Correct Answer - Option 2 :
\(\sqrt {\frac{8}{9}} c\)
Concept:
Time Dilation:
According to classical physics, time is an absolute quantity. But according to the special theory of relativity, Time is not an absolute quantity. It depends upon the motion of the frame of reference.
If the interval of time (say ticking of a clock) between two signals in an inertial frame S be t, then the time interval between these very two signals in another inertial frame S’ moving with respect to the first will be given by:
\(t' = \frac{t}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
v is speed of object, c is speed of light, t' is dilated time, t is original time.
Calculation:
Let speed of aircraft is v,
Given the dilated time is t' = 3t
or
\(\frac{t'}{t} = \frac{1}{3}\) -- (1)
t is time on earth, t is time on spacecraft.
\(t' = \frac{t}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
\(\implies \frac{t'}{t} = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
\(\implies \frac{1}{3} = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
Squaring
\(\implies \frac{1}{9} = \frac{1}{{ {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
\(\implies 1 - \frac{v^2}{c^2} = \frac{1}{9}\)
\(\implies \frac{v^2}{c^2} = 1- \frac{1}{9} = \frac{8}{9}\)
\(\implies v^2 = \frac{8}{9} c^2 \)
\(\implies v = \sqrt {\frac{8}{9}} c\)
So, the correct option is \(\sqrt {\frac{8}{9}} c\)
