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CSIR NET Physical Science Mock Test - 1 - UGC NET MCQ


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30 Questions MCQ Test - CSIR NET Physical Science Mock Test - 1

CSIR NET Physical Science Mock Test - 1 for UGC NET 2024 is part of UGC NET preparation. The CSIR NET Physical Science Mock Test - 1 questions and answers have been prepared according to the UGC NET exam syllabus.The CSIR NET Physical Science Mock Test - 1 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 - 1 below.
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CSIR NET Physical Science Mock Test - 1 - Question 1

A five digit number is formed with the digit 1,2,3,4 and 5 without repetition. Find the chance that the number is divisible by 5 ?

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

5! = 5*4*3*2*1 = 120
4! = 4*3*2*1 = 24
P = 24/120 = 1/5

CSIR NET Physical Science Mock Test - 1 - Question 2

If ‘S – T’ means ‘S is the wife of T’; ‘S + T’ means ‘S is the daughter of T’ and ‘S ÷ T’ means ‘S is the son of T’. What will ‘M + J ÷ K’ means?

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

‘M + J’ means ‘M is the daughter of J’; ‘J ÷ K’ means ‘J is the son of K’.

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CSIR NET Physical Science Mock Test - 1 - Question 3

Direction: Identify the diagram that best represents the relationship between the given classes. 

Metal, Oxygen, Copper

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 3

According to given information, copper is a metal and oxygen is a gas so oxygen is different from copper and metal.

CSIR NET Physical Science Mock Test - 1 - Question 4

Let S = [0, 1] ∪  [2, 3) and let f : S → R be defined by 

If T = {f(x) : x ∈ S}, then the inverse function f–1 : T → S

CSIR NET Physical Science Mock Test - 1 - Question 5

find the number of triangles in the given figure.

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 5

We may lebel the figure as shown.

  • The simplest triangles are BFG, CGH, EFM, FMG, GMN, GHN, HNI, LMK, MNK and KNJ i.e. 10 in number.
  • The triangles composed of three components each are FAK and HKD i.e. 2 in number.
  • The triangles composed of four components each are BEN, CMI, GLJ and FHK i.e. 4 in number.
  • The triangles composed of eight components each are BAJ and CLD i.e. 2 in number.
  • Thus, there are 10 + 2 + 4 + 2 = 18 triangles in the given figure.

Hence, option C is correct.

CSIR NET Physical Science Mock Test - 1 - Question 6

Directions: Study the given table carefully and answer the question that follow:
Number of candidates appeared and qualified for a test (in hundreds) in 6 different years from 5 different zones.
Note: Here App. means Appeared and Qual. Means qualified.

Q. From which zone was the total number of candidates who qualified the test, the second highest in the year?

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 6

Total number of candidates who qualified the test in 2009 and 2010 from various zones are:
P → (4.8 + 5.6) = 10.4
Q → (5.2 + 6.4) = 11.6
R → (6.8 + 7.4) = 14.2
S → (5.2 + 11.4) = 16.6
T → (6.9 + 9.4) = 16.3

CSIR NET Physical Science Mock Test - 1 - Question 7

PQRS is a circle and circles are drawn with PO, QO, RO and SO as diameters. Areas A and B are marked. A/B is equal to:

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 7

Such questions are all about visualization and ability to write one area in terms of others.

Here, Let the radius of PQRS be 2r 
∴ Radius of each of the smaller circles = 2r/2 = r

Area A can be written as:
A = π (2r)2 – 4 x π(r)2 (Area of the four smaller circles) + B (since, B has been counted twice in the previous subtraction)

A = 4πr2 - 4πr2 + B
A = B
A/B = 1

Choice (B) is therefore, the correct answer.
Correct Answer: 1

CSIR NET Physical Science Mock Test - 1 - Question 8

A Sales Executive gets a commission on total sales at 8%. If the sale is exceeded Rs.10,000 he gets an additional commission as a bonus of 4% on the excess of sales over Rs.10,000. If he gets the total commission of Rs.950, then the bonus he received is?

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 8

B. 50
Explanation: Commission up to 10000 = 10000 * 8/100 = 800 Ratio = 2x:x ; Commission = 2x, Bonus = x ; Bonus = 950 – 800 * 1/3 = 150 * 1/3 = 50

CSIR NET Physical Science Mock Test - 1 - Question 9

A dishonest dealer marks up the price of his goods by 20% and gives a discount of 10% to the customer. Besides, he also cheats both his supplier and his buyer by 100 grams while buying or selling 1 kilogram. Find the percentage profit earned by the shopkeeper.

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 9



CSIR NET Physical Science Mock Test - 1 - Question 10

The graph represents the average score of a Test player with respect to the number of Test matches he played. 
Determined the percentage change in average score b/w 10 - 40 match.

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 10

Formula used:

Calculation:
As shown in the graph,
Average score corresponding to 10 match (initial) = 64.5
Average score corresponding to 40 match (final) = 84.5
Therefore,

% change = 31 %
Hence, option (1) is correct.

