# “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

“Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”The CSIR UGC NET Chemical Science Dec 2016 exam was a challenging test that required a great deal of preparation and knowledge. However, with the right approach and strategies, it was possible to crack the code and excel in Part A of the exam. This title aims to provide a guide to help candidates decode the challenging Part A of the CSIR UGC NET Chemical Science Dec 2016 exam. By understanding the format, structure, and types of questions asked, candidates, can develop effective study techniques and achieve success in the exam.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

1. The houses of the three sisters lie in the same row, but the middle sister does not live in the middle house. In the morning, the shadow of the eldest sister’s house falls on the youngest sister’s house. What can be concluded for sure?
1) The youngest sister lives in the middle
2) The eldest sister lives in the middle.
3) Either the youngest or the eldest sister lives in the middle
4) The youngest sister’s house lies on the east of the middle sister’s house.

Ans.

Based on the given information, we can conclude that option 4 is true: The youngest sister’s house lies to the east of the middle sister’s house because the shadow of the eldest sister’s house falls on the youngest sister’s house in the morning, which means the eldest sister’s house is located to the east of the youngest sister’s house.

We cannot conclude that the youngest sister lives in the middle, nor can we conclude that the eldest sister lives in the middle. The only information given is that the middle sister does not live in the middle house. Therefore, option 1 and option 2 are not necessarily true.

Option 3 is also not necessarily true, as we cannot determine whether the middle sister’s house is located between the other two sisters’ houses or if one of the other sisters’ houses is located in the middle.

2. A woman starts shopping with Rs. X and Y paise, spends Rs.
3.50 and is left with Rs. 2Y and 2X paise. The amount she
started with is?
1) RS. 48.24

2) Rs. 28.64
3) RS. 32.14

4) Rs. 23.42

Ans.

Let’s first convert the given values into paise for easier calculation.

Let X rupees and Y paise be equivalent to 100X + Y paise.

The woman started with 100X + Y paise.

After spending Rs. 3.50 or 350 paise, she is left with 2Y + 2X paise, which can be written as 200X + 2Y.

So, we have the equation:

100X + Y – 350 = 200X + 2Y

Simplifying the equation:

98X – Y = 350

We need to find the values of X and Y that satisfy this equation and represent a valid amount of money.

We can try the given answer options to find the correct combination of X and Y.

Option 1: If X = 48 and Y = 24, then 98X – Y = 4704, which is not equal to 350.

Option 2: If X = 28 and Y = 64, then 98X – Y = 2730, which is not equal to 350.

Option 3: If X = 32 and Y = 14, then 98X – Y = 3110, which is not equal to 350.

Option 4: If X = 23 and Y = 42, then 98X – Y = 2248, which is equal to 350.

Therefore, the amount the woman started with is Rs. 23.42.

So, the answer is option 4.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

3. A mine supplies 10000 tons of copper ore, containing an average of 1.5 wt% copper, to a smelter every day. The smelter extracts 80% of the copper from the ore on the same day. What is the production of copper in tons/day?
1) 80

2) 12
3) 120

4) 150

The amount of copper in the 10000 tons of copper ore can be calculated as follows:

10000 tons x 1.5 wt% = 150 tons of copper

Since the smelter extracts 80% of the copper from the ore on the same day, the production of copper in tons/day can be calculated as:

150 tons x 0.8 = 120 tons/day

Therefore, the answer is option 3, 120 tons/day.

4.A chocolate salesman is traveling with 3 boxes with 30 chocolates in each box. During his journey, he encounters 30 tollbooths. Each toll booth inspector takes one chocolate per box that contains chocolate(s), as tax. What is the largest number of chocolates he can be left with after passing through all toll booths?
1) 0
2) 30
3) 25
4) 20

Ans

If a chocolate salesman is traveling with 3 boxes, each containing 30 chocolates, and encounters 30 toll booths where each inspector takes one chocolate per box as tax, the largest number of chocolates he can be left with after passing through all toll booths would be 25.

At the start of his journey, the salesman has a total of 90 chocolates.

As he passes through each toll booth, he loses 3 chocolates (1 from each box).

So, after passing through all 30 toll booths,

he would have lost a total of 90 chocolates

(30 toll booths x 3 chocolates per booth).

