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Physics Question on Dimensional Analysis

Pressure gradient has the same dimensions as that of

A

velocity gradient

B

potential gradient

C

energy gradient

D

None of these

Answer

None of these

Explanation

Solution

Pressure gradient = △p△x=N/m2m\frac{ \triangle p }{ \triangle x } = \frac{ N / m^2 }{ m } ∴ Dimensions of(△p△x)=[MLT−2]/[L2][L]\therefore \, Dimensions \, of \bigg( \frac{ \triangle p }{ \triangle x }\bigg) = \frac{ [ MLT^{ - 2} ] / [ L^2 ] }{ [ L ] } = [ML−2T−2][ ML^{ - 2} T^{ - 2} ] Dimension of velocity gradient = [−△v△x]=m/sm=[LT−1][l]\bigg[- \frac{ \triangle v }{ \triangle x }\bigg] = \frac{ m / s}{ m } = \frac{ [ LT^{ - 1} ] }{ [ l ] } = [M0L0T−1][ M^0 L^0 T^{ - 1 } ] Dimensions of potential gradient = (−△V△x)=△W/Q△x\bigg(- \frac{ \triangle V }{ \triangle x }\bigg) = \frac{ \triangle W / Q }{ \triangle x } = [MLT−2][L][AT][L]\frac{ [ MLT^{ - 2} ] [ L ] }{ [ AT ] [ L ] } = [MLT−3A−1][ MLT^{ - 3} A^{ - 1} ]b Energy gradient = △E△x=Nmm\frac{ \triangle E }{ \triangle x } = \frac{ Nm }{ m } ∴\therefore Dimensions of (△p△x)=[MLT−2]/[L][L]\bigg( \frac{ \triangle p }{ \triangle x }\bigg) = \frac{ [ MLT^{ - 2} ] / [ L] }{ [ L ] } = [MLT−2][ MLT^{ - 2} ] As observed from above results, we see that none of the dimensions are same as of pressure gradient.