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NEET PHYSICSMOTION IN A STRAIGHT LINEEasy

Question

Time taken by an object falling from rest to cover the height of h1h_1 and h2h_2 is respectively t1t_1 and t2t_2. Then the ratio of t1t_1 to t2t_2 is:

A

h1:h2h_1 : h_2

B

h1:h2\sqrt{h_1} : \sqrt{h_2}

C

h1:2h2h_1 : 2h_2

D

2h1:h22h_1 : h_2

Step-by-Step Solution

  1. Equation of Motion: For an object falling freely from rest (u=0u=0) under gravity (a=ga=g), the distance covered hh in time tt is given by the second equation of motion: h=ut+12gt2h = ut + \frac{1}{2}gt^2 h=0+12gt2    h=12gt2h = 0 + \frac{1}{2}gt^2 \implies h = \frac{1}{2}gt^2
  2. Relation between Time and Height: Rearranging the equation for time: t=2hgt = \sqrt{\frac{2h}{g}} Since gg is constant, tht \propto \sqrt{h}.
  3. Calculate Ratio:
  • For height h1h_1, time t1=2h1gt_1 = \sqrt{\frac{2h_1}{g}}
  • For height h2h_2, time t2=2h2gt_2 = \sqrt{\frac{2h_2}{g}}
  • Ratio t1t2=2h1/g2h2/g=h1h2\frac{t_1}{t_2} = \frac{\sqrt{2h_1/g}}{\sqrt{2h_2/g}} = \sqrt{\frac{h_1}{h_2}} .

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 LINEobjectfallingheightrespectively

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