Write the dimensions of a / b in the relation | vedclass

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(A) Given the relation $P = \frac{a - t^2}{bx}$.According to the principle of homogeneity of dimensions,the dimensions of $a$ must be equal to the dim

Vedclass · Accepted Answer

$ML^0 T^{-2}$

Vedclass · Answer

(A) Given the relation $P = \frac{a - t^2}{bx}$.According to the principle of homogeneity of dimensions,the dimensions of $a$ must be equal to the dimensions of $t^2$.Therefore,$[a] = [T^2]$.Now,the dimensions of the entire expression $\frac{a - t^2}{bx}$ must be equal to the dimensions of pressure $P$.$[P] = [ML^{-1} T^{-2}]$.So,$[P] = \frac{[a]}{[b][x]} \implies [ML^{-1} T^{-2}] = \frac{[T^2]}{[b][L]}$.Rearranging for $[b]$,we get $[b] = \frac{[T^2]}{[ML^{-1} T^{-2}][L]} = \frac{[T^2]}{[MT^{-2}]} = [M^{-1} T^4]$.Finally,the dimensions of $\frac{a}{b}$ are $\frac{[T^2]}{[M^{-1} T^4]} = [M^1 L^0 T^{-2}]$.

Write the dimensions of $a / b$ in the relation $P = \frac{a - t^2}{bx}$; where $P$ is the pressure,$x$ is the distance,and $t$ is the time.

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