Question

A meter scale that is moving with a speed v along its length appears to be a centimeter-scale to a stationary observer. Which of the following is correct?
1. v² = 0.99 c²
2. v² = 0.9999 c²
3. v = 0.99 c
4. v² = 0.9 c²

Answer 1

Correct Answer - Option 2 : v² = 0.9999 c²

Concept:

If we measure the length of anything moving relative to our frame, we find its length L to be smaller than the proper length L0 that would be measured if the object were stationary. 

At relativistic speeds, Close to the speed of light, distances measured are not the same when measured by different observers.

Length Contraction:

Length contraction is the decrease in the measured length of an object from its proper length when measured in a reference frame that is moving with respect to the object.

It is given by 

\(L = {L_0}\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} \)

where L0 is the length of the object in its rest frame, and L is the length in the frame moving with velocity v;

Calculation:

Given:

L₀ = 1 m = 100 cm and L = 1 cm

Using, \(L = {L_0}\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} \)

⇒  \(1 = 100\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} \)

⇒ \(\frac{1}{{1000}} - 1 - \frac{{{v^2}}}{{{c^2}}}\)

⇒ \(\frac{{{v^2}}}{{{c^2}}} - \frac{{9999}}{{10000}} = 0.9999\)

⇒ v² = 0.9999 c²