100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	A stone of mass ‘m’ is to be thrown to a height h1. What is the acceleration of the stone?2. With what minimum velocity should it be thrown.3. At what height does the KE and PE become equal?4. Find the velocity at that height
Answer» 1. ^–g or ^–9.8m/s. 2. v = 0, a = -g, S = h Substitute this values in V² = u² + 2as we get 0 = u² – 2gh u = \(\sqrt{2gh}\) 3. at \(\frac{h}{2}\)., KE and PE are equal. 4. V² = U² + 2aS = U² – 2g \(\frac{h}{2}\) (u² = 2gh) = U² – \(\frac{U^2}{2}\), V² = \(\frac{U^2}{2}\), V = \(\frac{U}{√2}\).

1.

A stone of mass ‘m’ is to be thrown to a height h1. What is the acceleration of the stone?2. With what minimum velocity should it be thrown.3. At what height does the KE and PE become equal?4. Find the velocity at that height

Answer»

1. ^–g or ^–9.8m/s.

2. v = 0, a = -g, S = h

Substitute this values in

V² = u² + 2as we get

0 = u² – 2gh

u = \(\sqrt{2gh}\)

3. at \(\frac{h}{2}\)., KE and PE are equal.

4. V² = U² + 2aS

= U² – 2g \(\frac{h}{2}\) (u² = 2gh) = U² – \(\frac{U^2}{2}\),

V² = \(\frac{U^2}{2}\),

V = \(\frac{U}{√2}\).

Discussion

2.	A graph paper is fitted on a board as shown in figure. Near to the graph paper a spring is placed. A pencil is attached to the end of the spring as shown in figure. The pencil is free to move on the graph paper. A stone of mass 50 gm is placed 1m above the spring. [Spring constant k = 98N/m]1. The energy possessed by the stone due to its height is called_______2. If this stone falls on the spring, find the length of mark that produced on the graph paper due to pencil [The change in P E of stone due to compression may be negleted]3. What will happen to the length of mark, if spring having smaller spring constant is used? Justify.
Answer» 1. Potential energy. 2. \(\frac{1}{2}\)kx² = mg \(\frac{1}{2}\) × 98 × x² = 50 × 10^-3 × 9.8 × 1 x² = 100 × 10⁴ x = 10cm. 3. The length of mark will be decreased. Compression of spring depends on spring constant.

2.

A graph paper is fitted on a board as shown in figure. Near to the graph paper a spring is placed. A pencil is attached to the end of the spring as shown in figure. The pencil is free to move on the graph paper. A stone of mass 50 gm is placed 1m above the spring. [Spring constant k = 98N/m]1. The energy possessed by the stone due to its height is called_______2. If this stone falls on the spring, find the length of mark that produced on the graph paper due to pencil [The change in P E of stone due to compression may be negleted]3. What will happen to the length of mark, if spring having smaller spring constant is used? Justify.

Answer»

1. Potential energy.

2. \(\frac{1}{2}\)kx² = mg

\(\frac{1}{2}\) × 98 × x² = 50 × 10^-3 × 9.8 × 1

x² = 100 × 10⁴

x = 10cm.

3. The length of mark will be decreased. Compression of spring depends on spring constant.

Discussion

3.	Given p(A) = 3/5 and p(B) = 1/5. Find P(A or B), if A & B are mutually exclusive events.
Answer» P(A ∪ B) = p(A) + p(B) (∵p(A ∩ B) = 0) = 3/5 + 1/5 = 4/5

3.

Given p(A) = 3/5 and p(B) = 1/5. Find P(A or B), if A & B are mutually exclusive events.

Answer»

P(A ∪ B) = p(A) + p(B) (∵p(A ∩ B) = 0)

= 3/5 + 1/5 = 4/5

Discussion

4.	In a group of 600 students in a school, 150 students were found to be taking tea, 225 and taking coffee and 100 were taking both tea and coffee. How may students were taking neither tea nor coffee.
Answer» Let set of students taking coffee = C & set of students taking tea = T Also set of students who take either Tea or Coffee = C ∪ T n (T) = 150 n(C) = 225 n(C ∩ T) = 100 ; n(C ∪ T) = ? n(C ∪ T) = n(C) + n(T) – n (C ∩ T) = 150 + 225 – 100 = 275 Number of students taking neither Tea nor Coffee = n(∪), n (C ∪ T) = 600 – 275 = 325

4.

