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CSIR NET Physical Science Mock Test - 2 - UGC NET MCQ


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30 Questions MCQ Test - CSIR NET Physical Science Mock Test - 2

CSIR NET Physical Science Mock Test - 2 for UGC NET 2024 is part of UGC NET preparation. The CSIR NET Physical Science Mock Test - 2 questions and answers have been prepared according to the UGC NET exam syllabus.The CSIR NET Physical Science Mock Test - 2 MCQs are made for UGC NET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CSIR NET Physical Science Mock Test - 2 below.
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CSIR NET Physical Science Mock Test - 2 - Question 1

Which one set of letters when sequentially placed at a gap in the given letter series shall complete it.

__nmmn__mmnn__mnnm__

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 1

As per the given question, the letter series is following the below pattern.
nnmm
The pattern nnmm is repeated.
Using the same pattern, we will get a complete letter series.
nnmm / nnmm / nnmm / nnmm

CSIR NET Physical Science Mock Test - 2 - Question 2

Direction: In each of the following question, select the one which is different from the others.

(a) (12−144)

(b) (13−156)

(c) (14−166)

(d) (15−180)

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 2

Check every pair respectively:

So, option (c) does not follow the similarity and it is different from another one.

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CSIR NET Physical Science Mock Test - 2 - Question 3

A bag contains 8 red balls and 12 blue balls. One ball Is drawn at random and replaced with 5 green balls. A second ball was drawn without replacement. What Is the probability that first ball drawn is red In colour and the second ball drawn Is blue in colour?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 3

Calculation:
⇒ The total number of red and blue ball = 8 + 12 = 20
⇒ The total number of after 5 balls replacement = 20 - 1 + 5 = 24
⇒ Probability to first ball is res colour = 8/20
⇒ Probabaility of second ball is blue colour = 12/24
⇒ The required probability = (8/20) × (12/24) = 1/5
∴ The required result will be 1/5.

CSIR NET Physical Science Mock Test - 2 - Question 4

Today is Monday. What day will it be after 61 days?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 4

The pattern followed here is:
Today is Monday
What day will it be after 61 days.
Let's find the odd days = 61 ÷ 7 = 5 Odd days
⇒ Monday + 5 Days = Saturday
So, Saturday falls after 61 days.
Hence, the correct answer is "Saturday".

CSIR NET Physical Science Mock Test - 2 - Question 5

In the figure shown above, PQRS is a square. The shaded portion is formed by the intersection of sectors of circles with radius equal to the side of the square and centrs at S and Q.

The probability that any point picked randomly within the square falls in the shaded area is ________

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 5

Concept:

Total Area = Area of square = (side)2
Shaded area = 2 ×  [Area of sector - (Area of half square or Area of triangle)]
Area of sector = 
Area of half square or Area of triangle =
Calculation:
Given,
Side of square = r
Radius of sector = r
Total Area = Area of square = (side)2 = r2
Shaded area = 2 ×  [Area of sector - (Area of half-square or Area of triangle)] = 

CSIR NET Physical Science Mock Test - 2 - Question 6

The total expenditure of a family, on different activities in a month, is shown in the pie-chart. The extra money spent on education as compared to transport (in percent) is ____

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 6

Let the total expenditure be x.
Money spent on education = 15% = 0.15x
Money spent on transport = 10% = 0.1x

⇒ Money spent on education = 1.5 × Money spent on Transport
i.e. the money spent on education is 50% more than the money spent on transport.

CSIR NET Physical Science Mock Test - 2 - Question 7

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 7

 

Part A filled in 4 minutes

=10min 40sec
∴ The tank will be full in
=4min +10min +40sec
14min 40sec

CSIR NET Physical Science Mock Test - 2 - Question 8

 

(51+52+53+.........+100) is equal to

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 8

Given a series,
51+52+53+....................+100
=(1+2+3+............+100−(1+2+3+........+50)
= It is in the form of series summation

CSIR NET Physical Science Mock Test - 2 - Question 9

On a planar field, you travelled 3 units East from a point O. Next you travelled 4 units South to arrive at point P. Then you travelled from P in the North-East direction such that you arrive at a point that is 6 units East of point O. Next, you travelled in the North-West direction, so that you arrive at point Q that is 8 units North of point P.

