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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

Water drops fall at regular intervals from a tap which is 5 m5 \text{ m} above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?

A

2.50 m

B

3.75 m

C

4.00 m

D

1.25 m

Step-by-Step Solution

  1. Analyze Time Intervals: Let the time interval between consecutive drops be tt.
  • The 3rd drop is just leaving, so its time of fall is 00.
  • The 2nd drop has been falling for one interval (tt).
  • The 1st drop touches the ground, meaning it has fallen for two intervals (2t2t) since the 3rd drop is just starting.
  1. Distance Equation: Using the equation of motion s=ut+12at2s = ut + \frac{1}{2}at^2 with u=0u=0 and a=ga=g .
  • Total height (HH) fallen by 1st drop in time 2t2t: H=12g(2t)2=12g(4t2)=2gt2=5 mH = \frac{1}{2}g(2t)^2 = \frac{1}{2}g(4t^2) = 2gt^2 = 5 \text{ m} From this, gt2=2.5 mgt^2 = 2.5 \text{ m}.
  1. Distance Fallen by 2nd Drop: The 2nd drop falls for time tt. Let this distance be h2h_2. h2=12gt2h_2 = \frac{1}{2}gt^2 Substituting gt2=2.5gt^2 = 2.5: h2=12(2.5)=1.25 mh_2 = \frac{1}{2}(2.5) = 1.25 \text{ m}
  2. Calculate Height Above Ground: The question asks for the height above the ground. Height=Hh2=5 m1.25 m=3.75 m\text{Height} = H - h_2 = 5 \text{ m} - 1.25 \text{ m} = 3.75 \text{ m}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEregularintervalsgroundleavinginstant

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