JEE PYQ: Motion in a Straight Line - Question ID cae7e45701c7 (JEE Main 2026)

Core Formula & Concept:

When water drops fall freely under gravity from rest, they execute uniformly accelerated motion with acceleration $g = 10\, \text{m/s}^2$. The key formulas we use are:

• Distance fallen in time $t$: $s = \tfrac12\,g\,t^2$.
• Velocity at time $t$: $v = g\,t$.

The problem states that drops fall at regular intervals. This means the time between the release of consecutive drops is constant. We denote this interval by $\Delta t$.

Step-by-Step Derivation:

Step 1 – Determine the time for the first drop to hit the floor

The first drop falls from rest through $h = 5\,$m. Using $s = \tfrac12\,g\,t^2$, $5 = \tfrac12 \times 10 \times t_1^2 \quad\Longrightarrow\quad t_1^2 = 1 \quad\Longrightarrow\quad t_1 = 1\, \text{s}.$

Step 2 – Find the time interval $\Delta t$ between drops

The first drop strikes the floor at $t = 1\,$s. At that same instant the sixth drop just begins to fall. Since drops are released at $t = 0,\, \Delta t,\, 2\Delta t,\, \dots$, the sixth drop is released at $t = 5\Delta t$. We are told $5\Delta t = 1\,$s, so $\Delta t = \frac{1}{5} = 0.2\, \text{s}.$

Step 3 – Locate the fourth drop at $t = 1\,$s

The fourth drop is released at $t = 3\Delta t = 0.6\,$s. At the instant $t = 1\,$s, this drop has been falling for $t_{\rm fall} = 1 - 0.6 = 0.4\, \text{s}.$ The distance it has fallen is $s_4 = \tfrac12\,g\,t_{\rm fall}^2 = \tfrac12 \times 10 \times (0.4)^2 = 0.8\, \text{m}.$ Since the total height is $5\,$m, the height of the fourth drop above the ground is $h_4 = 5 - 0.8 = 4.2\, \text{m}.$

Common Traps & Exam Tip:

1. Mis-counting drops. Students often confuse “the sixth drop begins to fall” with the sixth drop already in the air. Remember: if the first drop is released at $t=0$, the sixth is released at $t=5\Delta t$. 2. Incorrect time of fall for the fourth drop. One must subtract the release time of the fourth drop from the total time $t=1\,$s to find how long it has been falling. 3. Sign errors in height. Always subtract the distance fallen from the total height to get the remaining height above ground.

By carefully tracking the release times and applying $s = \tfrac12\,g\,t^2$, we arrive at the correct answer $4.2\,$m, which is option C.