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

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

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.

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.

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%

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

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

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.

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.
⇒ t_m ≠ t_n ⇒ 60/m ≠ 70/n ⇒ m/n ≠ 6/7
∴ Ratio of speeds should not be equal to 6/7.

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° 

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.

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.

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).

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 S₃ taken together
= 90000/600
= 150
∴ The total number of bottles sold by S₂ and S₃ is 150.

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%.

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.

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.

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

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

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.

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

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.

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^-

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

E₄ – E₁ = – 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

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

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^-.

P = K_HX(g)
X_(g) = Px1/K_H
Slope = 1/K_H
Hence the gas with highest K_H will have least slope i.e G1.

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

The order of bond is

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