CSIR NET Physical Science Mock Test - 3 - CSIR NET Physical Science MCQ

From the given information: A denotes '+', B denotes '×', C denotes '-', and D denotes ' ÷'
We substitute the values in the given equation and solve it using BODMAS rule:

124 D 4 C 86 D 2 A 61 A 17 B 3 = ?
124 ÷ 4 - 86 ÷ 2 + 61 + 17 × 3 = ?
31 - 43 + 61 + 17 × 3 = ?
31 - 43 + 61 + 51 = ?
143 - 43 = ?
100 = Answer

According to the logic,
First image is rotating by 90º in anti - clockwise direction.

Similarly, 
Fourth image is related to the third image in the same way as second image is related to the first image.

So,→ This image is following the logic explained above.
Hence, the correct answer is "Option (C)".

Family chart :-

Family Tree :

∴ Here, boy's mother is sister to Ranjitha's husband.
Hence, the correct answer is "Sister".

The least possible diagram is as follows:

Conclusions:
I. Some cockroaches are insects. - True (From the figure we can clearly see that some part of cockroaches intersects with insects).
II. Some lizards are cockroaches - False (There is no direct relation between lizards and cockroaches. It may be possible but not definite)
​Hence, the correct answer is "Only conclusion I follow".

The number which represents the words of English and Persian that are included in Hindi words is shown below:

Hence, '13' is the correct answer.

Straight Angle: A straight angle is an angle equal to 180 degrees. It is called straight because it appears as a straight line.
Calculation:
∠ABE + ∠ABC = 180° [∵ Straight Angle]
⇒ 120° + x = 180°
⇒ x = 180° - 120° 
⇒ x = 60° 
∠ACD + ∠ACB = 180° [∵ Straight Angle]
⇒ 130° + y = 180° 
⇒ y = 180° - 130° 
⇒  y = 50° 
∴ x = 60°  and y = 50° 

Given:
A bent can row 15 km per hour downstream
Upstream = 9 km/h

Formula Used:
Upstream = x − y
Downstream = x + y

Calculations:
Let the speed of the boat be x km/h
The speed of still water = y km/h
According to the formula,
⇒ x − y = 9 .......(1)
⇒ x + y = 15  ......(2)
On adding the eq (1) and eq (2), we get
⇒ x − y + x + y = 9 + 15
⇒ 2x = 24
⇒ x = 12 km/h
Hence, The speed of the boat in still water is 12 km/h.

Calculation:
In a gathering of 30 people, 20 people knew each other and 10 people didn't know each other.
And, people who don't know each other shake hands
So, from the question, 10 people shake their hands.
So, total handshakes between 10 people = 
Total handshakes between 10 people = 45

Given:
Arithmetic mean of 12 observations = 40
The smallest and the largest observations are 20 and 60 respectively.
Formula Used:
Arithmetic mean = (Sum of observations)/Number of observations
Calculation:
Arithmetic mean of 12 observations = 40
Thus, sum of 12 observations = 40 × 12 = 480
After removing 20 and 60 from the series, then sum of remaining 10 observations = 480 - 20 - 60 = 400
Thus, Arithmetic mean = (Sum of observations)/Number of observations
Arithmetic mean = 400/10 = 40
∴ The arithmetic mean of the remaining observations will not change.

Given:

Calculation:

Similarly

Now,

Equation of line passing through (a, b, c) and parallel to vector <u, v, w> is :
Midpoint O of the line passing through A (a, b, c) and B(x, y, z) is given by:

Calculation:
Given plane 3x - 5y + 4z = 5 . . .(1)
Here normal vector to the plane, <3 ,-5.4>
The equation of the line passing through the point (1, 2, -1) and perpendicular to the plane (1) is

Taking one at a time in equation (2), we get
Point on the line : (3r + 1, - 5r + 2, 4r - 1).
This point lies on the plane,
3(3r + 1) - 5(-5r + 2) + 4(4r - 1) = 5
or,  9r + 25r + 16r + 3 - 10 - 4 = 5
or, 50r = 16
or, 
So point on the plane is
Let (x, y, z) be the image of the point with respect to the plane 3x - 5y + 4z = 5.
Thenwill be the mid-point of the line joining (x, y, z) and (1, 2,-1)
Then we get

Hence, the required point is 

Therefore in 


Therefore in 

The responsivity is given as,

In which I_OP is the current due to electron hole pair and P_in is the power generated of incident light.
From question, the conversion efficiency of the solar cell = 10%

