If earth were to shrink to `(1)/(8)th` of its present volume, what would be the new length of the day in hour ?

Explanation

To solve the problem of finding the new length of the day if the Earth were to shrink to \( \frac{1}{8} \) of its present volume, we can follow these steps:

### Step 1: Understand the relationship between volume and radius
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
If the volume of the Earth shrinks to \( \frac{1}{8} \) of its original volume, we can express this as:
\[ V' = \frac{1}{8} V \]
Substituting the volume formula, we have:
\[ \frac{4}{3} \pi r'^3 = \frac{1}{8} \left( \frac{4}{3} \pi r^3 \right) \]

### Step 2: Simplify the equation
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ r'^3 = \frac{1}{8} r^3 \]

### Step 3: Find the new radius
Taking the cube root of both sides, we find:
\[ r' = \frac{1}{2} r \]

### Step 4: Use the conservation of angular momentum
Angular momentum \( L \) is conserved in the absence of external torques. The angular momentum before shrinking is equal to the angular momentum after shrinking:
\[ L_i = L_f \]
For a solid sphere, angular momentum is given by:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The moment of inertia for a solid sphere is:
\[ I = \frac{2}{5} m r^2 \]
Thus, we have:
\[ \frac{2}{5} m r^2 \omega_i = \frac{2}{5} m (r')^2 \omega_f \]
Cancelling \( \frac{2}{5} m \) from both sides gives:
\[ r^2 \omega_i = (r')^2 \omega_f \]

### Step 5: Substitute for \( r' \)
Substituting \( r' = \frac{1}{2} r \) into the equation:
\[ r^2 \omega_i = \left(\frac{1}{2} r\right)^2 \omega_f \]
This simplifies to:
\[ r^2 \omega_i = \frac{1}{4} r^2 \omega_f \]
Cancelling \( r^2 \) from both sides (assuming \( r \neq 0 \)):
\[ \omega_i = \frac{1}{4} \omega_f \]

### Step 6: Relate the new angular velocity to the length of the day
The original angular velocity \( \omega_i \) corresponds to a period of 24 hours (the length of one day):
\[ \omega_i = \frac{2\pi}{T_i} = \frac{2\pi}{24 \text{ hours}} \]
From the relation \( \omega_i = \frac{1}{4} \omega_f \), we can express \( \omega_f \) as:
\[ \omega_f = 4 \omega_i \]
Thus, the new period \( T_f \) is given by:
\[ T_f = \frac{2\pi}{\omega_f} = \frac{2\pi}{4 \omega_i} = \frac{1}{4} T_i = \frac{1}{4} \times 24 \text{ hours} = 6 \text{ hours} \]

### Conclusion
The new length of the day, if the Earth shrinks to \( \frac{1}{8} \) of its present volume, would be **6 hours**.

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