Question

Write the dimension of $a/b$ in the relation $P = \dfrac{{a - {t^2}}}{{bx}}$ ; where $P$ is pressure, $x$ is the distance and $t$ is the time.
(A) ${M^{ - 1}}{L^0}{T^{ - 2}}$
(B) $M{L^0}{T^{ - 2}}$
(C) $M{L^0}{T^2}$
(D) $ML{T^{ - 2}}$

Answer 1

Hint : Whenever quantities are operating on each other through addition or subtraction, they must be of the same unit. In an equation, the unit of the left hand side must be equal to the unit on the right hand side

Complete step by step answer
Given a relation between pressure time, and distance and two unknown variables, $a$ and $b$ . We are to find the dimension of $\dfrac{a}{b}$ .
To proceed, we must note that when two quantities are added, or one is subtracted from another, these two quantities must be of the same unit, thus dimension. Hence, $a$ in which ${t^2}$ is subtracted from must also have a dimension of ${t^2}$ which is ${T^2}$ .
Also, the dimension of the left hand side of the equation must be equal to the dimension of the right hand side. Hence, by dimension
$M{L^{ - 1}}{T^{ - 2}}$ = $M{L^{ - 1}}{T^{ - 2}} = \dfrac{{{T^2}}}{{\left[ b \right]L}}$ (since the dimension of Pressure is $M{L^{ - 1}}{T^{ - 2}}$ and according to question $x$ has a dimension of distance.
Furthermore, by calculating for $\left[ b \right]$ , we invert the equation, multiply both sides by ${T^2}$ and divide by $L$
We have,
${M^{ - 1}}L{T^2} \times \dfrac{{{T^2}}}{L} = \left[ b \right]$
$\Rightarrow \left[ b \right] = {M^{ - 1}}{T^4}$
Hence, the dimension of $\dfrac{a}{b}$ is
$\dfrac{{\left[ a \right]}}{{\left[ b \right]}} = \dfrac{{{T^2}}}{{{M^{ - 1}}{T^4}}} = M{T^{ - 2}}$
$\Rightarrow \dfrac{{\left[ a \right]}}{{\left[ b \right]}} = M{L^0}{T^{ - 2}}$
Hence, the correct option is B.

Note
Alternatively, since only the dimension of $\dfrac{a}{b}$ is to be found, it is quite unnecessary to find the dimension of $b$ .
Now, we know that the dimension $a$ is equal to that of ${t^2}$ , then the numerator is the dimension of $a$ .
Multiplying both sides by $x$ and using their dimensional equation, we have
$Px = \dfrac{{a - {t^2}}}{b}$
$\Rightarrow M{L^{ - 1}}{T^{ - 2}} \times L = \dfrac{{\left[ a \right]}}{{\left[ b \right]}}$
Hence, we simply have
$\dfrac{{\left[ a \right]}}{{\left[ b \right]}} = M{L^0}{T^{ - 2}}$
This is identical to the solution in the step by step procedure.