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NEET PHYSICSGRAVITATIONMedium

Question

The escape velocity of a body on the earth's surface is 11.2 km/s11.2 \text{ km/s}. If the same body is projected upward with a velocity 22.4 km/s22.4 \text{ km/s}, the velocity of this body at an infinite distance from the centre of the earth will be:

A

11.22 km/s11.2\sqrt{2} \text{ km/s}

B

zero

C

11.2 km/s11.2 \text{ km/s}

D

11.23 km/s11.2\sqrt{3} \text{ km/s}

Step-by-Step Solution

  1. Conservation of Energy: According to the principle of conservation of energy, the total energy of the body at the surface equals its total energy at infinity. Let vv be the projection velocity, vev_e be the escape velocity, and vfv_f be the final velocity at infinity. Ki+Ui=Kf+UfK_i + U_i = K_f + U_f 12mv2GMmR=12mvf2+0\frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2 + 0
  2. Relating to Escape Velocity: We know that the potential energy at the surface can be expressed in terms of escape velocity using the relation 12mve2=GMmR\frac{1}{2}mv_e^2 = \frac{GMm}{R}. Substituting this into the energy equation: 12mv212mve2=12mvf2\frac{1}{2}mv^2 - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_f^2 vf=v2ve2v_f = \sqrt{v^2 - v_e^2}
  3. Calculation: Given ve=11.2 km/sv_e = 11.2 \text{ km/s} and v=22.4 km/s=2vev = 22.4 \text{ km/s} = 2v_e. vf=(2ve)2ve2=4ve2ve2=3ve2=ve3v_f = \sqrt{(2v_e)^2 - v_e^2} = \sqrt{4v_e^2 - v_e^2} = \sqrt{3v_e^2} = v_e\sqrt{3} vf=11.23 km/sv_f = 11.2\sqrt{3} \text{ km/s}

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.

PHYSICSGRAVITATIONescapevelocityearthssurfaceprojected

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