Understand the setup

• Two blocks are in contact on the incline: $m_1=1\,\text{kg}$ and $m_2=2\,\text{kg}$.
• Block $m_1$ is frictionless with the plane.
• Block $m_2$ has friction with the plane: $\mu_s=\mu_k=\mu=0.3$
• We need the friction on block $m_2$ for different angles $\theta$.

Since the blocks are together, we first check whether they can remain at rest.

Condition for complete equilibrium of the two-block system

Along the incline, the total downward component of weight is $(m_1+m_2)g\sin\theta$

The only external friction available to oppose this is on block $m_2$: $f_{\max}=\mu m_2 g\cos\theta$

For the system to remain at rest, $(m_1+m_2)g\sin\theta \le \mu m_2 g\cos\theta$ $3\sin\theta \le 0.6\cos\theta$ $\tan\theta \le 0.2$

Using the given values, this means equilibrium is possible for $\theta \lesssim 11.5^\circ$

So:

• For $\theta=5^\circ$ and $10^\circ$: system can stay at rest.
• For $\theta=15^\circ$ and $20^\circ$: system cannot stay at rest, so it slides down.

Case 1: $\theta=5^\circ$ and $10^\circ$ (rest possible)

If both blocks are at rest, then for the combined system friction must balance the total downslope weight component: $f=(m_1+m_2)g\sin\theta$

So for

• $P: \theta=5^\circ \Rightarrow f=(m_1+m_2)g\sin\theta$ → List II item 2
• $Q: \theta=10^\circ \Rightarrow f=(m_1+m_2)g\sin\theta$ → List II item 2

Case 2: $\theta=15^\circ$ and $20^\circ$ (sliding occurs)

When the system slides, friction on $m_2$ is kinetic friction: $f=\mu m_2 g\cos\theta$

So for

• $S: \theta=20^\circ \Rightarrow f=\mu m_2 g\cos\theta$ → List II item 3

Special check for $\theta=15^\circ$

We must verify whether at $15^\circ$ the two blocks still move together or separate in behavior.

Suppose they move together with acceleration $a$ down the incline. For the two-block system, $(m_1+m_2)g\sin\theta - \mu m_2 g\cos\theta = (m_1+m_2)a$ $a = g\sin\theta - \frac{\mu m_2}{m_1+m_2}g\cos\theta$

For block $m_1$, the only force along incline is its weight component minus contact force $N$ from $m_2$: $m_1 g\sin\theta - N = m_1 a$ Hence, $N = m_1(g\sin\theta-a)$

Substitute $a$: $N = m_1\left[g\sin\theta - \left(g\sin\theta - \frac{\mu m_2}{m_1+m_2}g\cos\theta\right)\right]$ $N = \frac{m_1\mu m_2}{m_1+m_2}g\cos\theta >0$

So contact is maintained; they move together.

Therefore friction on $m_2$ at $15^\circ$ is also kinetic friction: $f=\mu m_2 g\cos\theta$ That corresponds to List II item 3.

Final matching

• $P(5^\circ) \to 2$
• $Q(10^\circ) \to 2$
• $R(15^\circ) \to 3$
• $S(20^\circ) \to 3$

Hence the correct option is $\boxed{\text{D}}$

Comparison with stored answer

Stored correct answer: D

My derived answer: D

So they agree.