Test: CSIR-NET Chemical Sciences Mock Test - 2 - UGC NET MCQ

TEAR, RATE, TARE
3 words can be made.

8th  letter from left is H and 7th  letters from right is T.
Between H and T there are 11 letters between them.

VENTURESOME
SERMON - use all letters from Venturesome.

Bases are those substances that turn red litmus paper into the blue.
It is bitter in taste and gives hydroxyl ions (OH⁻) in an aqueous solution. For example, ammonia gas turns red litmus paper blue because it is a base and has a pH of 11.6.

Area of triangle

The given sequence will follow the pattern 3, 4 4, 3 ..........
These are the difference between two consecutive numbers of the sequence.
So.

From the given information, we can draw the Venn diagram as follows:

From the above Venn diagram it is clear that Poonam is good at horse riding and shooting not swordsmanship.

Given:
Marked price (M.P) = Rs.3600
Concept used:
Discount% = M.P × D%
Effective discount (D) = (D₁) + (D₂) - (D₁ × D₂)/100
Calculation:
Discount (D₁) = 3600 × 35%
⇒ 36 × 35 = Rs.1260
Effective discount = 30 + 5 - (30 × 5)/100
⇒ 35 - 1.5 = 33.5%
Discount (D₂) = 3600 × 33.5% = Rs.1206
Difference between two discount = (1260 - 1206) = Rs.54
∴ The correct answer is Rs.54.

Concept used:
The sum of interior angles of any triangle = 180° 
Calculation:
In the figure, PRT and UQS forms a triangle.
⇒ ∠P + ∠R + ∠T = 180°
⇒ ∠U + ∠Q + ∠S = 180° 
On adding the above two, we get
⇒ ∠P + ∠Q + ∠R + ∠S + ∠T + ∠U = 180° + 180° = 360° 
∴ ∠P + ∠Q + ∠R + ∠S + ∠T + ∠U = 360°

Given:
Speed of the scooter of thief = 50 km/h
Speed of the scooter of owner = 80 km/h
Formula Use:
Speed = Distance/Time
Calculation:
We know,
Speed of the scooter of thief = 50 km/h
Since the thief is discovered at 2:30 P.M
The distance travel by thief in 3 : 30 hour = 7/2 × 50 = 175 km
Speed of the scooter of owner = 80 km/h
So, the relative speed of the owner = (80 - 50) km/h
⇒ 30 km/h
175 km, is the distance that has to be covered by the owner to catch the thief.
We have,
Speed = Distance / Time
⇒ 30 = 175 / Time
⇒ Time = 175/30 = 35/6 = 5 hr 50 min
Therefore, the time required to catch the thief will be 5 hr 50 min.
The owner will catch the thief at = 2:30 P.M + 5 hr 50 min = 8 : 20 P.M
So, The owner will catch the thief at  8 : 20 P.M
Hence, the correct answer is  8 : 20 P.M

Given: 
12 years ago, Sudhir and Ashish = 1 : 2
Current age of Sudhir and Ashish = 5: 7
Concept used:
We will be using the concept of ratio.
Calculation:
Let the ages of Sudhir and Ashish 12 years ago = 1x and 2x respectively,
According to the question,
(1x + 12) / (2x + 12) = 5/7
⇒ 7x + 84 = 10x + 60
⇒ 3x = 24
⇒ x = 8
∴ Present age of Sudhir  = 8 + 12 = 20 year
The age of Sudhir after five years from now = 20 + 5 = 25
The age of Sudhir after five years from now is 25 years.

Let total weight of goods = 1000 gm 
Let the CP of 1000 gm is Rs. 1000 
∴ CP of 1 gm is Rs. 1 
Shopkeeper uses 20% less weight
Weight uses by shopkeeper = 1000 - 1000 × 20% = 800 gm
CP for shopkeeper = 800 × 1 = 800
SP for shopkeeper = 1000 
Profit = 1000 - 800 = 200 
Profit percentage = (200/1000) × 100 = 25%

Given:
The frustum of a right circular cone has base diameter 10 cm, and on the top the diameter is 6 cm, and its height is 5 cm. (usе π = 22/7)
Formula used:
The volume of a frustum of a cone:

V = 1/3 π(R² + r² + Rr)h
Where:
h is the height of the frustum.
R is the radius of the larger base (the top base) of the frustum.
r is the radius of the smaller base (the bottom base) of the frustum.
Calculation:
According to the question,
R = 10/2 = 5 cm
r = 6/2 = 3 cm
h = 5
Now, on substituting the value of the respective variable in the required formula, we get,

Therefore, "256.67 cm²" is the required answer.

The original value was 5K. When the student multiplies by 1/5, the result is 0.2K, which is 96% less than the original value. So, x = 96%.
When the student multiplies by 1/25, the result is 0.04K, which is 99.2% less than the original value. So, y = 99.2%.
The percentage increase of y over x is (y - x)/x × 100 = (99.2 - 96)/96 * 100 = 3.33%.
Therefore, the answer is 3.33%. or 3(1/3)%.

Fingerprint region of IR spectra appears between (1400-600) cm^-1.

SBR has high load bearing capacity, high abrasion resistance, swells in oil and solvent and has low oxidation resistance.

It denotes that how much a reaction occurs.

Number of moles = mass / molecular weight.

Brunauer, Emmett and Teller proposed a theory in 1938 which is known as B.E.T. theory.

According to Faraday’s law of electrolysis, the amount of any substance deposited at the electrode is directly proportional to the quantity of electricity passed. It is given by:
m = ( Q / F ) * ( M / Z )
where, m = Mass of the substance liberated at an electrode in grams
Q = Total electric charge passed through the substance in Coulombs
F = Faraday constant
M = Molar mass of the substance in grams per mol
Z = Valency number of ions of the substance
For Faraday’s first law, M, F, and z are constants, so that the larger the value of Q the larger m will be.

Meso compounds are internally compensated form whereas racemic mixtures are externally compensated modification. All the other options are correct.

Cumene hydroperoxide act as a catalyst in the preparation of Buna-S (Butadiene styrene).

It is a very sensitive instrument as it could accurately measure the weight of a particular compound of upto 0.1 mg.

Reactions which are bimolecular yet following first order kinetics are called pseudo unimolecular reaction. Pseudo means pretentious. Inversion of cane sugar is an example of pseudo first order reaction.

The unit of extent of a reaction is Moles reacting.

The main cause of molecular aggregation is inter-molecular interactions. Inter-molecular interactions stem from the electric properties of atoms. Being the cause of molecular aggregation, inter-molecular forces are at the roots of chemistry and are the fabric of the world.

(1) KI + CuSO₄ → I₂
(2) I₂ + Na₂S₂O₃ → NaI + Na₂S₂O₆
(a) Na₂S₂O₃ is oxidized to Na₂S₂O₆
(b) Cu₂I₂ is formed as CuI₂ disproportionate.

Chromatography involves two mutually immiscible phases (stationary and mobile phases). The mixture is dissolved in a fluid called the mobile phase, which carries it through a structure holding another material called the stationary phase.