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Question: The Force \(F\) is given as \(F = at + b{t^2}\) where \(t\) is time. What are the dimensions of \(a\...

The Force FF is given as F=at+bt2F = at + b{t^2} where tt is time. What are the dimensions of aa and bb
A. [M][L][T]−1[M][L]{[T]^{ - 1}} and [M][L][T]0[M][L]{[T]^0}
B. [M][L][T]−3[M][L]{[T]^{ - 3}} and [M][L]2[T]4[M]{[L]^2}{[T]^4}
C. [M][L][T]−4[M][L]{[T]^{ - 4}} and [M][L][T]1[M][L]{[T]^1}
D. [M][L][T]−3[M][L]{[T]^{ - 3}} and [M][L][T]−4[M][L]{[T]^{ - 4}}

Explanation

Solution

Hint: Only quantities with similar dimensions can be equated or added. The dimension of Force is [M]1[L]1[T]−2{[M]^1}{[L]^1}{[T]^{ - 2}} and that of time is [M]0[L]0[T]1{[M]^0}{[L]^0}{[T]^1}. The dimension of atat and that of bt2b{t^2} should have the same dimension as that of force.

Complete step-by-step answer:
Equations of any kind have to satisfy dimensional consistency. This means that all individual terms in an equation should have the same dimension. For example, if you have an equation AB=C+DEAB = C + \dfrac{D}{E}, the dimensions of ABAB, CC and DE\dfrac{D}{E} should be equal.
This is because objects with the same dimensions can only be equated, added or subtracted. It makes no sense to subtract 2cm2cm from 3kg3kg

Here in the question, FF is given as at+bt2at + b{t^2}, where t is time and has the dimension: [M]0[L]0[T]1{[M]^0}{[L]^0}{[T]^1}
Now we know from our discussion that FF, atat, and bt2b{t^2} should have the same dimension.
The dimension of FF is [M]1[L]1[T]−2{[M]^1}{[L]^1}{[T]^{ - 2}}
So the dimension of atat should also be [M]1[L]1[T]−2{[M]^1}{[L]^1}{[T]^{ - 2}}
substituting the dimension of t as [M]0[L]0[T]1{[M]^0}{[L]^0}{[T]^1}, we get
[a]×[M]0[L]0[T]1=[M]1[L]1[T]−2[a] \times {[M]^0}{[L]^0}{[T]^1} = {[M]^1}{[L]^1}{[T]^{ - 2}}
So we get [a]=[M]1[L]1[T]−3[a] = {[M]^1}{[L]^1}{[T]^{ - 3}}

Similarly, for b,
The dimension of bt2b{t^2} should also be [M]1[L]1[T]−2{[M]^1}{[L]^1}{[T]^{ - 2}}.
[b]×([M]0[L]0[T]1)2=[M]1[L]1[T]−2[b] \times {\left( {{{[M]}^0}{{[L]}^0}{{[T]}^1}} \right)^2} = {[M]^1}{[L]^1}{[T]^{ - 2}}
[b]×[M]0[L]0[T]2=[M]1[L]1[T]−2[b] \times {[M]^0}{[L]^0}{[T]^2} = {[M]^1}{[L]^1}{[T]^{ - 2}}
So we get [b]=[M]1[L]1[T]−4[b] = {[M]^1}{[L]^1}{[T]^{ - 4}}
These answers correspond to option D and hence, it is the right answer.

Additional Information:
Dimensions of commonly used quantities are listed below. Remembering them would be beneficial.
Force [M]1[L]1[T]−2{[M]^1}{[L]^1}{[T]^{ - 2}}
Work [M]1[L]2[T]−2{[M]^1}{[L]^2}{[T]^{ - 2}}
spring constant [M]1[T]−2{[M]^1}{[T]^{ - 2}}
Moment of Inertia [M]1[L]2{[M]^1}{[L]^2}
Electric Field [M]1[L]1[T]−3[A]−1{[M]^1}{[L]^1}{[T]^{ - 3}}{[A]^ {- 1}}
Charge [T]1[A]1{[T]^1}{[A]^1}

Note: Though a correct equation should be dimensionally consistent. A dimensionally consistent equation need not be correct. Factors like the presence or absence of dimensionless constants, logarithms, sines, etc cannot be verified using dimensional analysis.