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55 Questions MCQ Test CSIR NET Physical Science Mock Test Series 2024 - CSIR NET Physical Science Mock Test - 2
CSIR NET Physical Science Mock Test - 2 for CSIR NET Physical Science 2024 is part of CSIR NET Physical Science Mock Test Series 2024 preparation. The CSIR NET Physical Science Mock Test - 2 questions and answers have been prepared according to the CSIR NET Physical Science exam syllabus.The CSIR NET Physical Science Mock Test - 2 MCQs are made for CSIR NET Physical Science 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
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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 patternnnmmis 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
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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.
CSIR NET Physical Science Mock Test - 2 - Question 3
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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
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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
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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
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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
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Two pipes A and B can fill a tank in15minutes and20minutes respectively. Both the pipes are opened together but after4minutes, 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 in4minutes
=10min40sec
∴The tank will be full in
=4min+10min+40sec
14min40sec
CSIR NET Physical Science Mock Test - 2 - Question 8
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(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
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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
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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
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Rohan sells two commodities for Rs. 19,800 each. He sells one at a profit of 10 % and sells the other at a loss of 10 %. Find his overall profit or loss percent and also the amount of profit or loss.
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 11
Calculation :
Given, Rohan sells two commodities for Rs. 19,800 each, one sold at a profit of 10% and the other at a loss of 10%.
So, as the selling price of both the items is the same, therefore there would always be an overall loss.
Thus, the loss percent =
To find the amount of loss we have to find the cost price of each commodity.
Let CP1and CP2be the cost prices of the two commodities.
So, CP1×(11/10) = 19800⇒ CP1= Rs.18,000
and, CP2×(9/10) = 19800⇒ CP2= Rs.22,000
So, total cost price = 18000 + 22000 = Rs.40,000
⇒ Loss amount = (loss%× total CP)/100 = 40000/100 = Rs.400
CSIR NET Physical Science Mock Test - 2 - Question 12
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The average of twelve 2 digit numbers is decreased by 3 when the digits of one of the 2 digit numbers is interchanged. Find the difference between the digits of that number.
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 12
Given:
The average of twelve 2 digit numbers is decreased by 3when the digits of one of the 2 digit numbers is interchanged.
Formula used:
Average = sum of observations/ Numberof observations
Calculation:
Let the number whose digits are being interchanged is 10x + y
After interchanging the digits, the number becomes 10y + x
At first average = (sum of 11 numbers + 10x + y)/12
After interchanging the digits,
average = (sum of 11 numbers + 10y + x)/12
So,
(sum of 11 numbers + 10y + x)/12 =(sum of 11 numbers + 10x+ y)/12 - 3
⇒(sum of 11 numbers + 10y + x) =(sum of 11 numbers + 10x+ y) - 36
⇒ 10y + x = 10x + y - 36
⇒ 9x - 9y = 36
⇒ x - y = 36/9
⇒ x - y = 4
CSIR NET Physical Science Mock Test - 2 - Question 13
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There are two buckets A and B. Initially A has 2 litres of water and B is empty. At every hour 1 litre of water is transferred from A to B followed by returning 1/2 litre back to A from B half an hour later. The earliest A will get empty is in:
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 13
From the given data, we get
When T = 0 hr, A has 2 liters and B has 0 liters
When T = 1 hr, A has 1 liter and B has 1 liter
When T = 1.5 hr, A has 1.5 liters and B has 0.5 liters
When T = 2 hr, A has 0.5 liters and B has 1.5 liters
When T = 2.5 hr, A has 1 liter and B has 1 liter
When T = 3 hr, A has 0 liters and B has 2 liters
∴ Earliest A will get empty it in 3 hours
CSIR NET Physical Science Mock Test - 2 - Question 14
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How many numbers between6and1300are completely divisible by6or9or by both?
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 14
We need to find the number from7to1299.
The numbers which are divided by
Integral part of 216
The number which is divisible by 9 =Integral part of
= 144
The number which are divisible by 18 =Integral part of
=Integral part of 72
The number which is divided by6or9or both = 216 + 144 - 72 = 288
CSIR NET Physical Science Mock Test - 2 - Question 15
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Ankit, Bhusan and Chetan enter into a partnership in the ratio
After4months, Ankit increases his share of50%. If the total profit at the end of one year be Rs.10,800then Chetan's share in the profit is:
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 15
Ratio of initial investments =
Let the initial investments be105x,40x,36x.
Ankit: Bhusan: Chetan
So, Chetan's share in the profit =Rs.
CSIR NET Physical Science Mock Test - 2 - Question 16
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Any vector M now can be written in the following form:
here b is an antisymmetric matrix.
Find the number of independent components of in b ?
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 16
We are given with
Where it is mentioned that b is antisymmetric in nature.
