36. A metal plate of area 10*3 cm*2 rests on a layer of
oil 6 mm thick. A tangential force 10*-2N is applied
on it to move it with a constant velocity of 6 cms*-1
The coefficient of viscosity of the liquid is
(1) 0.1 poise
(2) 0.5 poise
(3) 0.7 poise
(4) 0.9 poise
Answers
Given:
A metal plate of area 10^3 cm^2 rests on a layer of oil 6 mm thick .A tangential force 10^-2 N is applied on it to move it with a constant velocity of 6 cm/s.
To find:
Coefficient of viscosity of liquid.
Calculation:
Let coefficient of viscosity be \etaη .
So, we know that the general formula for force applied to plates in a viscous liquid is:
\boxed{ \sf{force = \eta \times A \times \dfrac{dv}{dx} }}
force=η×A×
dx
dv
"A" refers to area of plate, dv/dx refers to velocity gradient.
Putting all the available values in SI units:
\sf{ = > force = \eta \times {10}^{ - 1} \times \dfrac{0.06}{6 \times {10}^{ - 3} } }=>force=η×10
−1
×
6×10
−3
0.06
\sf{ = > force = \eta \times {10}^{ - 1} \times \dfrac{6 \times {10}^{ - 2} }{6 \times {10}^{ - 3} } }=>force=η×10
−1
×
6×10
−3
6×10
−2
\sf{ = > force = \eta \times \cancel{{10}^{ - 1} } \times \dfrac{6 \times \cancel{ {10}^{ - 2} }}{6 \times \cancel{ {10}^{ - 3}} } }=>force=η×
10
−1
×
6×
10
−3
6×
10
−2
\sf{ = > force = \eta \times \dfrac{6 }{6 } }=>force=η×
6
6
\sf{ = > {10}^{ - 2} = \eta \times \dfrac{6 }{6 } }=>10
−2
=η×
6
6
\sf{ = > {10}^{ - 2} = \eta \times \dfrac{ \cancel6 }{ \cancel6 } }=>10
−2
=η×
6
6
\sf{ = > \: \eta = {10}^{ - 2} \: poise }=>η=10
−2
poise
So, final answer is :
\boxed{ \red{ \large{\rm{ \: \eta = {10}^{ - 2} \: poise }}}}
η=10
−2
poise