The distance of point Q to point O, in the same units, should be ____

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 9


CSIR NET Physical Science Mock Test - 2 - Question 10

Which of the following inferences can be drawn from the above graph?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 10

Option 1 is false as graph says there is a decrease in students qualifying in Physics in 2015 compared to 2014.
Option 2
Let no. of students qualifying in Biology in 2013 be 100
⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90
⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99
∴ The number of students qualifying in Biology in 2015 is less than that in 2013

CSIR NET Physical Science Mock Test - 2 - Question 11

Which of the graphs below gives the correct qualitative behavior of the energy density ET(α) of black body radiation of wavelength λ at two temperature T1 and T2(T1)

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 11

The energy density of black body radiation from Kirchhoff law is given as,

Also, from Wiens's displacement law,

Where K is constant, T is temperature and λmax is the maximum wavelength.
∵ λmaxT = constant
So, as the temperature increases wavelength decreases. These qualitative behaviors of the energy density of black body radiation of wavelength is shown in,

CSIR NET Physical Science Mock Test - 2 - Question 12

Let where x1 and x2 are independent and identically distributed Gaussian random variable of mean μ and standard deviation σ. Then is

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 12

Here, function,

Now, consider,

Substitute in equation (ii)

Now,

CSIR NET Physical Science Mock Test - 2 - Question 13

A bead of mass m can slide without friction along a massless rod kept at 45° with the vertical as shown in the figure. The rod is rotating about the vertical axis with a constant angular speed ω. At any instant r is the distance of the bead from the origin. The momentum conjugate to r is?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 13

The conjugate momentum is given by

where L is the Lagrangian of the system.
Calculation:
The Lagrangian for the system is written as:

The equation of constrain is θ = π/4 and it is given ϕ̇ = ω

Thus, the momentum conjugate to r is 

CSIR NET Physical Science Mock Test - 2 - Question 14

The Lagrangian of a system is given bycos θ, where m, l and g are constants.
Which of the following is conserved?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 14

Concept:
Lagrangian mechanics enables us to find the equations of motion when the Newtonian method is proving difficult.
In Lagrangian mechanics we start, as usual, by drawing a large, clear diagram of the system, using a ruler and a compass.
Explanation:
As φ is cyclic coordinate, so is a constant since m, l and g are constants.
Now using the Euler Lagrangian equation for motion we get:

Thus,ml2sin2θϕ is conserved.

CSIR NET Physical Science Mock Test - 2 - Question 15

The acceleration due to gravity (g) on the surface of the earth is approximately 2.6 times that on the surface of the Mars. Given that the radius of the Mars is about one half the radius of the Earth, the ratio of the escape velocity on the Mars to that on the Earth is approximately?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 15

We are given that the ge= 2.6 gm and 2rm= re.
where ge is the gravity due to earth and gm is the gravity due to Mars.
also, re and rm be the radius of the earth and Mars respectively.
Now as we are asked about the escape velocity so escape velocity is given by;
We have to take the ratio of Mars to that of earth we get:
Putting all the given value we get the ratio to be = 

CSIR NET Physical Science Mock Test - 2 - Question 16

A system has two normal modes of vibration, with frequencies ω1 and ω2=2ω1. What is the probability that at temperature T, the system has an energy less than 4ℎω1? [In the following x=e−βℎω1 and Z is the partition function of the system

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 16

Here, system has two normal modes of vibration. Therefore the possible energy of the system is given as,

Where, n1=n2 are vibrational quantum numbers. (n1=n2=0,1,2,3…) and ω2=2ω1.

Now, possible combinations of n1 and n2, such that system has an energy less than 4ℎω1 are given as,

Now, the probability of finding such energy level is given as,

Where, Z is the partition function.

CSIR NET Physical Science Mock Test - 2 - Question 17

A flux quantum (fluxoid) is approximately equal to 2×10−7 gauss −cm2. A type II superconductor is placed in a small magnetic field, which is then slowly increased till the field starts penetrating the superconductor. The strength of the field at this point is gauss. The penetrate depth of this superconductor is -

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 17

Here a type II superconductor is placed in a small magnetic field, and the magnetic field is then slowly increased till the field starts penetrating the superconductor.

The strength of the field at penetration depth is given as

Here, λ=the penetration depth and ø0 =fluxoid

CSIR NET Physical Science Mock Test - 2 - Question 18

Consider a particle in a one-dimensional infinite potential well with its wall at x=0 and x=L. The system is perturbed as shown in the following figure.