CSIR NET Physical Science Mock Test - 1 - Question 11

Out of the two bar graphs provided below, one shows the amounts (in Lakh Rs.) invested by a Company in purchasing raw materials over the years and the other shows the values (in Lakh Rs.) of finished goods sold by the Company over the years.

The value of sales of finished goods in 1999 was approximately what percent of the sum of amount invested in Raw materials in the years 1997, 1998 and 1999?

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 11

CSIR NET Physical Science Mock Test - 1 - Question 12

If A exceeds B by 40%, B is less than C by 20%, then find A ∶ C

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 12

Given: 
 A exceeds B by 40% and  B is less than C by 20%,
Calculation:
Let,  B = 100
A exceeds B by 40%

B less than C by 20%

Now,  A : C  = 140 : 125 
⇒ A :C  = 28 : 25
∴ Option (C) is the correct answer.

CSIR NET Physical Science Mock Test - 1 - Question 13

The approximation of  cos θ ≈ 1  is valid up to 3 decimal places as long as θ is less than: 

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 13

Here the approximation of  cos θ ≈ 1 is possible only for those value of  θ , is  cos θ=1 .
The McLaurin series expansion of cosθ  is given as,

( ∵ Neglecting higher order terms)

CSIR NET Physical Science Mock Test - 1 - Question 14

Which of the following functions cannot be the real part of a complex analytic function of z = x + iy?

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 14

Let u be the real part of a complex analytic function of  w=u+iv , therefore for harmonic function;

So, x2y cannot be the real part.

CSIR NET Physical Science Mock Test - 1 - Question 15

Two objects of masses 2 kg and 3 kg are moving towards each other with velocities of 4 m/s and 2 m/s respectively. After the collision, the 2 kg object moves off with a velocity of 1 m/s in the opposite direction to its initial velocity. What is the velocity of the 3 kg object after the collision?

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 15

In a two-body collision, the law of conservation of momentum states that the total momentum of the system before the collision is equal to the total momentum of the system after the collision.
It is provided by the fact that there are no external forces acting on the system.
Let m1 and m2 be the masses of the two objects, and u1 and u2 be their initial velocities.
After the collision, let v1 and v2 be their final velocities.
Then, applying the law of conservation of momentum, we have:
m1u1 + m2u2 = m1v1 + m2v2
Substituting the given values, we get:
(2 kg) (4 m/s) + (3 kg) (-2 m/s) = (2 kg) (-1 m/s) + (3 kg) v2
Solving for v2, we get:
v2 = (2 kg) (4 m/s) + (3 kg) (-2 m/s) + (2 kg) (1 m/s) / (3 kg)
v2 = 4/3 m/s.
Therefore, the velocity of the 3 kg object after the collision is (a) 4/3 m/s.

CSIR NET Physical Science Mock Test - 1 - Question 16

A loop made of copper wire is bent in the shape as shown in the figure below. The radius of the smaller semi-circular segment is ‘a’ and that of the larger semi-circular segment is ‘b’. If a current I flows in the loop, the magnetic field at the centre ‘0’ of the semi-circular segment is

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 16

The magnetic field at the center of the circular loop is given as :
The magnetic field due to the semi-circular arc will be half :
Consider the magnetic field going inside as positive and magnetic field coming out at point O as negative.
So, the net magnetic field due to the two arcs is given as: 

CSIR NET Physical Science Mock Test - 1 - Question 17

The minimum energy of a collection of 6 non – interacting electrons of spin  and mass m placed in a one-dimensional infinite square well potential of width L is 

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 17

For one-dimensional infinite square well potential, the energy of nth state is given as,

The ground state can occupy two electrons, the first excited state occupied two electrons and the second excited state can occupy two electrons. Therefore total energy is, 

CSIR NET Physical Science Mock Test - 1 - Question 18

The potential of a diatomic molecule as a function of the distance between the atoms is given by V(r) = -a/ r6 + b/r12.The value of the potential at equilibrium separation between the atoms is:

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 18

The potential of a diatomic molecule as a function of the distance r between the atoms is given as,

At equilibrium point (r=r0), the first derivative of the potential will be minimum i. e., zero.

From equation (i)

CSIR NET Physical Science Mock Test - 1 - Question 19

With z=x+iy, which of the following function f(x,y) is not a (complex) analytic function of z?

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 19

Since we know that for analytic function Cauchy Riemann condition must be satisfied. For function Cauchy Riemann condition is not satisfied, therefore function is not analytic function of z.
Now, for

Now, is not analytic function and (2+z) is analytic function. Therefore product of these two functions is not be analytic.
For other options:

Analytic function

Analytic function

CSIR NET Physical Science Mock Test - 1 - Question 20

If and C is the circle of the unit radius in the plane defined by z = 1, with the centre on the z-axis, then the value of integral is:

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 20


Given vector is, 

From equation (1),

CSIR NET Physical Science Mock Test - 1 - Question 21

The first few terms in the Taylor series expansion of the function f(x) = sin x around x = π/4 are -

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 21

The Taylor series expansion is given as,

The first terms in the Taylor series expansion of the function f(x) = sin x around x = π/4 are -

CSIR NET Physical Science Mock Test - 1 - Question 22

The entropy S of a thermodynamic system as a function of energy E is given by the following graph,

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 22

As we know, the change in the entropy is given as,

From figure, in phase B, the entropy is constant. Therefore temperature of the phase B is much higher to maintain the entropy constant. In phase A, increasing of entropy with energy is more than increase of entropy in phase C with energy. Therefore from equation (1), TC > TA Therefore the temperature of the phase A, B and C denoted by TA, TB and TC, respectively, satisfy the TB > TC > TA.

