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Question: In the equation \(F = A\sin B{x^2} + \dfrac{C}{t}{e^{Dt}}\), \(F\), \(x\) and \(t\) are force, posit...

In the equation F=Asin⁡Bx2+CteDtF = A\sin B{x^2} + \dfrac{C}{t}{e^{Dt}}, FF, xx and tt are force, position and time respectively, then give the dimensions of ACB\dfrac{A}{{CB}}.

Explanation

Solution

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=Asin⁡Bx2+CteDtF = A\sin B{x^2} + \dfrac{C}{t}{e^{Dt}}.
Using 2nd2ndlaw of dimensional analysis we have:
Asin⁡Bx2A\sin B{x^2} and CteDt\dfrac{C}{t}{e^{Dt}} both have dimensions of force.
We know that the dimensional formula of Force =[M][L][T]−2 = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}}
Thus, the dimensions of the Asin⁡Bx2A\sin B{x^2} will be of force.
Now from 1st1st law we have to equate Asin⁡Bx2A\sin B{x^2} to [M][L][T]−2\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.
[M][L][T]−2=Asin⁡Bx2\left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}} = A\sin B{x^2}
[A]=[M][L][T]−2−−−(i)\left[ A \right] = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}} --- (i)
And,
Bx2=[M]0[L]0[T]0 [B][L]2=[M]0[L]0[T]0 [B]=[L]−2−−(ii)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 CteDt\dfrac{C}{t}{e^{Dt}} we know that its dimensions are the same as force.
So, Dimensions of Ct\dfrac{C}{t} are the same as force because eDt{e^{Dt}} is a dimensionless quantity.
Now,
Ct=[M][L][T]−2 ⇒[C][T]=[M][L][T]−2 ⇒[C]=[M][L][T]−1−−(iii)\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 [A]\left[ A \right],[B]\left[ B \right]and [C]\left[ C \right]
From equation (i)\left( i \right),(ii)(ii) and (iii)(iii) we get,
⇒ACB=[M][L][T]−2[[[M][L][T]−1[L]−2]] ⇒ACB=[L]2[T]−1\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 ACB=[L]2[T]−1\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.