Question

The time period of the inertial oscillation at a location R is 1.5 times that of a particle moving at a speed of 0.5 m/s at a location S (87°E, 45°S). Which of the following is the latitude of location R? (Round off to the nearest integer)

• A. 45°N
• B. 67°S
• C. 28°N
• D. 50°S

Hint:

The Coriolis parameter depends on the latitude, with higher latitudes having a larger Coriolis effect. The time period of inertial oscillations is inversely related to this parameter.

Answer 1

The Correct Option is C

The time period of the inertial oscillation \( T \) is related to the latitude of the location. For a particle moving at a velocity \( v \), the time period of the inertial oscillation is given by: \[ T = \frac{2\pi}{f} \] where \( f \) is the Coriolis parameter, given by: \[ f = 2 \cdot \Omega \cdot \sin(\phi) \] where \( \Omega \) is the angular velocity of the Earth and \( \phi \) is the latitude. The time period is inversely proportional to the Coriolis parameter. Given that the time period at location R is 1.5 times the time period at location S, we have: \[ \frac{T_R}{T_S} = 1.5 \] Since the time period is inversely proportional to the Coriolis parameter, this gives: \[ \frac{f_S}{f_R} = 1.5 \] Using the formula for \( f \), we can substitute the latitudes of locations R and S into the equation. After calculating the latitudes, we find that the latitude of location R is 28°N.