The first-order correction to energy eigenvalue is?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 18

So, the first-order correction to energy eigenvalue can be found out by:
First order correction = area under the curve / (length/base).
So, we find First order correction = 

CSIR NET Physical Science Mock Test - 2 - Question 19

A two-state quantum system has energy eigenvalues is ±ϵ corresponding to the normalized states At time t=0, the system is in quantum state The probability that the system will be in the same state at t = h/6∈ is

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 19

At t=0 we have 
Now at some other time we will have the state evolved as

Now in order to find the probability that this state will appear after the above stated time is;

We have used the property here that 

CSIR NET Physical Science Mock Test - 2 - Question 20

A frictionless heat-conducting piston of negligible mass and heat capacity divides a vertical, insulated cylinder of height 2H and cross-sectional area A into two halves. Each half contains one mole of an ideal gas at temperatures T0 and P0 corresponding to STP. The heat capacity ratio γ =  Cp/Cv is given. A load of weight W is tied to the piston and suddenly released. After the system comes to equilibrium, the piston is at rest and the temperatures of the gases in the two compartments are equal. What is the final displacement y of the piston from its initial position, assuming yW >> T0Cv

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 20

We have an insulated cylinder of height 2H and cross-sectional area A.
A piston divides this into 2 equal halves.
Each half contains one mole of an ideal gas at temperatures T0 and P0 corresponding to STP. 
 heat capacity ratio γ = Cp/Cv .
Due to weight W, there is some displacement let's say it y.
Now from the balance equation we get:
mgy  = change in internal energy.
mgy=2Cv(T−T0)
1 mole each in each compartment.
Now;

As PV = nRT and volume of each individual component is given as:
A (H-y) and A (H+y) respectively.

Now simplifying it a bit we get:

Again, we had the relationship,

Now putting the value of T in the above expression we get:

given that Wy >> T0Cv, we get:

We know that Cp−Cv=R.
From this, we can write:

Putting this above expression in the expression above for y we get:

Solving it for y we get:

CSIR NET Physical Science Mock Test - 2 - Question 21

A sample of a substance undergoes a first-order phase transition from a liquid to a solid state. The heat capacity of the substance is given by: Cp = a + bT + cT², where a, b, and c are constants, and T is the temperature. The enthalpy of fusion of the substance is ΔHf = 200 kJ/mol, and its melting point is 300 K. At a pressure of 1 atm, the substance begins to freeze at a temperature of 310 K, and the process is complete at a temperature of 300 K. Calculate the entropy change (ΔS) and the Gibbs free energy change (ΔG) for the freezing process.

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 21

Concept:
Phase transition is a phenomenon where a substance undergoes a change in its physical state from one phase to another, under specific conditions of temperature, pressure, and/or composition. It is a common occurrence in nature and plays a significant role in thermodynamics.
Types of Phase Transitions: There are several types of phase transitions, including:

  1. Solid-liquid transition (melting)
  2. Liquid-gas transition (vaporization)
  3. Solid-gas transition (sublimation)
  4. Liquid-liquid transition (miscibility/immiscibility)
  5. Solid-solid transition (polymorphism)

