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Test: CSIR-NET Chemical Sciences Mock Test - 6 - CSIR NET Chemical Science MCQ


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30 Questions MCQ Test CSIR-UGC (NET) Chemical Science Mock Test Series 2024 - Test: CSIR-NET Chemical Sciences Mock Test - 6

Test: CSIR-NET Chemical Sciences Mock Test - 6 for CSIR NET Chemical Science 2024 is part of CSIR-UGC (NET) Chemical Science Mock Test Series 2024 preparation. The Test: CSIR-NET Chemical Sciences Mock Test - 6 questions and answers have been prepared according to the CSIR NET Chemical Science exam syllabus.The Test: CSIR-NET Chemical Sciences Mock Test - 6 MCQs are made for CSIR NET Chemical Science 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: CSIR-NET Chemical Sciences Mock Test - 6 below.
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Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 1

Salaries of Ravi and Sumit are in the ratio 2:3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40:57. What is Sumit's salary?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 1

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, 
(2x+4000) / (3x+4000) = 40 / 57
⇒ 57 × (2x + 4000) = 40 × (3x+4000)
⇒ 6x = 68,000
⇒ 3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 2

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is 

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 2

Let the numbers be x and (100-x).
Then,x(100 − x ) = 5 * 495
⇒ x2 − 100 x + 2475 = 0
⇒ (x-55) (x-45) = 0
⇒ x = 55 or x = 45
The numbers are 45 and 55 
Required difference = (55-45) = 10

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Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 3

A, B and C enter into a partnership and their shares are in the ratio 1/2 : 1/3 : 1/4. After 2 months, A withdraws half of his capital and after 10 months, a profit of Rs. 378 is divided among them. What is B's share ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 3

Ratio of initial investments = 1/2 : 1/3 : 1/4 = 6 : 4 : 3.
Let their initial investments be 6x, 2x and 3x respectively.
A : B : C = (6x * 2 + 3x * 10) : (4x * 12) : (3x * 12) = 42 : 48 : 36 = 7 : 8 : 6.
B's share =  Rs. (378 * 8/21) = Rs. 144.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 4

A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 4

Let sum=Rs.x
C.I. when compounded half yearly = 
C.I. when compounded annually = 

⇒ > x = 20000

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 5

The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 5

Average of 20 numbers = 0. 
Sum of 20 numbers = (0 * 20) = 0. 
It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 6

What least number must be subtracted from 13601, so that the remainder is divisible by 87 ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 6


Required number is 29.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 7

405 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20% of the total number of children. How many sweets did each child recieve ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 7

Let the total number of children be x.
Then, x⋆(20% of x) = 405
⇒ x∗20×/100 = 405
(1/5)x= 405
⇒ x = 45
Number of sweets received by each child =20%
of 45 = 9

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 8

The ratio of the outer and the inner circumferences of a circular path is 11 ∶ 7. If path is 20 metres wide, then what is the radius of the inner circle? 

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 8

Given:
Outer circumference: Inner circumference = 11 : 7
Breadth of the path between concentric circles = 20 m
Calculation:
Let outer radius = R
Inner radius = r
Outer circumference : Inner circumference = 11 : 7
⇒ 2 × π × R : 2 × π × r = 11 : 7
⇒ R : r = 11 : 7
The breadth of the path between concentric circles = outer radius - inner radius
⇒ (11 - 7) units = 20 m
⇒ 4 units = 20 
⇒ 1 unit = 5 m
⇒ 7 units = 7 × 5 = 35 m 
∴ The correct answer is 35 m.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 9

A dishonest shopkeeper professes to sell pulses at the cost price, but he uses a false weight of 950gm for a kg. Find his gain percent.

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 9

The shopkeeper claims to sell 1 kg of pulses, but actually sells only 950 grams.

Let's assume the cost price (CP) of 1 kg of pulses is 100 units of currency.

Since he uses a weight of 950 grams instead of 1000 grams, the effective cost price for 950 grams is:

However, the shopkeeper sells 950 grams for the price of 1 kg, which is 100 units. Thus, the selling price (SP) for 950 grams is 100 units.

