According to Newton viscus force is given by F=-η A (d v/d x) where η= coefficient of viscosity, so dimensions of η will be :

Question

According to Newton viscus force is given by F=-η A (d v/d x) where η= coefficient of viscosity, so dimensions of η will be : (A)  [ML-1T-2] (B)  [MLT-2] (C)  [ML-1T-1] (D)  [M-1L2T-2]

Tardigrade Admin · Accepted Answer

F=-η A (d v/d x) ∴ η=-(F/A) (d x/d v) Writing the dimensions [η] =([ MLT -2]/[ L 2]) ([ L ]/[ LT -1]) =[ ML -1 T -1]

Q. According to Newton viscus force is given by
$F = - η A \frac{d v}{d x}$
where $η =$ coefficient of viscosity, so dimensions of $η$ will be :

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

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

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

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

$F = - η A \frac{d v}{d x}$
$∴ η = - \frac{F}{A} \frac{d x}{d v}$
Writing the dimensions
$[η] = \frac{[ M L T ^{- 2} ]}{[ L ^{2} ]} \frac{[ L ]}{[ L T ^{- 1} ]}$
$= [M L^{- 1} T^{- 1}]$

Q. According to Newton viscus force is given by F=−ηAdxdv​ where η= coefficient of viscosity, so dimensions of η will be :

[ML−1T−2]

[MLT−2]

[ML−1T−1]

[M−1L2T−2]