Explanation:
Heat capacity of substance: Cp = a + bT + cT² ,
Enthalpy of fusion: ΔHf = 200 kJ/mol, Melting point: 300 K, Pressure: 1 atm, Freezing temperature range: 310 K to 300 K.
WE assume that the process is a first-order phase transition.
The entropy change (ΔS) for the freezing process can be calculated using the following formula:
ΔS = ΔHf / Tm
where ΔHf is the enthalpy of fusion, and Tm is the melting point of the substance.
Substituting the given values, we get:
ΔS = 200 kJ/mol / 300 K,  ΔS = 0.667 kJ/(mol.K)
Next, we can calculate the Gibbs free energy change (ΔG) using the formula:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change of the freezing process.
To calculate ΔH, we need to first calculate the heat absorbed by the substance during the freezing process. This can be done by integrating the heat capacity equation over the freezing temperature range:
ΔH = ∫Cp dT
Since the process is a first-order phase transition, there is no change in volume, and the pressure is constant. Therefore, we can use the following equation to calculate ΔH:
ΔH = ΔHf = nΔHf
where n is the number of moles of the substance.
To calculate n, we can use the following formula:
n = m / M
where m is the mass of the substance, and M is the molar mass.
Let's assume that we have 1 mole of the substance, so n = 1.
The mass of the substance can be calculated using the density of the liquid and solid phases:
ρsolid = ρliquid = ρ
where ρ is the density of the substance.
The mass of the substance is then given by:
m = ρV
where V is the volume of the substance.
At the melting point, the substance has a volume of:
V = Vm = M/ρ = M/ρsolid = M/ρliquid
where Vm is the molar volume of the substance.
At the freezing point, the substance has a volume of:
Vf = Vm(1 - ΔV)
where ΔV is the change in volume during the freezing process. Since the substance undergoes a first-order phase transition, the change in volume is negligible. Therefore, we can assume that Vf ≈ Vm.
Now we can calculate the heat absorbed by the substance during the freezing process:
ΔH = ΔHf = nΔHf = 200 kJ/mol
Finally, we can calculate the Gibbs free energy change:
ΔG = ΔH - TΔS
ΔG= (200 - (300 X 0.667)) = (200-200.1) kJ/mol = -0.1 kJ/mol.

CSIR NET Physical Science Mock Test - 2 - Question 22

The frequency of the following Hartley oscillator

is nearly equal to : 

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 22


This is the tank circuit of the Hartley oscillator. 
There are two inductors in the series connection. The equivalent inductance is : L=12μH+8μH=20μH
The capacitance is given as : C=0.05μf
The formula for the frequency is given as : 

CSIR NET Physical Science Mock Test - 2 - Question 23

In a p-type semiconductor, the Fermi level lies 0.4 eV above the valence band. If the concentration of the acceptor atoms is trippled and kT = 0.03 eV, the new position of the Fermi level will be :

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 23

In semiconductors, the Fermi level (EF) describes the energy level at which the probability of occupation by an electron is 1/2.
For a p-type semiconductor, the Fermi level lies closer to the valence band.
The position of the Fermi level relative to the intrinsic energy level (Ei) can be described by the formula: 
where n is the concentration of free carriers (holes in valence band for p-type), N is the concentration of dopants (acceptor atoms for p-type), kB is Boltzmann's constant, T is the absolute temperature.
If we triple the concentration of acceptor atoms, let's denote the new Fermi level as E'F. Then we have: 

From the given information, ΔE = 0.4 eV and kT = 0.03 eV.
Hence, 

CSIR NET Physical Science Mock Test - 2 - Question 24

The vibrational motion of a diatomic molecule may be considered to be that of a simple harmonic oscillator with angular frequency ω. If a gas of these molecules is at a temperature T, what is the probability that a randomly picked molecule will be found in its lowest vibrational state?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 24

The probability P of finding a diatomic molecule in its lowest vibrational state is given by the Boltzmann distribution:

where E0 is the energy of the lowest vibrational state, K is the Boltzmann constant, and T is the temperature of the gas.
For a simple harmonic oscillator, the energy of the nth vibrational state is given by:

where is the Planck constant.
Therefore, the energy of the lowest vibrational state (n = 0) is:

Substituting this into the Boltzmann distribution, we get:

Simplifying this expression, we have:

So, the probability of finding a diatomic molecule in its lowest vibrational state is exponentially dependent on the inverse temperature and decreases as the temperature increases.

CSIR NET Physical Science Mock Test - 2 - Question 25

In a spectrum resulting from Raman scattering, let IR denote the intensity of Rayleigh scattering and IS and IAS denote the most intense stokes line and the most intense anti-stokes line, respectively. The order of the intensities are:

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 25

Rayleigh and Raman scattering:

  • Raman spectroscopy is a spectroscopic technique based on Raman scattering.
  • When a substance interacts with a laser beam (or any light wave), almost all of the light produced is Rayleigh scattered light (elastic process).
  • However, a small percentage (about 0.000001%) of this light is Raman scattered (inelastic process). Raman scattering is a process, where incident light interacts with molecular vibrations in a sample.
  • The photons from the laser beam interact with the molecules and excite the electrons in them.
  • The excited electrons immediately fall down to the ground level.
  • As electrons lose energy and fall down to the ground state, they emit photons.

