Kinematics

JEE Main level, single correct, medium

175 of 308

Question 175 of 308

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th{4}^{th} second after its fall to the next droplet is 34.3m34.3m. At what rate the droplets are coming from the tap ? (Take g=9.8ms−2g=9.8m{s}^{-2})

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Answer: (C) 11 drop/second

In 4thsec⁡{4}^{th}\sec, 1st{1}^{st} drop will travel
⇒12×(9.8)×(4)2=78.4m\Rightarrow \dfrac{1}{2}\times \left(9.8\right)\times {\left(4\right)}^{2}=78.4m
∴2nd∴{2}^{nd} drop would have travelled
⇒78.4−34.3=44.1m\Rightarrow 78.4-34.3=44.1m
Time for 2nd{2}^{nd} drop
12(9.8)t2=44.1\dfrac{1}{2}\left(9.8\right){t}^{2}=44.1
t=3sec⁡t=3\sec
∴∴ Each drop have time gap of 1sec⁡1\sec.
∴1∴1 drop per second.

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