Special Theory of Relativity MCQ Quiz in मल्याळम - Objective Question with Answer for Special Theory of Relativity - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

The total energy of a particle at rest can be calculated using Einstein's mass-energy equivalence formula:

\(E=mc^2\ \)

where, E is the total energy, m is the rest mass of the particle, and c is the speed of light.

Calculation:

Total Energy in Electron Volts

Given:

• Rest mass of the particle \(m = 5 \text{kg}\ \).
• Speed of light \( c = 3 \times 10^8 \text{m/s} \ \).
• Conversion factor: \(1 \text{ J} = 6.242 \times 10^{12} \text{ eV} \ \).

Using the formula \(E=mc^2\ \):

\( E = 5 , \text{kg} \times (3 \times 10^8 \text{m/s})^2 \\E = 5 \times 9 \times 10^{16} \text{kg} \cdot \text{m}^2/\text{s}^2 \\ E = 45 \times 10^{16} \text{J} \)

To convert the energy from joules to electron volts:

\( E_{\text{eV}} = 45 \times 10^{16} \text{J} \times 6.242 \times 10^{12} \text{eV/J} ) \\ E_{\text{eV}} = 45 \times 6.242 \times 10^{28} \text{eV}\\ E_{\text{eV}} \approx 280.89 \times 10^{28-9} \text{eV} \approx 2.81 \times 10^{20} \text{eV} \)

∴ The correct answer is option A.

Concept:

When a source of light is moving relative to an observer, the rate at which photons are received by the observer can be affected by the Doppler effect. For a source moving towards the observer, the observed frequency of the emitted photons increases (blue shift), while for a source moving away, it decreases (red shift). The relativistic Doppler effect for a source moving towards the observer is given by:

\( f' = f \sqrt{\frac{1 + β}{1 - β}} \ \)

where, f' is the observed frequency, f is the emitted frequency, and \( β = \frac{v}{c}\ \) is the velocity of the source as a fraction of the speed of light.

Calculation:

Number of Photons per Second Received by the Observer

Given:

• The star is emitting light at a rate of 1000 photons per second ( f = 1000 photons/second).
• The star is moving towards Earth at a velocity v = 0.7c  and β = 0.7.

Using the relativistic Doppler effect formula:

\(f' = 1000 \sqrt{\frac{1 + 0.7}{1 - 0.7}} = 1000 \sqrt{\frac{1.7}{0.3}} \\ f' = 1000 \sqrt{5.67} \approx 1000 \times 2.38 = 2380 \ \text{photons per second} \)

Since none of the given options exactly matches the calculated value, let's recheck and simplify the Doppler effect without underestimating intermediate values:

Simplifying:

\(f' \approx 1000 \times \sqrt{\frac{1.7}{0.3}} = 1000 \times \sqrt{5.67} = 1000 \times 2.38 \approx 2380 \text{ photons per second}\ \)

∴ The correct answer closely approximates to option C.

Concept:

According to the theory of Special Relativity, the speed of light in a vacuum is always c, regardless of the relative motion of the source and the observer. This postulate holds true even if the source and the observer are moving at significant fractions of the speed of light.

Calculation:

Speed of Light as Measured by the Spaceship

Given that the spaceship is traveling at 0.6c and it measures a red flashlight beam (directed backwards) with respect to the space station, the speed of light remains constant at c in all inertial frames of reference, according to Special Relativity.

Regardless of the motion of the spaceship or the space station, the speed of the flashlight beam as seen from the spaceship will remain c.

∴ The correct answer is option B.

Concept:

When a clock moves at a high speed, time dilation occurs according to the theory of Special Relativity. The time measured by the moving clock (proper time) will be less than the time measured by the stationary clock (coordinate time). The difference in the time measurements can be calculated using the Lorentz factor \(\gamma \ \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \ \). For small velocities compared to the speed of light (non-relativistic speeds), the time dilation can be approximated using the binomial expansion for the Lorentz factor.

Calculation:

Time Difference Due to Time Dilation

Given the jet's speed v = 900 km/h, which is \(v = 900 \times \frac{1000}{3600} \text{m/s} = 250 \text{m/s}\), we need to determine the time difference between the moving atomic clock and the stationary clock after 10 hours.

