Question

The velocity of water in a river is 18km/hr near the surface. If the river is 5m deep, find the shearing stress between the horizontal layers of water. The coefficient 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 1

Hint: This problem can be solved directly using the formula for Shear Stress. Convert unit of viscosity from poise to Pa.s. and unit of velocity from $ { km }/{ hr }$ to $ { m }/{s}$. And then substitute those values into the formula for Shear Stress.

Formula used:
$Shear\quad Stress= \mu \dfrac { dv }{ dx }$

Complete step-by-step answer:
Given: Velocity of water (v)= $18{ km }/{ hr }$
            Coefficient viscosity of water ($\mu$)= ${ 10 }^{ -2 }$poise
            Distance (x)= 5m
Formula for Shear Stress is given by,
$Shear\ Stress= \mu \dfrac { dv }{ dx }$...(1)
To balance the dimensions, convert the unit of viscosity from poise to Pa.s
1 poise= 0.1 Pa.s
$\therefore { 10 }^{ -2 } poise= { 10 }^{ -3 }Pa.s$
Now, substituting values in equation.(1) we get,
$\therefore \tau = { 10 }^{ -3 } \times \dfrac { 18\times 1000 }{ 3600\times 5 }$
$\therefore \tau = { 10 }^{ -3 }\dfrac { N }{ { m }^{ 2 } }$
Hence, the correct option is C i.e. $ { 10 }^{ -3 }{ N }/{ { m }^{ 2 } }$

So, the correct answer is “Option C”.

Note: Make sure you balance the dimensions. The options have a unit of ${ N }/{ { m }^{ 2 } }$ while the answer which we get is in terms of poise. So conversion of the unit of viscosity becomes mandatory.
So either you can convert the units initially, substitute the values and get the answer or the alternate method is to substitute the values first and then do the conversion of units.