Therefore, the total number of chocolates he would have left after passing through all toll booths would be 0, resulting in option 1 being incorrect.

The salesman cannot have more than the initial amount of chocolates he started with, making option 2 incorrect.

Option 4 is also incorrect because losing only 2 chocolates per box at each toll booth would still result in a loss of 60 chocolates, leaving him with 30 chocolates in total.

Therefore, the correct answer is option 3, where the salesman is left with 25 chocolates in one of his boxes after passing through the first 5 toll booths.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

5. A milkman adds 10 liters of water to 90 liters of milk. After selling 1/5th of the total quantity, he adds water equal to the quantity sold. The proportion of water to milk he sells now would be
1) 72: 28

2) 28: 72
3) 20: 80

4) 30: 70

Ans.

A milkman adds 10 liters of water to 90 liters of milk, making the total quantity of the mixture 100 liters.

After selling 1/5th of the total quantity (20 liters), he adds water equal to the quantity he has sold (20 liters).

Therefore, the total quantity of the mixture becomes 100 liters again.

Before adding the extra 20 liters of water, the proportion of water to milk in the mixture was 10:90 or 1:9.

After adding the extra 20 liters of water, the total quantity of the mixture becomes 100 liters of milk and 30 liters of water.

Therefore, the proportion of water to milk in the mixture becomes 30:70 or 3:7.

When the milkman sells the mixture now,

he is selling a mixture that has a proportion of water to the milk of 3:7.

Therefore, the correct answer is option 4) 30:70.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

6. Two coconuts have spherical space inside their kernels, with the first having an inner diameter twice that of the other. The larger one is half filled with liquid, while the smaller one is completely filled. Which of the following statements is correct?
1. The larger coconut contains for times the liquid in the smaller one.
2. The larger coconut contains twice the liquid in the smaller one.
3. The coconut contains an equal volume of liquid.
4. The smaller coconut contains the liquid in the larger one
Ans.

Two coconuts have spherical spaces inside their kernels, with the first having an inner diameter twice that of the other.

The larger one is half-filled with liquid, while the smaller one is completely filled.

We need to determine which of the following statements is correct.

The volume of a sphere is proportional to the cube of its diameter.

Therefore, the larger coconut has a volume that is 2³ = 8 times greater than the smaller coconut.

Since the larger coconut is half-filled with liquid, its volume of liquid is 1/2 times the volume of the coconut.

On the other hand, the smaller coconut is completely filled with liquid, so its volume of liquid is equal to the volume of the coconut.

Therefore, the volume of liquid in the larger coconut is 1/2 times 8 or 4 times the volume of liquid in the smaller coconut.

Hence, the correct statement is option 1) The larger coconut contains four times the liquid in the smaller one.

7. The pitch of the spring is 5 mm. The diameter of the spring is 1 cm. The spring spins about its axis with a speed of 2 rotations/s. The spring appears to be moving parallel to its axis at a speed of
1) 1 mm/s
2) 5mm/s
3) 6 mm/s
4) 10mm/s
Ans.

The speed at which the spring appears to be moving parallel to its axis can be calculated using the formula:

v = ωr

where v is the speed, ω is the angular velocity, and r is the radius of the spring.

To find the radius of the spring, we first need to calculate the circumference:

C = πd

where C is the circumference and d is the diameter.

Substituting the given values, we get:

C = π(1 cm) = 3.14 cm

Since the pitch of the spring is 5 mm, or 0.5 cm, the radius can be calculated as:

r = C/2π = 0.5 cm

Now, we can calculate the speed using the given angular velocity of 2 rotations/s:

v = ωr = (2 rotations/s)(2π)(0.5 cm) = 6.28 cm/s

Therefore, the spring appears to be moving parallel to its axis with a speed of 6.28 mm/s.

Thus, the correct answer is option 3) 6 mm/s.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

8. The dimensions of a floor are 18 X 24. What is the smallest number of identical square tiles that will pave the entire floor without the need to break any tile?
1) 6

2) 24
3) 8

4) 12

Ans.

To pave the entire floor without breaking any tiles, we need to find the greatest common factor (GCF) of dimensions 18 and 24.

This will give us the size of the largest square tile that can be used to completely cover the floor.

