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MHT CET 2025 (20 Apr, Shift 1) · Escape Velocity

Escape Velocity question from MHT CET 2025 (20 Apr, Shift 1)

Q13MHT CET 2025 · 20 APR · SHIFT 1Medium

Escape Velocity

A body is projected vertically from earth's surface with (13)rd\left(\frac{1}{3}\right)^{\text{rd}} of escape velocity. The maximum height reached by the body is ( R=R = radius of earth)

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Answer: (b)

  1. For v=Kvev = Kv_e: h=RK21−K2h = \frac{RK^2}{1-K^2}.
  2. K=13⇒h=R/98/9K = \frac{1}{3} \Rightarrow h = \frac{R/9}{8/9}.
  3. h=R8h = \frac{R}{8}.
  4. So the answer is (b) R8\frac{R}{8}.
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