In a group of 600 students in a school, 150 students were found to be taking tea, 225 and taking coffee and 100 were taking both tea and coffee. How may students were taking neither tea nor coffee.

Answer»

Let set of students taking coffee = C & set of students taking tea = T

Also set of students who take either Tea or Coffee = C ∪ T

n (T) = 150 n(C) = 225

n(C ∩ T) = 100 ; n(C ∪ T) = ?

n(C ∪ T) = n(C) + n(T) – n (C ∩ T)

= 150 + 225 – 100 = 275

Number of students taking neither Tea nor Coffee

= n(∪), n (C ∪ T) = 600 – 275 = 325

Discussion

5.	Name the parameter which is a measure of degree of elasticity of a body.
Answer» Coefficient of restitution, which is a measure of degree of elasticity of a body.

Discussion

6.	A function f is defined by f(x) = 2x – 5.Write down the values of(i) f(0),(ii) f(7),(iii) f(–3)
Answer» The given function is f(x) = 2x – 5. Therefore, (i) f(0) = 2 × 0 – 5 = 0 – 5 = –5 (ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9 (iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11

6.

A function f is defined by f(x) = 2x – 5.Write down the values of(i) f(0),(ii) f(7),(iii) f(–3)

Answer»

The given function is f(x) = 2x – 5.

Therefore,

(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11

Discussion

7.	Find the domain and range of the following real function: (i) f(x) = –\|x\| (ii) () = √(9 − 2)
Answer» (i) f(x) = –\|x\|, x ∈ R We know that \|\| = { , ≥ 0; −, < 0 ∴ () = −\|\| = { −, ≥ 0 ; , < 0 Since f(x) is defined for x ∈ R, the domain of f is R. It can be observed that the range of f(x) = –\|x\| is all real numbers except positive real numbers. ∴ The range of f is (−∞, 0]. (ii) () = √(9 − ²) Since √(9 − ²) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3]. For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3. ∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

7.

Find the domain and range of the following real function: (i) f(x) = –|x| (ii) () = √(9 − 2)

Answer»

(i) f(x) = –|x|, x ∈ R

We know that

|| = { , ≥ 0; −, < 0

∴ () = −|| = { −, ≥ 0 ; , < 0

Since f(x) is defined for x ∈ R, the domain of f is R.

It can be observed that the range of

f(x) = –|x| is all real numbers except positive real numbers.

∴ The range of f is (−∞, 0].

(ii) () = √(9 − ²)

Since √(9 − ²) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3]. For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

Discussion

8.	A = {x: x is a natural number less than 8}1. Write in roster form.2. Write a subset of A containing all even numbers in A.3. Which of the following could not be the number of elements of power set of a set [2, 8, 10, 16]?
Answer» 1. A = {1, 2, 3, 4, 5, 6, 7} 2. {2, 4, 6} or {2, 4, 6, 7} 3. 10. (since other are powers of 2.)

8.

A = {x: x is a natural number less than 8}1. Write in roster form.2. Write a subset of A containing all even numbers in A.3. Which of the following could not be the number of elements of power set of a set [2, 8, 10, 16]?

Answer»

1. A = {1, 2, 3, 4, 5, 6, 7}

2. {2, 4, 6} or {2, 4, 6, 7}

3. 10. (since other are powers of 2.)

Discussion

9.	Let A and B be two sets such that n( A) = 20, n(A ∪ B) = 42, n(A ∩ B) = 4. Find1. n(B)2. n(B – A)3. n(A – B)
Answer» 1. n(A ∪ B) = n(A) + n(B) – n(A ∩ B) ⇒ 42 = 20 + n(B) – 4 ⇒ n(B) = 26 2. n(B – A) = n(B) – n(A ∩ B) = 26 – 4 = 22 3. n(A – B) = n(A) – n(A ∩ B) = 20 – 4 = 16.

9.

Let A and B be two sets such that n( A) = 20, n(A ∪ B) = 42, n(A ∩ B) = 4. Find1. n(B)2. n(B – A)3. n(A – B)

Answer»

1. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ 42 = 20 + n(B) – 4

⇒ n(B) = 26

2. n(B – A) = n(B) – n(A ∩ B) = 26 – 4 = 22

3. n(A – B) = n(A) – n(A ∩ B) = 20 – 4 = 16.

