Question

If 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 increase

Answer 1

(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
$$\begin{gathered} l=\frac{l_o}{\gamma }=l_o\sqrt{1-\frac{v^2}{c^2}} \\ \mathrm{where}\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}={\left(1-\frac{v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{v^2}{2c^2}+...>1\mathrm{as}v<c \end{gathered}$$
If the speed is doubled, its multiplying factor
$$\begin{gathered} \gamma \text{'}=\frac{1}{\sqrt{1-{\displaystyle \frac{4v^2}{c^2}}}}={\left(1-\frac{4v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{2v^2}{c^2}+...>2\gamma \\ \mathrm{and}m=\gamma \text{'}{m}_o>2\gamma m_o,l=\frac{l_o}{\gamma \text{'}}<\frac{l_o}{2\gamma } \end{gathered}$$
Hence, the mass will increase but more than double and length will decrease but not exactly half of the original values.