GATE PH 2007 • Q61 MCQ • 2 Marks

Section 2: Classical Mechanics

Special theory of relativity

Subtopic: Length Contraction & Time Dilation

Question Statement:

A relativistic particle travels a length of $3 \times 10^{-3}$ m in air before decaying. The proper lifetime of the particle is $10^{-11}$ s. The speed of the particle is approximately

Options:

  • (A) $0.71 c$
  • (B) $0.86 c$
  • (C) $0.95 c$
  • (D) $0.99 c$

Official Answer Key:

A

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2007 Q.61)** **1. Relativistic Time Dilation**: - The lab frame lifetime is dilated: $\Delta t = \gamma \tau_0 = \frac{\tau_0}{\sqrt{1 - v^2/c^2}}$ - Distance traversed in lab frame: $d = v \Delta t = \frac{v \tau_0}{\sqrt{1 - v^2/c^2}}$ **2. Solving for Speed Ratio $\beta = v/c$**: - Substituting $d = 3 \times 10^{-3}$ m, $\tau_0 = 10^{-11}$ s, $c = 3 \times 10^8$ m/s: $\frac{d}{c \tau_0} = \frac{3 \times 10^{-3}}{3 \times 10^{-3}} = 1$ $\frac{\beta}{\sqrt{1 - \beta^2}} = 1 \implies \beta^2 = 1 - \beta^2 \implies 2\beta^2 = 1 \implies \beta = \frac{1}{\sqrt{2}} \approx 0.707$ - Therefore, $v \approx 0.71 c$. **3. Official Verification & Conclusion**: - **Official Master Answer**: **Option (A)**: $0.71 c$.

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