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NEET PHYSICSGRAVITATIONMedium

Question

The earth is assumed to be a sphere of radius RR. A platform is arranged at a height RR from the surface of the earth. The escape velocity of a body from this platform is fvefv_e, where vev_e is its escape velocity from the surface of the earth. The value of ff is:

A

2\sqrt{2}

B

12\frac{1}{\sqrt{2}}

C

13\frac{1}{3}

D

12\frac{1}{2}

Step-by-Step Solution

  1. Escape Velocity Formula: The escape velocity vev_e from the surface of the Earth (radius RR) is given by ve=2GMRv_e = \sqrt{\frac{2GM}{R}}.
  2. Escape Velocity at Altitude: The escape velocity from a point at a distance rr from the center of the Earth is vesc=2GMrv_{esc} = \sqrt{\frac{2GM}{r}}.
  3. Given Condition: The platform is at a height h=Rh = R from the surface. Therefore, the distance from the center is r=R+h=R+R=2Rr = R + h = R + R = 2R.
  4. Calculation: Substitute r=2Rr = 2R into the escape velocity formula: vplatform=2GM2R=122GMRv_{platform} = \sqrt{\frac{2GM}{2R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} Since ve=2GMRv_e = \sqrt{\frac{2GM}{R}}, we have: vplatform=12vev_{platform} = \frac{1}{\sqrt{2}} v_e
  5. Finding f: Comparing this with the given expression vplatform=fvev_{platform} = f v_e, we find: f=12f = \frac{1}{\sqrt{2}}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from GRAVITATION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSGRAVITATIONassumedsphereradiusplatformarranged

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