NEET-XII-Physics
47: The Special Theory of Relativity
- #2If the speed of a rod moving at a relativistic speed parallel to its length is doubled,
(a) the length will become half of the original value
(b) the mass will become double of the original value
(c) the length will decrease
(d) the mass will increasedigAnsr: c,dAns :
(c) the length will decrease
(d) the mass will increase
If the speed of a rod moving at a relativistic speed v parallel to its length, its mass
`` m=\gamma {m}_{o}=\frac{{m}_{o}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}``
and its length
`` l=\frac{{l}_{o}}{\gamma }={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}``
`` \,\mathrm{\,where\,}\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}={\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{{v}^{2}}{2{c}^{2}}+...>1\,\mathrm{\,as\,}v<c``
If the speed is doubled, its multiplying factor
`` \gamma \text{'}=\frac{1}{\sqrt{1-{\displaystyle \frac{4{v}^{2}}{{c}^{2}}}}}={\left(1-\frac{4{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{2{v}^{2}}{{c}^{2}}+...>2\gamma ``
`` \,\mathrm{\,and\,}m=\gamma \text{'}{m}_{o}>2\gamma {m}_{o},l=\frac{{l}_{o}}{\gamma \text{'}}<\frac{{l}_{o}}{2\gamma }``
Hence, the mass will increase but more than double and length will decrease but not exactly half of the original values.
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