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`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
The problem states that the Earth shrinks such that its density becomes 8 times its current value. We need to find the new duration of a day in hours.

### Step 2: Conservation of Mass
Since mass is conserved, we can express the mass of the Earth before and after the change in density. The mass (M) can be expressed as:
\[ M = \text{Density} \times \text{Volume} \]

Let the initial density be \(\rho\) and the initial radius be \(r_1\). The initial volume \(V_1\) is given by:
\[ V_1 = \frac{4}{3} \pi r_1^3 \]
Thus, the initial mass is:
\[ M = \rho \cdot \frac{4}{3} \pi r_1^3 \]

After the Earth shrinks, the new density becomes \(8\rho\) and let the new radius be \(r_2\). The new volume \(V_2\) is:
\[ V_2 = \frac{4}{3} \pi r_2^3 \]
So the new mass is:
\[ M = 8\rho \cdot \frac{4}{3} \pi r_2^3 \]

### Step 3: Set the Masses Equal
Since the mass is conserved:
\[ \rho \cdot \frac{4}{3} \pi r_1^3 = 8\rho \cdot \frac{4}{3} \pi r_2^3 \]

We can cancel \(\rho\) and \(\frac{4}{3} \pi\) from both sides:
\[ r_1^3 = 8 r_2^3 \]

### Step 4: Solve for the Radii
Taking the cube root of both sides gives:
\[ r_1 = 2 r_2 \]

### Step 5: Conservation of Angular Momentum
The angular momentum before and after the change must also be conserved. The angular momentum \(L\) is given by:
\[ L = I \omega \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

For a solid sphere, the moment of inertia \(I\) is:
\[ I = \frac{2}{5} M r^2 \]

For the initial state:
\[ I_1 = \frac{2}{5} M r_1^2 \]
And the angular velocity \(\omega_1\) is:
\[ \omega_1 = \frac{2\pi}{T_1} \]

For the new state:
\[ I_2 = \frac{2}{5} M r_2^2 \]
And the angular velocity \(\omega_2\) is:
\[ \omega_2 = \frac{2\pi}{T_2} \]

### Step 6: Set Up the Angular Momentum Equation
Setting the angular momentum before and after equal gives:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Substituting the expressions for \(I\) and \(\omega\):
\[ \frac{2}{5} M r_1^2 \cdot \frac{2\pi}{T_1} = \frac{2}{5} M r_2^2 \cdot \frac{2\pi}{T_2} \]

### Step 7: Simplify the Equation
Canceling common terms:
\[ r_1^2 \cdot \frac{1}{T_1} = r_2^2 \cdot \frac{1}{T_2} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = \frac{r_2^2}{r_1^2} \]

### Step 8: Substitute the Radius Ratio
From earlier, we found \(r_1 = 2r_2\), so:
\[ \frac{T_2}{T_1} = \frac{r_2^2}{(2r_2)^2} = \frac{1}{4} \]

### Step 9: Calculate the New Period
If \(T_1 = 24\) hours (the current duration of a day), then:
\[ T_2 = \frac{T_1}{4} = \frac{24}{4} = 6 \text{ hours} \]

### Final Answer
Thus, the new duration of the day will be **6 hours**.

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