CSIR NET Chemical Science Exam  >  CSIR NET Chemical Science Tests  >  CSIR-UGC (NET) Chemical Science Mock Test Series 2024  >  Test: CSIR-NET Chemical Sciences Mock Test - 5 - CSIR NET Chemical Science MCQ

Test: CSIR-NET Chemical Sciences Mock Test - 5 - CSIR NET Chemical Science MCQ


Test Description

30 Questions MCQ Test CSIR-UGC (NET) Chemical Science Mock Test Series 2024 - Test: CSIR-NET Chemical Sciences Mock Test - 5

Test: CSIR-NET Chemical Sciences Mock Test - 5 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 - 5 questions and answers have been prepared according to the CSIR NET Chemical Science exam syllabus.The Test: CSIR-NET Chemical Sciences Mock Test - 5 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 - 5 below.
Solutions of Test: CSIR-NET Chemical Sciences Mock Test - 5 questions in English are available as part of our CSIR-UGC (NET) Chemical Science Mock Test Series 2024 for CSIR NET Chemical Science & Test: CSIR-NET Chemical Sciences Mock Test - 5 solutions in Hindi for CSIR-UGC (NET) Chemical Science Mock Test Series 2024 course. Download more important topics, notes, lectures and mock test series for CSIR NET Chemical Science Exam by signing up for free. Attempt Test: CSIR-NET Chemical Sciences Mock Test - 5 | 75 questions in 180 minutes | Mock test for CSIR NET Chemical Science preparation | Free important questions MCQ to study CSIR-UGC (NET) Chemical Science Mock Test Series 2024 for CSIR NET Chemical Science Exam | Download free PDF with solutions
Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 1

If log 27 = 1.431, then the value of log 9 is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 1

log 27 = 1.431
log(33) = 1.431
3 log 3 = 1.431
log 3 = 0.477
log 9 = log(32)= 2 log 3 = (2 x 0.477) = 0.954

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 2

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 2

Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 3

Two dice are tossed. The probability that the total score is a prime number is:

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 3

Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = n(E)/n(S) = 15/36 = 5/12.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 4

If selling price is doubled, the profit triples. Find the profit percent ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 4

Let the C.P be Rs.100 and S.P be Rs.x, Then 
The profit is (x-100)
Now the S.P is doubled, then the new S.P is 2x
New profit is (2x-100)
Now as per the given condition; 
⇒ 3(x-100) = 2x-100
By solving, we get x = 200
Then the Profit percent = (200-100)/100 = 100
Hence the profit percentage is 100%

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

If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle.

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

let length = x and breadth = y then  
2(x+y) = 46
⇒ x+y = 23  
x2+y2 = 172 = 289  
now (x+y)2 = 23² 
⇒ x2+y2+2xy= 529 
⇒ 289+ 2xy = 529
⇒ xy = 120  
area = xy = 120 sq.cm

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

An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

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

Angle traced by the hour hand in 6 hours=(360/12)*6

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 7

In a certain code language COMPUTER is written as RFUVQNPC. How will MEDICINE be written in that code language?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 7

There are 8 letters in the word.
The coded word can be obtained by taking the immediately following letters of word, expect the first and the last letters of the given word but in the reverse order. That means, in the coded form the first and the last letters have been interchanged while the remaining letters are coded by taking their immediate next letters in the reverse order.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 8

Two bikes are moving with speeds m km/h and n km/h towards a crossing along two perpendicular roads. If their distances from the crossing be 60 meters and 70 meters respectively at an instant of time then they do not collide if their speeds are such that

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 8

Since the bikes should not be collided, the time taken by bikes to cover the distance to reach the junction should not be the same.
⇒ tm ≠ tn ⇒ 60/m ≠ 70/n ⇒ m/n ≠ 6/7
∴ Ratio of speeds should not be equal to 6/7.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 9

In the given figure, if AB || CD || EF, then the measure of ∠CEF is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 9

Given
AB || CD || EF
∠ B = 70° and ∠ BCE = 30° 
Calculation
AB || CD and BC is a transversal,
⇒ ∠ B = ∠ BCD [Alternate interior angles]
⇒ 70° = ∠ BCE + ∠ ECD
⇒ 70° = 30° + ∠ ECD
⇒ ∠ ECD = 70° - 30° = 40° 
EF || CD and EC is a transversal
⇒ ∠ CEF + ∠ ECD = 180° [Sum of angles on same side of transversal is 180°]
⇒ ∠ CEF + 40° = 180° 
⇒ ∠ CEF = 180° - 40° = 140° 
∴ ∠ CEF = 140° 

