Test: CSIR-NET Chemical Sciences Mock Test - 1 - CSIR NET Chemical Science MCQ

Saving  = 50% of (100 - 40)% of (100 - 30)% of Rs. 18,400 = Rs. (50/100 * 60/100 * 70/100 * 18400) = Rs. 3864.

= 21%

Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.
Remaining Work = 7/10, which was done by A,B and C in the initial number of days. 
Number of days required for this initial work = 7 days. 
Thus, the total numbers of days required = 4 + 7 = 11 days.

Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.

Let S be the sample space. Then, n(S)= number of ways of drawing 3 balls out of 
Let E= event of getting all the 3 red balls. 

We need to find the probability that at least one dice shows a number less than 5.
P₁ = 1 – Probability that none of the dice shows the number that is less than 5
P = probability that both dice does not have less than 5 that is can have 5 or 6

= 8/9

Given:
In an exam, there were 5 questions. 10% of students solved all questions, 10% did not solve any question and 15% of the remaining students solved 1 question and 16% of total students solved 4 questions 24% of students solved 2 questions and 140 students solved 3 questions.
Calculation:
Let the total number of students be x.
10% of students solved all questions, 10% did not solve any question.
Number of students who solved all questions = 10% of x = 0.10x
Number of students who solved 0 question = 10% of x = 0.10x
15% of the remaining solved single question.
Remaining students = x – 0.10x – 0.10x = 0.80x = 4x/5
Number of students who solved 1 question = 15% of 4x/5 = 3x/25
Number of students who solved 4 questions = 16% of x = 0.16x = 4x/25
Number of students who solved 2 questions = 24% of x = 0.24x = 6x/25
Number of students who solved 3 questions = 140
Total number of students = Number of students who solved 0 question + Number of students who solved 1 question + Number of students who solved 2 questions + Number of students who solved 3 questions + Number of students who solved 4 questions + Number of students who solved all questions
⇒ x = x/10 + 3x/25 + 6x/25 + 140 + 4x/25 + x/10
⇒ x = 18x/25 + 140
⇒ x – 18x/25 = 140
⇒ 7x/25 = 140
⇒ x = 500 students
∴ Total students are equal to 500.

Given data:

∠P : ∠Q : ∠R = 3 : 2 : 1
PR ⊥ RS
⇒ ∠PRS = 90°
Concept used:
The sum of all angles of a triangle is 180°.
Exterior angle = Sum of remaining two interior angles
⇒ ∠PRT = ∠QPR + ∠PQR
Calculation:
∠P : ∠Q : ∠R = 3 : 2 : 1
∠P = 3x
∠Q = 2x
∠R = x
∠P + ∠Q + ∠R = 180°
⇒ 3x + 2x + x = 180°
⇒ 6x = 180°
⇒ x = 30°
⇒ ∠P = 3 × 30°
⇒ 90°
⇒ ∠Q = 2 × 30°
⇒ 60°
⇒ ∠R = 1 × 30°
⇒ 30°
∠ PRT = ∠QPR + ∠PQR
⇒ 90° + 60°      ----(∵ ∠PRS = 90°)
⇒ ∠PRT = 150°
∠PRT = ∠PRS + ∠TRS      ----(Exterior angle sum)
⇒ 150° = 90° + ∠TRS
⇒ ∠TRS = 150° - 90°
⇒ ∠TRS = 60°
∴ The measure of ∠TRS is 60°.

⇒ (1 + m)(1 + n) = 4 ………… (i)
⇒ (1 + n)(1 + p) = 9 …………….. (ii)
⇒ (1 + p)(1 + m) = 16 ………………. (iii)
Now [(i) ÷ (ii)] (iii)
⇒ (m + 1)² = 4/9 16
⇒ m = 5/3
put the Value of m in (i)
⇒ n = 1/2
put the Value of n in (ii)
⇒ p = 5
Hence, 12mnp = 12 × 5/3 × 5 × 1/2 = 50

Work = 1/Efficiency
Calculation:
⇒ Pipe A can fill the tank in 6 hours, and Pipe A and B together can fill the tank in 12 hours.
⇒ Pipe B can empty the tank = 1/(1/12 – 1/6) = -12 hours
⇒ In 8 hours A will fill tank = 8/6 = 4/3
⇒ A will fill the extra water = 1 – 4/3 = -1/3
⇒ This extra water will be removed by pipe B in = (-1/12)/(-1/3) = 1/4
⇒ B will be closed after  4 hours
∴ The required result will be 4 hours.

