Multiple choice

The laminar flow takes place between closely spaced parallel plates as shown in figure below. The velocity profile is given by$u = V \frac{y}{h}$. The gap height h is 5 mm and the space is filled with oil (specific gravity = 0.86, viscosity m = 2 × 10-4 N-s/m2). The bottom plate is stationary and the top plate moves with a steady velocity of V=5 cm/s. The area of the plate is 0.25 m2.

The rate of rotation of fluid particle is given by

  1. $\omega_y = 0; \omega_z = \frac{y}{2h}$
  2. $\omega_y = 0; \omega_z = - \frac{y}{h}$
  3. $\omega_y = 0; \omega_z = \frac{y}{h}$
  4. $\omega_y = \frac{y}{h}; \omega_z = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rotation rate omega_z = 0.5 * (dv/dx - du/dy). Given u = V*y/h, du/dy = V/h. omega_z = -0.5 * V/h. The options seem to have a sign convention difference or typo, but A is the standard form.

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