Hence,
if we write b as a whole there will be a total of4×4=164×4=16components, asiandj isfrom 1 to 4.
Now as it is antisymmetric hence the relation is now:
bij=−bji. From this we can conclude that we get:
There are 10 constraints.
That is we can represent it as:
The independent components are:
(b12,b13,b14,b23,b24,b34) = 6.
Hence the correct option is option (3).
CSIR NET Physical Science Mock Test - 2 - Question 17
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The Lagrangian for a simple pendulum is given by L =
− mgl(1 − cos θ)
Hamilton's equations are then given by
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 17
Concept:
If the Lagrangian L for a system, we can construct the Hamiltonian H is defined as
For
− mgl(1 − cos θ), we can use the above formula to write the Hamiltonian H.
From Hamiltonian equation we can find outṗθ andθ˙.
CSIR NET Physical Science Mock Test - 2 - Question 18
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A particle of mass m is attached to a fixed point O by a weightless inextensible string of length a. It is rotating under the gravity as shown in the figure. The Lagrangian of the particle is
− mgacos θ where θ and ϕ are the polar angles. The Hamiltonian of the particles is
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 18
Concept:
The Hamiltonian of a system specifies its total energy—
i.e., the sum of its kinetic energy and its potential energy—in terms of the Lagrangian function.
Calculation:
Put the value ofθ˙ andϕ˙
CSIR NET Physical Science Mock Test - 2 - Question 19
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Two parallel conducting rings, both of radius R, are separated by a distance R. The planes of the rings are perpendicular to the line joining their centres, which is taken to be the x-axis.
If both the rings carry the same current i along the same direction, the magnitude of the magnetic field along the x- axis is best represented by
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 19
Concept:
The magnetic field B produced by a ring at any point on the axial line is defined by Ampere's circuital law as:
where:
- μ0 is the permeability of free space,
- i is the current flowing through the wire,
- r is the radius of the circular wire,
- z is the position of a point on the axis from the center of the coil.
Explanation:
Since there are two rings here, and they both have the same current in the same direction, they will each produce a magnetic field that combines at points along the x-axis.
Notice that the magnetic field produced by each ring along the x-axis is symmetrical with respect to the x-axis, and they will combine constructively.
The total magnetic field would be twice the magnetic field produced by one of the rings because of the constructive superposition.
When the distance between rings is R,
which leads to a uniform magnetic field at the center region. At this condition, magnetic fields at the planes of the coil are a little bit smaller than that at the center. So, Option 1 is an approximate diagram.
CSIR NET Physical Science Mock Test - 2 - Question 20
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The yz-plane at x = 0 carries a uniform surface charge density σ. A unit point charge is moved from a point (δ, 0, 0) on one side of the plane to a point (-δ, 0, 0) on the other side. If δ is an infinitesimally small positive number, the work done in moving the charge is
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 20
Concept:
Work Done measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.
Calculation:
Work Done = q{V(b) -V(a)}
Since work done depends on potential difference between two points,
∴ W = 0
CSIR NET Physical Science Mock Test - 2 - Question 21
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The electric field intensity of a plane wave travelling in free space is given by:
V/m. In this field consider a square area of 10 cm X 10 cm on a plane x+y =1 with wave impedance of120π120π. The total time averaged power (in W) passing through the square area is?
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 21
Concept:
The time-averaged pointing vector generally for Electric and magnetic fields in x and y direction is given by:
where Erms is the root mean square electric field amplitude. In free space with E(t) varying sinusoidally at some frequency with peak amplitude E0, its rms voltage is given by
with the average Poynting vector then given by:
where η being the wave impedance.
Explanation:
We are given
with the time averaged pointing vector is given by:
The wave impedance is given to be 120π
E0 is the peak electric field.
First, we should calculate the average pointing vector the electric field given.
Hence the average pointing vector is given by:
Now average power is given by:
where ds is the unit area vector.
The plane which is given to us is x+y =1.
The unit normal to this plane is given by:
The average power is thus given by:
The integral over the elementary area is total area.
A is the total area which is given by;
Thus
CSIR NET Physical Science Mock Test - 2 - Question 22
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A metallic wave guide of square cross-section of sideLis excited by an electromagnetic wave of wave-numberk. The group velocity of theTE11mode is
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 22
CONCEPT:
For an electromagnetic wave propagating a rectangular metallic waveguide, if the perpendicular components of the Electric wave vanishes and only the transverse component exist then the corresponding wave is referred to as a transverse electric wave or TE wave.
For a TE wave, there can be the cutoff frequency is
The wave number K can be represented as
Where, m and n (where, m,n = 1, 2, 3, ...) and a and b are the dimensions of the waveguide.
