Question

Dimensional formula of eta

Answer 1

Answer:

F = tangential force , area , R = distance between the layers , velocity . Kinematic viscosity is divided by density is the formula of eta .

I hope this helps you please mark it as brainliest answer.

Answer 2

$\: \: \: \: \: \: \bf{ \large{ \underline{ \: \: \: Defination \: of \: \eta : \: \: \: }}}$

If A = 1 and Δv/Δz = 1, then $\eta$= F . Thus the coefficient of viscosity of a liquid is defined as the viscous force per unit area of content between two layers having a unit with velocity gradients between them.

$\bigstar \textbf{ \underline{Dimention of coefficient of viscosity }}$

$\sf{ \eta = \dfrac{F}{A \bigg(\dfrac{Δv}{Δz}\bigg) } } \\ \\ \rightarrow \sf{F = [ MLT^{ - 2} ]} \\ \rightarrow \sf{A =[L^{2} ] } \\ \rightarrow \sf{\bigg(\dfrac{Δv}{Δz}\bigg) =[T^{ - 1} ] } \\ \\ \bf{ \eta = \dfrac{F}{A \bigg(\dfrac{Δv}{Δz}\bigg) } } \\ \\ \sf{ \eta = \frac{[M L T^{ - 2} ]}{[M^{0} L^{2} T^{0} ][M^{0} L^{0} T^{ - 1} ] } } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \eta = [M L^{ - 1} T^{ - 1} ] }}$

$\bf{ SI \: unit \: of \: \eta }= \sf{kg {m}^{ - 1}{s}^{ - 1}}$