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

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

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

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

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.

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

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

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

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.

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.

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)

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.

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.

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: 

For one-dimensional infinite square well potential, the energy of n^th 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, 

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)

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

Given vector is, 

From equation (1),

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 -

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 T_A, T_B and T_C, respectively, satisfy the T_B > T_C > T_A.

Given electric field is

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

Here, from question, Separation of mirror (d) = 20cm = 20 x 10-2 m and Refractive index (n_o) = 1 are given.
Now, the values of the separation between the modes  is given as,

And width of each mode of laser cavity is,

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 f_c is the lower limit of frequency in this white region where as  f_h  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:

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

For Acoustical branch: 

Velocity of sound is

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.

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

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 α_n 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 

As we know that energy is given as

The g - factor corresponding to J = 1