IIT JAM Physics MCQ Test 9 - Physics MCQ

 Let lo be the length of the rod in the frame in which it is at rest and s' is the frame which is moving with a speed 0.8c in a direction making an angle 60° with x-axis. The components of lo along and perpendicular to the direction of motion are lo cos 60° and lo sin 60° respectively.
Now length of the rod along the direction of motion 
= lo cos 60°_/1-(0.8) 2/c2
= lo/2×0.6
= 0.3 lo.
Length of the rod perpendicular to the direction of motion.
= lo sin 60°
=_/3/2 lo
Length of moving rod
l = [(0.3lo)2+{lo_/3/2} 2] 1/2
= 0.916 lo.
Percentage contraction
= lo-0.916lo/lo×100
= 8.4%.

X = ( x - vt ) A

T = ( t - vx/c^2 )

 Y = y ,  Z = z

X = ( x - vt) / ( 1 - v^2 / c^2)^½

   = 100 - ( - 0.8 * 3 * 10^5 km/s *5 * 10^-4) / (1 - 0.8 ^2 )^½

X = 367km

T = ( t - vx/c^2) / (1-v^2 /c^2 )

   = 5 * 10 - ( - 0.8 * 3 * 10^5 * 100 km) / (1 - 0.8 ^2 )^½

    = 1.28 ms

y=10 , z=1

Let L be the length of the line as seen in the first spaceship. The horizontal and vertical extents of the line are each L/√2. In the second ship the horizontal extent is shortened to  while the vertical extent is unchanged. Then

or 
and θ = 17.3°

Let 0-6c be the velocity of the rod along X-axis, then we have 

As the contraction takes place only along the direction of motion, in this case the contraction will take place only along X-axis; while the component of rod along Y-axis will remain the same.
If I_x’ is the X component of the rod, when in motion, then we have

The Y-component of the rod, when in motion, will be the same, i.e.,

The length of the rod in motion.

If θ is the inclination of the rod with X-axis in S’, then

or tan θ = 0.72
θ = tan^-1 (0.72) 

We know 


These type of event are happened at different place but at same time .so,

i.e, 
 [for condition of like space]

 Which is not possible.

Given rest mass energy  
And kinetic energy T = 9.69 mev.
We know from einstein theory
E = mc² = KE. + rest mass energy 
mc² = T + m₀c²

Let us suppose that the particle moving with speed -0.9c is at rest in system S. The it may be supposed that laboratory (S’) is moving with velocity + 0.9c relative to S along + ve X-axis, 
i.e. v = 0.9c
∴ The velocity of the particle relative to S’, i.e. the relative speed of the particles is given by

= 0.995c.

A spaceship is moving with velocity less than velocity of light v < c. So this is the case of relativistic mechanics and to analysis the motion . We can use Application of Lorentz transformation equation. So,
According to “ time dealation “ A clocks running slower than a stationary clock by a factor  because observed time  Where  proper time interval.
According to Length contraction it’s Length shrunk in the direction of relative  motion by a factor of  because observed length  Where L₀ is proper length.
Similarly mass in motion is m  where m_o is rest mass.
So, it’s mass increased by a factor of 
So all of the above statements are true for a spaceship in space.

Speed of photon = C 
Speed of astronaut = 0.8C
Let the photon and astronauts super spaceship moving along positive and negative direction of x - axis respectively.
Let the electron moving with velocity - 0.8C.
We know from addition of velocity the x - component of velocity in moving from is

Energy incident per sec = Number of incident photon per sec x 
Or 
h = 6.62 x 10^-34 J.s c = 
∴ N = 4 x 10¹¹ photons per sec.

The wavelength of visible light is between 400 and 700 nm. Thus we can take a typical visible wavelength to be
 ...(i)
The energy of a single photon is  ...(ii)
Before we evaluate this, it is useful to note that the product he enters into many calculations and is a useful combination to remember. Since h has the dimension energy * time, he has the dimension energy * length and is conveniently expressed in eV-nm, as follows 

or he = 1240 eV.nm ...(iii)
Putting numbers into (2), we find the energy for a typical visible photons to be
 ...(iv)
On the atomic level this energy is significant, but by everyday standards it is extremely small. When we look at the moon, the energy entering our eye per second is given by IA, where I is the intensity  and A is the area of the pupil  if we take the diameter of the pupil to be about 6mm). Thus the number of photons entering our eye per second is number of photons per second  photons per second This is such a large number that the restriction to integer numbers of photons is quite unimportant even for this weak source.

