A block of mass $m_1 = 1\ \text{kg}$ another mass $m_2 = 2\ \text{kg}$, are placed together (see figure) on an inclined plane with angle of inclination $\theta$. Various values of $\theta$ are given in List-I. The coefficient of friction between the block $m_1$ and plane is always zero. The coefficient of static and dynamic friction between the block $m_2$ and the plane are equal to $\mu = 0.3$. In List-II expressions for the friction on block $m_2$ are given. Match the correct expression of the friction in List-II with the angles given in List-I, and choose the correct option. The acceleration due to gravity is denoted by $g$. [Useful information: $\tan(5.5) \approx 0.1$; $\tan(11.5) \approx 0.2$; $\tan(16.5) \approx 0.3$] List-I: (P) $\theta=5$, (Q) $\theta=10$, (R) $\theta=15$, (S) $\theta=20$. List-II: (1) $m_2 g \sin\theta$, (2) $(m_1+m_2) g \sin\theta$, (3) $\mu m_2 g \cos\theta$, (4) $\mu (m_1+m_2) g \cos\theta$.

Step 1: Analyze the forces acting on the blocks.

The block $m_1$ is smooth ($\mu_1 = 0$) and placed up-slope, while $m_2$ is rough ($\mu = 0.3$) and placed down-slope.

Step 2: Determine the contact force between the blocks.

Since $m_1$ has no friction, the normal contact force $N_c$ exerted by $m_1$ on $m_2$ along the incline is equal to the component of its weight down the slope.

$$N_c = m_1 g \sin\theta$$

Step 3: Find the friction force required for equilibrium.

For both blocks to remain at rest, the friction force $f$ on $m_2$ must balance the total down-slope gravitational force of both blocks.

$$f = m_2 g \sin\theta + N_c$$ $$f = (m_1 + m_2) g \sin\theta$$

Step 4: Determine the maximum angle for which the blocks remain at rest.

The static friction force cannot exceed its maximum possible value.

$$f \le \mu m_2 g \cos\theta$$ $$(m_1 + m_2) g \sin\theta \le \mu m_2 g \cos\theta$$ $$\tan\theta \le \frac{\mu m_2}{m_1 + m_2}$$

Step 5: Calculate the threshold angle.

Substitute the given values $m_1 = 1\,\text{kg}$, $m_2 = 2\,\text{kg}$, and $\mu = 0.3$.

$$\tan\theta \le \frac{0.3 \times 2}{1 + 2}$$ $$\tan\theta \le 0.2$$

Using the given approximation $\tan(11.5^\circ) \approx 0.2$, the blocks remain at rest if:

$$\theta \le 11.5^\circ$$

Step 6: Match the angles in List-I with the friction expressions in List-II.

For $\theta \le 11.5^\circ$, the blocks are at rest, and the friction is given by expression (2).

$$f = (m_1 + m_2) g \sin\theta$$

For $\theta > 11.5^\circ$, the blocks slide, and the kinetic friction is given by expression (3).

$$f = \mu m_2 g \cos\theta$$

Evaluating each case:

• For (P) $\theta = 5^\circ$, the blocks are at rest, so $f$ matches (2).
• For (Q) $\theta = 10^\circ$, the blocks are at rest, so $f$ matches (2).
• For (R) $\theta = 15^\circ$, the blocks slide, so $f$ matches (3).
• For (S) $\theta = 20^\circ$, the blocks slide, so $f$ matches (3).

Answer: P-2, Q-2, R-3, S-3 (Option 4)