1. Magnetic flux and induced current during entry

While the loop is partially entering the region $y\ge 0$, the area inside the magnetic field increases.

If $x$ is the length of the loop that has entered the field, then $\Phi = B_0(Lx).$ Since $\dot x = v$, the induced emf is $\mathcal E = \frac{d\Phi}{dt} = B_0L v.$ Hence the induced current is $I=\frac{\mathcal E}{R}=\frac{B_0Lv}{R}.$

By Lenz's law, the magnetic force opposes the upward motion.

2. Retarding force on the loop

Only the horizontal segment inside the field experiences a net vertical magnetic force during entry. Its magnitude is $F = I L B_0 = \frac{B_0Lv}{R}\,LB_0 = \frac{B_0^2L^2}{R}v.$ This force is downward, so $M\frac{dv}{dt}=-\frac{B_0^2L^2}{R}v.$ Given $K=\frac{B_0^2L^2}{RM},$ we get $\frac{dv}{dt}=-Kv.$

So during entry, $v(t)=v_0 e^{-Kt}.$

3. Position during entry

Let $y(t)$ be the upward displacement of the loop into the field from the instant the top edge touches $y=0$. Then $\frac{dy}{dt}=v_0 e^{-Kt}.$ Integrating, $y(t)=\int_0^t v_0e^{-K\tau}d\tau=\frac{v_0}{K}\left(1-e^{-Kt}\right).$

The loop completely enters when $y=L.$ So complete entry is possible only if the maximum possible displacement satisfies $y_{\max}=\frac{v_0}{K}\ge L \quad\Rightarrow\quad v_0\ge KL.$

4. Check option A

Given $v_0=1.5KL$. Then $y_{\max}=\frac{1.5KL}{K}=1.5L > L.$ So the loop does not stop before entering completely; it does enter fully.

Therefore, A is false.

5. Check option B

When the complete loop is inside the uniform magnetic field, the flux through the loop is constant: $\Phi = B_0L^2 = \text{constant}.$ Thus, $\mathcal E = -\frac{d\Phi}{dt}=0,$ so no induced current flows: $I=0.$ Hence magnetic force on the loop is zero.

Therefore, B is true.

6. Check option C

Given $v_0=\dfrac{KL}{10}$. Then the maximum displacement is $y_{\max}=\frac{v_0}{K}=\frac{L}{10}.$ So the loop stops before fully entering.

To find when it comes to rest: from $v(t)=v_0e^{-Kt},$ velocity becomes zero only asymptotically as $t\to\infty$. It does not become exactly zero at any finite time.

The given time $t=\frac{1}{K}\ln\left(\frac52\right)$ is not a stopping time.

Therefore, C is false.

7. Check option D

Given $v_0=3KL$. For complete entry, set $L=\frac{v_0}{K}(1-e^{-Kt}).$ Substitute $v_0=3KL$: $L=\frac{3KL}{K}(1-e^{-Kt})=3L(1-e^{-Kt}).$ So $\frac13=1-e^{-Kt}$ $e^{-Kt}=\frac23$ $t=\frac{1}{K}\ln\left(\frac32\right).$ Thus D is true.

8. Final conclusion

Correct options are: $\boxed{B,\ D}$

This matches the stored correct answer.