NEET-XII-Physics
47: The Special Theory of Relativity
- #4A person standing on a platform finds that a train moving with velocity 0.6c takes one second to pass by him. Find (a) the length of the train as seen by the person and (b) the rest length of the train. (a) the length of the train as seen by the person and (b) the rest length of the train.Ans : Given:
Velocity of train, v = 0.6c
Time taken, t = 1 s (a) 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: (a) 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-athe length of the train as seen by the person andAns : Length observed by the observer, L= vt
⇒ L = 0.6 × 3 × 108
= 1.8 × 108 m
- #4-bthe 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: