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Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: The Special Theory of Relativity

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Special Theory of Relativity

Assume that Devlok (heaven) moves constantly at a speed of 0.9999c...

Assume that Devlok (heaven) moves constantly at a speed of

0.9999c

relative to the earth. In the earth's frame of reference, how much time elapses on the earth for one day to pass on Devlok?

Options

A $70.7$ days
B $0.014$ days
C $100$ days
D $141.4$ days

Correct answer

A. $70.7$ days

Step-by-step solution

The time dilation formula is given by

\Delta t = \dfrac{\Delta t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}

Here, the proper time

\Delta t_0 = 1

day and the velocity

v = 0.9999c

.

Calculating the term inside the square root:

1 - \dfrac{v^2}{c^2} = (1 - 0.9999)(1 + 0.9999) = 0.0001 \times 1.9999 \approx 2 \times 10^{-4}

Taking the square root:

\sqrt{1 - \dfrac{v^2}{c^2}} \approx \sqrt{2 \times 10^{-4}} = 1.414 \times 10^{-2}

Substituting this back into the time dilation formula:

\Delta t = \dfrac{1}{1.414 \times 10^{-2}} = \dfrac{100}{1.414} = 70.7

days

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