Question

Consider a tornado in cyclostrophic balance. The tangential wind speed at a radial distance of 500 m from the center of the tornado is _________ \(\text{m/s}\), if the pressure gradient at that location in the radial direction is 5 N/m^3. Assume the density of air to be 1 kg/m^3 . (Round off to the nearest integer).

Hint:

Cyclostrophic balance occurs when the centrifugal force due to the tangential wind equals the radial pressure gradient force. Use \( v = \sqrt{r \cdot \frac{dP}{dr} \cdot \rho} \) for the tangential wind speed.

Answer 1

Correct Answer: 49

Cyclostrophic balance equation is:
\[ \frac{v^2}{r} = \frac{1}{\rho} \frac{dP}{dr} \] Where:
- \( v \) is the tangential wind speed (unknown)
- \( r = 500\ \text{m} \) (radial distance from the center of the tornado)
- \( \rho = 1\ \text{kg/m}^3 \) (density of air)
- \( \frac{dP}{dr} = 5\ \text{N/m}^3 \) (pressure gradient)
Rearrange for \( v \):
\[ v = \sqrt{r \cdot \frac{dP}{dr} \cdot \rho} \] Substitute the given values:
\[ v = \sqrt{500 \times 5 \times 1} \] \[ v = \sqrt{2500} = 50\ \text{m/s} \] Thus, the tangential wind speed is:
\[ \boxed{50\ \text{m/s}} \]