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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 body is rolling down an inclined plane.If KE of rotation is 40% of KE in a translatory
- The force of repulsion between two electrons at a certain distance is F.
- Which of the following pairs is wrong
- For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing
- The electric potential at a point in free space due to a charge Q coulomb is Q x 10¹¹ v
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]