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 10

The sum of any three consecutive natural numbers where first number is even number is always divisible by?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 10

The number is always divisible by 3.
∵ The sum of smallest 3 consecutive number = 2 + 3 + 4 = 9
Taking another example;
82 + 83 + 84 = 249.
∵ 249 is a multiple of 3, thus it is divisible by 3.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 11

A clock takes 5 seconds to strike 5 times at 5 O'clock. How long will it take to strike 9 times at 9 O'clock ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 11

But we know that, there should be an interval of time between two successive strikes.
Then, between 5 strikes, there must have 4 intervals of time.
And it is given that it takes 5 seconds for that 4 intervals.
Hence, 5 strikes → 4 intervals → 5 seconds
So, each interval is 5454 seconds long.
Now we have to find out the time taken to strike 9 times at 9 O'clock.
As per the given concept, between 9 strikes, there must have 8 intervals of time.
That is : 9 strikes → 8 intervals.
Then the time taken for 9 strikes = Number of intervals between them × Time taken for one interval = 8 × 5454 = 2 × 5
= 10 seconds
Hence, 10 seconds is the correct answer.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 12


The trends of two quantities over five years are shown in the graph. Which of the following are valid inferences?
A. The mean values of the quantities are nearly equal
B. The variations in the two quantities are nearly equal
C. Quantity 1 varies less over the given period as compared to Quantity 2 

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 12

The mean or Average of a set of experiments is the sum of the values divided by the number of values.
It is clear from the graph that quantity I is lesser in every part of the graph.
So mean values will be the same.
The correct answer is an option (3).

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 13

The bar chart given below shows the total revenue (in Rs. '000) earned by 5 shops by selling steel bottles.

The price of each bottle is Rs. 600 in all the shops. How many bottles have been sold by S2 and S3 taken together?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 13

Total Revenue of shop S2 and S3 by selling steel bottles = 54000 + 36000
= 90000
The price of each bottle is Rs 600 in all the shops.
Number of bottles sold by S2 and Staken together
= 90000/600
= 150
∴ The total number of bottles sold by S2 and S3 is 150.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 14

Study the following graph and answer the question that follows.

The actual production of ACs in March is what percentage more than the average target production of ACs over the period of 5 months (correct to the nearest integer)? 

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 14

Total targeted production of AC in 5 months = 60 + 30 + 80 + 90 + 100
⇒ 360
Average = 360/5 = 72
Actual production of AC in March = 100
Difference = 100 - 72 = 28
Percentage = 28/72 × 100
⇒ 38.88% ≈ 39%
∴ Required answer is 39%.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 15

At what angle are the hands of a clock at 15 minutes past 4 o'clock?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 15

Given:
Hands of a clock are at 15 minutes past 4 o’clock.
Concepts used:
In 12 hours the hour hand of a clock moves 360°
So, in one hour the hour hand of a clock moves = 360 /12 = 30°.
Also, in 1 minute the hour hand of a clock moves = 30 / 60 = 0.5°.
In 1 hour or 60 minutes, the minute hand of a clock moves 360°.
So, in 1 minute the minute hand of a clock moves = 360 / 60 = 6°.
Solution:
At 4 o’clock the hour hand of the clock will be making = 4 × 30 = 120°.
Angle made by the hour clock at 15 minutes past 4 o’clock = 120 + 0.5 × 15 = 127.5°.
At 15 minutes, the minute hand of the clock will be making = 15 × 6 = 90°.
Thus, the angle made by the hands of clock at 15 minutes past 4 o’clock
= Angle made by hour hand – angle made by minute hand
= 127.5° – 90°
= 37.5°.
∴ 37.5° is the angle made by the hands of a clock at 15 minutes past 4 o'clock.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 16

Consider the following reactions and choose an option correctly indicating the chemical property of water:

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 16

 

H2O(l) + H2O (l) →H3O+ (aq)+ OH-(aq) Autoprotolysis or self-ionization reaction where 2 molecules of solvent react to give cation and anion characteristic of that solvent.
Al4C3 (s)+ 12H2O(l) →4Al(OH)3 (aq)+3CH4 (g) Hydrolysis
2Na(s) +2H2O (l) →2NaOH (aq) +H2 (g) Oxidizing agent as it is getting reduced to H2 gas.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 17

