This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Einstein’s Special Theory of Relativity”.
1. As an object approaches the speed of light, it’s mass becomes _____________
a) Zero
b) Double
c) Remains Same
d) Infinite
View Answer
Explanation: We know, for an object in motion m=\(\frac{m_0}{\sqrt{1-\frac{V^2}{c^2}}}\)
Thus, as v = c, the quantity in denominator becomes zero. Hence the mass of the object becomes infinite.
2. If the sun radiates energy at the rate of 4 x 1026 Js-1, what is the rate at which its mass is decreasing?
a) 5.54 x 109 kgs-1
b) 4.44 x 109 kgs-1
c) 3.44 x 109 kgs-1
d) 2.44 x 109 kgs-1
View Answer
Explanation: As we know from Einstein’s mass energy relation, E = mc2.
Therefore, ΔE= Δmc2
ΔE = 4 x 1026 Js-1, c = 3 x 108 ms-1.
Δm = 4 x 1026 Js-1/ 9 x 1016m2s-2
Δm = 4.44 x 109 kgs-1.
3. The orbit of mercury is changing slightly due to the sun’s gravity.
a) True
b) False
View Answer
Explanation: According to Einstein’s theory of relativity, a massive object distorted space and time. Thus, due to the curvature of space-time around the massive sun, the orbit of mercury is shifting gradually over time.
4. According to Einstein’s Special Theory of Relativity, laws of physics can be formulated based on ____________
a) Inertial Frame of Reference
b) Non-Inertial Frame of Reference
c) Both Inertial and Non-Inertial Frame of Reference
d) Quantum State
View Answer
Explanation: One of the postulates of Einstein’s Special Theory of Relativity states that all the inertial frames are equivalent to the formulation of laws of physics. Thus, it is suitable for all the inertial frames.
5. For Einstein’s relation, E2 – p2c2 = _____________
a) moc2
b) mo2c4
c) moc4
d) mo2c6
View Answer
Explanation: We know, E = mc2 and momentum, p = mv
Now, E2 – p2c2 = m2c4 – m2v2c2
Now, we know, m = \(\frac{m_0}{\sqrt{1-\frac{V^2}{c^2}}}\)
Therefore, E2 – p2c2 = m2c4(1-v2/c2)
=c4m2(1-v2/c2)
Using, m2(1-v2/c2) = mo2 we get
E2 – p2c2 = mo2c4.
6. A frame of reference has four coordinates, x, y, z, and t is referred to as the_____________
a) Inertial frame of reference
b) Non-inertial frame of reference
c) Space-time reference
d) Four-dimensional plane
View Answer
Explanation: Such a frame, having four coordinates of x, y, z and time t is called the space-time frame. It plays a major role in Einstein’s special theory of relativity.
7. A man, who weighs 60 kg on earth, weighs 61 kg on a rocket, as measured by an observer on earth. What is the speed of the rocket?
a) 2.5 X 108 m/s
b) 2.5 X 107 m/s
c) 5.5 X 107 m/s
d) 5.5 X 108 m/s
View Answer
Explanation: Now, as we know m = \(\frac{m_0}{\sqrt{1-\frac{V^2}{c^2}}}\)
Here, m = 61 kg and m0 = 60 kg
Therefore, 61/60 = 1/\({\sqrt{1-\frac{V^2}{c^2}}}\)
v2/c2 = 1 – (60/61)2 = 121/3600
v = 11 X c/60 = 5.5 X 107 m/s.
8. The momentum of a photon having energy 1.00 X 10-17 J is ____________
a) 2.33 X 10-26 kg m/s
b) 3.33 X 10-26 kg m/s
c) 4.33 X 10-26 kg m/s
d) 5.33 X 10-26 kg m/s
View Answer
Explanation: Now, the rest mass of a photon is zero.
Therefore, it’s momentum, p = E/c
= 1.00 X 10-17/3 X 108
= 3.33 X 10-26 kg m/s.
9. According to Einstein’s special theory of relativity, which of these objects should be the heaviest?
a) A
b) B
c) C
d) D
View Answer
Explanation: As we know, m = \(\frac{m_0}{\sqrt{1-\frac{V^2}{c^2}}}\)
Using this formula on all of the objects we get,
For object A: m = 1.127 m0
For object B: m = 2.197 m0
For object C: m = 3.034 m0
For object D: m = 0.70 m0
Hence, Object C has the highest mass.
10. Which effect is shown by the following figure?
a) Gravitational Redshift
b) Gravitational Blueshift
c) Gravitational Lensing
d) Gravitational force
View Answer
Explanation: Gravitational lensing refers to the distribution of matter between the light source and the observer such that it can bend the light coming from the source to the observer. According to Einstein’s theory of special relativity, it happens because a massive object can alter the space-time frame around itself.
Sanfoundry Global Education & Learning Series – Engineering Physics.
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