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NEET-XII-Physics

47: The Special Theory of Relativity

with Solutions - page 2
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  • #6
    The rest distance between Patna and Delhi is 1000 km. A nonstop train travels at 360 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?
    Ans : Given:
    Rest distance between Patna and Delhi, L0 = 1000 km = 106 m
    Speed of train, v = 360 km/h`` =\frac{360\times 5}{18}=100\,\mathrm{\,m\,}/\,\mathrm{\,s\,}`` (a) Apparent distance between Patna and Delhi is given by
    `` L\text{'}={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}``
    `` L\text{'}={10}^{6}\sqrt{1-{\left(\frac{100}{3\times {10}^{8}}\right)}^{2}}``
    `` L\text{'}={10}^{6}\sqrt{1-\frac{{10}^{4}}{9\times {10}^{16}}}``
    `` L\text{'}={10}^{6}\sqrt{1-\frac{1}{9}\times {10}^{-12}}``
    `` L\text{'}=56\times {10}^{-9}\,\mathrm{\,m\,}``
    Change in length = 56 nm
    So, the distance between Patna and Delhi in the train frame is 56 nm less than 1000 km. (b) Actual time taken by the train is given by
    t = `` \frac{{L}_{0}}{v}=\frac{1000\times {10}^{3}}{100}={10}^{4}\,\mathrm{\,s\,}=\frac{500}{3}\,\mathrm{\,min\,}``
    Change in time, `` ∆t=\frac{∆L}{v}=\frac{L\text{'}}{v}=\frac{56\times {10}^{-9}}{100}=0.56\times {10}^{-9}\,\mathrm{\,s\,}=0.56\,\mathrm{\,ns\,}``
    So, 0.56 nm less than `` \frac{500}{3}`` min elapse in the train frame between Patna and Delhi.
    Page No 458:
  • #6-a
    What is the distance between Patna and Delhi in the train frame?
    Ans : Apparent distance between Patna and Delhi is given by
    `` L\text{'}={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}``
    `` L\text{'}={10}^{6}\sqrt{1-{\left(\frac{100}{3\times {10}^{8}}\right)}^{2}}``
    `` L\text{'}={10}^{6}\sqrt{1-\frac{{10}^{4}}{9\times {10}^{16}}}``
    `` L\text{'}={10}^{6}\sqrt{1-\frac{1}{9}\times {10}^{-12}}``
    `` L\text{'}=56\times {10}^{-9}\,\mathrm{\,m\,}``
    Change in length = 56 nm
    So, the distance between Patna and Delhi in the train frame is 56 nm less than 1000 km.
  • #6-b
    How much time elapses in the train frame between Patna and Delhi?
    Ans : Actual time taken by the train is given by
    t = `` \frac{{L}_{0}}{v}=\frac{1000\times {10}^{3}}{100}={10}^{4}\,\mathrm{\,s\,}=\frac{500}{3}\,\mathrm{\,min\,}``
    Change in time, `` ∆t=\frac{∆L}{v}=\frac{L\text{'}}{v}=\frac{56\times {10}^{-9}}{100}=0.56\times {10}^{-9}\,\mathrm{\,s\,}=0.56\,\mathrm{\,ns\,}``
    So, 0.56 nm less than `` \frac{500}{3}`` min elapse in the train frame between Patna and Delhi.
    Page No 458: