Question

The sea surface height concentric isolines (\(L1\) and \(L2\) in cm) and the distance between them (\(dx\) in km) for three different eddies at the same latitude are given in the figure.

Which one of the following orders is correct about the magnitudes of the geostrophic currents within the isolines?

• A. (i)>(ii)>(iii)
• B. (i)>(iii)>(i)
• C. (ii)>(i)>(iii)
• D. (iii)>(ii)>(i)

Hint:

For geostrophic current strength, always compare \(\Delta h / \Delta x\). Larger sea surface slope → stronger geostrophic flow.

Answer 1

The Correct Option is C

Step 1: Geostrophic current formula.
Geostrophic speed is proportional to the horizontal pressure gradient, here represented by the slope of sea surface height: \[ V_g \propto \frac{\Delta h}{\Delta x} \] where \(\Delta h\) is the height difference between isolines, and \(\Delta x\) is their separation.

Step 2: Compute slope for each case.
- Case (i): \(L1=20,\ L2=30,\ \Delta h=10~\mathrm{cm}=0.1~\mathrm{m},\ \Delta x=200~\mathrm{km}=2\times 10^5~\mathrm{m}\). \[ \frac{\Delta h}{\Delta x} = \frac{0.1}{2\times 10^5} = 5\times 10^{-7}. \] - Case (ii): \(L1=10,\ L2=20,\ \Delta h=10~\mathrm{cm}=0.1~\mathrm{m},\ \Delta x=300~\mathrm{km}=3\times 10^5~\mathrm{m}\). \[ \frac{\Delta h}{\Delta x} = \frac{0.1}{3\times 10^5} = 3.3\times 10^{-7}. \] - Case (iii): \(L1=5,\ L2=15,\ \Delta h=10~\mathrm{cm}=0.1~\mathrm{m},\ \Delta x=100~\mathrm{km}=1\times 10^5~\mathrm{m}\). \[ \frac{\Delta h}{\Delta x} = \frac{0.1}{1\times 10^5} = 1.0\times 10^{-6}. \]

Step 3: Compare values.
- (iii): \(1.0\times 10^{-6}\) (largest)
- (i): \(5.0\times 10^{-7}\) (second)
- (ii): \(3.3\times 10^{-7}\) (smallest) So the order is: (iii)>(i)>(ii). Final Answer:
\[ \boxed{(iii)>(i)>(ii)} \]