CSIR NET Physical Science Mock Test - 1 - Question 23

An electromagnetic wave is travelling in free space (of permittivity ε0) with electric field The average power (per unit area) crossing planes parallel to 4x +3y = 0 will be:

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 23

Given electric field is

Therefore, the magnetic field,
Now, the pointing vector,
Therefore, the average power per unit area crossing planes parallel to is

CSIR NET Physical Science Mock Test - 1 - Question 24

Consider a  laser cavity consisting of two mirrors of reflectivity R1 = 1 and R2 = 0.98. The mirrors are separated by a distance d = 20 cm and the medium in between has a refractive index n0 = 1 and absorption coefficient a = 0. The values of the separation between the modes and the width  of each mode of the laser cavity are -

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 24

Here, from question, Separation of mirror (d) = 20cm = 20 x 10-2 m and Refractive index (no) = 1 are given.
Now, the values of the separation between the modes  is given as,

And width of each mode of laser cavity is,

CSIR NET Physical Science Mock Test - 1 - Question 25

Assume that the noise spectral density at any given frequency, in a current amplifier is independent of frequency. The bandwidth of measurement is changed from 2 Hz to 30 Hz. The ratio (B/A) of the RMS noise current before (A) and after (B) the bandwidth modification?

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 25

In communications, noise spectral density (NSD), noise power density, noise power spectral density, or simply noise density (N0) is the power spectral density of noise or the noise power per unit of bandwidth. It has dimension of power over frequency, whose SI unit is watt per hertz (equivalent to watt-second or joule). It is commonly used in link budgets as the denominator of the important figure-of-merit ratios, such as carrier-to-noise-density ratio as well as Eb/N0 and Es/N0.

If the noise is one-sided white noise, i.e., constant with frequency, then the total noise power N integrated over a bandwidth B is N = BN0 (for double-sided white noise, the bandwidth is doubled, so N is BN0/2). This is utilized in signal-to-noise ratio calculations.

For thermal noise, its spectral density is given by N0 = kT, where k is the Boltzmann constant in joules per kelvin, and T is the receiver system noise temperature in kelvins.
The noise amplitude spectral density is the square root of the noise power spectral density, and is given in units such as

Explanation:
In the general opamp noise spectrum, we have two regions one is pink region.
This is when the spectrum is dependent on frequency whereas,
the other one is where the spectrum is independent of frequency which we call as white noise.
Now we are given with the condition that the noise spectral density at any given frequency, in a current amplifier is independent of frequency. 
So, we have the rms noise spectral density to be:

where fc is the lower limit of frequency in this white region where as  f is the higher frequency limit.
We are given with the difference in lower and higher frequency before the modification to be 2Hz.
Therefore:

(B) which is after modification gives: 
The ratio is given as B/A:

CSIR NET Physical Science Mock Test - 1 - 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 - 1 - 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 - 1 - Question 27

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 - 1 - Question 27

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 - 1 - Question 28

If the reaction are governed by weak interactions, then the particle Z in the reaction, Y → Z + X is

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 28

In reaction:
using conservation laws 
Now, for we have Thus, 

CSIR NET Physical Science Mock Test - 1 - Question 29

A particle of mass m is confined in an infinite square well with a shelf of height V0 extending half way across so that :

Using WKB approximation, the allowed energies En (n = 1, 2, 3,....) for this particle are, (given are the energies of particle in an infinite well with no shelf) :

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 29

The Schrodinger equation is given by 
The above equation can be written as where 
Now, we know that is the value for any wavefunction.
So, the solution of above equation using WKB Approximation, is 
Calculation:



We are finding the energies of the particle from a to-a in an infinite potential well.
Using WKB Approximation, is 
Now, we will substitute value of p(x) in above integration equation and taking limits from 0 to a and a to a/2 and V = V0 for first term and  V = 0 for second term, we get then,

After integration with respect to x, and applying limits, we get,


Taking andto right hand side, we get, 


Squaring both sides, and putting 


Take term αn to left side, we get,

Again squaring both sides, we get,

Cancelling terms and find the value of E, we get,


Now, substitute value of in above equation, we get,



So, the correct answer is 

CSIR NET Physical Science Mock Test - 1 - Question 30

The spectral line corresponding to an atomic transition from J=1 to J=0 states splits in a magnetic field of 1KG into three components separated by 1.6×10−3. If the zero field spectral line corresponding to 1849, what is the g− factor corresponding to the J=1 state? ( You may use )

Detailed Solution for CSIR NET Physical Science Mock Test - 1 - Question 30

As we know that energy is given as

The g - factor corresponding to J = 1

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