A homogenous rod of length `l=etax` and mass `M` is lying on a smooth horizontal floor. A bullet of mass m hits the rod at a distance `x` from the middle of the rod at a velocity `v_(0)` perpendicular to the rod and comes to rest after collision. If the velocity of the farther end of the rod just after the impact is in the opposite direction of `v_(0)`, then:

Options

A. `etagt3`
B. `etalt3`
C. `etagt2`
D. `etalt6`

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the System
We have a homogeneous rod of length \( l = \eta x \) and mass \( M \) lying on a smooth horizontal floor. A bullet of mass \( m \) strikes the rod at a distance \( x \) from the middle of the rod with a velocity \( v_0 \) perpendicular to the rod.

### Step 2: Determine the Impact
When the bullet hits the rod, it comes to rest after the collision. The rod will start rotating about its center of mass due to the impact. The distance from the center of the rod to the point of impact is \( x \).

### Step 3: Analyze the Motion of the Rod
After the impact, the rod will have both translational and rotational motion. The center of mass of the rod will move, and the rod will also rotate about its center of mass.

### Step 4: Calculate the Linear Momentum Change
The impulse imparted by the bullet is equal to the change in linear momentum of the rod. Since the bullet comes to rest, the change in momentum for the bullet is:
\[ \Delta p = mv_0 \]
This impulse will cause the rod to gain some linear momentum.

### Step 5: Calculate the Angular Impulse
The angular impulse about the center of mass due to the bullet striking at a distance \( x \) from the center is given by:
\[ \text{Angular Impulse} = \text{Impulse} \times \text{Distance} = mv_0 \cdot x \]

### Step 6: Relate Linear and Angular Quantities
Let \( \omega_0 \) be the angular velocity of the rod after the impact. The moment of inertia \( I \) of the rod about its center is:
\[ I = \frac{1}{12}ML^2 = \frac{1}{12}M(\eta x)^2 \]
The relationship between linear velocity \( V \) of the center of mass and angular velocity \( \omega_0 \) is:
\[ V = \omega_0 \cdot \frac{L}{2} = \omega_0 \cdot \frac{\eta x}{2} \]

### Step 7: Set Up the Equations
From the conservation of linear momentum:
\[ mv_0 = MV \]
Substituting for \( V \):
\[ mv_0 = M\left(\omega_0 \cdot \frac{\eta x}{2}\right) \]

From the angular impulse:
\[ mv_0 \cdot x = I \cdot \omega_0 \]
Substituting for \( I \):
\[ mv_0 \cdot x = \left(\frac{1}{12}M(\eta x)^2\right) \cdot \omega_0 \]

### Step 8: Solve for \( \omega_0 \)
From the angular impulse equation:
\[ \omega_0 = \frac{12mv_0 \cdot x}{M(\eta x)^2} \]

### Step 9: Substitute and Simplify
Now substituting \( \omega_0 \) back into the linear momentum equation:
\[ mv_0 = M\left(\frac{12mv_0 \cdot x}{M(\eta x)^2} \cdot \frac{\eta x}{2}\right) \]
This simplifies to:
\[ mv_0 = \frac{12mv_0 \cdot \eta x}{2(\eta x)^2} \]
Cancelling \( mv_0 \) (assuming \( v_0 \neq 0 \)):
\[ 1 = \frac{6\eta}{\eta^2} \]
Rearranging gives:
\[ \eta^2 - 6\eta < 0 \]
Factoring gives:
\[ \eta(\eta - 6) < 0 \]
Thus, \( \eta < 6 \).

### Conclusion
The condition for \( \eta \) is:
\[ \eta < 6 \]