NEET 2021 · PhysicsEscape energyPrevious Year Question
A particle of mass m is projected with velocity u = k·vₑ (k < 1) from the earth's surface (vₑ = escape velocity). The maximum height above the surface reached by the particle is:
Answer: (B) Rk²/(1−k²). Correct Answer (B): Rk²/(1−k²) Solution: ½mu² − GMm/R = −GMm/(R+H).
- A.Rk²/(1+k)
- B.Rk²/(1−k²)✓
- C.R(k/(1−k))²
- D.R(k/(1+k))²
Correct Answer
(B) Rk²/(1−k²)
Solution & Explanation
Correct Answer (B): Rk²/(1−k²) Solution: ½mu² − GMm/R = −GMm/(R+H). With u² = k²(2GM/R): (GM/R)(k²−1) = −GM/(R+H) ⇒ R+H = R/(1−k²) ⇒ H = Rk²/(1−k²).
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