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

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

Check every pair respectively:

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

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.

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".

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)] = 

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.

Part A filled in 4 minutes

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

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

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

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 λ_max is the maximum wavelength.
∵ λ_maxT = constant
So, as the temperature increases wavelength decreases. These qualitative behaviors of the energy density of black body radiation of wavelength is shown in,

Here, function,

Now, consider,

Substitute in equation (ii)

Now,

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 

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.

We are given that the g_e= 2.6 g_m and 2r_m= r_e.
where g_e is the gravity due to earth and g_m is the gravity due to Mars.
also, r_e and r_m 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 = 

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.

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

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 = 

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 

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/C_v .
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=2C_v(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 C_p−C_v=R.
From this, we can write:

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

Solving it for y we get:

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.

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 : 

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, 

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.

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 I_R 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 I_R, 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 I_R > I_S > I_AS.

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

For Acoustical branch: 

Velocity of sound is

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) : 

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.

Initially, we had the number of U as N₀ and Pb to be 0.
After a time t we have the number of U as N_ut and Pb to be N_pbt.
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.

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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