Key idea: motion in the lift frame

The water comes out of the side hole due to pressure at the hole. In the lift frame, the effective gravity is:

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

depending on whether the lift accelerates upward or downward.

The horizontal range of the water jet inside the lift depends on:

• the speed of efflux from the hole,
• the time taken to fall to the floor.

When the lift is at rest

Let the vertical depth of the hole below the water surface be $h$ and the vertical height of the hole above the floor be $y$.

Then by Torricelli’s theorem,

$v = \sqrt{2gh}$

Time to hit the floor:

$t = \sqrt{\frac{2y}{g}}$

Hence horizontal distance:

$d = vt = \sqrt{2gh}\,\sqrt{\frac{2y}{g}} = 2\sqrt{hy}$

So for the given case, this equals $1.2\,\text{m}$.

Case P: Lift accelerating vertically up

Effective gravity becomes:

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

Then

$v = \sqrt{2(g+a)h}$

and

$t = \sqrt{\frac{2y}{g+a}}$

Therefore,

$d = vt = \sqrt{2(g+a)h}\,\sqrt{\frac{2y}{g+a}} = 2\sqrt{hy}$

Thus the distance remains unchanged.

So,

$P \to 1$

Case Q: Lift accelerating vertically down with acceleration less than $g$

Effective gravity is:

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

Then

$v = \sqrt{2(g-a)h}, \qquad t = \sqrt{\frac{2y}{g-a}}$

Hence,

$d = vt = \sqrt{2(g-a)h}\,\sqrt{\frac{2y}{g-a}} = 2\sqrt{hy}$

Again unchanged.

So,

$Q \to 1$

Case R: Lift moving vertically up with constant speed

Constant speed means no acceleration, so the lift frame is inertial. Effective gravity is just $g$.

Therefore the distance remains the same:

$R \to 1$

Case S: Lift falling freely

In free fall,

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

So there is no hydrostatic pressure difference in the water. Therefore water does not come out of the hole.

So,

$S \to 4$

Final matching

$P \to 1, \quad Q \to 1, \quad R \to 1, \quad S \to 4$

This corresponds to Option C.