Discussion

10.	Write the following in set builder form.1. [0, 10]2. [-2, 7)3. (3, 4)
Answer» 1. {x: x∈ R, 0 ≤ x ≤ 10 } 2. { x: x ∈ R, -2 ≤ x < 7 } 3. {x: x ∈ R, 3 < x < 4 }

10.

Write the following in set builder form.1. [0, 10]2. [-2, 7)3. (3, 4)

Answer»

1. {x: x∈ R, 0 ≤ x ≤ 10 }

2. { x: x ∈ R, -2 ≤ x < 7 }

3. {x: x ∈ R, 3 < x < 4 }

Discussion

11.	If cos∅ = 1/2 then the value of cosec∅ is –(A) 2(B) 2/√3(C) √3/2(D) 1/√3
Answer» Correct answer is (B) 2/√3

Discussion

12.	the mode of the data 21, 26, 22, 29, 23, 29, 26, 29, 22, 23 is?
Answer» Answer: 23 and 29

Discussion

13.	In an online test, sami scores 18 points for correct answers and -3 points for incorrect answers. Find his score.
Answer» Total = Correct answers - Incorrect answers Total = 18 - 3 Total = 15 Therefore Sam scored a total of 15 points.

13.

In an online test, sami scores 18 points for correct answers and -3 points for incorrect answers. Find his score.

Answer»

Total = Correct answers - Incorrect answers

Total = 18 - 3

Total = 15

Therefore Sam scored a total of 15 points.

Discussion

14.	Consider the set Q of rational numbers. Let * be the operation on Q defined by a * b = a + b - ab. The identity element under * is
Answer» An identity relation is one in which every element of a set is related to itself only. a * b = a + b - ab . As in identity relation 'a' is related to 'a', so the correct option will be the one which gives the value of the relation = 'a'. So, equating a + b - ab = a, we get b(1 - a) = 0. Now putting the values of a, we find b and the option in which a = b, will be the answer. For a = 0, b = 0, so the correct option. For a = 1, b(1 - 1) = 0 ⇒ b can have multiple values. For a = 2, b(1 - 2) = 0 ⇒ b = 0 but a = 2

14.

Consider the set Q of rational numbers. Let * be the operation on Q defined by a * b = a + b - ab. The identity element under * is

Answer»

An identity relation is one in which every element of a set is related to itself only.

a * b = a + b - ab .

As in identity relation 'a' is related to 'a', so the correct option will be the one which gives the value of the relation = 'a'.

So, equating a + b - ab = a, we get b(1 - a) = 0.

Now putting the values of a, we find b and the option in which a = b, will be the answer.

For a = 0, b = 0,

so the correct option. For a = 1, b(1 - 1) = 0 ⇒ b can have multiple values.

For a = 2, b(1 - 2) = 0 ⇒ b = 0 but a = 2

Discussion

15.	what is diamond made up of like water is made of H2O
Answer» diamond is an allotrope of carbon who's formula is { C} Diamond is made up of carbon It is the earth's hardest substances