. Wavelength I=660nm
Solar cell area A=30cm²
Therefore generated power is,

The conversion efficiency or quantum efficiency is given as,

Substituting the values of λ,Pin  and Q.E in above equation,

For disappearing interference patterns, the path difference for bright fringes is equal to the path difference for dark fringes.
- For bright fringe, path difference,
- For dark fringe, path difference 
- From equation (1), 
Substitute the value of m in equation (2)

Bose - Einstein distribution function is given as,

Here, μ is the chemical potential and 
For Bose - Einstein Condensation, all Bose particles occupy the ground state. Therefore,

taking log both sides, therefore


Substitute the value of β in above equation,

For Bose - Einstein condensation, T must be tends to zero,

For three dimensional harmonic oscillation,

For ground state, n = 0

Here  the energy dispersion of a two dimensional electron system is given as,
E=ℎku
Now, the number of states is given as,

Now, let hk = p
⇒ E = pu
Differentiating both sides of the equation E = pu, we get
dE = udp

Now, the density of the sates can be given as,

Therefore density of state is directly proportional to E.

At the boundary

For a spin- 3/2 particle, the degeneracy is:
2s+1 = 2. (3/2) + 1 = 4.
For 8 particles the ground state energy is given by:

The first excited state is given by:

The minimum excitation energy is given by 

We have 
Now <ψ|ψ> =1. Hence normalized.
Now we have to find the 
Now using the relation that we get:

In two dimensional system, the number of allowed k-states in range k and k + dk is

Given dispersion relation is 

Assume energy at ground state is 0 and energy at first excited state is ∈.
The partition function is Z = 1 + 2e^-β∈
Energy = 
Specific heat, 

The truth table is given by:

We are given with,
E_gGoN = 3.5 eV and E_glnN = 1.5 eV
Band Gap energy of Gax In1-x N is E ∝ x.
For blue light of wavelength 400nm, the band gap


Thus equating slopes we get;

Suppose we have separation between the energy levels of a two-level atom is 2 eV.
4 × 1020 atoms are in the ground state and 7 × 1020 atoms are pumped into the excited state just before lasing starts.
Lets, say N₁=4 × 10²⁰, N₂= 7× 10²⁰.
Then,
N₂ - N₁ = 3 × 10²⁰
We know that the energy of the laser pulse will be:

Applying the given data we have:
Energy of laser pulse, 
⇒ E = 48 J.

Concept:
Each energy level is capable of holding a spin-up electron as well as a spin-down electron, i.e. in each orbital there are two different spin states which are degenerate (some would call these spin orbitals).

Explanation:
We know that total spin of a hydrogen atom is due to the contribution of the spins of the electron and the proton. 
Also, we know that:

The degeneracy of quantum level is 2F + 1.
Thus, for this case we have the ratio to be:

The reciprocal lattice is a lattice of points in momentum space that is dual to the direct lattice of points in real space.
It is defined as the set of all vectors G that satisfy the equation.
G⋅R=2πn,
where, R is any vector in the direct lattice,
and n is an integer.
In two dimensions, the reciprocal lattice is also a two-dimensional lattice.
For a square lattice with lattice constant a, the direct lattice is defined by the vectors.

The reciprocal lattice vectors are given by;

To find the reciprocal lattice vector corresponding to the (1,1) direction in the direct lattice,
we take a linear combination of the reciprocal lattice vectors.
Thus, the reciprocal lattice vector corresponding to the (1,1) direction in the direct lattice 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:

We have;

Z = 51.
Using the nuclear shell model, we have:
(s_1/2)² (p_3/2)⁴ (p_1/2)² (d_5/2)⁶ (s_1/2)² (d_3/2)⁴ (f_7/2)⁸ (p_3/2)⁴ (f_5/2)⁶ (p_1/2)² (g_9/2)¹⁰ (g_7/2)¹ 
⇒ j = 7/2 and l = 4. Thus spin and parity = 

N = 51:
(s1/2)² (p3/2)⁴ (p1/2)² (d5/2)⁶ (s1/2)² (d3/2)⁴ (f7/2)⁸ (p3/2)⁴ (f5/2)⁶ (p1/2)² (g9/2)¹⁰ (g7/2)¹
⇒ j = 7/2 and l = 4. Thus spin and parity = 

The parity or space inversion operation converts a right-handed coordinate system to left-handed:
x → −x, y → −y, z → −z.
This is a case of a noncontinuous operation, i.e. the operation cannot be composed of infinitesimal operations.
The symmetry operations we require that π is unitary, i.e.
π†π=1
Explanation:
No parity change; ΔJ = 2, 3
For E_l = type, Δπ = (-1)^l  (for no parity change l = 2)
For M_l type, Δπ = (-1)l⁺¹  (for no parity change l = 3)
ΔJ = 2, No parity change → E2; ΔJ = 3,
No parity change → M3.