A person in lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance $d$ of $1.2\ \text{m}$ from the person. In the following, state of the lift's motion is given in List-I and the distance where the water jet hits the floor of the lift is given in List-II. Match the statements from List-I with those in List-II and select the correct answer using the code given below the lists. (P) Lift is accelerating vertically up (1) $d = 1.2\ \text{m}$ (Q) Lift is accelerating vertically down with an acceleration less than the gravitational acceleration (2) $d > 1.2\ \text{m}$ (R) Lift is moving vertically up with constant speed (3) $d < 1.2\ \text{m}$ (S) Lift is falling freely (4) No water leaks out of the jar

Step 1: Define the effective acceleration due to gravity.

The effective acceleration due to gravity $g_{\text{eff}}$ inside the lift depends on its state of motion.

For a lift accelerating vertically upward with acceleration $a$:

$$g_{\text{eff}} = g + a$$

For a lift accelerating vertically downward with acceleration $a < g$:

$$g_{\text{eff}} = g - a$$

For a lift moving with constant speed:

$$g_{\text{eff}} = g$$

For a lift in free fall:

$$g_{\text{eff}} = 0$$

Step 2: Calculate the horizontal distance of the water jet.

Let $h$ be the height of the water column above the hole, and $H$ be the height of the hole above the floor of the lift.

The horizontal velocity of efflux $v$ is given by Torricelli's law.

$$v = \sqrt{2 g_{\text{eff}} h}$$

The time of flight $t$ for the water to reach the floor is determined by the vertical motion.

$$t = \sqrt{\frac{2H}{g_{\text{eff}}}}$$

The horizontal distance $d$ traveled by the jet is the product of velocity and time.

$$d = v \cdot t$$ $$d = \sqrt{2 g_{\text{eff}} h} \cdot \sqrt{\frac{2H}{g_{\text{eff}}}}$$ $$d = 2\sqrt{hH}$$

The distance $d$ is independent of $g_{\text{eff}}$, meaning it remains constant at $1.2\,\text{m}$ as long as $g_{\text{eff}} > 0$.

Step 3: Match the states of motion.

For case (P), the lift is accelerating vertically up, so $g_{\text{eff}} = g + a > 0$.

$$d = 1.2\,\text{m}$$

For case (Q), the lift is accelerating vertically down with $a < g$, so $g_{\text{eff}} = g - a > 0$.

$$d = 1.2\,\text{m}$$

For case (R), the lift is moving vertically up with constant speed, so $g_{\text{eff}} = g > 0$.

$$d = 1.2\,\text{m}$$

For case (S), the lift is falling freely, so $g_{\text{eff}} = 0$.

$$v = 0$$

Answer: (P) $\to$ (1), (Q) $\to$ (1), (R) $\to$ (1), (S) $\to$ (4)