>All the rules for dimensional analysis mentioned above must be followed otherwise the chances of mistakes can be increased.
>Silly mistakes in calculations must be avoided to get the correct answer.