A rod of mass 𝑚 and length 𝐿, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed 𝑣 strikes the rod horizontally at a distance 𝑥 from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $$\omega$$ about the pivot. The maximum angular speed $$\omega_M$$ is achieved for $$x = x_M$$. Then

The bullet-rod collision is very quick, so the external impulse torque about the pivot is negligible. Hence the angular momentum about the pivot is conserved during the impact.

Step 1 Initial angular momentum of the bullet
The bullet of mass $$m$$ travels horizontally with speed $$v$$ and hits the rod at a perpendicular distance $$x$$ from the pivot.
Magnitude of angular momentum about the pivot:
$$L_i = m v x$$

Step 2 Moment of inertia of the combined system after the impact
• Uniform rod of mass $$m$$, length $$L$$, about one end: $$I_{\text{rod}} = \frac{1}{3} m L^{2}$$.
• Bullet embedded at distance $$x$$ from the pivot: $$I_{\text{bullet}} = m x^{2}$$.
Total moment of inertia:
$$I = I_{\text{rod}} + I_{\text{bullet}} = \frac{1}{3} m L^{2} + m x^{2}$$

Step 3 Angular speed immediately after the collision
Conserving angular momentum, $$L_i = I \omega$$:
$$m v x \;=\; \Bigl(\frac{1}{3} m L^{2} + m x^{2}\Bigr)\,\omega$$
Cancelling $$m$$ and rearranging,
$$\omega = \frac{v x}{\dfrac{1}{3} L^{2} + x^{2}} = \frac{3 v x}{L^{2} + 3 x^{2}}$$

This matches Option A.

Step 4 Finding the value of $$x$$ that maximises $$\omega$$
Write $$\omega(x)=\dfrac{3 v x}{L^{2}+3x^{2}}$$.
Differentiate with respect to $$x$$ and set to zero:
$$\frac{d\omega}{dx} = \frac{3v\,(L^{2}+3x^{2}) - 3v x\,(6x)}{(L^{2}+3x^{2})^{2}} = \frac{3v\,(L^{2} - 3x^{2})}{(L^{2}+3x^{2})^{2}} = 0$$
Hence $$L^{2} - 3x^{2} = 0 \;\Rightarrow\; x = \frac{L}{\sqrt{3}}$$

This gives Option C.

Step 5 Maximum angular speed $$\omega_M$$
Substitute $$x_M = \dfrac{L}{\sqrt{3}}$$ into the expression for $$\omega$$:
Numerator: $$3 v x_M = 3 v \frac{L}{\sqrt{3}} = v L \sqrt{3}$$.
Denominator: $$L^{2} + 3 x_M^{2} = L^{2} + 3 \left(\frac{L^{2}}{3}\right) = L^{2} + L^{2} = 2L^{2}$$.
Therefore
$$\omega_M = \frac{v L \sqrt{3}}{2 L^{2}} = \frac{v}{2L}\sqrt{3}$$

This corresponds to Option D.

Step 6 Checking the remaining option
Option B gives $$\omega = \dfrac{12 v x}{L^{2}+12x^{2}}$$, which does not agree with the derived expression, so it is incorrect.

Final result
Correct options: Option A, Option C, Option D.