First, we convert 10 hours into seconds:

\(t = 10 \text{ hours} = 10 \times 3600 = 36000 \text{ seconds} \ \)

Next, we calculate the Lorentz factor \(\gamma \ \) for the jet's speed:

Since, v is much smaller than c, we can use the binomial expansion approximation for the Lorentz factor:

\( \gamma \approx 1 + \frac{v^2}{2c^2}\ \)

Substituting \(v = 250 \text{ m/s} \ \) and \( c = 3 \times 10^8 \text{ m/s}\ \):

\(\frac{v^2}{2c^2} = \frac{(250)^2}{2 \times (3 \times 10^8)^2} = \frac{62500}{2 \times 9 \times 10^{16}} = \frac{62500}{18 \times 10^{16}} = 3.47 \times 10^{-12} \ \)

Therefore,

\( \gamma \approx 1 + 3.47 \times 10^{-12} \ \)

Now we calculate the time difference \( \Delta t \ \) between the moving clock and the stationary clock:

\(\Delta t = t - t_{\text{moving}} = t - \frac{t}{\gamma} \approx t - t (1 - \frac{v^2}{2c^2}) = t \left( \frac{v^2}{2c^2} \right) \ \ \)

Substituting \( t = 36000 \text{ seconds} \ \) and the \( \frac{v^2}{2c^2}\ \) factor:

\( \Delta t \approx 36000 \times 3.47 \times 10^{-12} \text{ seconds} = 1.25 \times 10^{-7} \text{ seconds} \ \ \)

∴ The correct answer is closest to option D.

Concept:

The lifetime of a muon as measured in the Earth's frame is extended due to time dilation. The time dilation effect in Special Relativity states that the observed lifetime \((t)\ \) in the Earth's frame is related to the proper lifetime (\( t_0\ \)) by the Lorentz factor (\(\gamma \ \)), where ( \(t = \gamma t_0 \ \)). The distance d traveled by the muon is given by \((d = vt)\ \), where v is the velocity of the muon.

Calculation:

Determining if the Muon Reaches the Ground

Given that the muon is moving at (v = 0.998c) and its proper lifetime \( t_0 = 2.2 \text{ microseconds}=2.2 \times 10^{-6} \text{seconds}\ \) , we first calculate the Lorentz factor \( \gamma\ \). The Lorentz factor is given by:

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

Substituting v = 0.998c:

\( \gamma = \frac{1}{\sqrt{1 - (0.998)^2}} = \frac{1}{\sqrt{1 - 0.996004}} = \frac{1}{\sqrt{0.003996}} \approx 15.82 \ \)

Next, we calculate the dilated lifetime t observed in the Earth's frame:

\( t = \gamma t_0 = 15.82 \times 2.2 \times 10^{-6} \text{ seconds} \approx 3.48 \times 10^{-5} \text{ seconds}\ \)

Now, we find the distance d traveled by the muon during its dilated lifetime:

\(d = vt = 0.998c \times 3.48 \times 10^{-5} \text{ seconds} \ \)

Substituting \( c = 3 \times 10^8 \text{ m/s} \ \):

\( d = 0.998 \times 3 \times 10^8 \text{ m/s} \times 3.48 \times 10^{-5} \text{ seconds} \approx 1.04 \times 10^4 \text{ meters} \ \)

Since this distance (approximately 10,400 meters) is greater than the initial altitude of 5000 meters, the muon will indeed reach the ground before decaying.

∴ The correct answer is option B.

Concept:

The annihilation of an electron and a positron produces two photons. The total energy of each particle before the collision is the sum of its rest mass energy and kinetic energy, which must be conserved and converted into the photons' energy. Using the concept of relativistic energy, the total energy (E) of a particle moving at a speed (v) is given by

\(E = \gamma m c^2 \ \),

where,  \(\gamma \ \) is the Lorentz factor  \( \left(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right). \)

Calculation:

Energy Conservation in Particle Annihilation

Given that the rest mass of the electron and positron is \(m = 9.11 \times 10^{-31} \text{kg} \ \) and their speed prior to the collision is v = 0.99c:

First, we calculate the Lorentz factor \(\gamma \ \) for the speed 0.99c:

\(\gamma = \frac{1}{\sqrt{1 - (0.99)^2}} = \frac{1}{\sqrt{1 - 0.9801}} = \frac{1}{\sqrt{0.0199}} \approx 7.09 \ \)

The total energy of each particle before annihilation is:

