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A ballon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h.The time taken by the stone to reach the ground is
Options
(a) 4√(h/g)
(b) 2√(h/g)
(c) √(2h/g)
(d) √(g/h)
Correct Answer:
2√(h/g)
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Question Type: A
(1)
Difficulty Level: Medium
(3)
Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A radiactive substance emits 100 beta particles in the first 2s and 50 beta
- In a compound microscope, the focal length of the objective is 2.5 cm and of eye lens
- In an astronomical telescope in normal adjustment a straight black line of length
- The period of oscillation of a mass M suspended from a spring of negligible mass is T
- The dimensional formula for Young’s modulus is
Question Type: A (1)
Difficulty Level: Medium (3)
Topics: Motion in Straight Line (93)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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The velocity of the balloon at the height h is
v = √(2ah) = √(2gh/8) = √(gh)/2
Initial velocity of the stone at height h is u = √(gh)/2 upwards
h = ut + gt²/2
put the value of u in the above relation and rearrange the terms to obtain,
(√(gH)/2 )t + gt²/2 – h = 0
(√(gH))t + gt² – 2h = 0
The time taken by the stone to reach the ground can be obtained by solving the above quadratic.
t = 2√[h/g]