IIT JAM Physics MCQ Test 9

30 Questions MCQ Test Mock Test Series for IIT JAM Physics | IIT JAM Physics MCQ Test 9

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Calculate the percentage contraction of a rod moving with a velocity 0.8c in a direction inclined at 60o to its own length.


 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%.


As measured by S a flashbulb goes off at x = 100 km, y = 10 km, z = 1 km at t = 0.5 ms. What are the coordinates x' y' z' and t’ of this event as determined by a second observer, S, moving relative to S at - 0 . 8 c along the common xx ’ axis?


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


The insignia painted on the side of spaceship is a circle with a line across it at 45° to the vertical. As the ship shoots past another ship in space, with a relative speed of 0.95c, the second ship observes the insignia. What angle does the observed line make to the vertical?


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

and θ = 17.3°


A thin rod has proper length I0. If the rod is moving at 0-6c is a direction at 30o to its own length, calculate its length and inclination.


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 Ix’ 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) 


Let two events A and B occur at two space points  and difference vector be  then the condition for xμ to be like space is


We know 

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

 [for condition of like space]

 Which is not possible.


Rest mass energy of an election is 0. 51 MeV. A moving election has a kinetic energy of 9.69 MeV. The ratio of the mass of the moving electron to its mass is


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


Two particles are travelling in opposite directions with speed 0.9c relative to laboratory. Calculate their relative speed.


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 in space will have


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 mo 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.


An astronaut moves in a super spaceship travelling at a speed of 0.8C. The astronaut observes a photon approaching him from space. The speed of photon with respect to the astronaut is


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


The power incident on a detector of light is 100 nW. Determine the number of photons per second incident on the detector if the wavelength is 800 nm.


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


What is the energy of a typical visible photon? About how many photons enter the eye per second when one looks at a weak source of light such as the moon, which produces light of intensity of about 3 * 10 -4 watts/m2?


The wavelength of visible light is between 400 and 700 nm. Thus we can take a typical visible wavelength to be
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
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.


In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?


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


For the X-ray spectrum due to transition between n = 2 and n = 1 states for large nuclear charge Ze, we have frequencies   respectively . Mosley’s law implies which one of the following equations?


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

Adding Eqs. (iv) and (vi)



What is the uncertainty in the location of a photon of wavelength 3000 Å if this wavelength is known to an accuracy of one part in a million?


The momentum of the photon is given by

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

from which


The least possible uncertainty in the position of an electron moving with velocity  2.8 x 109 cms -1 is


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]

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

*Multiple options can be correct

In the photoelectric effect, electromagnetic radiation is incident upon the surface of a metal. Which of the following are true statement about the photoelectric effect?


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.

*Multiple options can be correct

Which one of the following relations are false for Pauli matrices 


For Pauli matrices,  

*Multiple options can be correct

The matrices are called Paulis spin matrices, then which one of the relations are incorrect?


We know pauli spin matrices are

and they obey the following commutation relation-

So, A, B are never correct option.

Let us take 

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


so that  commutes with each component  

*Multiple options can be correct

Consider the following statements and choose the correct answer


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

*Multiple options can be correct

An o16 nucleus is spherical and has a charge radius R and a volume  According to the empirical observations, the volume of 54 X128 nucleus assumed to be spherical is V ’ and radius R'. Then


R = R0A1/3


*Answer can only contain numeric values

A μ- meson moves with a speed of 0.998 c and possesses an average lifetime of  at rest, find the distance (in km) travelled by μ- meson, before decay?


Distance travelled by it before decay

*Answer can only contain numeric values

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?


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

*Answer can only contain numeric values

One milliwatt of light of wavelength 4560 is incident on a cesium surface. Calculate the photoelectric current (in microampere) liberated assuming a quantum efficiency of 0-5%, given Planck’s constant h = 6.62 * 1O-34 J-s and velocity of light c = 3 * 108 m/sec.


The energy of each photon of incident light is

Number of photons in one milliwatt source

(... Power of source = 1 milliwatt = 10-3W) 
Number of photons released

∴ Photoelectric current

*Answer can only contain numeric values

The wavelength of -line characteristic X-rays emitted by an element is 0.32. The wavelength of -line (in A°) emitted by the same element will be _______ .


*Answer can only contain numeric values

Find the energy release (in MeV), if two nuclei can fuse together to form 42He nucleus. The binding energy per nucleon of 2H and 4He is 1.1 meV and 7.0 MeV respectively.


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 = 

*Answer can only contain numeric values

The binding energy of 17CI35 nucleus is 298 MeV. Find its atomic mass (in a.m.u.). The mass of hydrogen atom (1H1) is 1 -008143 a.m.u. and that of a neutron is 1.008986 a.m.u. Given 1 a.m.u. = 931 MeV.


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.

*Answer can only contain numeric values

There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half-life of neutrons is 700 seconds, what fraction of neutrons will decay before they travel a distance of 10m?
Given mass of neutron = 1.675 * 10-27 kg.


= 0.0327 x 1.6 x 10-19 J
= 0.0327 x 1.6 x 10-19
On solving, we get 
v =2.5 x 103 m /s.
∴ Fraction of neutron decayed

= 0.000048.

*Answer can only contain numeric values

A piece of burnt wood of mass 20g is found to have a C14 activity of 4 decays/s. How long has the tree that this wood belonged to been dead (in years)?
Given T1/2 of C14 = 5730 years.


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).
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 

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

= 0.58 x 1011 s
= 1842 years

*Answer can only contain numeric values

Find the time (in years) (as measured by clock at rest on a rocket) taken by a rocket to reach a distant star and return to the earth with a constant velocity

if the distant of the star is 4 light years.


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.

*Answer can only contain numeric values

The rest mass of the electron is 9.023 *10-28 gm. Calculate the energy equivalent in electron volts.


Rest mass of the electron m0 = 9.028 *10-28 gm.
∴ Energy equivalent =  = 81.252 x10-8 ergs 
Now one electron volt = 1.6 * 10-12ergs 
∴ Energy in electron volt
 = 0.5078 -106 eV = 0.5078 MeV
(1 MeV = one million electron volts)

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