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MCQMediumJEE 2026Motion in a Straight Line

JEE Physics 2026 Question with Solution

Water drops fall from a tap on the floor, 5 m5 \, \text{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 the ground, at the instant when the first drop strikes the ground, is _____ m\text{m}. (g=10 m s−2g = 10 \, \text{m s}^{-2})

  • A

    4.04.0

  • B

    3.83.8

  • C

    4.24.2

  • D

    2.52.5

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The floor is 5 m5 \, \text{m} below the tap, and drops fall at equal time intervals. The first drop strikes the ground when the sixth drop begins to fall. Also, g=10 m s−2g = 10 \, \text{m s}^{-2}.

Find: The height of the fourth drop from the ground at the instant the first drop strikes the ground.

Use uniformly accelerated motion for each drop starting from rest.

s=12gt2s = \frac{1}{2}gt^2

For the first drop:

5=12⋅10⋅t25 = \frac{1}{2}\cdot 10 \cdot t^2t2=1t^2 = 1t=1 st = 1 \, \text{s}

So, the first drop reaches the ground in 1 s1 \, \text{s}.

Between the first and sixth drops, there are 55 equal intervals. Let the interval be τ\tau.

5τ=15\tau = 1τ=0.2 s\tau = 0.2 \, \text{s}

The fourth drop starts falling after 3τ3\tau from the first drop:

3τ=3×0.2=0.6 s3\tau = 3 \times 0.2 = 0.6 \, \text{s}

Hence, when the first drop hits the ground at t=1 st = 1 \, \text{s}, the fourth drop has been falling for

t4=1−0.6=0.4 st_4 = 1 - 0.6 = 0.4 \, \text{s}

Distance fallen by the fourth drop:

s4=12gt42s_4 = \frac{1}{2}gt_4^2s4=12⋅10⋅(0.4)2=0.8 ms_4 = \frac{1}{2}\cdot 10 \cdot (0.4)^2 = 0.8 \, \text{m}

Therefore, its height above the ground is

h=5−0.8=4.2 mh = 5 - 0.8 = 4.2 \, \text{m}

Therefore, the correct option is C and the height of the fourth drop from the ground is 4.2 m4.2 \, \text{m}.

Common mistakes

  • Counting the number of intervals between the first and sixth drops as 66 instead of 55. This is wrong because intervals are the gaps between releases, not the number of drops. Count release gaps carefully: first to sixth means 55 equal intervals.

  • Using the full 1 s1 \, \text{s} as the falling time of the fourth drop. This is wrong because the fourth drop starts later, after 3τ3\tau. First find its release time, then subtract from 1 s1 \, \text{s}.

  • Taking 0.8 m0.8 \, \text{m} as the answer directly. This is wrong because 0.8 m0.8 \, \text{m} is the distance fallen by the fourth drop from the tap, not its height above the ground. Subtract the fallen distance from 5 m5 \, \text{m}.

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