# Engineering Physics Questions and Answers – Lorentz Transformation Equations

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This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Lorentz Transformation Equations”.

1. A rod of length 5 m is moving at a speed of 0.6 c. To an observer sitting perpendicular to the direction of motion, the length appears to be _______________
a) 5 m
b) 4 m
c) 3 m
d) 2 m

Explanation: In Lorentz Length transformation, there is no change in the dimensions of the objects in the direction perpendicular to the direction of motion. Thus, to the observer, the length remains the same.

2. If an object reaches the speed of light, it’s length changes to ___________
a) Infinite
b) Double of the value
c) Half of the value
d) Zero

Explanation: We know, l = l0$$\sqrt{1-\frac{v^2}{c^2}}$$. Thus, as v tends to c, the length of the rod in motion reduces to zero, which is physically impossible.

3. A 20-year-old person goes at a high speed in a rocket on his birthday. when he comes back to earth after 1 earth year, he would be ___________
a) 1 year older
b) 2 years older
c) A few months older
d) Same age

Explanation: When a person is going around at high speed, time dilation takes place. For that person, the time starts running slowly. Thus, as 1 earth year has passed away but for that person, it must have been only a few months.

4. The length of a rod seems shorter to an observer when it moves in a specific direction. What change would he observe when the direction of rod changes by 180o?
a) The rod becomes even smaller
b) The length of the rod increases
c) The length of the rod remains the same
d) The rod has the length equal to its proper length

Explanation: As we know, by Lorentz contraction, l = l0$$\sqrt{1-\frac{v^2}{c^2}}$$.
Thus, as it deals with the square of velocity, v, there is no effect on the length of the rod by changing the direction of the rod from v to -v.

5. An object of length 1 m is moving at speed 0.5c. To an observer at rest relative to the object, the length of the object seems to be ___________
a) 0.86 m
b) 0.5 m
c) 1 m
d) 0.14 m

Explanation: The observer at rest relative to the object does not notice any kind of contraction of length of the object. It is so because the scale with which he measures will also get contracted by the same amount.

6. How fast does a rocket have to move relative to an observer for its length to be contracted to 95% of its original length?
a) 0.5 c
b) 0.4 c
c) 0.3 c
d) 0.2 c

Explanation: l = 0.95lo, v =?
l = l0$$\sqrt{1-\frac{v^2}{c^2}}$$
Putting values, we get,
0.95lo = l0$$\sqrt{1-\frac{v^2}{c^2}}$$
v2/c2 = 0.0975
v ≈ 0.3 c.

7. A cube with sides of proper length lo is viewed from a reference frame moving with velocity v, parallel to an edge of the cube. The expression for the volume of the cube for the observer is __________
a) Lo3
b) Lo3$$\sqrt{1-\frac{v^2}{c^2}}$$
c) Lo3$$(1-\frac{v^2}{c^2})^{3/2}$$
d) Lo3$$(1-\frac{v^2}{c^2})$$

Explanation: The rest volume of the cube is Lo3. However, as the observer is moving at a high speed, one of the edges of the cube contracts. The rest of the edges, however, retain their proper length.
Thus, the edge of the cube reduces to l0$$\sqrt{1-\frac{v^2}{c^2}}$$
Volume = Lo3$$\sqrt{1-\frac{v^2}{c^2}}$$.

8. A rod of length 1m moves with a speed of 0.5 c. How much length contraction takes place?
a) 50 %
b) 14 %
c) 10 %
d) 35 %

Explanation: We know, l = l0$$\sqrt{1-\frac{v^2}{c^2}}$$
Therefore, as v = 0.5 c and lo = 1 m we get,
L = 0.86 m
% contraction = (1 – 0.86)/1 X 100 %
= 14%.

9. A particle with a lifetime of 2 X 10-6 s moves through the laboratory with a speed of 0.9 c. It’s lifetime, as measured by an observer in the laboratory, is ___________
a) 2 X 10-6 s
b) 3.2 X 10-6 s
c) 4.6 X 10-6 s
d) 5.4 X 10-6 s

Explanation: Δt0 = 2 X 10-6 s, v = 0.9c, Δt =?
Δt = $$\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
= 2 X 10-6 s/0.43
= 4.6 X 10-6s.

10. All the given particles have a lifetime of 1 microsecond. Which of them will survive the longest?

a) A
b) B
c) C
d) D

Explanation: As we know, Δt = $$\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Thus, as v approaches c, Δt approaches infinity.
Hence, the one with the maximum speed would survive the longest, as the time would be dilated maximum for it.

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