CSIR NET Physical Science Mock Test - 4 - CSIR NET Physical Science MCQ

Svedberg unit (S/ Sv) is a unit for sedimentation coefficients. It offers a measure of size of the particle based on sedimentation rate under acceleration. It calculates how fast the particle settles to the bottom of the solution. Svedberg unit is a measure of time (exactly 10⁻¹³ seconds).

The protective layer around your teeth, called enamel, is the strongest substance that our body produces. In fact, tooth enamel is stronger than any other substance on earth except for diamonds. It is comprised largely of calcium and phosphate mineral crystals, similar to the composition of our bones.

It gets discolored in air because of presence of Hydrogen sulphide in the air Brass constitute zinc and copper and these two elements are Hydro sulphide is a colourless gas and if mixed with others, will get colourless. Because brass is metal it get react with sulphide.

The fruit content in both the fresh fruit and dry fruit is the same.
Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100 kg
Dry fruit has 20% water.so remaining 80% is fruit content. Let weight if dry fruit be y kg.
Fruit % in freshfruit = Fruit % in dryfruit
Therefore, 
We get, y = 40 kg.

Total Zink =1/4×120+2/7×91+34
=30+26+34=90
Total Copper =3/4×120+5/7×91
=90+65=155
Ratio =90:155=18:31

Note that we are talking about Smart Cities Mission. Thus, anything related to rural areas is beyond the scope. Whereas, in option (C), we focus only on urban development. Therefore, option (C) will be an effect of the statement.

The total number of male employees working in the sells department 
Required

The image embedded in given figure is as shown below:Hence,

The pattern here is:
1) The triangle is moving from one corner to another and the point is moving in an anticlockwise direction. Also, the middle image is moving in a clockwise direction.
2) we can say that the missing image will look the same as that in figure 1.
Thus, options 2 and 3 are eliminated because the triangle is in the wrong place and facing in the wrong direction too.
And, option 4 is eliminated because the middle image is placed wrong. 

Here, we make a table as per the given data:

The number of single officers who have a salary above Rs. 50,000 but do NOT have a post-graduate degree means a common region of the rectangle and circle only.

Here, The number of single officers who have a salary above Rs. 50,000 but do NOT have a post-graduate degree is 10.
Hence, the correct answer is "10".

Given:
A = B + 2.5kg
C = B + 10.25kg
A + B + C = 48.75 kg
Solution:
B + 2.5 + B + B + 10.25 = 48.75
3B = 48.75 - 2.5 - 10.25
B = 36/3
B = 12 kg
A = 12 + 2.5 = 14.50 kg
Hence, the correct option is 4

Given:
LCM = 126
HCF = 9
Concept used:
LCM = Lowest Common Factor
It is the product of the greatest power of each prime factor, involved in the numbers.
LCM × HCF = Product of two numbers
HCF = Highest Common Factor
It is the product of the smallest power of each common prime factor in the numbers.
Solution:
Let, the other number = m
126 x 9 = 18 x m
m = (126 x 9)/18
m = 63
Other number = 63
Hence, the correct option is 1.

Concept:
To find the mean using the direct method, for grouped data,

where xi is the classmark and fi are the frequencies. 
Classmark is the average of the lower limit and upper limit of any class interval.
Calculation:

We know that,

⇒ x̅ = 315/25 = 12.60
Hence, the mean is 12.60.

Calculation:
The total number of computers sold by dealer A on Monday, Tuesday, and Friday = 75 + 80 + 78 = 233
The total number of computers sold by dealer B on Wednesday, Thursday, and Saturday = 60 + 80 + 75 = 215
Hence, difference between them = 233 - 215 = 18
∴ The difference between the total number of computers sold by dealer A on Monday, Tuesday and Friday and total number of computers sold by dealer B on Wednesday, Thursday and Saturday is 18.

Given:
∠BQR = 50°
Calculation:
∠BQR = ∠QAB  (∵ PQR is a tangent)
By the property of alternate segment theorem
⇒ ∠QAB = 50°      ----- (i)
∠BQR = ∠QBA  (∵ AB || PR, Alternate interior angles)
⇒ ∠QBA = 50°       ----- (ii) 
In Δ ABQ,
From (i) and (ii)
⇒ ∠QAB + ∠QBA + ∠AQB = 180° (∵ Sum of all the interior angles of the triangle )
⇒ 50° + 50° + ∠AQB = 180°
⇒ ∠AQB = 180° - 100° = 80° 

Given:

The Boltzmann's entropy is

From combined I and II law of thermodynamics, we have

So, the area of the rectangle is 2 square units.

For b. c. c lattice, primitive vertices are given as, 

Where, a is lattice constant. 
Now, according to Bloch theorem,

where

Which means, function is periodic w.r.t. translational vector 

Other functions are not periodic

For 
Proton is a stable particle, therefore it cannot decay. So the given process is forbidden. 
For 
From the process, the charge is not conserved (0 →+2), therefore this process is also forbidden. 
For 
Charge is conserved (+1 = +1) and spin is also conserved (0 = 0). Isospin is 1. 
Therefore this reaction is allowed through E. M. interaction. 
For  
Here, in reaction γ is involved, therefore, it is allowed for E. M. interaction.

Here, for a classical model, a scalar meson consists of a quark and an antiquark bound by a potential is given, For determination of minimum value of r, differentiation of equation (i) with respect to r is zero.

From question, b=100MeVfm and a=200MeVfm⁻¹.

Now, substitute the value of r,a and b in equation (i),

From the question, the masses of the quark and antiquark are negligible. Therefore from energy conservation,

Here, the entropy S is related to the internal energy U and volume V is given as,

According first law of thermodynamics,

Substitute the value of U from equation (1), we have

Now, Gibb's potential is given as,
G=U−TS+PV
Substitute the values of U,T and V from equation (1), (2) and (3) we get,

Here, a photon splits an ∝ - particle at rest into a tritium and a proton. The reaction may be given as,

Therefore, the energy of the photon will be minimum only if the tritium and created at rest.

 We know that the Ladder angular momentum is L₊= L_x +i L_y.
Now we have the relation that <l,m' |L± |l,m  > = 
Now for our case
We further simplify it and get:

We have where A is a normalization constant.
Now first we have to find the normalization constant.
So, we can find it using 
Using the above expression, we get:

So, we get: 
We are given a statehose probability in we get:

We have
Now 
Using the above identity, we get:

where,
For critical condition 
Putting the expressions for n we have:

From equation (i) and (ii) 

put the value of in equation (i)

We have:

This wave function is normalized as if we do:

The expectation value of the energy in this state is:

We can find the E's first we get:

corresponding all wavefunctions.
Now, we can find the expectation value to be:

Solving it we will get:

<E>=−5.1eV.

When there is minima or maxima in the graph of resistivity vs temperature,
Then it indicates that there will be two different temperatures for which the material will show the same resistivity.
That kind of resistivity variating materials are not suitable candidates for sensors as they can be misleading.
Hence the only resistivity profile that is suitable for the temperature sensor is given by the one that shows some monotonic behavior.

For the figure (a).
Applying KVL in the input circuit we get:
−10.7+10k×IB+0.7=0
Hence I_B=1mA
Current gain is given as:

Therefore, I_C=0.1A
Now applying KVL in the output circuit we get:
−10+0.1×2mA+V_CE=0
Hence V_CE=−190V.
The output junction voltage is given by 
V_BC=V_BE+V_EC=0.7+190=190.7V
Hence both the input as well as output junctions are forward biased.
So the figure (a) is in saturation region.
Similarly,
For the figure (b).
Applying KVL in the input circuit we get:
−5+I_B×10K+0.7=0
Hence I_B=0.43mA
Current gain is given as:

Therefore, I_C=4.3mA
Now applying KVL in the output circuit we get:
−10+4.3+V_CE=0
Hence V_CE=5.7V
The output junction voltage is given by 
V_BC=V_BE+V_EC=0.7−5.7=−5V
Hence the input junction here is forward biased but the output one is reverse biased.
So the figure (b) is in active region.

CONCEPT:

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 shorter wavelength. This type of Raman scatter is called “Anti-Stokes”.