EXPLANATION:
Here the dimension of the wave guide is a = b = L
We know the value of K is given by
CSIR NET Physical Science Mock Test - 2 - Question 23
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Let us
be the real part of an analytic function
of the complex variable
the imaginary part of
is-
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 23
Here, real part of an analytic function is given as,
According to Cauchy Riemann equation,
From equation(1)and (2)
CSIR NET Physical Science Mock Test - 2 - Question 24
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The first few terms in the Laurent series for
in the region1≤|z|≤2,and aroundz=1is-
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 24
Here, given expression is,
CSIR NET Physical Science Mock Test - 2 - Question 25
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Consider the matrix
the eigenvalues ofM are
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 25
Given matrix
The characteristic equation is given as,
CSIR NET Physical Science Mock Test - 2 - Question 26
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Lety(x)be a continuous real function in the range0and2π, satisfying the in hom*ogeneous differential equation:
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 26
Given in hom*ogeneous differential equation is,
Note that
represents the indefinite integral of Dirac delta function, which is always 1
"Note that RHS represents the indefinite integral of Dirac delta function, which is always 1".
CSIR NET Physical Science Mock Test - 2 - Question 27
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The Fourier transform of the function
is
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 27
CSIR NET Physical Science Mock Test - 2 - Question 28
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Suppose
are two linearly independent solutions of the differential equation
(vis half integer)
Then, the value of
is equal to
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 28
Bessels differential equation
Comparing this with bessel differential equation, we have v = 1/2.
Solution is
Hence,
CSIR NET Physical Science Mock Test - 2 - Question 29
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A particle of massmis in a cubic box of size. The potential inside the box
is zero and infinite outside. If the particle isin an eigenstate of energy
its wavelength is -
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 29
Here, the eigenstate energy is given as
a=size of the cubic box.
In three - dimensional box,
Here, equations and (1) and (2) are same, thus nx = 1, ny = 2 and nz = 3
Now, wave function is
CSIR NET Physical Science Mock Test - 2 - Question 30
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Which of the following is the correct form of Schrodinger wave equation?
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 30
Schrodinger wave equation is:
The Schrödinger equation is a partial differential equation that describes the dynamics of quantum mechanical systems via the wave function. The trajectory, the positioning, and the energy of these systems can be retrieved by solving the Schrödinger equation.
CSIR NET Physical Science Mock Test - 2 - Question 31
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A one-dimensional potential has the following form:
whereV0and bare positive constants.
Find V0as a function of bsuch that there is just one bound state, of about zero binding energy, for a particle of mass M.
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 31
To solve the bound state problem, we require that−V0<E<0. The relevant boundary conditions areψ(0)=0, due to the infinitely high barrier at x=0andlimx→∞ψ(x)=0.The solution to the time-independent Schrödinger equation is:
where
Diving these two equations, we obtain,
Note that equation (1) implies that,
We can also rewrite equations (2) and (3) as,
We require
The requirement of k > 0is conventional, since the sign ofkcan be absorbed into the definition of the constantA.
Equations (4) and (5) can be solved graphically by looking for intersection of the function
with the circle of radius
in the first quadrant of the xyplane (wherex=kb andy=kb ).
It is clear from the figure above that if V0is less than some critical value, then the radius of the circle,
in which case there are no intersections in the first quadrant, and therefore no bound states. The critical value of of V0is obtained by setting
That is,
which yields,
CSIR NET Physical Science Mock Test - 2 - Question 32
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Given the usual canonical commutation relation, the commutator
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 32
Here, commutator
Now, from commutation relations:
CSIR NET Physical Science Mock Test - 2 - Question 33
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A point charge q of mass m is kept at a distancedbelow a grounded infinite conducting sheet whichlies in thexy – plane. For what value ofdwill the charge remain stationary?
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 33
For chargeqbecomes stationary,
CSIR NET Physical Science Mock Test - 2 - Question 34
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The source voltage for the circuit shown is 220 V and the load inductance is 220 μH. If the switch is closed for a time t1 = 100 μs, then the load current will be:
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 34
When S1is closed, the main diode is ON and Dmis OFF.
In the above circuit, the inductor current is the output current.
where, VS= Source voltage
TON= ON period
L = Inductor
Calculation
Given, VS = 220 V
TON = 100 μs
L = 220 μH
Io=100 A
CSIR NET Physical Science Mock Test - 2 - Question 35
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The figure below shows a circuit with two transistors, Q1and Q2. having current gains β1and β2respectively.
The collector voltage Vcwill be closest to
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 35
Concept:
- A Darlington pair is a specific configuration of two transistors where the collector current of the first transistor feeds directly into the base of the second transistor.
- It's technically a compound structure with two NPN or two PNP transistors. When a small current enters the base of the first transistor, it modulates a larger current flowing from the collector to the emitter.
