A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When

Q: Why does the effective gravity cancel out in the range formula?

The velocity of efflux is proportional to $\sqrt{g_{eff}}$ because a higher gravity pushes the water out faster. However, the time of flight is proportional to $1/\sqrt{g_{eff}}$ because a higher gravity pulls the water down to the floor faster. When you multiply velocity and time to get the range, these two opposing effects cancel each other out perfectly!

Q: What happens during free fall? Why doesn't water leak out?

During free fall, the effective gravity inside the lift is exactly zero. This means there is no hydrostatic pressure gradient in the water, so there is no force to push the water out of the hole. Both the water and the jar fall with the same acceleration $g$, so they remain at rest relative to each other.

Q: Does this independence of range hold true for any shape of the jar?

Yes, as long as the hole is small enough that the velocity of efflux follows Torricelli's Law, the shape of the jar does not affect the cancellation of $g_{eff}$.

Q: What if the lift accelerates downwards with an acceleration greater than $g$?

If $a > g$ downwards, the effective gravity becomes negative (directed upwards). The water would rise to the top of the jar, and no water would leak out of the bottom hole.

Q: Is the atmospheric pressure assumed to be constant inside the lift?

Yes, the atmospheric pressure is assumed to be uniform and constant inside the lift, so it does not create any pressure difference between the top surface of the water and the hole.

Question

A person in a 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 $1.2	ext{ m}$ 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.

\begin{array}{l|l}
\hline
	extbf{List I} & 	extbf{List II} \
\hline
	ext{A. Lift is accelerating vertically up.} & 	ext{p. } d = 1.2	ext{ m} \
	ext{B. Lift is accelerating vertically down with an acceleration less than } g. & 	ext{q. } d > 1.2	ext{ m} \
	ext{C. Lift is moving vertically up with constant speed.} & 	ext{r. } d < 1.2	ext{ m} \
	ext{D. Lift is falling freely.} & 	ext{s. No water leaks out of the jar} \
\hline
\end{array}

Options:

(A) A-q, B-r, C-q, D-s

(B) A-q, B-r, C-p, D-s

(C) A-p, B-p, C-p, D-s

(D) A-q, B-r, C-p, D-p

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing inside a lift, holding a water jar with a small hole at the bottom.

When the lift is at rest, the water jet shoots out horizontally and lands on the floor at a distance of $1.2	ext{ m}$.

Now, what happens when the lift starts moving?

Does the water jet land closer, further, or does it stop flowing altogether?

To answer this, we must look at the physics of fluid flow and kinematics from the perspective of an observer inside the lift.

Let's define our parameters:
- Let $H$ be the height of the water level above the hole.
- Let $h$ be the height of the hole from the floor of the lift.
- Let $g_{eff}$ be the effective acceleration due to gravity inside the lift.

### The Master Equation

First, let's find the velocity of efflux $v$ using Torricelli's Law.

According to Torricelli's Law, which is derived from Bernoulli's Principle, the speed at which water leaves a small orifice is:

$$v = \sqrt{2 g_{eff} H}$$

Once the water leaves the hole horizontally, it undergoes projectile motion.

Vertically, it is in a state of free fall from a height $h$ under the influence of the effective gravity $g_{eff}$.

Using the kinematic equation for vertical motion:

$$h = \frac{1}{2} g_{eff} t^2 \implies t = \sqrt{\frac{2h}{g_{eff}}}$$

Now, the horizontal distance $d$ traveled by the water jet before hitting the floor is simply the horizontal velocity multiplied by the time of flight:

$$d = v \cdot t$$

Substituting our expressions for $v$ and $t$:

$$d = \left(\sqrt{2 g_{eff} H}ight) \cdot \left(\sqrt{\frac{2h}{g_{eff}}}ight)$$

$$d = \sqrt{2 g_{eff} H \cdot \frac{2h}{g_{eff}}} = 2\sqrt{Hh}$$

### The Magic of Cancellation

Notice something absolutely beautiful here!

The effective gravity $g_{eff}$ cancels out of the equation completely!

This means that the horizontal range $d$ of the water jet is **completely independent** of the acceleration of the lift, as long as water actually flows out of the jar (i.e., $g_{eff} > 0$).

Therefore, in any state of motion where $g_{eff} > 0$, the horizontal distance $d$ will remain exactly equal to its value at rest, which is $1.2	ext{ m}$.

### Analyzing the Cases

Let's evaluate each case from List I:

- **Case A: Lift is accelerating vertically up.**
  In this case, a downward pseudo force acts on the water, making the effective gravity:
  $$g_{eff} = g + a > 0$$
  Since $g_{eff} > 0$, water flows out, and the range remains $d = 1.2	ext{ m}$. 
  Thus, **A matches with p**.

- **Case B: Lift is accelerating vertically down with $a < g$.**
  Here, an upward pseudo force acts on the water, making the effective gravity:
  $$g_{eff} = g - a > 0$$
  Since $g_{eff} > 0$, water flows out, and the range remains $d = 1.2	ext{ m}$. 
  Thus, **B matches with p**.

- **Case C: Lift is moving vertically up with constant speed.**
  Since the speed is constant, the acceleration is zero, so:
  $$g_{eff} = g > 0$$
  Water flows out normally, and the range is $d = 1.2	ext{ m}$. 
  Thus, **C matches with p**.

- **Case D: Lift is falling freely.**
  When the lift falls freely, its downward acceleration is exactly $a = g$. 
  The effective gravity inside the lift becomes:
  $$g_{eff} = g - g = 0$$
  In this state of weightlessness, there is no hydrostatic pressure difference to push the water out of the hole. 
  Both the water and the jar fall together at the same rate, so **no water leaks out of the jar**.
  Thus, **D matches with s**.

### Conclusion

Combining all our matches, we get:
- $A 	o p$
- $B 	o p$
- $C 	o p$
- $D 	o s$

This perfectly corresponds to option (c).

Select Answer:

Visualized Solution

Understanding the Setup

Velocity of Efflux ( $v$ )

Time of Flight ( $t$ )

Horizontal Range ( $d$ )

Independence of $g_{e f f}$

Case A: Accelerating Vertically Up

Case B: Accelerating Vertically Down

Case C: Moving Up with Constant Speed

Case D: Free Fall

Matching and Conclusion

What if $a > g$ Downwards?

The Sigma Insight: Flow of Fluid

Analyzing the Setup

The Master Equation

The Magic of Cancellation

Analyzing the Cases

Conclusion

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