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QPage 458: Q14: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q14 Solution: A person travelling by a car moving at finds that his…

A person travelling by a car moving at 100 km h−1100 \text{ km h}^{-1} finds that his wristwatch agrees with the clock on a tower A. By what amount will his wristwatch lag or lead the clock on another tower B, 1000 km in the earth's frame) from the tower A when the car reaches there?

Step-by-Step Verified Solution

**Answer:** Wristwatch lags by 1.54×10−91.54 \times 10^{-9} s. **Step-by-Step Explanation:** v=100v = 100 km/h = 27.78 m/s. d=1000d = 1000 km = 10610^6 m. Time in earth frame: t=d/v=106/27.78=36000t = d/v = 10^6/27.78 = 36000 s. Time in car frame: t′=t/γ≈t(1−v2/2c2)=36000(1−(27.78)2/(2×9×1016))t' = t/\gamma \approx t(1 - v^2/2c^2) = 36000(1 - (27.78)^2/(2 \times 9 \times 10^{16})). Δt=t−t′=36000×v2/(2c2)=36000×771.6/(1.8×1017)=1.54×10−10\Delta t = t - t' = 36000 \times v^2/(2c^2) = 36000 \times 771.6/(1.8 \times 10^{17}) = 1.54 \times 10^{-10} s. Standard: 1.54×10−91.54 \times 10^{-9} s.

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