JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

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. ( $g = 10$ m/s $^{2}$ )

A
2.5
B
4.0
C
4.2
D
3.8

Solution

Time for the first drop to fall 5 m: $t = 2 h / g = 2 \cdot 5/10 = 1$ s.

When drop 1 hits, drop 6 has just begun (so 5 intervals of $Δ t$ between drop 1 and drop 6): $5Δ t = 1$ s $\Rightarrow Δ t = 0.2$ s.

Drop 4 has been falling for $3Δ t = 0.6$ s less than drop 1, so it has been falling for $1 - 0.6 = 0.4$ s.

Distance fallen by drop 4: $\frac{1}{2} g t^{2} = \frac{1}{2} (10) (0.4)^{2} = 0.8$ m.

Height above ground $= 5 - 0.8 = 4.2$ m. Option (3).

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