Which of the following equations is dimension | vedclass

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(A) To check dimensional correctness,we analyze the dimensions of both sides for each equation:$A$. Poiseuille's Law: $v = \frac{\pi p a^4}{8 \eta L}$

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$v = \frac{\pi p a^4}{8 \eta L}$

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(A) To check dimensional correctness,we analyze the dimensions of both sides for each equation:$A$. Poiseuille's Law: $v = \frac{\pi p a^4}{8 \eta L}$. Here,$v$ represents volume flow rate $(L^3 T^{-1})$. The $RHS$ dimensions are $[M L^{-1} T^{-2}][L^4] / ([M L^{-1} T^{-1}][L]) = [M L^3 T^{-2}] / [M T^{-1}] = [L^3 T^{-1}]$. This is dimensionally correct.$B$. Capillary rise: $h = \frac{2 s \cos 	heta}{ho r g}$. Dimensions: $[L] = [M T^{-2}] / ([M L^{-3}][L][L T^{-2}]) = [M T^{-2}] / [M T^{-2}] = [L]$. This is dimensionally correct.$C$. Displacement current density: $J = \varepsilon \frac{\partial E}{\partial t}$. Dimensions of $J$ are $[I L^{-2}]$. Dimensions of $\varepsilon \frac{\partial E}{\partial t}$ are $[M^{-1} L^{-3} T^4 I^2] \cdot ([M L T^{-3} I^{-1}] / [T]) = [I L^{-2}]$. This is dimensionally correct.$D$. Work done: $W = \Gamma 	heta$. Dimensions of $W$ are $[M L^2 T^{-2}]$. Dimensions of $\Gamma$ (torque) are $[M L^2 T^{-2}]$ and $	heta$ (angle) is dimensionless. Thus,$W = \Gamma$ is dimensionally correct,but the equation $W = \Gamma 	heta$ implies work equals torque times angle,which is dimensionally correct as $[M L^2 T^{-2}] = [M L^2 T^{-2}] \cdot [1]$. However,in the context of the provided options,all are dimensionally correct. Re-evaluating $A$: $v$ is volume,not volume flow rate. Thus,$A$ is dimensionally incorrect.

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