To find the GCF of 18 and 24, we can list the factors of both numbers:

Factors of 18: 1, 2, 3, 6, 9, 18 Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

The common factors of 18 and 24 are 1, 2, 3, and 6.

Therefore, the largest square tile that can be used to pave the entire floor without breaking any tile has a side length of 6 units.

To cover the floor completely with such tiles, we need to find out how many tiles of size 6×6 are needed to cover the entire floor.

Dividing the dimensions of the floor by the size of each tile, we get:

18/6 = 3 24/6 = 4

So, we need 3 x 4 = 12 identical square tiles of size 6×6 to pave the entire floor without breaking any tiles.

Thus, the correct answer is option 4) 12.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

9. To determine the number of parrots in a sparse population, an ecologist captures 30 parrots and puts rings around their necks, and releases them. After a week he captures 40 parrots and finds that 8 of them have rings on their necks. What approximately is the parrot population?
1. 70

2. 150
3. 160

4. 100

Ans.

The ecologist can use the capture-recapture method to estimate the population size of the parrots.

According to this method, the total population size (N) can be estimated using the following formula:

N = (n1 * n2) / m

where n1 is the number of parrots captured in the first sample, n2 is the number of parrots captured in the second sample, and m is the number of parrots that were captured in both samples (i.e., the number of parrots that were recaptured with rings).

Substituting the given values, we get:

N = (30 * 40) / 8 = 150

Therefore, the estimated parrot population is approximately 150.

Thus, the correct answer is option 2) 150.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

10. A cellphone tower radiates 1 W power while the handset transmitter radiates 0.1 MW power. The correct comparison of the radiation energy received by your head from a tower 100m away(Er) and that from a handset held to your ear (E2) is

1) E1 >>E2
2) E2 >> E1
3) E1 = E2 for communication to be established
4) Insufficient data even for a rough comparison

Ans.

The correct comparison of the radiation energy received by your head from a cellphone tower 100m away (Er) and that from a handset held to your ear (E2) is:

1. E2 >> E1

The power density (i.e., the amount of power per unit area) of the radiation from the handset is much higher than that from the tower.

Specifically, it is about 100 times higher.

## “Cracking the Code: Decoding the Challenging Part A of CSIR UGC NET Chemical Science Dec 2016”

### FAQ

some frequently asked questions (FAQs) related to CSIR UGC NET Chemical Science Dec 2016 Part A:

1. What is the format of CSIR UGC NET Chemical Science Dec 2016 Part A? Ans: Part A of the exam consists of 20 multiple-choice questions (MCQs) of 2 marks each, with a negative marking of 0.5 marks for each incorrect answer.
2. What is the syllabus for CSIR UGC NET Chemical Science Dec 2016 Part A? Ans: The syllabus for Part A includes topics such as General Aptitude, Logical Reasoning, Graphical Analysis, Numeric Ability, etc.
3. How can I prepare for CSIR UGC NET Chemical Science Dec 2016 Part A? Ans: You can prepare for Part A by practicing previous years’ question papers, studying basic concepts of the topics mentioned in the syllabus, taking online mock tests, etc.
4. Can I use a calculator during the exam? Ans: No, you are not allowed to use a calculator during the exam.
5. Is there any sectional cut-off for Part A of CSIR UGC NET Chemical Science Dec 2016? Ans: No, there is no sectional cut-off for Part A. However, you need to secure an overall minimum cut-off score to qualify for the exam.

### SUMMARY

Part A of the CSIR UGC NET Chemical Science Dec 2016 exam was a challenging section that required a thorough understanding of the fundamentals of chemistry. It consisted of 20 multiple-choice questions, each carrying 2 marks.

The questions were designed to test the candidate’s knowledge of topics such as organic, inorganic, and physical chemistry, as well as analytical and environmental chemistry.

The questions were of varying difficulty, with some requiring simple recall of facts while others demanded a deeper understanding of concepts and their applications. Time management was crucial, as candidates had only 3 hours to complete the entire exam, including Part A.

To succeed in Part A of the CSIR UGC NET Chemical Science Dec 2016 exam, candidates needed to have a strong foundation in chemistry, as well as an ability to think critically and analyze problems.

By utilizing effective study techniques, such as regular practice and review of past exam papers, candidates could improve their chances of success in this challenging section of the exam.

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