Q89 / 120 · VITEEE 2008 · Physics

An observer 'A' sees an asteroid with a radioactive element moving by at a speed $=0.3 \mathrm{c}$ and measures the radioactivity decay time to be $\mathrm{T}_{\mathrm{A}^{\prime}}$. Another observer ' $\mathrm{B}$ ' is moving with the asteroid and measures its decay time as $\mathrm{T}_{\mathrm{B}}$. Then $\mathrm{T}_{\mathrm{A}}$ and $\mathrm{T}_{\mathrm{B}}$ are related as below

  • A
    $\mathrm{T}_{\mathrm{B}} < \mathrm{T}_{\mathrm{A}}$
  • B
    $\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}}$
  • Correct answer
    C
    $\mathrm{T}_{\mathrm{B}}>\mathrm{T}_{\mathrm{A}}$
  • D
    Either (A) or (C) depending on whether the asteroid is approaching or moving away from A

Solution

Due to time dilation the interval between two events at the same point in a moving frame appears to be longer by a factor $\gamma\left(\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)$ to an observer in a stationary frame. Time dilation is independent of the direction of velocity and depends only on its magnitude.

Hence, the correct answer is option: 'C'.

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