CBSE-XI-Physics
47: The Special Theory of Relativity
- #14A person travelling by a car moving at 100 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?Ans : Given:
Velocity of the car, v = 100 km/h = `` \frac{1000}{36}`` m/s-1
Distance between tower A and tower B, s = 1000 km
If ∆t be the time interval to reach tower B from tower A, then
`` ∆t=\frac{v}{s}=\frac{1000}{100}=10\,\mathrm{\,h\,}`` = 36000 s
`` ∆t\text{'}=\frac{∆t}{\sqrt{1-{v}^{2}/{c}^{2}}}``
`` =\frac{36000}{\sqrt{1-{\left({\displaystyle \frac{1000}{36\times 3\times {10}^{8}}}\right)}^{2}}}``
`` ∆t\text{'}=36000{\left[1-{\left(\frac{1000}{36\times 3\times {10}^{8}}\right)}^{2}\right]}^{-\frac{1}{2}}∆t``
Now,
∆t - ∆t' = 0.154 ns
∴ Time will lag by 0.154 ns.
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