The kinetic energy of photoelectron is given as 
KE = hv-W
where v → frequency o f incident photon 
W → work function of the substance 
h→ Planck constant

Now, hv₀ = 8 eV, KE = 2eV 
So, work function = (8 - 2) eV 
W = 6 eV
Now, if _hv0 = 8 eV
⇒ 1.25 hv₀ = 10 eV
Thus, hv = 10 e V , W = 6 eV
⇒ Kinetic energy = hv - W = 10 eV - 6 eV = 4 eV

Mosley’s law states that square root of frequency for Kα -lines is directly proportional to the atomic number of the element i.e,

or 
Adding Eqs. (iv) and (vi)

⇒ 

The momentum of the photon is given by

The uncertainty in the photon momentum is (working with magnitudes only):

from which

Let  Axileaabe the least uncertainty in the position of the electron and  the maximum uncertainty in the momentum of the electron. Then

[where m₀ be the rest mass of electron]

or 
Given 
v = 2.8 x 10⁹cm^-1 = 2.8 * 10⁷ ms^-1
m₀ = 9.1 * 10^-31 kg 
c = 3 x ¹⁰8 ms^-1
∴  = 0.046 x 10^-10 m = 0.046

In the photoelectric effect, by conservation of energy

where is the work function of the metal. Furthermore,

where is the work function of the metal. Furthermore,

Thus the stopping potential V₀ is directly proportional to the incident light frequency v²

In fact this is one way of determining Planck’s constant.

For Pauli matrices,  

We know pauli spin matrices are

and they obey the following commutation relation-

and 
So, A, B are never correct option.

Let us take 

But 
SO ]
We know that
[ab, c] = a [b, c] + [a, c] b.
So, 


Similarly,

so that  commutes with each component  

All statements -A,B,C are possible but D are not possible because fission of nucleus is achieved by neutron not by protons.

R = R₀A^1/3

⇒ 
and 

Distance travelled by it before decay


= 9.46km

Now, as we know
m =
Here, m = 61 kg and m₀ = 60 kg
Therefore, 61 / 60 =
v² / c² = 1 – (60 / 61)² = 121 / 3600

v = 11 X c / 60 = 5.5 X 10⁷ m/s

Since number nucleo ns in nucleus is 4, hence B.E. for  = 28.0 MeV, similarly B.E. for ²H nucleus (combination of one proton and one neutron) = 2.2 MeV.
Mass of  nucleus = 2 [Mess of proton] + 2 [Mass of neutron) - 28.0 MeV. 
Mass of  nucleus = (Mass of proton) + [Mass of neutron) - 2.2 MeV.
In fusion reaction energy released, ΔE = 

34.980094
The ₁₇CI³⁵ atom has 17 protons and 18 neutrons in its nucleus.
M ass o f 1 7 ₁H¹ atom i.e., 17 protons and 17 electrons = 17 * 1.008143
= 17-138431 a.m.u.
Mass of 18 neutrons = 18 * 1.008986
= 18.161748 a.m.u.
Total = 35.300179 a.m.u.
Mass defect Δm = 298/931 = 0.320085 a.m.u.
The atomic mass of ₁₇CI³⁵ would be the sum of the masses of the constituent particles minus the mass equivalent of the binding energy of the nucleus. Hence Atomic mass of ₁₇CI³⁵ = 35.300179 - 0-320085
= 34.980094 a.m.u.

0.000048
Here; 
= 0.0327 x 1.6 x 10^-19 J
or 
= 0.0327 x 1.6 x 10^-19
On solving, we get 
v =2.5 x 10³ m /s.
Time, 
Now 
∴ Fraction of neutron decayed

= 0.000048.

The decay constant of C¹⁴ is


= 3.83 x 10^-12 s^-1
To find the number C¹⁴ nuclei in 20 g of burnt wood , we first calculate the number of C¹² nuclei in 20 g of carbon (burnt wood).
Thus 
Now, assuming that the ratio of C¹⁴ to C¹² is 1.3 x 10^-12,the number of C¹⁴ nuclei in 20 g before decay is

We thus have for the initial activity of the sample 

= 4.979 decays/s
= decays/s
The age of the sample can now be calculated from the relation 

or 
or 
It is given that R = 4 decays/s and we have calculated R₀ = 5 decays/s.
Thus 

= 0.58 x 10¹¹ s
= 1842 years

Here 
Distance of the star from the earth = 4 light years.
Total distance travelled by the rocket = 8 light years.
Time taken by the rocket to travel 8 light years as measured by a clock on the earth is given by

Let t₀ be the time measured by the clock on the rocket, we have

Putting the values of t and v² / c², we have

= 0.08 years.

According to Einstein's mass-energy equivalence theory, E₀=m₀c², where E₀ is the rest mass energy, m₀=9.11×10⁻³¹kg be the rest mass of the electron and c be the velocity of light.