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Question 11 Mark

If the speed of a rod moving at a relativistic speed parallel to its length is doubled,

The length will become half of the original value.
Mass will become double of the original value.
The length will decrease.
The mass will increase.

Answer

The length will decrease.
The mass will increase.

Explanation:

If the speed of a rod moving at a relativistic speed v parallel to its length, its mass:

$\text{m}=\curlyvee\text{m}_0=\frac{\text{m}_0}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

And its length,

$\text{l}=\frac{\text{l}_0}{\curlyvee}=\text{l}_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$

where, $\curlyvee=\frac{1}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}=\Big(1-\frac{\text{v}^2}{\text{c}^2}\Big)^{-\frac{1}{2}}=1+\frac{\text{v}^2}{2\text{c}^2}+\ \cdots>1$ as $\text{v}<\text{c}$

If the speed is doubled, its multiplying factor,

$\curlyvee'=\frac{1}{\sqrt{1-\frac{4\text{v}^2}{\text{c}^2}}}$

$=\Big(1-\frac{4\text{v}^2}{\text{c}^2}\Big)^{-\frac{1}{2}}$

$=1+\frac{2\text{v}^2}{\text{c}^2}+\ \cdots>2\curlyvee$

And $\text{m}=\curlyvee\text{m}_0>2\curlyvee\text{m}_0,\text{l}-\frac{\text{l}_0}{\curlyvee}<\frac{\text{l}_0}{2\curlyvee}$

Hence, the mass will increase but more than double and length will decrease but not exactly half of the original values.

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Question 21 Mark

Mark the correct statements:

Equations of special relativity are not applicable for small speeds.
Equations of special relativity are applicable for all speeds.
Nonrelativistic equations give exact result for small speeds.
Nonrelativistic equations never give exact result.

Answer

Equations of special relativity are applicable for all speeds.

Nonrelativistic equations never give exact result.

Explanation:

According to special relativity, if a particle is moving at a very high speed v, its mass:

$\text{m}=\gamma\text{ m}_0,$length $\text{l}=\frac{\text{l}_0}{\gamma}, $ change in time $\triangle\text{t}=\gamma\ \triangle\text{t}_0$

where, $\gamma=\frac{1}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}\text{i}\text{f}\ \text{v}<<\text{c}$

$\Rightarrow\gamma\cong1$

that is at non relativistic speed (small speed), $\text{m}\cong\text{m}_0,\text{l}\cong\text{l}_0,\triangle\text{t}\cong\triangle\text{t}_0$

where $\text{m}_0,\text{l}_0$ and $\triangle\text{t}_0$ are the rest mass, length and time interval respectively. Therefore, relativistic equations are applicable for all speeds. But

$\gamma=\Big(1-\frac{\text{v}^2}{\text{c}^2}\Big)^{-\frac{1}{2}}$

$\Rightarrow\gamma=1+\frac{\text{v}^2}{\text{2c}^2}+\cdots$ (expanding binomially)

$\frac{\text{v}^2}{2\text{c}^2}+\cdots=\text{K}<<1$ if $\text{v}<<\text{c}$ but silll $\text{K}>0$

Hence. non relativistic equations. in which $\curlyvee$ factor is taken to be exactly 1 never give exact results.

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MCQ 31 Mark

The magnitude of linear momentum of a particle moving at a relativistic speed $v$ is proportional to:

A
$\text{v}$
B
$\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$
C
$1-\frac{\text{v}^2}{\text{c}^2}$
✓
None of these.

Answer

Correct option: D.

None of these.

Linear momentum of a particle moving at a relativistic speed $v$ is given by:
$\text{P}=\frac{\text{m}_0\text{v}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$
Here, $m_o$ is the rest mass of the particle.
So, linear momentum is proportional to $\frac{\text{v}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$
Hence, none of the above options is correct.

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MCQ 41 Mark

An experimenter measures the length of a rod. In the cases listed, all motions are with respect to the lab and parallel to the length of the rod. In which of the cases the measured length will be minimum?

A
The rod and the experimenter move with the same speed $v$ in the same direction.
✓
The rod and the experimenter move with the same speed $v$ in opposite directions.
C
The rod moves at speed $v$ but the experimenter stays at rest.
D
The rod stays at rest but the experimenter moves with the speed $v.$

Answer

Correct option: B.

The rod and the experimenter move with the same speed $v$ in opposite directions.

