In the equation,pressure P = frac{c - t^{2}}{ | vedclass

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(A) Given the equation $P = \frac{c - t^{2}}{DS}$.According to the principle of homogeneity of dimensions,the dimensions of $c$ must be equal to the d

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$[L^{0} M^{-1} T^{2}]$

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(A) Given the equation $P = \frac{c - t^{2}}{DS}$.According to the principle of homogeneity of dimensions,the dimensions of $c$ must be equal to the dimensions of $t^{2}$.Therefore,$[c] = [t^{2}] = [T^{2}]$.Now,the dimension of pressure $P$ is $[P] = [M L^{-1} T^{-2}]$.Given $S$ is distance,so $[S] = [L]$.Substituting these into the equation $P = \frac{c}{DS} - \frac{t^{2}}{DS}$,we consider the term $\frac{c}{DS}$:$[P] = \frac{[c]}{[D][S]} \Rightarrow [M L^{-1} T^{-2}] = \frac{[T^{2}]}{[D][L]}$.Solving for $[D]$:$[D] = \frac{[T^{2}]}{[M L^{-1} T^{-2}][L]} = \frac{[T^{2}]}{[M T^{-2}]} = [M^{-1} T^{4}]$.We need to find the dimensions of $\left(\frac{D}{c}\right)$:$\left[\frac{D}{c}\right] = \frac{[M^{-1} T^{4}]}{[T^{2}]} = [M^{-1} T^{2}] = [L^{0} M^{-1} T^{2}]$.

In the equation,pressure $P = \frac{c - t^{2}}{DS}$,where $S$ and $t$ represent the distance and time respectively. The dimensions of $\left(\frac{D}{c}\right)$ are

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