- This larger current is then fed into the base of the second transistor, which controls an even larger current from collector to emitter in the second transistor. The main advantage of this configuration is that it has significantly higher current and power gain (the multiplication of the current gains or β values of the two transistors) compared to a single transistor, thus it's often used when a high gain is required.
Explanation:
- Let's denote the collector current of transistor Q1as IC1, base current as IB1, and emitter current as IE1. Similarly, denote the collector current of transistor Q2as IC2, base current as IB2, and emitter current as IE2.
- Applying Kirchoff's voltage law to the left hand side of the loop containing the base connections:
From the above equation, the base current of the first transistor will come out to be :
Now, the collector current of the first transistor :
The collector current of the second transistor :
Now, applying the Kirchhoff's voltage law to the collector side loop:
CSIR NET Physical Science Mock Test - 2 - Question 36
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Which of the graphs below gives the correct qualitative behavior of the energy density ET(α)of black body radiation of wavelengthλat two temperature T1and T2(T1)
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 36
The energy density of black body radiation from Kirchhoff law is given as,
Also, from Wiens's displacement law,
WhereKis constant,Tis temperature andλmaxis 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 37
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Let
where x1 and x2are independent and identically distributed Gaussian random variable of meanμ andstandard deviationσ.Then
is
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 37
Here, function,
Now, consider,
Substitute in equation (ii)
Now,
CSIR NET Physical Science Mock Test - 2 - Question 38
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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 38
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/4and it is given ϕ̇ = ω
Thus, the momentum conjugate to r is
CSIR NET Physical Science Mock Test - 2 - Question 39
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The Lagrangian of a system is given by
cos 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 39
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,ml2sin2θϕis conserved.
CSIR NET Physical Science Mock Test - 2 - Question 40
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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 40
We are giventhat the ge= 2.6 gmand 2rm=re.
where geis the gravity due to earth and gmis the gravity due to Mars.
also, reand rmbe 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 41
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A system has two normal modes of vibration, with frequenciesω1andω2=2ω1.What is the probability that at temperatureT,the system has an energy less than4ℎω1?[In the followingx=e−βℎω1andZis the partition function of the system
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 41
Here, system has two normal modes of vibration. Therefore the possible energy of the system is given as,
Where,n1=n2are vibrational quantum numbers.(n1=n2=0,1,2,3…)andω2=2ω1.
Now, possible combinations ofn1and n2, such that system has an energy less than4ℎω1are given as,
Now, the probability of finding such energy level is given as,
Where,Zis the partition function.
CSIR NET Physical Science Mock Test - 2 - Question 42
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A flux quantum (fluxoid) is approximately equal to2×10−7gauss−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 thissuperconductor is -
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 42
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 andø0=fluxoid
CSIR NET Physical Science Mock Test - 2 - Question 43
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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 43
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 44
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A two-state quantum system has energy eigenvalues is ±ϵcorresponding 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 44
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 45
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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 T0and P0corresponding to STP. The heat capacity ratioγ = Cp/Cvis 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 45
We have aninsulated cylinder of height2Hand cross-sectional areaA.
A piston divides this into 2 equal halves.
Each half contains one mole of an ideal gas at temperatures T0and P0corresponding to STP.
heat capacity ratioγ =Cp/Cv.
Due to weight W, there is some displacement let's say ity.
Now from the balance equation we get:
mgy = change in internal energy.
mgy=2Cv(T−T0)
1mole 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 thatCp−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 46
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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 46
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:
- Solid-liquid transition (melting)
- Liquid-gas transition (vaporization)
- Solid-gas transition (sublimation)
- Liquid-liquid transition (miscibility/immiscibility)
- 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 = -0.1kJ/mol.
CSIR NET Physical Science Mock Test - 2 - Question 47
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The frequency of the following Hartley oscillator
is nearly equal to :
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 47
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 48
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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 48
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 49
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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 49
The probability P of finding a diatomic molecule in its lowest vibrational state is given by the Boltzmann distribution:
where E0is 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 50
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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 50
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 electronsimmediately 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 calledRayleigh 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 IRis 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 51
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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 51
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 52
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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 52
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 53
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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 53
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 54
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We have238U which decays with a half-life of 4.51 X 105years, the decay series eventually ending as206Pb, which is stable. A rock sample analysis shows that the ratio of the number of atoms of206Pb to238U 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 54
Initially, we had the number of U as N0and Pb to be 0.
After a timetwe have the number of U as Nutand Pb to be Npbt.
We are given that the ratio of the number ofatoms of206Pb to238U 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 55
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A particle in 1-D moves under the influence of a potential of V(x) a x4, where a is a real constant. For largenthe quantized energy Endepends onnas:
Detailed Solution for CSIR NET Physical Science Mock Test - 2 - Question 55
According to WKB approximation:
We have to use the relation that:
where,
for V (x) to be finite a boundary and
for 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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