| ⇦ | 
| ⇨ |
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.
Related Questions: - Charge density for intrinsic semiconductor will be
- The voltage of clouds is 4×10⁶ volt with respect to ground. In a lightening
- The primary of a transformer when connected to a dc battery of 10 volt draws a current
- Heat is not being exchanged in a body. If its internal energy is increased, then
- A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm
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
- Charge density for intrinsic semiconductor will be
- The voltage of clouds is 4×10⁶ volt with respect to ground. In a lightening
- The primary of a transformer when connected to a dc battery of 10 volt draws a current
- Heat is not being exchanged in a body. If its internal energy is increased, then
- A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm
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
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
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]