If the radius of earth is reduced to three-fourth of its present value without change in its mass then value of duration of the day of earth will be ______ hours 30 minutes.
Correct Answer: 13
Approach Solution - 1
By conservation of angular momentum:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Moment of inertia of a sphere:
\[ I = \frac{2}{5} M R^2 \]
Angular velocity relation:
\[ \omega = \frac{2 \pi}{T} \]
\[ \frac{2}{5} M R^2 \cdot \frac{2 \pi}{T_1} = \frac{2}{5} M \left( \frac{3}{4} R \right)^2 \cdot \frac{2 \pi}{T_2} \]
Simplifying:
\[ \frac{1}{T_1} = \frac{9}{16} \cdot \frac{1}{T_2} \]
\[ T_2 = \frac{16}{9} \cdot T_1 \]
Substituting $T_1 = 24$ hours:
\[ T_2 = \frac{9}{16} \cdot 24 = \frac{27}{2} = 13 \, \text{hours} \, 30 \, \text{minutes}. \]
Approach Solution -2
Step 1: Apply Conservation of Angular Momentum
Initial angular momentum \( L \) must equal final angular momentum \( L' \):
\( L = L' \rightarrow I \cdot \omega = I' \cdot \omega' \).
Since radius reduces to \( \frac{3}{4}r \), moment of inertia becomes \( I' = \frac{2}{5} m \left(\frac{3}{4}r\right)^2 = \frac{9}{16}I \).
So, \( I \cdot \omega = \frac{9}{16}I \cdot \omega' \).
Thus, \( \omega' = \frac{16}{9} \omega \).
Step 2: Relating Angular Velocity and Time Period
The period \( T \) of Earth's rotation is related to angular velocity by:
\( \omega = \frac{2\pi}{T} \).
Therefore,
\( \omega' = \frac{2\pi}{T'} = \frac{16}{9} \omega \).
Hence, \( \frac{2\pi}{T'} = \frac{16}{9} \cdot \frac{2\pi}{T} \).
It leads to:
\( T' = \frac{9}{16} T \).
Given \( T = 24 \) hours, then:
\( T' = \frac{9}{16} \times 24 = 13.5 \) hours.
Final Check
The new duration of Earth's day is 13.5 hours.
Hence, the answer is 13 hours 30 minutes.
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