Let the amount invested in scheme I be 'Rs x', then the amount invested in scheme II be 'Rs. (100000 - x)'.
Case - A:
Profit from scheme I is = 10%.
∴ the amount = 10% of x = 0.1x
Profit from scheme II = 12%.
∴ the amount = 12% of (100000 - x) = 0.12 × (100000 - x)
Total amount = 0.1x + 0.12 × (100000 - x)
If the profit percentage gets interchanged:
Case - B:
Profit from scheme I is = 12%.
∴ the amount = 12% of x = 0.12x
Profit from scheme II = 10%.
∴ the amount = 10% of (100000 - x) = 0.1 × (100000 - x).
Total amount = 0.12x + 0.1 × (100000 - x)
According to problem:
[0.1x + 0.12 × (100000 - x)] - [0.12x + 0.1 × (100000 - x)] = 120
⇒ 0.1x + 12000 - 0.12x - 0.12x - 10000 + 0.1x = 120
⇒ 0.2x - 0.24x = 120 - 2000
⇒ -0.04x = 1880

The ratio of investment will be:

Given: 
In a certain time a boat can cover 168 km downstream and in the same time it can cover 96 km upstream.
Speed of boat in still water is 11 km/h.
Formula used:
Downstream speed = Speed of the boat in still water + Speed of the stream
Upstream speed = Speed of the boat in still water – Speed of the stream
Calculation:
Let the time taken by boat to cover 168 km downstream = ‘t’ hours
So, the time taken by boat to cover 96 km upstream = ‘t’ hours
The downstream speed of boat = (168/t) km/h
And the upstream speed of boat = (96/t) km/h
According to the question:
⇒ [(168/t) + (96/t)]/2 = 11
⇒ (168 + 96)/2t = 11
⇒ 264/2t = 11
⇒ t = 12 hours
∴the speed of current = [(168/12) - (96/12)]/2 = (14 - 8)/2 = 3 km/h

When X is unit digit, The number,
⇒ 10Y + X
When Y is unit digit, the number,
⇒ 10X + Y
∴ The sum of the numbers,
⇒ 10Y + X + 10X + Y
⇒ 11(X + Y)
For the sum to be a perfect square the value of (X + Y) = 11

If an ion has 2 electrons in K shell, 8 electrons in L shell and 6 electrons in M shell, then number of S electrons present in that element is 6 because of 2 electrons in 1s, 2 electrons in 2s and 2 electrons in 3s.
The maximum number of electrons in a subshell is given by 2(2l +1).
If electron has magnetic number –1, then it cannot be present in s-orbital because only possible magnetic number for s is 0.
Only one radial node is present in 3p orbital.

Due to nascent oxygen, it can act as a permanent bleaching agent

Oxidation state of hydrogen remains the same. Oxygen in one molecule of H₂O₂ is oxidised to O₂ (–1 → 0) whereas in another molecule is reduced to H₂O (–1 → –2) Thus, H₂O₂ acts both as an oxidizing and reducing agent.

Actual number of moles in 98mg of H₂SO₄ = 1x10^-3 mol
Number of molecules in 10^-3 mol of 98 mg of H₂SO₄= 6.023 x1020
Since 3.01 x 1020 are removed from 98 mg of H₂SO₄, so half of the molecules are left.
Numbre of moles left in 98 mg of H₂SO₄=
(1─0.5)x10^-3
= 0.5x10^-3 moles

Molecular mass of butane (C₄H₁₀) is= 4(12)+10 = 58
No. Of moles of C₄H₁₀ = 29/58
No. Of molecules = (29/58)*(6.022*10²³)
No. Of atoms = (29/58)*(6.022*10²³)*14 = 4.21*10²⁴

Nitrogen in pyridine has sp² hybridization because the lone pair of electron present on it is involved in delocalization.

Mass of 1F atom=  (since At.wt. of F = 19)
Mass of 1N atom = 
Mass of 1 H atom = 
Mass of 1 O atom= 
i > iii > ii > iv

KOH is a strong base and HCOOH is a weak acid.
C = (160 x 0.1 – 40 x 0.2)/200 = 0.04M
PH = 1/2 (pK_a - logC) = 1/2 (4 – log 2 - log 0.04) = 2.55
POH = 14 – 2.55 = 11.4

The relation between octahederal and tetrahederal splitting is:
∆_t = (4/9)∆₀

Momentum and kinetic energy are related by:

By de-broglie relation:
λ = h/p = h/
KE of electron = q x V = 1.6 x 10^-19 x 400 = 6.4 x 10^-17 J

m = 9.1 x 10^-31Kg
therefore, wavelength = 0.06nm

The energy emitted or absorbed by a black body is discontinuous in the form of small packets of energy called quantum.In  case of light, this quantum is called photon.The energy of radiation is directly proportional to frequency(v).
So the energy is given by: E=h\v  ---->(1)
Also, v =c/λ
Given,the energies of two photons are in the ratio 3 : 2, their wavelengths will be in the ratio of 2 : 3,because according to Planck’s quantum theory

So, λ1 : λ2 = 2:3
Hence option (2) is correct.

The process of separating a crystalloid from a colloid by filtration is called Dialysis.
Dialysis is the separation of suspended colloidal particles from dissolved ions or molecules of small dimensions (crystalloids) by means of their unequal rates of diffusion through the pores of semipermeable membranes.

Fac – Mer isomerism is shown by complexes of form Ma₃b₃

Meq. of H₃PO₃ = Meq. of Ca(OH)₂
⇒ V × 0.25 × 2 = 25 × 0.03 × 2 (H₃PO₃ is dibasic acid)