Question

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 cms . The coefficient of viscosity of the liquid. dont spam plzzz​

Answer 1

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 $\eta$.

So, we know that the general formula for force applied to plates in a viscous liquid is:

$\boxed{ \sf{force = \eta \times A \times \dfrac{dv}{dx} }}$

"A" refers to area of plate, dv/dx refers to velocity gradient.

Putting all the available values in SI units:

$\sf{ = > force = \eta \times {10}^{ - 1} \times \dfrac{0.06}{6 \times {10}^{ - 3} } }$

$\sf{ = > force = \eta \times {10}^{ - 1} \times \dfrac{6 \times {10}^{ - 2} }{6 \times {10}^{ - 3} } }$

$\sf{ = > force = \eta \times \cancel{{10}^{ - 1} } \times \dfrac{6 \times \cancel{ {10}^{ - 2} }}{6 \times \cancel{ {10}^{ - 3}} } }$

$\sf{ = > force = \eta \times \dfrac{6 }{6 } }$

$\sf{ = > {10}^{ - 2} = \eta \times \dfrac{6 }{6 } }$

$\sf{ = > {10}^{ - 2} = \eta \times \dfrac{ \cancel6 }{ \cancel6 } }$

$\sf{ = > \: \eta = {10}^{ - 2} \: poise }$

So, final answer is :

$\boxed{ \red{ \large{\rm{ \: \eta = {10}^{ - 2} \: poise }}}}$

Answer 2

0.1poise

η=FA(dv/dy)

η=10−210−3×10−4(6×10−46×10−3)