Water droplets are coming from an open tap at a particular rate. The spacing between a droplet obser

Q: Why is the initial velocity $u = 0$?

The drops are falling from an open tap, which means they detach and start their free fall from rest. They are not being thrown downwards.

Q: Why do we subtract the spacing to find the second drop's distance?

The first drop has been falling for a longer time, so it is lower (has covered more distance) than the second drop. The spacing is the difference in their positions, so $s_2 = s_1 - \text{spacing}$.

Q: Does the mass of the water droplet affect its falling speed?

No, in the absence of air resistance, all objects fall at the same rate with acceleration $g$, regardless of their mass. This is a fundamental principle of kinematics.

Q: What does the time difference between the two drops represent?

The time difference represents the interval at which consecutive drops are released from the tap. If drop 1 falls at $t=0$ and drop 2 falls at $t=1$, the interval is 1 second.

Q: Could we use relative motion to solve this?

Yes, the relative acceleration between the two drops is zero (since both accelerate at $g$), so their relative velocity is constant. However, finding their individual times using the ground frame is much more straightforward here.

Question

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th second after its fall to the next droplet is $34.3 	ext{ m}$. At what rate, the droplets are coming from the tap ? (Take, $g = 9.8 	ext{ m/s}^2$)

Options:

(A) 3 drops / 2 s

(B) 2 drops / s

(C) 1 drop / s

(D) 1 drop / 7 s

Sigma Percentile · Accepted Answer

### The Rhythm of Falling Drops

Imagine a leaky tap, steadily releasing water drops one after another. As each drop falls, gravity pulls it downwards, accelerating it continuously. Because the first drop has been falling for a longer time than the second drop, it is moving faster. This difference in speed means that the gap between any two consecutive drops doesn't stay constant—it keeps increasing as they fall!

In this problem, we are given a snapshot of this dynamic system. We know the exact time the first drop has been falling, and we know the physical distance separating it from the drop that followed it. Our mission is to work backwards and figure out the rhythm of the tap.

### The Master Equation of Free Fall

Since the drops detach from the tap and fall freely under gravity, their initial velocity is zero ($u = 0$). The distance $s$ covered by a freely falling body in time $t$ is governed by the second equation of motion:

$$s = ut + \frac{1}{2}gt^2$$

Substituting $u = 0$, this simplifies beautifully to:

$$s = \frac{1}{2}gt^2$$

This equation tells us that the distance fallen is directly proportional to the square of the time.

### Pinpointing the First Drop

We are told that the first drop has been falling for $t = 4 	ext{ s}$. Let's calculate exactly how far it has traveled from the tap. We'll call this distance $s_1$.

$$s_1 = \frac{1}{2} 	imes 9.8 	imes (4)^2$$

$$s_1 = 4.9 	imes 16$$

$$s_1 = 78.4 	ext{ m}$$

So, at this exact moment, the first drop is $78.4 	ext{ m}$ below the tap.

### Locating the Second Drop

The problem states that the spacing between this first drop and the *next* droplet (the one that fell immediately after it) is $34.3 	ext{ m}$.

Because the second drop hasn't been falling as long, it hasn't traveled as far. To find its distance from the tap, which we'll call $s_2$, we simply subtract this gap from the first drop's distance:

$$s_2 = s_1 - \Delta s$$

$$s_2 = 78.4 - 34.3$$

$$s_2 = 44.1 	ext{ m}$$

### Unveiling the Time Interval

Now that we know the second drop has fallen exactly $44.1 	ext{ m}$, we can use our master equation again to find out *how long* it has been falling. Let's call its time $t'$.

$$s_2 = \frac{1}{2}g(t')^2$$

$$44.1 = \frac{1}{2} 	imes 9.8 	imes (t')^2$$

$$44.1 = 4.9 	imes (t')^2$$

Dividing both sides by $4.9$:

$$(t')^2 = \frac{44.1}{4.9} = 9$$

Taking the square root gives us:

$$t' = 3 	ext{ s}$$

### The Final Conclusion

We have uncovered the timeline! The first drop has been falling for $4 	ext{ s}$, and the second drop has been falling for $3 	ext{ s}$.

The difference between their fall times is exactly the time interval between their release from the tap:

$$	ext{Time Interval} = 4 	ext{ s} - 3 	ext{ s} = 1 	ext{ s}$$

This means the tap releases one drop every second. Therefore, the rate at which the droplets are coming from the tap is **1 drop / s**.

Animated Solution for Physics - Kinematics: Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th second after its fall to the next droplet is $34.3 m$ . At what rate, the droplets are coming from the tap ? (Take, $g = 9.8 m/s^{2}$ )

Select Answer:

Visualized Solution

$t_{1} = 4 s, t_{2} = t^{'}$

$s = u t + \frac{1}{2} g t^{2}$

$s_{1} = \frac{1}{2} \times 9.8 \times 4^{2}$

$s_{1} = 78.4 m$

$s_{2} = s_{1} - Δ s$

$s_{2} = 44.1 m$

$44.1 = \frac{1}{2} \times 9.8 \times (t^{'})^{2}$

$t^{'} = 3 s$

\text{Rate} = 1 \text{ drop/s}

\text{What if?}

The Sigma Insight: Motion in a Straight Line

The Rhythm of Falling Drops

The Master Equation of Free Fall

Pinpointing the First Drop

Locating the Second Drop

Unveiling the Time Interval

The Final Conclusion

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