\( E_{\text{total}} = \gamma m c^2 \ \)

Substituting the values, we get:

\(E_{\text{total}} = 7.09 \times 9.11 \times 10^{-31} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \ \)

Simplifying the constants, we convert this energy to MeV. where, \(mc^2 \approx 0.511 \ MeV\ \) per particle for an electron or positron at rest:

\( E_{\text{total}} = 7.09 \times 0.511 \text{ MeV} \approx 3.62 \text{ MeV} \ \)

Since each particle (electron and positron) is annihilated, its total energy is converted into the photon energy. The total energy for both particles is:

\( E_{\text{photons}} = 3.62 \text{ MeV} + 3.62 \text{ MeV} = 7.24 \text{ MeV}\ \)

This energy is shared equally by the two photons produced:

\( E_{\text{each photon}} = \frac{7.24}{2} \approx 3.62 \text{ MeV}\ \)

∴ The correct answer is option D.

Concept:

Lorentz transformations are used to relate the space and time coordinates of two observers in uniform relative motion within the framework of special relativity. To determine which differential operator remains invariant, we need to consider Lorentz invariance. A quantity is Lorentz invariant if it has the same value in all inertial frames of reference.

For a differential operator to be invariant under Lorentz transformation, it must retain its form (remain unchanged) when transforming the space and time coordinates between different inertial frames.

We recognize that the D'Alembertian operator, which is used in the wave equation and in the field equations of electromagnetism, is invariant under Lorentz transformations. The D'Alembertian operator in four-dimensional spacetime is:

\( \Box = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \ \)

This operator is invariant under Lorentz transformations because it corresponds to the invariant spacetime interval in four-dimensional spacetime. It applies the same way regardless of the observer's inertial frame.

Let's analyze the options:

\(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} - \frac{1}{c^2} \frac{\partial}{\partial t}\\\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} + \frac{1}{c^2} \frac{\partial^2}{\partial t^2}\\ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}\\\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\)

From these, option 3 matches the D'Alembertian operator form, which is known to be Lorentz invariant.

Hence, the correct answer is option 3.

Explanation:

Given:

\(t=48 \ \text{h}\)

One complete rotation of earth in its rest frame is \(24\text{ h}\)

 From the concept of time dilation:
\(t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}\\ \Rightarrow \frac{24}{\sqrt{1 - \frac{v^2}{c^2}}} =48\)

\(\Rightarrow \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{2} \Rightarrow 1 - \frac{v^2}{c^2} = \frac{1}{4}\)
\(\Rightarrow \frac{v^2}{c^2} = \frac{3}{4} \Rightarrow v = \frac{\sqrt{3}}{2} c = 0.85c\)

The value of f is 0.85.

Explanation:

Initial Conditions :
    The speeds \(v_{1,0}\) and\( v_{2,0}\) refer to the velocities of two particles in a given reference frame \(F_0\) .
    The formula for the speed of\( F_2\) with respect to \(F_0 \) is derived from the relativistic velocity addition formula.
The relativistic velocity addition formula states:
   
  \( v = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}\)
   
   Where c is the speed of light.

The speed \(v_{2,0}\) is calculated as:
   
   \(v_{2,0} = v_{1,0} + v_{2,1} = \frac{c}{2} + \frac{4c}{5}\)
   
   Substituting the values:
   
  \( v_{2,0} = \frac{5c}{10} + \frac{8c}{10} = \frac{13c}{10}\)
   
   The speed \(v_{2,0}\) with respect to \(F_0\) is expressed using:
   
   \(v_{2,0} = \frac{v_{1,0} + v_{2,1}}{1 + \frac{v_{1,0} v_{2,1}}{c^2}}\)
   
   Which leads to:
   
  \( v_{2,0} = \frac{\frac{c}{2} + \frac{4c}{5}}{1 + \frac{\frac{c}{2} \cdot \frac{4c}{5}}{c^2}} = \frac{\frac{13c}{10}}{1 + \frac{2c^2}{10c^2}} = \frac{\frac{13c}{10}}{\frac{14}{10}} = \frac{13c}{14}\)
  Now,
   
  \( v_{p} = \frac{c(5 + 8)}{2} = \frac{c \cdot 13}{10}\)
   
   This gives:
   
   \(v_{p} = 0.93c \quad \text{(where } f \approx 0.93 \text{)}\)
  The value of f is 0.93.