Topic: Fluid Mechanics
A thin plane Lamina, of Area A and weight W, slides down a fixed plane inclined at an angle $\alpha$ and maintains a uniform gap $\epsilon$ from the surface of the plane, the gap being filled with oil of constant viscosity $\mu$. The terminal velocity of the plane is:
- A. $W\epsilon \cos\alpha/\mu A$
- B. $W\epsilon \sin\alpha/\mu A$
- C. $\mu \epsilon \cos\alpha / AW$
- D. $\mu \epsilon \sin\alpha / AW$
Answer: B. $W\epsilon \sin\alpha/\mu A$
Explanation: The downward component of the weight of the lamina along the inclined plane is $W\sin\alpha$.
The viscous drag force resisting the motion is given by Newton's law of viscosity: $F_v = \tau A = \mu \left(\frac{V}{\epsilon}\right) A$, where V is the terminal velocity.
At steady terminal velocity, downward force = drag force:
$W\sin\alpha = \mu \left(\frac{V}{\epsilon}\right) A \Rightarrow V = \frac{W\epsilon \sin\alpha}{\mu A}$.