JEE PYQ: Motion in a Straight Line - Question ID 2a0f5f54f7f1 (JEE Main 2021)

Let us solve the problem systematically, ensuring clarity at every step.

The problem involves uniformly accelerated motion under gravity. The key concepts and formulas are:

• Equation of motion for free fall: $s = ut + \frac{1}{2} g t^2$ where:
  • $s$ = displacement (distance fallen from the nozzle),
  • $u$ = initial velocity (0 for drops released from rest),
  • $g$ = acceleration due to gravity ($9.8\ \text{m/s}^2$ downward),
  • $t$ = time elapsed since release.
• Time to fall from height $H$: Since $u = 0$, the time $T$ to fall a distance $H$ is: $H = \frac{1}{2} g T^2 \implies T = \sqrt{\frac{2H}{g}}$
• Regular interval of drops: The drops are released at a fixed time interval $\Delta t$. This means:
  • Drop 1 is released at $t = 0$.
  • Drop 2 is released at $t = \Delta t$.
  • Drop 3 is released at $t = 2\Delta t$.

The problem states: When the first drop strikes the floor, the third drop begins to fall. This gives a crucial relationship between $\Delta t$ and the total fall time $T$ of the first drop.

Step 1: Compute total fall time of first drop

The first drop falls from height $H = 9.8\ \text{m}$. Using $H = \frac{1}{2} g T^2$: $9.8 = \frac{1}{2} \times 9.8 \times T^2 \implies T^2 = 2 \implies T = \sqrt{2} \approx 1.414\ \text{s}$

Step 2: Relate drop interval to total fall time

When the first drop hits the floor at $t = T$, the third drop is just beginning to fall. This means the third drop is released at $t = 2\Delta t = T$. Hence: $2\Delta t = T \implies \Delta t = \frac{T}{2} = \frac{\sqrt{2}}{2} \approx 0.707\ \text{s}$

Step 3: Determine release time of second drop

The second drop is released at $t = \Delta t = \frac{T}{2}$. At the instant the first drop hits the floor ($t = T$), the second drop has been falling for: $t_{\text{fall}} = T - \Delta t = T - \frac{T}{2} = \frac{T}{2}$

Step 4: Compute distance fallen by second drop

Using $s = \frac{1}{2} g t_{\text{fall}}^2$: $s = \frac{1}{2} \times 9.8 \times \left(\frac{T}{2}\right)^2 = \frac{1}{2} \times 9.8 \times \frac{T^2}{4} = \frac{9.8 \times 2}{8} = \frac{19.6}{8} = 2.45\ \text{m}$ This is the distance the second drop has fallen from the nozzle.

Step 5: Locate position of second drop from the floor

The nozzle is $9.8\ \text{m}$ above the floor. The second drop has fallen $2.45\ \text{m}$ from the nozzle, so its height above the floor is: $9.8 - 2.45 = 7.35\ \text{m}$

Conclusion: The position of the second drop from the floor when the first drop strikes the floor is $7.35\ \text{m}$, which corresponds to option D.

Students often make these mistakes:

• Misinterpreting the drop numbering: Some assume the first drop is released at $t = \Delta t$, leading to incorrect timing. Always clarify: first drop at $t=0$, second at $t=\Delta t$, third at $t=2\Delta t$.
• Incorrect fall time calculation: Using $T = \sqrt{\frac{H}{g}}$ instead of $T = \sqrt{\frac{2H}{g}}$ is a common error. Remember: $s = \frac{1}{2} g t^2$, so $t = \sqrt{\frac{2s}{g}}$.
• Confusing distance fallen with height from floor: The question asks for position from the floor, not distance fallen. Subtract the fallen distance from total height.
• Assuming equal spacing in space: Drops are released at equal time intervals, not equal spatial intervals. Their positions are not equally spaced.

Exam Tip: Always draw a timeline marking release and impact events. Label each drop’s release time and compute fall duration carefully.