(D) The shearing stress $\tau$ is given by the formula: $\tau = \eta \left( \frac{dv}{dx} \right)$.Here,$\eta = 10^{-3} \ SI$ units is the coefficient

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(D) The shearing stress $\tau$ is given by the formula: $\tau = \eta \left( \frac{dv}{dx} \right)$.Here,$\eta = 10^{-3} \ SI$ units is the coefficient of viscosity.The velocity gradient $\frac{dv}{dx}$ is approximated as $\frac{v}{h}$,where $v = 10 \ ms^{-1}$ is the velocity at the surface and $h = 20 \ m$ is the depth.Substituting the values: $\tau = 10^{-3} \times \left( \frac{10}{20} \right)$.$\tau = 10^{-3} \times 0.5 = 0.5 \times 10^{-3} \ Nm^{-2}$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

Water in a river $20 \ m$ deep is flowing at a speed of $10 \ ms^{-1}$. The shearing stress between the horizontal layers of water in the river in $Nm^{-2}$ is (Coefficient of viscosity of water $= 10^{-3} \ SI$ units).

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A
$1 \times 10^{-2}$
B
$0.5 \times 10^{-2}$
C
$1 \times 10^{-3}$
D
$0.5 \times 10^{-3}$

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Water in a river $20 \ m$ deep is flowing at a speed of $10 \ ms^{-1}$. The shearing stress between the horizontal layers of water in the river in $Nm^{-2}$ is (Coefficient of viscosity of water $= 10^{-3} \ SI$ units).

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