Test: CSIR-NET Chemical Sciences Mock Test - 3 - CSIR NET Chemical Science MCQ

Total no of student = 10000
Total number of students passed in all five subjects = 5583
Total number of students passed in four subjects = total number of students failed in one subject only = 75+145+140+200+157 = 717
Therefore, number of students passed in at least four subjects is 5583 + 717 = 6300.

Surveys can be done in various ways. Some of the methods of conducting surveys are personal interviews, telephonic interviews, mailing questionnaires, through schedules, etc. Whereas archive is a historical record of an event which can be utilised for data collection but it is not appropriate for a survey.

Brinjal is one of the most common vegetables grown throughout the country for it and green or white pendulous fruit.
It is a member of the Solanaceae family and is closely related to tomato and potato.

Present age of father= x years
Present age of son= y years
(x+y)/2 = 27
x + y = 54 ------- 1
After 18 years,
father’s age = (x + 18) years
Son’s age = (y + 18) years
So,
x + 18 = 2(y+18)
x + 18 = 2y + 36
x - 2y = 18    ---- 2
Put value of 1 in 2,
54-y-2y = 18
3y = 36
y = 12, x = 42
present age of father = 42 years
present age of son = 12 years

Calcium oxalate (in archaic terminology, oxalate of lime) is a calcium salt of oxalic acid with the chemical formula CaC₂O₄.
It is a chemical compound that forms envelope-shaped crystals, known in plants as raphides.
A major constituent of human kidney stones, calcium oxalate is also found in beerstone, a scale that forms on containers used in breweries.

Given: Ratio between the angles of the triangle = 3 : 4 : 5
Let the angles of the triangle are 3x, 4x and 5x, then
3x + 4x + 5x = 180
⇒ x = 15°
Thus, biggest angle = 5 × 15
= 75°

A thesis is different from a research article in many aspects. Firstly, thesis is more scientific and formal; hence, it will cover all the important aspects of a research study. However, this is not true about a research article as it presents some particular idea or ideas about a research study. Writing a thesis is also a relatively longer process than writing a research article. So, all of them are true.

As per the instruction given in the question, the diagram is drawn below and all the angles are marked.

Now, ∠ A = ∠B = ∠C = 60º

• The tiangle is an equilateral triangle as all three angles of the traingle is 60º. 
• As AB = 20 km, CA and BA will also be 20 km.

So,the distance between point C and A is 20 km.

It is given that the decay rate is proportional to the amount of substance. If the amount of the substance is N then, 

Here, λ is the decay constant.
The half life of any radioactive substance is given as: 
Given: The half life of the substance is : t_1/2 = 1years.
Substituting in the half life formula : 

Considering the Binary Number System:

101 = (1×2⁶)+(1×2⁴)+(0×2⁴)+(0×2³)+(1×2²)+(0×2¹)+(1×2⁰)

The total number of alphabets is 26. Now, as per the series given in the question, the complete series is given below:

Substituting the values in the equation:

x = −1

• A's rating is the average of V and K. It means that  A has a place between V and K. 
• Now, let us suppose that the ratings are:  V = 4, A = 3, and K =2.
• Now, it is given that T is above A. Supposing that rating of T = 5.
• The average of S and A is equal to a minimum of V and K. So, that suggests that the rating of S should be significantly less. Let us suppose it is S = 1.

Summarising the ratings from higher to lower order.

Note: The rating of V and K might be different but in the question, only the third position is to be obtained. So, the position of the rest players doesn't matter.

• Let the total number of people be: A
• Let the total number of chairs be: B
• The total number of occupied chairs is: 7B/9
• The total number of people seated: 9A/13

Now total number of unoccupied chairs is: B−(7B/9) = 36
2B = 324
B = 162
Hence, the total number of chairs is: 162
Now, the total number of chairs and the people occupying it will be the same.

∴ A = 182
Now, the number of chairs needed to occupy the rest members: 182−162 = 20

It is given that  B²−A² = 103 which implies (B−A)(B+A)=103. Now, it is given that A and B are consecutive numbers, so their difference will be 1.
(B−A)=1 and  (B+A)=103
Adding both the equations, B is obtained as: 52. Hence, A will be 51. C and D will be 53 and 54 respectively.
So, 
D²−C² = (54)²−(53)²
D²−C² = 2916−2809
D²−C² = 107

Case 1: Consider 100 students attempting the exam and the exam is of 100 marks.

• Now it is given that no two students have the same marks. Each student will have a different score.
• So, if the median of the marks achieved is considered, it will be : X = (49+50)/2 = 49.5
• Now, in this case, the cut-off is half of the median: X/2 = 25.25
• Here, 90 students pass and 10 fail.

Case 2: Now, here the median is double of case 1: X = 49.5.

• As it is given that no students can have the same score. So, the students scoring from 50 to 100 have passed the exam.
• So, in total 51 students passed the exam.
• Out of 100,51 students clear the exam which states that approx 50% of students failed the exam.

Let the amount of nitrogen added be x, and amount of 20% nitrogen solution be P. Nitrogen balance: x + 0.2(P) = 0.4(100), Water balance: 0.8(P) = 0.6(100), ⇒ P = 75 Kg, ⇒ x = 0.4(100) – 0.2(75), ⇒ x = 25 Kg.

Exit components, ammonia: 0.05*0.6 = 0.03, air: 0.95, ⇒ yammonia = 0.03/0.98 = 0.0306, ⇒ Total exit pressure = 10/0.0306 = 326.8 mm Hg.

The reaction is 4C₃H₇O₄ + 11O₂ → 12CO₂ + 14H₂O, ⇒ Limiting reagent is O₂. Let 100 moles of feed was supplied, ⇒ after the reaction, moles of CO₂ = 24, moles of H₂O = 28, moles of O₂ = 22, moles of C₃H₇O₄ = 38, moles of N₂ = 10, ⇒ percentage of O₂ = 18%

Q = ∆H – V∆P = (250 – 100) – 10(10 – 5) = 100 J/Kg.

Amount of water added to the vessel in 10 seconds = (8 – 5)*10 = 30 liters, => Final amount of water = 10 + 30 = 40 liters.

Since in steady-state the rate of flow in is equal to the rate of flow out, so final amount of fluid in the vessel is same as the initial amount.

Moles of bone dry air = 6, ⇒ humidity = 5/6*100 = 83%.

Density of the substance = 100*2.2 = 220 lb/m³, Specific gravity = (100*2.2)/100 = 2.2.

A dew point is a temperature at which condensation starts.

It takes a finite time for an irreversible process to complete.

Pressure of gas = Pressure due to height difference of liquid = d*g*h, ⇒ 5 = 10*10*h ⇒ h = 0.05 m = 50 mm.

New partial pressure of N₂ = Old partial pressure of N₂ = 0.8P.

U-tube manometer is used to measure the pressure of gas only.

In air conditioning process, moist air is heated and then cooled.

Flow rate of B = 5 – 10/5 = 3 mole/s.