If radius of earth becomes n times its present value, without change in mass, then duration of day becomes

Options

A. `(24)/(n^(2))`
B. `24n^(2)`
C. `24(1-(1)/(n^(2)))`
D. `24(1-n^(2))`

Answer

To solve the problem of how the duration of a day changes if the radius of the Earth becomes \( n \) times its present value without changing its mass, we can follow these steps:

### Step 1: Understand the relationship between angular momentum and moment of inertia
The angular momentum \( L \) of a rotating body is given by the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Calculate the initial moment of inertia
For a solid sphere (like the Earth), the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the Earth and \( r \) is its radius. Initially, we have:
\[ I_1 = \frac{2}{5} m r^2 \]

### Step 3: Calculate the final moment of inertia
If the radius of the Earth becomes \( n \) times its present value, the new radius is \( nr \). The moment of inertia now becomes:
\[ I_2 = \frac{2}{5} m (nr)^2 = \frac{2}{5} m n^2 r^2 \]

### Step 4: Apply the conservation of angular momentum
Since the mass of the Earth does not change and there are no external torques acting on it, angular momentum is conserved:
\[ L_1 = L_2 \]
This gives us:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Substituting the expressions for \( I_1 \) and \( I_2 \):
\[ \frac{2}{5} m r^2 \omega_1 = \frac{2}{5} m n^2 r^2 \omega_2 \]
We can cancel out the common terms:
\[ \omega_1 = n^2 \omega_2 \]

### Step 5: Relate angular velocity to the period of rotation
The period \( T \) of rotation is related to angular velocity by the formula:
\[ T = \frac{2\pi}{\omega} \]
Thus, the initial period \( T_1 \) is:
\[ T_1 = \frac{2\pi}{\omega_1} \]
And the final period \( T_2 \) is:
\[ T_2 = \frac{2\pi}{\omega_2} \]

### Step 6: Substitute for \( \omega_2 \) in terms of \( \omega_1 \)
From the conservation of angular momentum, we have:
\[ \omega_2 = \frac{\omega_1}{n^2} \]
Substituting this into the expression for \( T_2 \):
\[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{\frac{\omega_1}{n^2}} = 2\pi \cdot \frac{n^2}{\omega_1} = n^2 \cdot T_1 \]

### Step 7: Calculate the new duration of the day
Given that the current duration of the day \( T_1 \) is 24 hours:
\[ T_2 = n^2 \cdot 24 \text{ hours} \]

### Conclusion
The new duration of the day becomes \( n^2 \times 24 \) hours.

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