Question

The zonal gradient of meridional current and the meridional gradient of zonal current is \(-0.3 \times 10^{-3}~\mathrm{s^{-1}}\) and \(0.3 \times 10^{-3}~\mathrm{s^{-1}}\), respectively, at a location P (87°E, 15°N). Which one of the following best explains the nature of the flow?

• A. The flow is non-divergent in nature.
• B. The flow is non-rotational in nature.
• C. The flow is counter-clockwise in nature.
• D. The flow is clockwise in nature.

Hint:

Remember: \(\zeta = \partial v/\partial x - \partial u/\partial y\). Positive vorticity = counter-clockwise; Negative vorticity = clockwise (in the Northern Hemisphere).

Answer 1

The Correct Option is B

Step 1: Recall vorticity formula.
Relative vorticity of a horizontal flow field is: \[ \zeta = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}, \] where \(u\) is the zonal (eastward) velocity and \(v\) is the meridional (northward) velocity.

Step 2: Substitute given gradients.
- Zonal gradient of meridional current: \(\frac{\partial v}{\partial x} = -0.3 \times 10^{-3}~\mathrm{s^{-1}}\).
- Meridional gradient of zonal current: \(\frac{\partial u}{\partial y} = 0.3 \times 10^{-3}~\mathrm{s^{-1}}\). So, \[ \zeta = (-0.3 \times 10^{-3}) - (0.3 \times 10^{-3}) = -0.6 \times 10^{-3}~\mathrm{s^{-1}}. \]

Step 3: Interpret sign of vorticity.
- Positive \(\zeta\) = counter-clockwise rotation (Northern Hemisphere).
- Negative \(\zeta\) = clockwise rotation. But note: in this question, the given values suggest divergence structure as well. Let’s carefully re-check. Actually, the correct expression for vorticity is: \[ \zeta = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}. \] Plugging in: \[ \zeta = (-0.3 \times 10^{-3}) - (0.3 \times 10^{-3}) = -0.6 \times 10^{-3}. \] This is negative, which corresponds to clockwise motion.

Step 4: Correction and final interpretation.
Therefore, the flow is actually clockwise. Final Answer:
\[ \boxed{\text{The flow is clockwise in nature.}} \]