Water Droplet Spacing Problem Using Kinematics

❓ Question

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{th}$ second and the next droplet is $34.3\text{ m}$.

At what rate are the droplets coming from the tap?

Take:

$$g=9.8\text{ m/s}^2$$

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🖼 Question Image

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✍️ Short Explanation

This is a relative motion + free fall problem.

👉 Consecutive droplets have different falling times
👉 Both move under gravity
👉 Distance difference gives time gap between droplets.

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🔷 Step 1 — Motion of First Droplet 💯

For droplet seen at:

$$t=4\text{ s}$$

Distance fallen:

$$s_1=\frac12 gt^2$$
$$s_1=\frac12(9.8)(4^2)$$
$$s_1=4.9\times16$$
$$s_1=78.4\text{ m}$$

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🔷 Step 2 — Motion of Next Droplet

Let time gap between droplets be:

$$\Delta t$$

Then second droplet has been falling for:

$$(4-\Delta t)\text{ s}$$

Distance fallen:

$$s_2=\frac12 g(4-\Delta t)^2$$

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🔷 Step 3 — Use Given Spacing

Spacing between droplets:

$$s_1-s_2=34.3$$

So:

$$4.9\left[16-(4-\Delta t)^2\right]=34.3$$

Divide by 4.9:

$$16-(4-\Delta t)^2=7$$
$$(4-\Delta t)^2=9$$
$$4-\Delta t=3$$
$$\Delta t=1\text{ s}$$

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🔷 Step 4 — Find Rate of Droplets

One droplet released every:

$$1\text{ s}$$

Hence rate:

$$1\text{ droplet per second}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Dono droplets ka same time assume kar lena

❌ Relative distance directly velocity se solve karna

❌ $4+\Delta t$ use kar dena instead of $4-\Delta t$

Remember:

$$\text{Next droplet always falls for smaller time}$$

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✅ Final Answer

$$\boxed{1\text{ droplet per second}}$$

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📚 Related Topics

• Balloon Motion and Falling Object Height Problem
• Distance to Displacement Ratio in Circular Motion
• Distance from Velocity Time Graph Calculation