Discussion

16.	What are the two types of ecosystem?
Answer» 1. Natural Ecosystem. 2. Artificial Ecosystem. Terrestrial Ecosystem. Aquatic Ecosystem. Terrestrial Ecosystems Terrestrial ecosystems are exclusively land-based ecosystems. There are different types of terrestrial ecosystems distributed around various geological zones. They are as follows: Forest Ecosystems Grassland Ecosystems Tundra Ecosystems Desert Ecosystem Forest Ecosystem A forest ecosystem consists of several plants, animals and microorganisms that live in coordination with the abiotic factors of the environment. Forests help in maintaining the temperature of the earth and are the major carbon sink. Grassland Ecosystem In a grassland ecosystem, the vegetation is dominated by grasses and herbs. Temperate grasslands, savanna grasslands are some of the examples of grassland ecosystems. Tundra Ecosystem Tundra ecosystems are devoid of trees and are found in cold climates or where rainfall is scarce. These are covered with snow for most of the year. The ecosystem in the Arctic or mountain tops is tundra type. Desert Ecosystem Deserts are found throughout the world. These are regions with very little rainfall. The days are hot and the nights are cold. Aquatic Ecosystem Aquatic ecosystems are ecosystems present in a body of water. These can be further divided into two types, namely: Freshwater Ecosystem Marine Ecosystem Freshwater Ecosystem The freshwater ecosystem is an aquatic ecosystem that includes lakes, ponds, rivers, streams and wetlands. These have no salt content in contrast with the marine ecosystem. Marine Ecosystem The marine ecosystem includes seas and oceans. These have a more substantial salt content and greater biodiversity in comparison to the freshwater ecosystem. Structure of the Ecosystem The structure of an ecosystem is characterised by the organisation of both biotic and abiotic components. This includes the distribution of energy in our environment. It also includes the climatic conditions prevailing in that particular environment. The structure of an ecosystem can be split into two main components, namely: Biotic Components Abiotic Components The biotic and abiotic components are interrelated in an ecosystem. It is an open system where the energy and components can flow throughout the boundaries Biotic Components Biotic components refer to all life in an ecosystem. Based on nutrition, biotic components can be categorised into autotrophs, heterotrophs and saprotrophs (or decomposers). Producers include all autotrophs such as plants. They are called autotrophs as they can produce food through the process of photosynthesis. Consequently, all other organisms higher up on the food chain rely on producers for food. Consumers or heterotrophs are organisms that depend on other organisms for food. Consumers are further classified into primary consumers, secondary consumers and tertiary consumers. Primary consumers are always herbivores that they rely on producers for food. Secondary consumers depend on primary consumers for energy. They can either be a carnivore or an omnivore. Tertiary consumers are organisms that depend on secondary consumers for food. Tertiary consumers can also be an omnivore. Quaternary consumers are present in some food chains. These organisms prey on tertiary consumers for energy. Furthermore, they are usually at the top of a food chain as they have no natural predators. Decomposers include saprophytes such as fungi and bacteria. They directly thrive on the dead and decaying organic matter. Decomposers are essential for the ecosystem as they help in recycling nutrients to be reused by plants. Abiotic Components Abiotic components are the non-living component of an ecosystem. It includes air, water, soil, minerals, sunlight, temperature, nutrients, wind, altitude, turbidity, etc. Functions of Ecosystem The functions of the ecosystem are as follows: It regulates the essential ecological processes, supports life systems and renders stability. It is also responsible for the cycling of nutrients between biotic and abiotic components. It maintains a balance among the various trophic levels in the ecosystem. It cycles the minerals through the biosphere. The abiotic components help in the synthesis of organic components that involves the exchange of energy. So the functional units of an ecosystem or functional components that work together in an ecosystem are: Productivity – It refers to the rate of biomass production. Energy flow – It is the sequential process through which energy flows from one trophic level to another. The energy captured from the sun flows from producers to consumers and then to decomposers and finally back to the environment. Decomposition – It is the process of breakdown of dead organic material. The top-soil is the major site for decomposition. Nutrient cycling – In an ecosystem nutrients are consumed and recycled back in various forms for the utilisation by various organisms. Important Ecological Concepts <h3 style="font-family: inherit; font-weight: 40 Natural ecosystem Artificial ecosystem

16.

What are the two types of ecosystem?

Answer»

1. Natural Ecosystem.

2. Artificial Ecosystem.

Terrestrial Ecosystem.

Aquatic Ecosystem.

Terrestrial Ecosystems

Terrestrial ecosystems are exclusively land-based ecosystems. There are different types of terrestrial ecosystems distributed around various geological zones. They are as follows:

Forest Ecosystems
Grassland Ecosystems
Tundra Ecosystems
Desert Ecosystem

Forest Ecosystem

A forest ecosystem consists of several plants, animals and microorganisms that live in coordination with the abiotic factors of the environment. Forests help in maintaining the temperature of the earth and are the major carbon sink.

Grassland Ecosystem

In a grassland ecosystem, the vegetation is dominated by grasses and herbs. Temperate grasslands, savanna grasslands are some of the examples of grassland ecosystems.

Tundra Ecosystem

Tundra ecosystems are devoid of trees and are found in cold climates or where rainfall is scarce. These are covered with snow for most of the year. The ecosystem in the Arctic or mountain tops is tundra type.

Desert Ecosystem

Deserts are found throughout the world. These are regions with very little rainfall. The days are hot and the nights are cold.

