JEE PYQ: Motion in a Straight Line - Question ID 1d0bff9b5625 (JEE Main 2021)

In this problem, we analyze the motion of water droplets falling freely under gravity from an open tap. The key concept is uniformly accelerated motion (free-fall) with acceleration $g = 9.8 \, \text{m/s}^2$ downward.

The position $s(t)$ of a droplet at time $t$ after being released is given by: $s(t) = \frac{1}{2} g t^2$ This formula assumes the droplet starts from rest ($u = 0$) at $t = 0$.

The question involves two consecutive droplets. Let’s denote: - Droplet 1: released at time $t = 0$ - Droplet 2: released at time $t = T$, where $T$ is the time interval between successive droplets (i.e., the inverse of the drop rate).

We are told that at $t = 4 \, \text{s}$ after the first droplet is released, the spacing between the two droplets is $34.3 \, \text{m}$. This means: $\text{Distance between droplets at } t = 4 \, \text{s} = s_1(4) - s_2(4 - T) = 34.3 \, \text{m}$ where $s_1(4)$ is the distance fallen by the first droplet in 4 seconds, and $s_2(4 - T)$ is the distance fallen by the second droplet in $(4 - T)$ seconds (since it was released $T$ seconds later).

Step 1: Express positions of both droplets

For Droplet 1 (released at $t = 0$): $s_1(t) = \frac{1}{2} g t^2 \Rightarrow s_1(4) = \frac{1}{2} \times 9.8 \times 4^2 = 4.9 \times 16 = 78.4 \, \text{m}$

For Droplet 2 (released at $t = T$): At $t = 4 \, \text{s}$, it has been falling for $(4 - T)$ seconds. $s_2(4 - T) = \frac{1}{2} g (4 - T)^2$

Step 2: Use the given spacing condition

The spacing between the droplets at $t = 4 \, \text{s}$ is: $s_1(4) - s_2(4 - T) = 34.3$ Substitute the expressions: $78.4 - \frac{1}{2} \times 9.8 \times (4 - T)^2 = 34.3$ Simplify: $78.4 - 4.9 (4 - T)^2 = 34.3$ $4.9 (4 - T)^2 = 78.4 - 34.3 = 44.1$ $(4 - T)^2 = \frac{44.1}{4.9} = 9$ Take square root: $4 - T = \pm 3$ This gives two possibilities: - $4 - T = 3 \Rightarrow T = 1$ - $4 - T = -3 \Rightarrow T = 7$

Step 3: Interpret the solutions

$T = 1 \, \text{s}$: This means droplets are released every 1 second → rate = 1 drop per second.
$T = 7 \, \text{s}$: This would mean droplets are released every 7 seconds → rate = 1 drop per 7 seconds.

Step 4: Validate the physical meaning

At $t = 4 \, \text{s}$, if $T = 7 \, \text{s}$, then the second droplet was released at $t = 7 \, \text{s}$, which is after $t = 4 \, \text{s}$. This means at $t = 4 \, \text{s}$, the second droplet hasn't even been released yet — so it cannot have fallen any distance. But the problem states that the spacing is observed at $t = 4 \, \text{s}$, implying both droplets exist and are in motion. Hence, $T = 7 \, \text{s}$ is physically invalid in this context.

Therefore, the only valid solution is $T = 1 \, \text{s}$, corresponding to a drop rate of 1 drop per second.

Trap 1: Students often forget that the second droplet was released later and thus has been falling for less time. They mistakenly assume both droplets have fallen for 4 seconds.

Trap 2: Ignoring the physical meaning of negative time. The solution $T = 7 \, \text{s}$ leads to $4 - T = -3$, which implies the second droplet hasn't been released yet. This is not consistent with the observation of spacing at $t = 4 \, \text{s}$. Always check the domain of validity.

Exam Tip: When dealing with relative motion of objects released at different times, always define the time of release clearly and express positions as functions of time since release. Use the difference in positions to set up the equation, and always interpret the physical meaning of mathematical solutions.

Final Answer: Option C: 1 drop / second