If a rod is moving with speed v parallel to its length $l_0$ and the experimenter is at rest, its new length will be given as,
$\text{l}=\text{l}_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$
If a rod is moving with speed v parallel to its length $l_o$ and the experimenter is at rest, its new length will be given as,
$\text{l}=\text{l}_0\sqrt{1-\frac{(-\text{v})^2}{\text{c}^2}}=\text{l}_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$
f the rod and the experimenter both are moving with the same speed in the same direction, then $l = l_0$ while if they are moving with same speed in the opposite directions, the length of the rod will be given as,
$\text{l}=\text{l}_0\sqrt{1-\frac{\big(\text{v}-(-\text{v})\big)^2}{\text{c}^2}}=\text{l}_0\sqrt{1-\frac{4\text{v}^2}{\text{c}^2}}$
Where, $\text{v}< \text{c}$
As, $1-\frac{4\text{v}^2}{\text{c}^2}<1-\frac{\text{v}^2}{\text{c}^2}$
$\therefore\text{l}_0\sqrt{1-\frac{4\text{v}^2}{\text{c}^2}}<\text{l}_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}<\text{l}_0$
Therefore, the length will be minimum in the case when both are travelling in opposite direction.

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MCQ 51 Mark

A charged particle is projected at a very high speed perpendicular to a uniform magnetic field. The particle will:

✓
Move along a circle.
B
Move along a curve with increasing radius of curvature.
C
Move along a curve with decreasing radius of curvature.
D
Move along a straight line.

Answer

Correct option: A.

Move along a circle.

If a charged particle is projected at a very high speed perpendicular to a uniform magnetic field, its mass will increase from $m_0$ to $\text{m}=\frac{\text{m}_0}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$ and its radius will increase from.

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MCQ 61 Mark

An experimenter measures the length of a rod. Initially the experimenter and the rod are at rest with respect to the lab. Consider the following statements:

If the rod starts moving parallel to its length but the observer stays at rest, the measured length will be reduced.
If the rod stays at rest but the observer starts moving parallel to the measured length of the rod, the length will be reduced.

A
Is true but $B$ is false.
B
$B$ is true but $A$ is false.
✓
Both $A$ and $B$ are true.
D
Both $A$ and $B$ are false.

Answer

Correct option: C.

Both $A$ and $B$ are true.

If a rod is moving with speed $v$ parallel to its length $l_o$ and the observer is at rest, its new length will be given as:
$\text{l}=\text{l}_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$
As, $\text{V}<\text{c}$
$\therefore\frac{\text{v}^2}{\text{c}^2}<1$
$\Rightarrow\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}<1$
Therefore, $\text{l}<\text{l}_0$
If the rod is at rest and the observer is moving with speed $v$ parallel to measured length of the rod, the rod's length will be given as:
$\text{l}=\text{l}_0\sqrt{1-\frac{(-\text{v})^2}{\text{c}^2}}$
$\Rightarrow\text{l}=\text{l}_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$
As, $\text{v}<\text{c}$
$\therefore\Big(\frac{\text{-v}^2}{\text{c}^2}\Big)^2<1$
$\Rightarrow\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}<1$
$\Rightarrow\text{l}<\text{l}_0$
Therefore, the length will be reduced in both the cases.

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Question 71 Mark

If the speed of a particle moving. at a relativistic speed is doubled, its linear momentum will:

Become double.
Become more than double.
Remain equal.
Become leas than double.

Answer

Becomes more than double.

Explanation:

If a particle is moving at a relativistic speed v, its linear momentum (p) is given as,

$\text{P}=\frac{\text{m}_0\text{v}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

$\Rightarrow\text{P}=\text{m}_0\text{v}\Big(1-\frac{\text{v}^2}{\text{c}^2}\Big)^{-\frac{1}{2}}$

Expanding binomially and neglecting higher terms we have,

$\text{P}\simeq\text{m}_\text{v}\Big(1+\frac{\text{v}^2}{\text{2c}^2}\Big)$

$\Rightarrow\text{P}\simeq\text{m}_0\text{v}+\frac{\text{m}_0\text{v}^2}{2\text{c}^2}$

If the speed is doubled, such that it is travelling with speed 2v, linear momentum will be given as,

$\text{P}'=\frac{\text{m}_0(2\text{v})}{\sqrt{1-\frac{4\text{v}^2}{\text{c}^2}}}$

$\Rightarrow\text{P}=2\text{m}_0\text{v}\Big(1-\frac{4\text{v}^2}{\text{c}^2}\Big)^{-\frac{1}{2}}$

Expanding binomially and neglecting higher terms we have,

$\text{P}'\simeq2\text{m}_0\text{v}\big(1+\frac{4\text{v}^2}{2\text{c}^2}\Big)$

$\Rightarrow\text{P}'\simeq2\text{m}_0\text{v}+\frac{4\text{m}_0\text{v}^3}{\text{c}^3}$

$\therefore\text{P}'\simeq2\text{P}+\frac{3\text{m}_0\text{v}^3}{\text{c}^3},\frac{3\text{m}_0\text{v}^3}{\text{c}^3},>0$

Therefore, p' is more than double of p.

