Test: CSIR-NET Chemical Sciences Mock Test - 4 - CSIR NET Chemical Science MCQ

Let the total profit be Rs. 100.
After paying to charity, A's share  = (95 x 3/5) = Rs. 57.
If A's share is Rs. 57, total profit = Rs. 100.
If A's share is Rs. 855, total profit  = (100/57 x 855) = 1500.

Given:
If each edge of a cube is increased by 50%.
We know that,
The surface area of cube = 6 side²
So,
According to the question,
Let the side of the cube be x.
Each side of the cube increased by 50%.
So,
= (x)(100+50))/100
= 1.5x
The surface area of the cube = 6x²
The new surface area of the cube (side =1.5x) = 6 x 2.25x²
Increase percentage in the surface area = ((6 x 2.25x²-6x²)/6x²) x 100
= 125%
∴ The percentage increase in the surface area is 125%.

A is the sister of B and B is the daughter of C.
So, A is the daughter of C. Also, D is the father of C.
So, A is the granddaughter of D. 

The fruit content in both the fresh fruit and dry fruit is the same.
Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100kg
Dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg.
Fruit % in freshfruit = Fruit% in dryfruit 
Therefore, (32/100) x 100 = (80/100 ) x y 
we get, y = 40 kg.

Let the number be x.
Then, ideally he should have multiplied by  x by 5/3. Hence Correct result was x * (5/3)= 5x/3. 
By mistake he multiplied x by 3/5 . Hence the result with error  = 3x/5 
Then, error = (5x/3 - 3x/5) = 16x/15 
Error %  = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 %

Total number of votes = 7500 
Given that 20% of Percentage votes were invalid
⇒ Valid votes = 80%
Total valid votes = 7500*(80/100) 
1st candidate got 55% of the total valid votes.
Hence the 2nd candidate should have got 45% of the total valid votes
⇒ Valid votes that 2nd candidate got = total valid votes x (45/100)
7500 x (80/100) x (45/100) = 2700

20% of a = b  ⇒ (20/100)a = b 
b% of 20 = (b/100) x 20 = (20a/100) x (1/100) x (20) = 4a/100 = 4% of a.

LCM of 18, 24 = 72
Area of the floor = 18 × 24 = 432
the smallest number of identical square tiles that pave the entire floor without the need to break any tile
⇒ (LCM²)/Area 
⇒ (72 × 72)/432 = 12 
We can rather check by the options, the square of which of the options have a multiple in 432.

There's an interesting pattern to be noticed when we write down the powers of 3.
The number of digits in the expansion of powers of 3 increases by 1 after every two expansions. For example,
3¹ = 3               (1 digit)
3² = 9               (1 digit)
3³ = 27             (2 digits)
3⁴ = 81             (2 digits)
3⁵ = 243           (3 digits)
And so on...
Using this knowledge, we can conclude that for 3¹⁵ and 3¹⁶ there will be 8 digits in the expansion.

x ⊕ y = 2 ⊙ 2

x² + y² = 2xy
(x - y)² = 0
x = y

Letter series:
F, E, D, C, B, A
Number series:
2
2 × 2 = 4
4 × 2 = 8
8 × 2 = 16
16 × 2 = 32
32 × 2 = 64.
Hence, the missing term is E4.

As per the given data,
Option 3 is the correct answer as numerical value is given.

Area of a triangle = Semi - perimeter × In - radius
⇒ Area of a triangle = (1/2) × Perimeter × In - radius
∴ The number is 1/2.

Chords subtending angles at the same point on the circumference of the circle are equal.
∴ AB : CD = 1 : 1.

Cost price of 1 liter alcohol = Rs. 75
Selling price of 1 liter alcohol = 75 + 50% of 75 = Rs. 112.5
Price of water = Rs. 0/L

∴ Ratio of alcohol to water = 2 : 1

The statement that describes C₆₀ is C₆₀ is soluble in benzene.
Because of the organic nature of C₆₀, it is soluble in a non-polar solvent, such as benzene and reacts with tert-butyllithium. It is made up of 12 five-membered and 20 six-membered rings and adjacent five and six-membered rings having a common edge.

The conversion requires reduction; however, the conditions necessary (LiAlH₄) would also reduce the ketone carbonyl. The ketone functionality is therefore protected as the cyclic acetal.

Reduction of the carboxylic acid may now be carried out

The number of possible isomers for [Ru(bpy)₂Cl₂] is 3.

At constant temperature the final pressure 
 =2 atmosphere

Partial pressure of C₂H₆ = Mole fraction × Total pressure
No. of Moles of C₂H₆ = (60/30)= 2
No. of Moles of CO =(28/28) = 1
Mole fraction of C₂H₆ = 
Partial pressure of C₂H₆ =(2/3) × 3 = 2 atm

Alcohols are organic compounds that contain a hydroxyl (-OH) functional group attached to a carbon atom. They exhibit certain characteristics and properties, but one of these characteristics is not valid for alcohols:
A: Higher alcohols are stronger and have a bitter taste
This statement is incorrect because higher alcohols are not necessarily stronger or have a bitter taste. The strength of an alcohol refers to its ability to act as a base and donate a proton (H+). It is determined by the stability of the resulting conjugate base. However, the strength of an alcohol does not increase with the length of its carbon chain (higher alcohols have longer carbon chains). Instead, it depends on factors such as the electron-donating or electron-withdrawing nature of any substituents present on the carbon atom bearing the hydroxyl group.
Similarly, the taste of alcohols is not solely determined by their carbon chain length. While some alcohols may have a bitter taste, this characteristic is not exclusive to higher alcohols. Taste is subjective and can vary depending on the specific alcohol compound and individual preferences.
The other characteristics mentioned are valid for alcohols:
B: Lower alcohols are stronger and have a bitter taste
Lower alcohols, such as methanol and ethanol, can exhibit a stronger taste and may have a bitter flavor. They also tend to have lower carbon chain lengths.
C: The boiling points of alcohols increase with increasing carbon chain length
The boiling points of alcohols generally increase with an increase in the carbon chain length. This is because longer carbon chains result in more extensive London dispersion forces between molecules, leading to stronger intermolecular attractions and higher boiling points.
D: The lower alcohols are soluble in water
Lower alcohols, particularly those with one to three carbon atoms (methanol, ethanol, and propanol), are generally soluble in water due to their ability to form hydrogen bonds with water molecules. This solubility decreases with increasing carbon chain length.

HOMO-σ2p_x

HOMO - π *2p_y

The energy of photon (E) = hv
W = hc/λ
Given       h = 6.6 × 10^-27
c = 3× 10¹⁰ cm/sec.
I = 3000 A° 3000× 10^-8 cm 

Argon have stable full filled electronic configuration. So that’s why Argon have the highest ionization potential.
Ar(Inert gas) — 3s² 3p⁶

E = Eº – (RT/nF)  lnQ
At equilibrium E_cell = 0
So equilibrium const. H_C = Q

The maximum to minimum values of the quantum number J are given by
J = (j₁ + j₂), (j₁ + j₂ – 1), (j₁ + j₂ – 2)...|j₁ – j₂|
∴ For j₁ = 2 & j₂ = 4, we have
J_max = 2 + 4 = 6
J_min = |2 – 4| = 2
Hence possible values are J = 6, 6 – 1, 6 – 2, 6 – 3, 6 – 4,
∴ J = 6, 5, 4, 3, 2.