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(A) The dimensional formula for the coefficient of viscosity $\eta$ is $[M^1 L^{-1} T^{-1}]$.Let the dimensional formula be expressed as $\eta = P^x A

Vedclass · Accepted Answer

$[PA^{-1}T^0]$

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(A) The dimensional formula for the coefficient of viscosity $\eta$ is $[M^1 L^{-1} T^{-1}]$.Let the dimensional formula be expressed as $\eta = P^x A^y T^z$.We know the dimensions of the fundamental quantities are:$[P] = [M^1 L^1 T^{-1}]$$[A] = [L^2]$$[T] = [T^1]$Substituting these into the equation:$[M^1 L^{-1} T^{-1}] = [M^1 L^1 T^{-1}]^x [L^2]^y [T^1]^z$$[M^1 L^{-1} T^{-1}] = M^x L^{x+2y} T^{-x+z}$Comparing the powers of $M, L,$ and $T$ on both sides:For $M: x = 1$For $L: x + 2y = -1 \implies 1 + 2y = -1 \implies 2y = -2 \implies y = -1$For $T: -x + z = -1 \implies -1 + z = -1 \implies z = 0$Thus,the dimensional formula is $[P^1 A^{-1} T^0]$.

If momentum $[P]$,area $[A]$,and time $[T]$ are taken as fundamental quantities,then the dimensional formula for the coefficient of viscosity is:

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