💧 Spacing Between Water Droplets – Motion Under Gravity | JEE Physics | Doubtify JEE

    

📌 Question:

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at the 4th second after its fall to the next droplet is 34.3 m. At what rate are the droplets coming from the tap?
(Take g = 9.8 m/s²)

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🖼️ Question Image:

💡 Concept Overview:

This question is a perfect example of motion under gravity with a time lag between successive drops. Such problems test your understanding of how objects behave when released one after another under constant acceleration.

In this case, we’re observing two water droplets – both falling freely under gravity – but released at different times. One droplet is seen at the 4th second, and we are asked to find the time interval between successive droplet releases that results in a 34.3 m distance between them.

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🧠 Physics Behind the Problem:

Let’s assume the tap releases droplets at a fixed time interval of T seconds.

That means the second droplet was released T seconds after the first droplet.

We’re observing the position of the first droplet after 4 seconds, and the second droplet after (4 – T) seconds, because it started falling later.

Using the formula for displacement under gravity:

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

Let’s find their positions:

• Position of 1st droplet at 4 sec =

$$s_1 = \frac{1}{2} \cdot 9.8 \cdot (4)^2 = 78.4 \, \text{m}$$

• Position of 2nd droplet at (4 – T) sec =

$$s_2=\frac{1}{2}\cdot 9.8\cdot (4-T{)}^2$$

Given that their spacing is:

$$s_1 - s_2 = 34.3$$

Substitute values:

$$78.4 - \frac{1}{2} \cdot 9.8 \cdot (4 - T)^2 = 34.3$$

Solving:

$$78.4 - 4.9(4 - T)^2 = 34.3 \Rightarrow 4.9(4 - T)^2 = 44.1 \Rightarrow (4 - T)^2 = 9 \Rightarrow 4 - T = 3 \Rightarrow T = 1 \, \text{second}$$

So, droplets are coming out at a rate of 1 drop per second. ✅

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🧠 Solution (Image):

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🎥 Video Solution:

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