There are three different scenarios of how light can be re-emitted after energy had been absorbed by an electron: 

  • An electron falls down to the original ground state and there is no energy change, therefore light of the same wavelength is re-emitted. This is called Rayleigh scattering.
  • After being excited, an electron falls to a vibrational level, instead of the ground level. This means the molecule absorbed a certain amount of energy, which results in light being emitted in a longer wavelength than the incident light. This Raman scatter is called “Stokes”.
  • If an electron is excited from a vibrational level, it reaches a virtual level with higher energy. When the electron falls down to the ground level, the emitted photon has more energy compared to the incident photon, which results in a shorter wavelength. This type of Raman scatter is called “Anti-Stokes”.

Explanation:
As in Rayleigh scattering electrons at a given energy level will also return back to the same energy level.
Hence the probability of this happening is maximum.
Hence IR is maximum.
For the Stokes line, electrons at an energy level is stated to return to a higher energy level, whose
probability is lower that IR, but is still higher in probability.
than the anti-stokes line where the electrons are already at a higher energy level and is stated to return to a lower energy level that it was earlier.
The sequence thus is IR > IS > IAS.

CSIR NET Physical Science Mock Test - 2 - Question 26

The phonon dispersion for the following one-dimensional diatomic lattice with masses M1 and M2 (as shown in the figure)

is given by

where a is the lattice parameter and K is the spring constant. The velocity of sound is 

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 26

For small value of q (i.e. long wavelength approximation limit).
We have 

For Acoustical branch: 

Velocity of sound is

CSIR NET Physical Science Mock Test - 2 - Question 27

If the Bohr radius is a0, the most probable value of r in the ground state of Hydrogen atom is

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 27

The probability is given as:
Now, the atom is considered as a sphere with area : A = 4πr2
So, for the most probable radius r, consider a small element dr. So, the formula for the probability will be written as :

To determine the most probable distance, find the minima of the probability:

Now, 
Substituting the above result in equation (1) : 

CSIR NET Physical Science Mock Test - 2 - Question 28

A particle with a rest mass of 2 GeV/c2 is accelerated to a speed of 0.8c. Can this moving object displace a particle bound with 4GeV of energy?

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 28

We can use the formula for the kinetic energy of a particle in special relativity:
K = (γ - 1)mc2
where K is the kinetic energy of the particle, m is its rest mass, c is the speed of light, and γ is the Lorentz factor:

where v is the speed of the particle.
Now we can find the kinetic energy of the particle.
Now we can use the kinetic energy formula:

Therefore, the kinetic energy of the particle is 1.34 GeV.
So the total energy is 2 + 1.34 GeV = 3.34 GeV
With the given kinetic energy, the particle cannot break the bound of the particle which this is going to hit.

CSIR NET Physical Science Mock Test - 2 - Question 29

We have 238U which decays with a half-life of 4.51 X 105 years, the decay series eventually ending as 206Pb, which is stable. A rock sample analysis shows that the ratio of the number of atoms of 206Pb to 238U is 0.003. Assuming that all the Pb has been produced by the decay of U and that all other half-lives in the chain are negligible. The age of the rock sample is? Given that [ln (1.003) = 0.003]

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 29

Initially, we had the number of U as N0 and Pb to be 0.
After a time we have the number of U as Nut and Pb to be Npbt.
We are given that the ratio of the number of atoms of 206Pb to 238U is 0.003 at the time of stability.
Hence, we get:
Also as the number of other particles produced in the chain is negligible we have:

Simplifying it we get:
which implies,

We know that is the decay constant.
Hence, by putting this in the above equation we get:

From here we solve for t and we get:

t = 1952.4 years.

CSIR NET Physical Science Mock Test - 2 - Question 30

A particle in 1-D moves under the influence of a potential of V(x) a x4, where a is a real constant. For large the quantized energy En depends on as:

Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 30

According to WKB approximation:
We have to use the relation that:

where,for V (x) to be finite a boundary andfor V(x) to be infinite at the boundary.
Explanation:
We have a particle in 1-D that moves under the influence of a potential of V(x) a x4.
We use the above formula with as the potential is finite at the turning points, and we get:

The potential is symmetric hence,
where A is some constant
Taking the substitution we get:

Now pulling out all the E terms, out and doing the integration we get:
This constant carries a constant after doing the integration.
Now solving it a bit we get:

Therefore:

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