Viscous Drag and Shear Stress on Floating Thin Square Plate

To determine which statements are correct, we analyze the viscous flow of the liquid beneath the floating plate.

1. Velocity Profile and Velocity Gradient

Since the depth $h$ of the liquid layer is very small compared to the lateral dimensions of the plate, the fluid flow can be considered laminar and one-dimensional (Couette-like flow).

Taking the bottom floor of the tank at $y = 0$ (where velocity $u = 0$) and the floating plate at $y = h$ (moving with velocity $u = u_0$), the linear velocity profile across the liquid depth is given by: $u(y) = u_0 \frac{y}{h}$

The velocity gradient inside the liquid layer is: $\frac{du}{dy} = \frac{u_0}{h}$

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2. Shear Stress in the Liquid

By Newton's Law of Viscosity, the tangential (shear) stress $\tau$ at any height $y$ in the liquid layer is given by: $\tau = \eta \frac{du}{dy} = \eta \frac{u_0}{h}$

where $\eta$ is the dynamic viscosity of the liquid.

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3. Viscous Resistive Force on the Plate

The total resistive force $F$ exerted by the viscous fluid on the bottom surface of the plate of area $A$ is: $F = \tau \cdot A = \frac{\eta A u_0}{h}$

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4. Evaluation of the Statements

• Option (A): The resistive force $F = \frac{\eta A u_0}{h}$ is inversely proportional to the liquid depth $h$ ($F \propto \frac{1}{h}$).
  Statement (A) is correct.

• Option (B): The resistive force $F$ is directly proportional to the surface area $A$ of the plate ($F \propto A$), and thus it depends on the area.
  Statement (B) is incorrect.

• Option (C): The shear stress on the floor of the tank ($y = 0$) is $\tau_{\text{floor}} = \frac{\eta u_0}{h}$, which increases linearly with the speed $u_0$.
  Statement (C) is correct.

• Option (D): The shear stress on the plate ($y = h$) is $\tau_{\text{plate}} = \frac{\eta u_0}{h}$, which varies linearly with the viscosity $\eta$ of the fluid.
  Statement (D) is correct.

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Final Answer:

The correct statements are A, C, and D.