Aquatic Ecosystem

Aquatic ecosystems are ecosystems present in a body of water. These can be further divided into two types, namely:

Freshwater Ecosystem
Marine Ecosystem

Freshwater Ecosystem

The freshwater ecosystem is an aquatic ecosystem that includes lakes, ponds, rivers, streams and wetlands. These have no salt content in contrast with the marine ecosystem.

Marine Ecosystem

The marine ecosystem includes seas and oceans. These have a more substantial salt content and greater biodiversity in comparison to the freshwater ecosystem.

Structure of the Ecosystem

The structure of an ecosystem is characterised by the organisation of both biotic and abiotic components. This includes the distribution of energy in our environment. It also includes the climatic conditions prevailing in that particular environment.

The structure of an ecosystem can be split into two main components, namely:

Biotic Components
Abiotic Components

The biotic and abiotic components are interrelated in an ecosystem. It is an open system where the energy and components can flow throughout the boundaries

Biotic Components

Biotic components refer to all life in an ecosystem. Based on nutrition, biotic components can be categorised into autotrophs, heterotrophs and saprotrophs (or decomposers).

Producers include all autotrophs such as plants. They are called autotrophs as they can produce food through the process of photosynthesis. Consequently, all other organisms higher up on the food chain rely on producers for food.
Consumers or heterotrophs are organisms that depend on other organisms for food. Consumers are further classified into primary consumers, secondary consumers and tertiary consumers.
- Primary consumers are always herbivores that they rely on producers for food.
- Secondary consumers depend on primary consumers for energy. They can either be a carnivore or an omnivore.
- Tertiary consumers are organisms that depend on secondary consumers for food. Tertiary consumers can also be an omnivore.
- Quaternary consumers are present in some food chains. These organisms prey on tertiary consumers for energy. Furthermore, they are usually at the top of a food chain as they have no natural predators.
Decomposers include saprophytes such as fungi and bacteria. They directly thrive on the dead and decaying organic matter. Decomposers are essential for the ecosystem as they help in recycling nutrients to be reused by plants.

Abiotic Components

Abiotic components are the non-living component of an ecosystem. It includes air, water, soil, minerals, sunlight, temperature, nutrients, wind, altitude, turbidity, etc.

Functions of Ecosystem

The functions of the ecosystem are as follows:

1. It regulates the essential ecological processes, supports life systems and renders stability.
2. It is also responsible for the cycling of nutrients between biotic and abiotic components.
3. It maintains a balance among the various trophic levels in the ecosystem.
4. It cycles the minerals through the biosphere.
5. The abiotic components help in the synthesis of organic components that involves the exchange of energy.

So the functional units of an ecosystem or functional components that work together in an ecosystem are:

Productivity – It refers to the rate of biomass production.
Energy flow – It is the sequential process through which energy flows from one trophic level to another. The energy captured from the sun flows from producers to consumers and then to decomposers and finally back to the environment.
Decomposition – It is the process of breakdown of dead organic material. The top-soil is the major site for decomposition.
Nutrient cycling – In an ecosystem nutrients are consumed and recycled back in various forms for the utilisation by various organisms.

Important Ecological Concepts

<h3 style="font-family: inherit; font-weight: 40

Natural ecosystem

Artificial ecosystem

Discussion

17.

How hard wood differ form soft wood

Answer»

Hardwood	Softwood
Comes from angiosperm trees that are not monocots; trees are usually broad-leaved. Has vessel elements that transport water throughout the wood; under a microscope, these elements appear as pores.	Comes from gymnosperm trees which usually have needles and cones. Medullary rays and tracheids transport water and produce sap. When viewed under a microscope, softwoods have no visible pores because of tracheids.
hardwoods are more likely to be found in high-quality furniture, decks, flooring, and construction that needs to last.	About 80% of all timber comes from softwood. Softwoods have a wide range of applications and are found in building components (e.g., windows, doors), furniture, medium-density fiberboard (MDF), paper, Christmas trees, and much more.
Examples of hardwood trees include alder, balsa, beech, hickory, mahogany, maple, oak, teak, and walnut.	Examples of softwood trees are cedar, Douglas fir, juniper, pine, redwood, spruce, and yew.
Most hardwoods have a higher density than most softwoods.	Most softwoods have a lower density than most hardwoods.

