If the earth shrinks such that its density becomes `8` times to the present values, then new duration of the day in hours will be

Options

A. 24
B. 12
C. 6
D. 3

Answer

To solve the problem of determining the new duration of the day if the Earth's density becomes 8 times its current value, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between mass, density, and volume**:
- The mass of the Earth is given by the formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
- Let the initial density be \(\rho_i\) and the initial volume be \(V_i\). The final density is \(\rho_f = 8 \rho_i\), and the final volume is \(V_f\).

2. **Express the volumes**:
- The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the initial volume \(V_i\) can be expressed as:
\[ V_i = \frac{4}{3} \pi r_i^3 \]
- The final volume \(V_f\) can be expressed as:
\[ V_f = \frac{4}{3} \pi r_f^3 \]

3. **Set up the conservation of mass equation**:
- According to the conservation of mass:
\[ \rho_i V_i = \rho_f V_f \]
- Substituting the expressions for density and volume:
\[ \rho_i \left(\frac{4}{3} \pi r_i^3\right) = 8 \rho_i \left(\frac{4}{3} \pi r_f^3\right) \]
- Canceling out \(\rho_i\) and \(\frac{4}{3} \pi\) gives:
\[ r_i^3 = 8 r_f^3 \]

4. **Solve for the radius**:
- Taking the cube root of both sides:
\[ r_i = 2 r_f \]

5. **Apply conservation of angular momentum**:
- The moment of inertia \(I\) for a sphere is given by:
\[ I = \frac{2}{5} m r^2 \]
- Using conservation of angular momentum:
\[ I_i \omega_i = I_f \omega_f \]
- Substituting for \(I\) and \(\omega\):
\[ \frac{2}{5} m r_i^2 \cdot \omega_i = \frac{2}{5} m r_f^2 \cdot \omega_f \]
- Canceling out common terms:
\[ r_i^2 \omega_i = r_f^2 \omega_f \]

6. **Substituting the radius**:
- Substitute \(r_i = 2 r_f\):
\[ (2 r_f)^2 \omega_i = r_f^2 \omega_f \]
- Simplifying gives:
\[ 4 r_f^2 \omega_i = r_f^2 \omega_f \]
- Canceling \(r_f^2\):
\[ 4 \omega_i = \omega_f \]

7. **Relate angular velocity to the period**:
- The angular velocity \(\omega\) is related to the period \(T\) by:
\[ \omega = \frac{2\pi}{T} \]
- Therefore:
\[ 4 \cdot \frac{2\pi}{T_i} = \frac{2\pi}{T_f} \]
- Canceling \(2\pi\) gives:
\[ 4 \cdot \frac{1}{T_i} = \frac{1}{T_f} \]

8. **Solving for the final period**:
- Rearranging gives:
\[ T_f = \frac{T_i}{4} \]
- The initial period \(T_i\) (the current duration of a day) is 24 hours:
\[ T_f = \frac{24 \text{ hours}}{4} = 6 \text{ hours} \]

### Final Answer:
The new duration of the day will be **6 hours**.