Question

A clock keeps correct time. With what speed should it be moved relative to an observer so that it may seem to lose 20 sec in 100 sec?

• A. \( 0.6 \times 10^8 \) m/sec
• B. \( 1.8 \times 10^8 \) m/sec
• C. \( 0.64 \times 10^8 \) m/sec
• D. \( 2.0 \times 10^8 \) m/sec

Hint:

Time dilation occurs in moving frames. The faster an object moves, the slower its time appears to an observer.

Answer 1

The Correct Option is B

Time dilation in special relativity is given by:
\[ \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \] Given \( \Delta t' = 100 \) sec and \( \Delta t = 80 \) sec:
\[ 100 = \frac{80}{\sqrt{1 - \frac{v^2}{c^2}}} \] Solving for \( v \):
\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{4}{5} \] \[ 1 - \frac{v^2}{c^2} = \frac{16}{25} \] \[ \frac{v^2}{c^2} = \frac{9}{25} \] \[ v = 0.6 c = 1.8 \times 10^8 \text{ m/sec} \] Thus, the correct answer is:
\[ Option 2: 1.8 \times 10^8 \text{ m/sec}. \]