Question

If the diameter of the Earth is increased by 4% without changing the mass, then the length of the day is ...... hours.
(Take the length of the day before the increment as 24 hours. Assume the Earth to be a sphere with uniform density.)

Hint:

The period of rotation of a planet is related to its radius by the formula \( T \propto R^{3/2} \) for a sphere with uniform density.

Answer 1

Correct Answer: 25.95

Step 1: Use the formula for the moment of inertia.
The Earth is treated as a solid sphere, and its moment of inertia is given by: \[ I = \frac{2}{5} M R^2 \] where \( M \) is the mass of the Earth and \( R \) is its radius. The rotational period \( T \) is related to the moment of inertia by the following relation: \[ T = \frac{2 \pi}{\omega} \] where \( \omega \) is the angular velocity, and \( \omega \) is inversely proportional to the moment of inertia.
Step 2: Relation between radius and period.
Since the moment of inertia depends on \( R^2 \), and the angular velocity \( \omega \) is inversely proportional to \( I \), the period \( T \) is proportional to \( R^{3/2} \). Thus, we have: \[ \frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2} \] Given that \( R_2 = 1.04 R_1 \), the new period is: \[ \frac{T_2}{24} = (1.04)^{3/2} \] \[ T_2 = 24 \times (1.04)^{3/2} \approx 23.19 \, \text{hours} \]
Step 3: Conclusion.
Thus, the new length of the day is approximately 23.19 hours.