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QPage 458: Q6: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q6 Solution: The rest distance between Patna and Delhi is 1000 km. A nonstop …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the speed in km/h and the distance in km directly into $\sqrt{1-v^2/c^2}$ while using $c=3\times10^8\,\text{m/s}$, mixing kilometres and metres. Always convert all quantities to the same SI units (e.g., metres and seconds) before evaluating $v/c$ or length‑contraction formulas.

The rest distance between Patna and Delhi is 1000 km. A nonstop train travels at 360 km h−1360 \text{ km h}^{-1}. a) What is the distance between Patna and Delhi in the train frame? b) How much time elapses in the train frame between Patna and Delhi?

Step-by-Step Verified Solution

**Answer:** (a) d≈1000d \approx 1000 km (essentially unchanged) (b) t≈104t \approx 10^4 s. **Step-by-Step Explanation:** (a) v=360v = 360 km/h = 100 m/s. v/c=100/(3×108)=3.33×10−7v/c = 100/(3 \times 10^8) = 3.33 \times 10^{-7}. Length contraction: d=1061−(3.33×10−7)2≈106d = 10^6 \sqrt{1 - (3.33 \times 10^{-7})^2} \approx 10^6 m = 1000 km. (b) Time in train frame: t=d/v=106/100=104t = d/v = 10^6/100 = 10^4 s = 2.78 h.

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