Two guns are pointed at each other one upwards at an angle of elevation of 30º and other at the same angle of depression, the muzzle being 30 m apart. If the charges leave the gun with velocities of 350 m/s and 300 m/s respectively. Find when will they meet?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 17


In ∆ABC,
sin 30 ° = AC/AB
AC = AB sin 30 ° = 30 × (12) = 15 m

Height risen by Ist shell
= H1 = 350 sin 30 ° t – (1/2)gt2
= 175 t – gt2/2
Height fallen by IInd shell

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 18

By passing excess Cl2(g) in boiling toluene, which one of the following compounds is exclusively formed?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 18

When chlorine gas is passed through boiling toluene, side chain substitution occurs forming benzotrichloride.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 19

Which of the following compounds is not coloured ?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 19

Complex compound having lack of unpaired electron is colourless. Among the given complexes, K4[Fe(CN)6] has no unpaired electron as CN- being strong field ligand causes pairing of electrons. Thus, it is not coloured.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 20

Iodoform can be prepared from all, except

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 20

Iodoform can be prepared from compounds
Containing ¾CH3CO group or CH3¾CHOH group.
Propan -1-ol does not contain may such group.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 21

Which one of the following represents the correct ratio of the energy of electron in ground state of H atom to that of the electron in the first excited state of Li+?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 21

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 22

The bivalent metal ion having maximum paramagnetic behavior among the first transition series elements is,

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 22


Higher the number of unpaired electrons, higher will be the paramagnetic behavior shown by the ion. Among the given ion, Mn2+ has the highest number of unpaired electrons, thus it will show maximum paramagnetic behavior.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 23

Which of the following is correct representation of bleaching powder?

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 23

Bleaching powder is simply represented by formula CaCl2 and its aqueous solution gives positive test for Ca2+, Cl- and Clo- thus, its correct representation is
Ca2+(ClO-)Cl-

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 24

What is the pH of a saturated solution of Mg(OH)2? Ksp = 1.8 x 10-11.

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 24

Concentration of OH- = [2s]
Concentration of Mgg2+ = [s]
Ksp = [s][2s]2 = 1.8 x 10-11
S = 1.65 x 10-4
[OH] = 2 x 1.65 x 10-4 = 3.3 x 10-4
PH = 14 – POH = 14 + log[OH] = 10.5

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 25

Electrons accelerated from rest by a potential difference of 12.75 V, are bombarded on a monoatomic hydrogen gas. Possible emission of spectral lines are -

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 25

E4 – E= – 0.85 eV – (–13.6 eV)
= 12.75 eV.
Possible emission of spectral lies
(i) First three lyman lines
(ii) First two Balmer lines
(iii) First paschen line

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 26

In order to oxidise a mixture of one mole of each of FeC2O4,Fe(C2O4)3,FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 required is:

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 26

Writing the oxidation reaction for each compound

 No oxidation because all elements are already in highest oxidation stake

Now, we can equate gram equivalent of KMnO4 & other reactants.
gm.eq. 
⇒  moles × valence factor =_______
x × 5 = (1 × 3) + (1 × 6) + (1x1)
x = 2

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 27

CO is a stronger ligand that Cl- because

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 27

The strength of a ligand is determined by the amount of crystal field energy. Since, CO causes more crystal field splitting than Cl- due to the presence of π bonds, so it has more crystal field splitting and thus is stronger ligand than Cl-.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 28

The variation of solubility of four different gases (G1, G2, etc.) in a given solvent with pressure at a constant temperature is shown in the plot.

The gas with the highest value of Henry’s Law constant is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 28

P = KHX(g)
X(g) = Px1/KH
Slope = 1/KH
Hence the gas with highest KH will have least slope i.e G1.

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 29

BrF5 hybridization is

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 29

 

sp3d2 hybridization having octahedral shape with one position occupied by lone pair (or square pyramidal) with five 4sp3d2-2p bonds.
The valance electrons in BrF5 = 7 + 5 × 7 = 42. The distribution of these electrons in BrF5

Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 30

The bond length of H2+,H2-, and H2 are in the following order

Detailed Solution for Test: CSIR-NET Chemical Sciences Mock Test - 5 - Question 30


The order of bond is

Thus, the order of bond length is
H2- > H2+ > H2

View more questions
9 tests
Information about Test: CSIR-NET Chemical Sciences Mock Test - 5 Page
In this test you can find the Exam questions for Test: CSIR-NET Chemical Sciences Mock Test - 5 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: CSIR-NET Chemical Sciences Mock Test - 5, EduRev gives you an ample number of Online tests for practice
Download as PDF