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NEET-XII-Physics

47: The Special Theory of Relativity

with Solutions - page 2
Qstn# iv-4-a Prvs-QstnNext-Qstn
  • #4-a
    the length of the train as seen by the person and (b) the rest length of the train.
    Ans : Length observed by the observer, L= vt
    ⇒ L = 0.6 × 3 × 108
    = 1.8 × 108 m (b) Let the rest length of train be L0. Then,
    `` L={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}``
    `` \Rightarrow 1.8\times {10}^{8}={L}_{0}\sqrt{1-\frac{{\left(0.6\right)}^{2}{c}^{2}}{{c}^{2}}}``
    `` \Rightarrow 1.8\times {10}^{8}={L}_{0}\times 0.8``
    `` \Rightarrow {L}_{0}=\frac{1.8\times {10}^{8}}{0.8}``
    `` \Rightarrow {L}_{0}=2.25\times {10}^{8}\,\mathrm{\,m\,}``
    Page No 458:
  • #4-b
    the rest length of the train.
    Ans : Let the rest length of train be L0. Then,
    `` L={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}``
    `` \Rightarrow 1.8\times {10}^{8}={L}_{0}\sqrt{1-\frac{{\left(0.6\right)}^{2}{c}^{2}}{{c}^{2}}}``
    `` \Rightarrow 1.8\times {10}^{8}={L}_{0}\times 0.8``
    `` \Rightarrow {L}_{0}=\frac{1.8\times {10}^{8}}{0.8}``
    `` \Rightarrow {L}_{0}=2.25\times {10}^{8}\,\mathrm{\,m\,}``
    Page No 458: