Test: CSIR-NET Chemical Sciences Mock Test - 9 - CSIR NET Chemical Science MCQ

Numbers are (2³ - 1), (3³ - 1), (4³ - 1), (5³ - 1), (6³ - 1), (7³ - 1) etc.
So, the next number is (8³ - 1) = (512 - 1) = 511.

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day.
100 years have 5 odd days.
75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days
1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. 
Jan   Feb   Mar  Apr  May  Jun  Jul  
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) 
Total number of odd days = (0 + 2) = 2.  
Required day was 'Tuesday'.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C₃ x 4C₂) 
= 210. 
Number of groups, each having 3 consonants and 2 vowels = 210. 
Each group contains 5 letters. 
Number of ways of arranging 5 letters among themselves = 5! = 120 
Required number of ways = (210 x 120) = 25200.

Let the average after 7th inning = x
Then average after 16th inning = x - 3
16(x-3)+87 = 17x 
 x = 87 - 48 = 39

Let the numbers 13a and 13b.
Then, 13a x 13b = 2028 
⇒ ab = 12. 
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.

A's 5 days work =50% 
B's 5 days work = 33.33\% 
C's 2 days work =16.66%[100−(50+33.33)]
Ratio of contribution of work of A, B and C = 50:33(1/3):16(2/3) =3:2:1
A's total share = Rs. 1500
B's total share = Rs. 1000
C's total share = Rs. 500
A's one day's earning = Rs.300 
B's one day's earning = Rs.200 
C's one day's earning = Rs.250

We have n(s)=52C₂52=52∗51/2∗1=1326
Let A = event of getting both black cards 
B= event of getting both queens
A ∩ B = event of getting queen of black cards 

 

The perimeter of Circle of radius r = 2πr
The perimeter of a square of side s = 4s
The perimeter of an equilateral triangle of side a = 3a
The perimeters of a circle, a square, and an equilateral triangle are equal.
2πr = 4s = 3a

• s = 3/4 a = 0.75 a
• r = 3/2π a = 0.477 a

Area of an equilateral triangle:
Area of square = s² = (0.75 a)² = 0.5625 a²
Area of Circle = 2πr² = π(0.477 a)² = 0.714 a²
A_circle > A_Square > A_trg

Given,
A family has several children.
Each boy in this family has as many sisters as brothers, but each girl has twice as many brothers as sisters.
By checking the given options.
1) 1 and 2
Here we see only one boy is there so he has not as many sisters as brothers, thus this option is gets eliminated.
2) 3 and 4
Here we see if one boy is selected he has two brothers which are not as many as sisters, thus this option is gets eliminated.
3) 6 and 3
Here we see if one boy is selected he has five brothers which are also not as many as sisters, thus this option is gets eliminated.
4) 4 and 3
Here we see if one boy is selected he has three brothers and three sisters, and if a girl selected she has two sisters and four brothers which is she has twice as many brothers as sisters.
Hence, the correct answer is 4 and 3.

In the case of the repeating year, we add 6 with the given year if the given year is a leap year + 1.
2020 is a leap year.
2021 = leap year + 1
so we add 6. 2021 + 6= 2027
The trick to remember:

2021 has the same calendar as that of 2027.
Hence, “2027 is the correct answer.

Given:
n is a natural number.
Concept Used:
If n is an even number, then n, n + 2 is divisible by 2.
If n is an odd number, then n + 1 is divisible by 2.
Calculations:
If n is an even number, then n, n + 2 is divisible by 2.
If n is an odd number, then n + 1 is divisible by 2.
⇒ Therefore, (n + 1)(n + 2) is divisible by 2.
If we take three consecutive numbers, then there will be divisible by 3.
So, from the above
⇒ n(n + 1)(n + 2) is divisible by 3.
⇒ and it is divisible by 2 also.
⇒ Therefore, It is divisible by 6 also.
∴ The three consecutive numbers are divisible by 6.

Let the radius of the semicircle is r as shown in the diagram given below.

Now, OA = OC = OB = r
AB = OA + OB = 2r

Since PA ∶ PB = 3 ∶ 2 and AB is diameter which is equal to 10 cm,
⇒ PA = 3/5 × 10 = 6 cm and PB = 10 - 6 = 4 cm
Since AO is the radius of the circle,
∴ OP = 6 - 5 = 1 cm
In the same way,
ND = 6 cm and NC = 4 cm
∴ NO = 6 - 5 = 1 cm
Now applying Pythagoras theorem in ΔOSN,
⇒ OS² = NS² + NO²
Since OS is the radius of the circle, OS = 5 cm
⇒ 25 = NS² + 1
⇒ NS = 2√6
∴ SM = (2√6 - 1) cm

Total production of suitcases of branch P in the periods of 2012 to 2016 = 60 + 50 + 35 + 55 + 50 = 250
The average production of suitcases of branch P in the periods of 2012 to 2016 = 250/5 = 50
Total production of suitcases of branch Q in the periods of 2012 to 2016 = 30 + 35 + 45 + 45 + 55 = 210
The average production of suitcases of branch P in the periods of 2012 to 2016 = 210/5 = 42
The average production of suitcases of branch P in the periods of 2012 to 2016 : The average production of suitcases of branch P in the periods of 2012 to 2016
⇒ 50 : 42
⇒ 25 : 21

Given:
The average of twelve 2 digit numbers is decreased by 3 when the digits of one of the 2 digit numbers is interchanged.
Formula used:
Average = sum of observations/ Number of observations
Calculation:
Let the number whose digits are being interchanged is 10x + y
After interchanging the digits, the number becomes 10y + x
At first average = (sum of 11 numbers + 10x + y)/12
After interchanging the digits, average = (sum of 11 numbers + 10y + x)/12
So, (sum of 11 numbers + 10y + x)/12 = (sum of 11 numbers + 10x + y)/12 - 3
⇒ (sum of 11 numbers + 10y + x) = (sum of 11 numbers + 10x + y) - 36
⇒ 10y + x = 10x + y - 36
⇒ 9x - 9y = 36
⇒ x - y = 36/9 
⇒ x - y = 4
∴ difference between two digits is 4.

The strength as oxidizing agents in acidic solution is in the order of 

Nylon is polymer having amide linkages, thus it is a polyamide.

4FeCrO₄ + 8Na₂CO₃ + 7H₂O → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
Chromite Ore
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + 2NaSO₄ + H₂O
Na₂Cr₂O₇ + 3C → Na₂Cr₂O₄ + 3CO
Na₂Cr₂O₄ + H₂O → Cr₂O₃ + 2NaOH
Cr₂O₃ + 2A₁ → 2Cr + Al₂O₃

More electropositive metals associated with Cu (Fe, Zn, Co, Ni) pass in solution state and less electropositive metals (Au, Ag) collect below the anode as anode mud.
Hence, option C is correct.

V₂O₅ is red or orange in color and it is an amphoteric oxide.

The diazotisation of -NH₂ group takes place in the presence of NaNO₂/HCland a coupling productis formed with the diazonium chloride and the active aryl nucleus.

The given reaction is Hoffmann Bromamide degradation reaction to synthesize an amine from amide as shown below:

The mechanism of this reaction can be represented as:
(i) RCONH2 + Br2  RCONHBr +HBr

From the mechanism ,it is evident that RCONHBr (step-II) and RNCO (step-IV) are formed as intermediates.

The repeating unit has the cis-configuration in natural rubber with chain extensions on the same side of the ethylene double bond. This is essential for elasticity. If the configuration is trans, the polymer is either a hard plastic or like gutta-percha.

2Cu₂O + Cu₂S → 6Cu + SO₂
Hence, option A is correct.

Basic character increases with increase in the electron donating (+I) group. N, N-dimethyl aniline is most basic owning to the restriction in the delocalization of the lone pair over the benzene ring owing to the presence of two methyl groups.

In basic medium KMnO₄ does not changes to Mⁿ⁺², it changes to MnO₂

The Gabriel phthalimide reaction can be used only for preparation of primary amines. It cannot be used to prepare secondary or tertiary amines. The compound given in option C is a tertiary amine and hence cannot be prepared by Gabriel phthalimide reaction. The reaction is as shown below:

Galvanisation is a process in which iron is protected by forming a layer of Zn on Fe.

Malachite green is a dye formed when benzaldehyde condenses with N,N-dimethyl aniline in presence anhydrous ZnCl₂.