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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.
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Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- 64 drops of mercury, each charged to a potential of 10 V, are combined
- For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing
- A super conductor exhibits perfect
- Two bodies of masses m and 4 m are moving with equal K.E. The ratio of their linear
- The phase difference between the instantaneous velocity and acceleration of a particle
Topics: Gravitation (63)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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