Discussion Forum



1)

The velocity of water in a river is 18 km/hr near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water.-The co-efficient of viscosity of water = 10-2 poise.

 


A) $10^{-1}N/m^{2}$

B) $10^{-2}N/m^{2}$

C) $10^{-3}N/m^{2}$

D) $10^{-4}N/m^{2}$

Answer:

Option B

Explanation:

$\eta = 10^{-2}$ poise

v = 18 Km/hr = $\frac{18000}{3600}$ = 5m/s

l = 5m

Strain rate = $\frac{v}{l}$

Coefficient of viscosity,

$\eta=\frac{Shearing stress}{strain rate}$

.'. Shearing stress = $\eta\times Strain rate$

= $10^{-2}\times\frac{5}{5}$ = 10-2 N/m2