Q60 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is $\_\_\_\_$ m.
$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
- Correct answerA4.2
- B2.5
- C4.0
- D3.8
Solution
Time for first drop to fall 5 m:
$5 = \frac{1}{2} \times 10 \times t^2$ gives $t = 1$ s.
Since the first drop strikes when the sixth begins, there are 5 intervals: each interval $T = \frac{1}{5} = 0.2$ s.
When drop 1 hits the ground (at $t = 1$ s), drop 4 has been falling for $t_4 = 1 - 3 \times 0.2 = 0.4$ s.
Distance fallen by drop 4: $s_4 = \frac{1}{2} \times 10 \times (0.4)^2 = 0.8$ m.
Height of drop 4 above ground: $5 - 0.8 = 4.2$ m.
$5 = \frac{1}{2} \times 10 \times t^2$ gives $t = 1$ s.
Since the first drop strikes when the sixth begins, there are 5 intervals: each interval $T = \frac{1}{5} = 0.2$ s.
When drop 1 hits the ground (at $t = 1$ s), drop 4 has been falling for $t_4 = 1 - 3 \times 0.2 = 0.4$ s.
Distance fallen by drop 4: $s_4 = \frac{1}{2} \times 10 \times (0.4)^2 = 0.8$ m.
Height of drop 4 above ground: $5 - 0.8 = 4.2$ m.
Hence, the correct answer is option: 'A'.