Question

A column of air mass extending from surface to a height of \(10~\mathrm{km}\) moving eastward along \(30^\circ\mathrm{N}\) strikes a north–south oriented mountain range. While crossing the range, the air mass acquires a relative vorticity of \(-3.65\times 10^{-5}~\mathrm{s^{-1}}\) at the top. If the air mass remains at the same latitude and conserves potential vorticity, the height of the mountain range is \(__________\) km. (Round off to the nearest integer) \textit{[Assume Earth’s angular velocity \(\Omega=7.3\times10^{-5}~\mathrm{s^{-1}}\) and initial relative vorticity is zero.]}

Hint:

With PV \(q=(f+\zeta)/H\) conserved at fixed latitude, column compression (\(H\downarrow\)) makes \(\zeta\) negative (anticyclonic) to keep \(q\) constant; stretching does the opposite.

Answer 1

Correct Answer: 5

Step 1: Potential vorticity (PV) conservation for a column.
For a shallow air column of thickness \(H\) at a fixed latitude, PV is \[ q=\frac{f+\zeta}{H}=\text{constant}, \] where \(f=2\Omega\sin\phi\) is the Coriolis parameter and \(\zeta\) is the relative vorticity.

Step 2: Initial and final states.
Latitude \(30^\circ\Rightarrow f=2\Omega\sin30^\circ=2(7.3\times10^{-5})(0.5)=7.3\times10^{-5}~\mathrm{s^{-1}}\). Initially: \(\zeta_0=0\), \(H_0=10~\mathrm{km}\Rightarrow q_0=f/H_0\). Over the mountain: \(\zeta_1=-3.65\times10^{-5}~\mathrm{s^{-1}}\), column thickness \(H_1\) (reduced by the terrain height).

Step 3: Apply PV conservation at same latitude.
\[ \frac{f}{H_0}=\frac{f+\zeta_1}{H_1} \Rightarrow H_1=H_0\,\frac{f+\zeta_1}{f} =10\,\frac{7.3-3.65}{7.3} =10\times \frac{3.65}{7.3} =10\times 0.5=5~\mathrm{km}. \]

Step 4: Mountain height.
Height \(=H_0-H_1=10-5=5~\mathrm{km}\). Final Answer:
\[ \boxed{5~\text{km}} \]