Now, we can calculate the gain:

The gain percent is given by:

So, the correct answer is: 5.26%

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 10

In grade 11 of a school, 40 students opted for Physics, 17 opted for Biology, and 20 opted for Chemistry. If the total number of students in grade 11 was 60, all of these students opted for at least one of the three subjects mentioned here, and exactly five of these students opted for all these three subjects, what is the probability that a randomly selected student of grade 11 of this school would have opted for exactly one of these three subjects?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 10

Concept used:
Probability = Number of students who opted for exactly one subject / Total number of students
Calculation:
Total number of students = 60
Number of students who opted two or more subjects = 40 + 17 + 20 - 60 - 5 = 12
So number of students who opted exactly one subject = 60 - 12 = 48
Required probability = 48/60 = 0.80
∴ The correct answer is 0.80

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 11

A train started at 9 AM from a station A with a speed of 72 km/hr. Another train after two hours started from the station B towards A with a speed of 90 km/ph. The two trains are expected to cross each other at 1.30 PM. At 12 noon because of the signals both the trains reduced their speeds by the same quantity and they crossed each other at 4.30 PM. The speed of the train, after 12 noon, that started from the station A, is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 11


AC in coverd in 2 hr by 1st train = 72 × 2 = 144 km
BC in coverd by both trains in (11 am to  1: 30pm = 2.5 hr) so speed will be added (opposite direction)
⇒ (72 + 90) × 2.5
⇒ 162 × 2.5 = 405 km
Distance covered after 12 noon = 405 - (72 + 90) = 243 km
Time taken in 243 km (12 pm to 4: 30 pm) = 4.5 hr
Let the reduction be 'y' km/h in each train
So, the speed of both train = 243/4.5 = 54 km
⇒ (72 - y) + (90 - y) = 54
⇒ 162 - 2y = 54
⇒ 2y = 108
⇒ y = 54 km
The speed of the train, after 12 noon, that started from the station A, = 72 - x = 72 - 54 = 18 km/h

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 12

The price of a commodity increased by 5% from 2010 to 2011, 8% from 2011 to 2012 and 77% from 2012 to 2013. What is the average price increase (approximate) from 2010 to 2013 ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 12

Given:
The price of a commodity increased by 5% from 2010 to 2011, 8% from 2011 to 2012 and 77% from 2012 to 2013. 
Formula used:
Average = Sum of observations/Number of observations
Calculation:
According to the question
The average price increases from 2010 to 2013 = (5 + 8 + 77)/3
⇒ (90/3)%
⇒ 30%
∴ The required average is 30%.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 13

Study the given graph and answer the following question.

The number of lecturers recruited in state A in the year 2017 was what percentage of the number of lecturers recruited in state B in the year 2019 ? (Correct to 2 decimal places)

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 13

Given:
No. of lecturers in state A in 2017 = 10
No. of lecturers in state B in 2019 = 22
Formula used:
Percentage = 
Calculations:
According to the formula,
⇒ Percentage = (10/22) x 100 = 45.45%
⇒ Hence, No. of lecturers in state A in 2017 is 45.45% of state B in 2019. 

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 14

A swimming pool is fitted with 3 pipes A, B, C to fill the pool. A and B together can fill the pool in half the time that is required for C to fill the pool. B takes 20 hours more than the time required for A and 14 hours more than the time required for C to fill the pool. Then the time (in hours) required for all the 3 pipes together to fill the pool is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 14

Given that A, B, and C are pipes that take a, b, and c hours respectively to fill the pool. According to the problem,
we have: b = a + 20 and b = c + 14
Subtracting these equations, we get:
a + 20 = c + 14 ⇒ a - c = 6 ---- (equation 1)
Also given that A and B together can fill the pool in half the time that is required for C to fill the pool, we can write:
a*b / (a + b) = c / 2
Given b = a + 20, replacing b in above equation we get:
a × (a+20) / (2a + 20) = c / 2
Solving this gives:
a² + 20a = (c + a) ×  c
As a - c = 6 from equation 1, replacing c with (a-6) in above equation we get:
a² + 20a = (a - 6) × a
Upon simplifying, we get:
a² + 20a = a² - 6a
This simplifies to:
26a = 60
Solving further we find:
a = 15 hours
Substituting a = 15 into equations b = a + 20 and b = c + 14 equations, we get:
b = 15 + 20 = 35 hours
c = b - 14 = 35 - 14 = 21 hours
To find out how long it would take for all three to fill the pool together, we add their rates of work. The rate of work is the reciprocal of the time taken.
The combined rate of work = 1/a + 1/b + 1/c = 1/15 + 1/35 + 1/21 = 7/105 + 3/105 + 5/105 = 15/105 = 1/7
So, they will fill the pool together in 1 / (1/7) = 7 hours.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 15

