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Question: The dimensions of \( \dfrac{a}{b} \) in the equation \( P = \dfrac{{a - {t^2}}}{{bx}} \) where \( P ...

The dimensions of ab\dfrac{a}{b} in the equation P=a−t2bxP = \dfrac{{a - {t^2}}}{{bx}} where PP is the pressure , xx is the distance and tt is time are:

A) M2LT−2{M^2}L{T^{ - 2}}

B) M1L0T−2{M^1}{L^0}{T^{ - 2}}

C) M1L3T−1{M^1}{L^3}{T^{ - 1}}

D) ML1T−3M{L^1}{T^{ - 3}}

Explanation

Solution

In this solution, we will determine using the relations of the dimensional formula for a and b using the rules of dimensional formula. For subtracting two terms, they must have the same dimensional formula.

Complete step by step answer:

We’ve been the formula for PP as

P=a−t2bxP = \dfrac{{a - {t^2}}}{{bx}}

Now we know that when subtracting two terms, they must have the same dimensional formula. So in the numerator in the formula for pressure, we know that

[a]=[t2]=T2[a] = [{t^2}] = {T^2}

Now to equate two terms, we know that both the terms must have the same dimensional formula. So, we can say that the dimensional formula of the term on the right side of the equation also must have the dimensions of pressure. We can write that as

[a−t2bx]=[P]\left[ {\dfrac{{a - {t^2}}}{{bx}}} \right] = [P]

Let us find the dimensional formula of pressure now. Since we know that P=FAP = \dfrac{F}{A} where FF is the force and AA is the area. Since the dimensional formula of [F]=[ma]=M1L1T−2[F] = [ma] = {M^1}{L^1}{T^{ - 2}} and dimensional formula of [A]=L2[A] = {L^2} , the dimensional formula of pressure will be

[P]=M1L1T−2L2=M1L−1T−2[P] = \dfrac{{{M^1}{L^1}{T^{ - 2}}}}{{{L^2}}} = {M^1}{L^{ - 1}}{T^{ - 2}}

So, coming back to the equation in question, we can now say that the dimensions of the left side will be equal to:

[a−t2bx]=M1L−1T−2\left[ {\dfrac{{a - {t^2}}}{{bx}}} \right] = {M^1}{L^{ - 1}}{T^{ - 2}}

Since the numerator has the dimensions of T2{T^2} , the dimensions of bb can be determined by

T2[bx]=M1L−1T−2\dfrac{{{T^2}}}{{[bx]}} = {M^1}{L^{ - 1}}{T^{ - 2}}

⇒M−1L1T4=[bx]\Rightarrow {M^{ - 1}}{L^1}{T^4} = [bx]

So, the dimension of b will be

[b]=M−1T4[b] = {M^{ - 1}}{T^4}

Then the dimensions of a/ba/b will be

[ab]=T2M−1T4\left[ {\dfrac{a}{b}} \right] = \dfrac{{{T^2}}}{{{M^{ - 1}}{T^4}}}

∴[ab]=M1T−2\therefore \left[ {\dfrac{a}{b}} \right] = {M^1}{T^{ - 2}}

Hence the dimensions of [ab]=M1L0T−2\left[ {\dfrac{a}{b}} \right] = {M^1}{L^0}{T^{ - 2}}. So, option (B) is correct.

Note:

There is no such formula for pressure; however , this question tests our proficiency in dealing with dimensional formulas. To answer such questions, we must be able to easily determine the dimensional formula of basic mechanical quantities.