Given data

$M=1.00\,\text{kg},\quad L=0.20\,\text{m},\quad m=0.10\,\text{kg},\quad u=5.00\,\text{m s}^{-1}$

The particle hits the rod at distance

$r=\frac{L}{2}=0.10\,\text{m}$

The rod is pivoted at one end and rotates freely about that pivot.

Moment of inertia of the rod about the pivot

For a uniform rod about one end,

$I=\frac{1}{3}ML^2=\frac{1}{3}(1.00)(0.20)^2=\frac{0.04}{3}=0.01333\,\text{kg m}^2$

Use conservation of angular momentum about the pivot

During collision, external impulse at pivot has zero moment about the pivot, so angular momentum about pivot is conserved.

Initial angular momentum of the small mass:

$L_i=mur=(0.10)(5.00)(0.10)=0.05\,\text{kg m}^2\text{/s}$

After collision, the particle returns back along the same path, so its final velocity is opposite in direction. Taking initial direction as positive,

$v_f=-v$

Hence final angular momentum of particle is

$L_{m,f}=-mvr$

Rod's angular momentum after collision:

$L_{rod}=I\omega$

Therefore,

$mur=-mvr+I\omega$

$0.05=-0.1\times v\times 0.1+0.01333\,\omega$

$0.05=-0.01v+0.01333\omega \qquad (1)$

Use conservation of kinetic energy

Collision is elastic, so total kinetic energy is conserved.

Initial KE:

$K_i=\frac12 mu^2=\frac12(0.10)(5.00)^2=1.25\,\text{J}$

Final KE:

$K_f=\frac12 mv^2+\frac12 I\omega^2$

So,

$1.25=\frac12(0.10)v^2+\frac12(0.01333)\omega^2$

$1.25=0.05v^2+0.006667\omega^2 \qquad (2)$

Solve equations

From (1):

$0.01333\omega=0.05+0.01v$

$\omega=\frac{0.05+0.01v}{0.01333}=3.75+0.75v$

Now test the options.

Option A: $v=4.30$ m/s

$\omega=3.75+0.75(4.30)=3.75+3.225=6.975\approx 6.98\,\text{rad/s}$

This matches the stated $\omega$.

Check energy:

$K_f=0.05(4.30)^2+0.006667(6.975)^2$

$=0.05(18.49)+0.006667(48.65)$

$=0.9245+0.3243=1.2488\approx 1.25\,\text{J}$

So Option A is correct.

Option B: $v=4.30$ gives $\omega=6.98$, not $3.75$ rad/s. Incorrect.

Option C: $v=10.0$ gives

$\omega=3.75+0.75(10)=11.25\,\text{rad/s}$

not $3.75$ rad/s. Incorrect.

Option D: $v=4.10$ gives

$\omega=3.75+0.75(4.10)=6.825\approx 6.83\,\text{rad/s}$

close to $6.80$, but now check energy:

$K_f=0.05(4.10)^2+0.006667(6.825)^2$

$=0.8405+0.3107=1.1512\,\text{J}$

which is not equal to $1.25$ J. So incorrect.

Final answer

The correct option is:

$\boxed{\text{A: } \omega=6.98\,\text{rad s}^{-1},\; v=4.30\,\text{m s}^{-1}}$

Comparison with stored correct answer

Stored correct answer: A

My derived answer also: A

Hence, they agree.