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Question: Calculate the dynamic viscosity of oil which is used for lubrication between a square plate of size ...

Calculate the dynamic viscosity of oil which is used for lubrication between a square plate of size 0.8m×0.8m0.8m\times 0.8m and an inclined plane with an angle of inclination 30o{{30}^{o}} as shown. The mass of the square plate is 30kg30kg and it slides down the inclined plane with a velocity of 0.3ms10.3m{{s}^{-1}}. The thickness of the oil film is 1.5mm1.5mm.
(A). μ=9.5poise\mu =9.5poise
(B). μ=11.7poise\mu =11.7poise
(C). μ=12.2poise\mu =12.2poise
(D). μ=13.2poise\mu =13.2poise

Explanation

Solution

Oil is used for lubrication between block and an inclined plane. According to Newton's second law, the block is moving due to the forces acting on it. Using the FBD diagram, we can resolve the forces. Drag will resist the motion of the block; hence it will act opposite to the force. The dynamic viscosity is related to velocity of the block, thickness of film and area of cross section.

Formulas used:
mgsinθ=drag(viscousforce)mg\sin \theta =drag(viscous\,force)
F=μvtAF=\mu \dfrac{v}{t}A

Complete step-by-step solution:

The above given figure is the FBD diagram for the block on the inclined plane. From the above figure we can see that,
mgsinθ=drag(viscousforce)mg\sin \theta =drag(viscous\,force)

The dynamic viscosity is the internal resistance provided by the fluid in the direction of flow. It is a constant quantity and the SI unit is Pascal-second (PsP-s).

Drag is the frictional force acting between a fluid and a surface moving through the fluid. It is the force acting on a body opposite to its motion relative to the fluid Its SI unit is Newton (NN). It is given by-
F=μvtAF=\mu \dfrac{v}{t}A
Here, μ\mu is the coefficient of viscosity
vv is the velocity of the body
tt is the thickness of fluid film
AA is the area of cross section

We substitute the given values in above equation to get,
F=μvtA mgsinθ=μ0.3×1021.5×101cm0.8×0.8×104cm2 30×103gm×10×1.5×1052×0.3×0.8×0.8=μ μ=11.7poise \begin{aligned} & F=\mu \dfrac{v}{t}A \\\ & \Rightarrow mg\sin \theta =\mu \dfrac{0.3\times {{10}^{-2}}}{1.5\times {{10}^{-1}}cm}0.8\times 0.8\times {{10}^{-4}}c{{m}^{2}} \\\ & \Rightarrow \dfrac{30\times {{10}^{-3}}gm\times 10\times 1.5\times {{10}^{5}}}{2\times 0.3\times 0.8\times 0.8}=\mu \\\ & \therefore \mu =11.7poise \\\ \end{aligned}

Therefore, the value of dynamic viscosity of the fluid is 11.7poise11.7poise.

Hence, the correct option is (B).

Note:
Dynamic viscosity is also known as the coefficient of viscosity. The dynamic viscosity acts between the layers of the fluid when they slide over each other. The drag is the frictional force acting on the block hence it is negative. We converted all units into CGS to calculate poise which is a CGS unit of dynamic viscosity.