| ⇦ | 
| ⇨ |
The escape velocity of a body from earth’s surface is Ve. The escape velocity of the same body from a height equal to 7R from earth’s surface will be
Options
(a) Ve/√2
(b) Ve/2
(c) Ve/2√2
(d) Ve/4
Correct Answer:
Ve/2√2
Explanation:
Escape velocity of body from the earth’s surface is Ve=√2gR
Escape velocity of a same body at a height h from the earth’s surface is,Ve’=√2gR²/R+h
Given h=7R
Ve’=√2gR²/R+7R ⇒ √2gR²/R[1+7]
=√2gR/8 ⇒ 1/2√2 √2gR
=1/2√2 Ve.
Related Questions: - A wire of length 1 m is placed in a uniform magnetic field of 1.5 tesla at an angle
- The excess pressure inside one soap bubble is three times that inside a second soap bubble
- The dimensions of Planck’s constant is same as that of
- The energy released by the fission of one uranium atom is 200 MeV.
- The sun light reaches us as white and not as its components because
Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A wire of length 1 m is placed in a uniform magnetic field of 1.5 tesla at an angle
- The excess pressure inside one soap bubble is three times that inside a second soap bubble
- The dimensions of Planck’s constant is same as that of
- The energy released by the fission of one uranium atom is 200 MeV.
- The sun light reaches us as white and not as its components because
Topics: Gravitation (63)
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
Leave a Reply