A and B run in opposite directions from a point P on the circumference of a circle with different but constant speeds. A runs in the clockwise direction. They meet for the first time at distance of 900m in the clockwise direction from P, and for the second time at a distance of 800m in the anticlockwise direction from P, and B is yet to complete one round. The circumference of the circle is:

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 15

By the time they meet for the second time, both of them, together, have completed exactly 2 rounds, since they are running at different speeds, one of them has completed more than one round and other is yet to complete a round. A runs 900m per meeting. Hence A ran 1800m by the time they met for the second time. B ran a total of 800 m by this time.
Cirumference = 1800 + 800 / 2 = 1300 m

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 16

What is the name of the reaction when acetamide changes into methylamine?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 16

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 17

Which type of compounds cannot exhibit geometrical isomerism?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 17

Triply bonded compounds cannot exhibit geometrical isomerism as the -C=C- bond in these compounds is linear.
Geometrical isomerism is a type of stereoisomerism in which the molecular formula and structure are same but the relative arrangement of atoms is different. Geometrical isomerism arises commonly in heteroleptic complexes. This type of isomerism arises due to the different possible geometric arrangements for the ligand.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 18

From the options given below, select the option which does not contain the correct isotopes of the elements.

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 18

The isotopes of Sulfur and Carbon are given incorrectly.
Isotopes of Carbon : 12C6,13C6,14C6
Isotopes of Sulfur: 32S16,33S16,34S16,35S16
Atoms of the same element that contains the same number of electrons but a different number of neutrons are termed Isotopes. Thus, isotopes have the same atomic number but different mass number.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 19

Which of the following is a macronutrient?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 19

macronutrient is a chemical element (e.g. potassium, magnesium, calcium) required in large amounts for plant growth.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 20

The electronegativity of sp2 hybridised atom will be

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 20

The electronegativity of sp2 hybridised atom will be 2.75. Fluorine is the most electronegative element.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 21

Which of the following can make difference in optical isomers?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 21

An optically active substance is one which can rotate the plane of polarisation of plane polarised light. if you shine a beam of polarised monochromatic light (light of only a single frequency – in other words a single colour) through a solution of an optically active substance, when the light emerges, its plane of polarisation is found to have rotated.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 22

The optimum DP value of cellulose is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 22

The optimum DP value of cellulose is 250. The number of monomeric units contained in the polymer is called degree of polymerisation (DP).

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 23

Reduction of nitroalkanes yields which compound?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 23

Reduction of nitroalkanes yields amines, as shown in below reaction.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 24

The temperature at which solid and liquid coexist in equilibrium is called

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 24

The temperature at which solid and liquid coexist in equilibrium is called melting point of solid or freezing point of liquid.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 25

Which of the following is not a secondary pollutant

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 25

Secondary air pollutants are poisonous substance formed from primary air pollutants. In bright sun light nitrogen, nitrogen oxides, hydrocarbons and O2 interact to produce more powerful photochemical oxidants like ozone (O3), peroxyacetyl nitrate (PAN), aldehydes, sulphuric acid, peroxides, etc.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 26

Which of the following is a not a five membered ring?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 26

Pyridine is a basic five membered heterocyclic organic compound with the chemical formula C5H5N. It is structurally related to benzene, with on (=CH–) group replaced by a nitrogen atom.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 27

Which source of water is free from hardness and surface impurities?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 27

Rain water is free from hardness and surface impurities. Surface water and sea water is very impure and saline.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 28

Why are aryl halides less reactive towards nucleophilic substitution reactions as compared to alkyl halides?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 28

Overlapping of sp2 orbital of carbon with p-orbital of halogen is one of the reasons. Due to conjugation double bond character in alkyl halide.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 29

If ΔH is the change in enthalpy and ΔE, the change in internal energy accompanying a gaseous reaction, then:

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 29

ΔH<ΔE only if the number of moles of products is less than the number of moles of the reactants.
If n< nr; Δn= np−n=− ve. where, n= No. of moles of products
n= No. of moles of reactants, n= No. of moles of gas
Therefore, ΔH < ΔE.

Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 30

In primary alkyl halides, carbon attached to the halogen atom is further attached to how many carbon atoms?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 6 - Question 30

As we can see below, carbon attached to the halogen atom is further attached to one carbon atom.
Example: CH3-X → Methyl halide
CH3-CH2-X → Ethyl halide
CH3-CH2-CH2-X → n-Propyl halide.

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