(C) The shearing stress $\tau$ is given by the formula: $\tau = \eta \frac{dv}{dy}$.Here,$\eta = 10^{-3}\,SI\,unit$ is the coefficient of viscosity.Th

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(C) The shearing stress $\tau$ is given by the formula: $\tau = \eta \frac{dv}{dy}$.Here,$\eta = 10^{-3}\,SI\,unit$ is the coefficient of viscosity.The velocity gradient $\frac{dv}{dy}$ can be approximated as $\frac{v}{h}$,where $v = 5\,m/s$ is the velocity and $h = 10\,m$ is the depth of the river.Substituting the values:$\tau = 10^{-3} \times \frac{5}{10}$$\tau = 10^{-3} \times 0.5$$\tau = 0.5 \times 10^{-3}\,N/m^2$.

Class 11 Physics Mechanical Properties of Fluids Newton's Law of Viscosity

Medium

$A$ river $10\,m$ deep is flowing at $5\,m/s$. The shearing stress between horizontal layers of the river is :- $(\eta = 10^{-3}\,SI\,unit)$

A
$10^{-3}\,N/m^2$
B
$0.8\times 10^{-3}\,N/m^2$
C
$0.5\times 10^{-3}\,N/m^2$
D
$1\,N/m^2$

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Mechanical Properties of Fluids

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Newton's Law of Viscosity

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$A$ river $10\,m$ deep is flowing at $5\,m/s$. The shearing stress between horizontal layers of the river is :- $(\eta = 10^{-3}\,SI\,unit)$

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