| ⇦ | 
| ⇨ |
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: - The oscillation of a body on a smooth horizontal surface is represented by the equation
- The radius of a ball is(5.2±0.2)cm.The percentage error in the volume of the ball
- Heat produced in a resistance according to which law?
- 1 a.m.u. is equivalent to
- In Young’s double slit experiment, the intensity of light coming from the first slit
Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The oscillation of a body on a smooth horizontal surface is represented by the equation
- The radius of a ball is(5.2±0.2)cm.The percentage error in the volume of the ball
- Heat produced in a resistance according to which law?
- 1 a.m.u. is equivalent to
- In Young’s double slit experiment, the intensity of light coming from the first slit
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