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Question 81 Mark

When a rod moves at a relativistic speed v, its mass:

Must incrase by a factor of $\curlyvee.$
May remain unchanged.
May increase by a factor other than $\curlyvee.$
May decrease.

Answer

Must incrase by a factor of $\curlyvee.$

Explanation:

If a rod is moving at a relativistic speed v, its mass is given by:

$\text{m}=\curlyvee\text{m}_0=\frac{\text{m}_0}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

Here,

$\curlyvee=\frac{1}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

Thus, its mass must increase by a factor of $\curlyvee.$

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Question 91 Mark

If a constant force acts on a particle, its acceleration will:

Remain constant.
Gradually decrease.
Gradually increase.
Be undefined.

Answer

Gradually decrease.

Explanation:

If a constant force is acting on a particle, the force will tend to accelerate the particle and increase its speed. But, due to increase in speed, the mass of the particle will increase by the relation $\text{m}=\frac{\text{m}_0}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

leading to decrease in acceleration as constant force is acting. Therefore, after sometime its acceleration will start decreasing gradually.

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MCQ 101 Mark

As the speed of a particle increases, its rest mass:

A
Increases.
B
Decreases.
✓
Remains the same.
D
Changes.

Answer

Correct option: C.

Remains the same.

Rest mass of a particle $m_0$ is universally constant. It doesn't vary with the frames as well as with the speed with which the particle is moving.

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Question 111 Mark

Which of the following quantities related to an electron has a finite upper limit?

Mass.
Momentum.
Speed.
Kinetic energy.

Answer

Speed.

Explanation:

If an electron is given a very high speed v, its mass:

$\text{m}=\curlyvee\text{m}_0=\frac{\text{m}_0}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

Momentum, $\text{P}=\text{mv}=\curlyvee\text{m}_0\text{v}=\frac{\text{m}_0\text{v}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

Kinetik energy, $\text{k}=\frac{1}{2}\text{mv}^2=\frac{1}{2}\curlyvee\text{m}_0\text{v}_2=\frac{1}{2}=\frac{1}{2}\frac{\text{m}_0\text{v}^2}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

$\curlyvee=\frac{1}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

At, $\text{v}=\text{c},\curlyvee=\frac{1}{\sqrt{1-\frac{\text{c}^2}{\text{c}^2}}}=\infty$

$\Rightarrow\text{at}\ \text{v}=\text{c},\text{m}=\text{p}=\text{k}=\infty$

Therefore, there's an upper bound for v to be always less than c but no upper limits for mass, momentum and kinetic energy of the electron.

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MCQ 121 Mark

Two events take place simultaneously at points $A$ and $B$ as seen in the lab frame. They also occur simultaneously in a frame moving with respect to the lab in a direction:

A
Parallel to $AB$.
✓
Perpendicular to $AB$.
C
Making an angle of $45^\circ$ with $AB$.
D
making an angle of $135^\circ$ with $AB.$

Answer

Correct option: B.

Perpendicular to $AB$.

Two events take place simultaneously at points $A$ and $B$ as seen in the Jab frame.
They also occur simultaneously in a frame moving with respect to the lab in a direction will be Perpendicular to $AB$.

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Question 131 Mark

A rod of rest length L moves at a relativistic speed. $\text{L}=\frac{\text{L}}{\curlyvee}.$ Let Its length:

Must be equal to L.
May be equal to L.
May be more than L' but less than L.
May be more than L.

Answer

May be equal to L.
May be more than L' but less than L.

Explanation:

If a rod of rest length L is moving at a relativistic speed v and its length is contracted to L', then:

$\text{L}'\frac{\text{L}}{\curlyvee}=\text{l}\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$

If $\curlyvee=\frac{1}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}},$ then $\text{v}<<\text{c},\curlyvee\cong1.$

$\Rightarrow\text{L}'\cong\text{L}$

But the length of the rod may be more than L' depending on the frame of the observer. However, it cannot be more than L because as the speed of the rod increases, its length contracts more and more due to increasing value of $\curlyvee.$

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