Question

A person is watching a rocket with an astronaut inside move by at a speed near the speed of light.
Which of the following statements are true?
A. The passage of time on the astronaut’s which is faster form the person’s perspective than form the astronaut’s
B. The passage of time on the astronaut’s watch is the same from the perspective of the person and the astronaut
C. The passage of time on the astronaut’s watch is faster from the perspective of the astronaut than form the perspective of the person
D. The passage of time on the person’s watch is faster, from his own perspective, as the rocket files by, than it was before the rocket flew by
E. The passage of time on the astronaut’s watch is faster, from his own perspective, as the flies by the person, than it was before he flew by the person

Answer 1

Hint:Einstein had given the theory of relative difference in the different quantities when a body moves with the speed more to the speed of light we can solve this problem by conditioning the astronaut and the person two different frames references.
Formula used:
${t^{ - 1}} = \dfrac{l}{{\sqrt {l - \dfrac{{{v^2}}}{{{c^2}}}} }}$.

Complete step by step answer:
Let us imagine the rocket and the astronaut in one frame of reference. And the person, observing the rocked into another frame of reference.
Now, using the theory of relativity given by Einstein, we can write
${t^1} = \dfrac{t}{{\sqrt {l - \dfrac{{{v^2}}}{{{c^2}}}} }}$ . . . (1)
Where,
${t^1}$ is the rest time observed by the person.
t is the rest time observed by the astronaut.
$v$ is the velocity of a rocket.
$c$ is the speed of light.
Now, check out the following observation.
Even though the speed of the rocket is closed to the speed of light, it would steel be less than the speed of light.
$ \Rightarrow v < c$
Divide both the sides by c
$ \Rightarrow \dfrac{v}{c} < 1$
By squaring both the sides, we get
$\dfrac{{{v^2}}}{{{c^2}}} < 1$
Multiplying both the sides by$ - ve$sing, we get
$ - \dfrac{{{v^2}}}{{{c^2}}} > - 1$ (since, multiplying by –ve sing changes the inequality)
Now, add 1 to both the sides of above inequality
$ \Rightarrow l - \dfrac{{{v^2}}}{{{c^2}}} > l - 1$
$ \Rightarrow l - \dfrac{{{v^2}}}{{{c^2}}} > 0$
‘Taking square root to both the sides, we get
$\sqrt {l - \dfrac{{{v^2}}}{{{c^2}}} > - 0} $
$\therefore \sqrt {l - \dfrac{{{v^2}}}{{{c^2}}}} $ exists
Now, we can observe that
$\sqrt {l - \dfrac{{{v^2}}}{{{c^2}}} < 1} $
Therefore, reciprocal of above equation
$\dfrac{l}{{\sqrt {l - \dfrac{{{v^2}}}{{{c^2}}}} }} > 1.$
Since, taking reciprocal changes the direction of inequality.
From this observation, we can say that, in equation (1)
${t^1} = \dfrac{t}{{\sqrt {l - \dfrac{{{v^2}}}{{{c^2}}}} }}$
Since, $\dfrac{l}{{\sqrt {l - \dfrac{{{v^2}}}{{{c^2}}}} }} > 1.$
${t^1}$must be greater than t
i.e. ${t^1} > t$
$ \Rightarrow $The time observed by the person will be greater than the time observed by the astronaut in this watch.
Therefore, the correct option is (A).

Note:In order to do this question, we must have known about the inequality because it helps us in comparing the things. Moreover, we can always understand the concept of relativity whenever a body moves with the velocity near about the speed of light.