Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{\text{th}}$ second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take $\mathrm{g}=9.8 \mathrm{~m}/\mathrm{s}^{2}$)

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Kinematics

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at

4^{th}

second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take

g = 9.8 m / s^{2}

)

(A)1 drop / 7 seconds
(B)3 drops / 2 seconds
(C)1 drop / second✓ Correct
(D)2 drops / second

Answer: (C)

Solution

Let the drops leave the tap at equal intervals

T

. At the instant described the drop in question has been falling for

4

s and the one behind it for

4 - T

s, so the gap between them is

s = \frac{g}{2} [4^{2} - (4 - T)^{2}]

Putting

s = 34.3

m and

g = 9.8 m/ s^{2}

gives

16 - (4 - T)^{2} = 7

, so

(4 - T)^{2} = 9

and

T = 1

s. The drops leave the tap at the rate of one drop per second.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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