Discussion

18.	Let \( y=e^{3 x} \) and \( f(x)=-4 e^{-3 x}+e^{6 x} \)i) Find \( f^{\prime}(x) \) and \( f^{\prime \prime}(x) \)ii) Show that \( f^{\prime \prime}(x)=36\left[y^{2}-\frac{1}{y}\right] \)
Answer» f(x) = -4e^-3x + e^6x f'(x) = 12e^-3x + 6e^6x f''(x) = -36e^-3x + 36e^6x = 36(- 1/e^3x + (e^3x)²) = 36(y² - 1/y) (∵ y = e^3x)

18.

Let \( y=e^{3 x} \) and \( f(x)=-4 e^{-3 x}+e^{6 x} \)i) Find \( f^{\prime}(x) \) and \( f^{\prime \prime}(x) \)ii) Show that \( f^{\prime \prime}(x)=36\left[y^{2}-\frac{1}{y}\right] \)

Answer»

f(x) = -4e^-3x + e^6x

f'(x) = 12e^-3x + 6e^6x

f''(x) = -36e^-3x + 36e^6x

= 36(- 1/e^3x + (e^3x)²)

= 36(y² - 1/y) (∵ y = e^3x)

Discussion

19.	H.C.F of \( \left(x^{2}-3 x+2\right) \) and \( \left(x^{2}-4 x+3\right) \) is
Answer» x²−3x+2=x 2 −2x−x+2 x(x−2)−1(x−2)=(x−2)(x−1) Now x² −4x+3=x 2 −3x−x+3 x(x−3)−1(x−3)=(x−3)(x−1) Thus, the only common factor is (x-1)

19.

H.C.F of \( \left(x^{2}-3 x+2\right) \) and \( \left(x^{2}-4 x+3\right) \) is

Answer»

x²−3x+2=x 2 −2x−x+2

x(x−2)−1(x−2)=(x−2)(x−1)

Now

x² −4x+3=x 2 −3x−x+3

x(x−3)−1(x−3)=(x−3)(x−1)

Thus, the only common factor is (x-1)

Discussion

20.	Describe the structure of chloroplast.
Answer» Chioroplast containing chlorophyll, caretenoids and helps in photosynthesis. Chioroplast is found in mesophyll tissue of leaves. They may be lens shaped, spherical, discoid and reticulate or ribbon like. It is covered by double membranes. The membrane encloses the stroma containing thylakoids in the grana. Stroma is liquid containing 70S ribosomes and enzymes and double stranded circular DNA.

Discussion

21.	Name the building blocks of proteins and explain the structural aspects of proteins.
Answer» Amino acids. Primary structure: The amino acids molecules arranged in a linear chain with peptide bond. They exist like rigid rod. Secondary structure: In this organization the chain is folded. They are folded at right handed side. Tertiary structure: The protein chain folded one over the other form woolen ball, form three dimensional structure. Quaternary structure: The polypeptide chain arranged in the form of strings of sphere or may form cube.

Discussion

22.	Name the building blocks of protein explain the structural aspects of protein.
Answer» The building blocks of proteins are amino acids, which are small organic molecules that consist of an alpha (central) carbon atom linked to an amino group, a carboxyl group, a hydrogen atom, and a variable component called a side chain.

Discussion

23.	Which of the following is the distance of a point P(x, y) from the origin?(a) x2 + y2 (b) x2 - y2(c) √(x2+y2)(d) √(x2-y2)
Answer» The following is the distance of a point P(x, y) from the origin√(x²+y²)

Discussion

24.	Which of the following is the first positive term of the given A.P. –11, –8, –5, .......?(a) 1 (b) –2 (c) 2 (d) 3
Answer» The first positive term of the given A.P. –11, –8, –5, 1

Discussion

25.	The symbol for circumcenter is
Answer» Circumcenter is equidistant to all the three vertices of a triangle. The circumcenter is the centre of the circumcircle of that triangle. Circumcenter is denoted by O (x, y).

Discussion

26.	Manjit wants to donate a rectangle plot of level for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50m and breadth is increased by 50m, then its area will remain same, but if length increased by 10 m and breadth is decreased by 20m, then its area will decrease by 5300 m2.with the help of above facts, answer the following question:The relation between x and y is given by :(a) x − y = 50, 2x − y = 550(b) x − y = 50, 2x + y = 550(c) x + y = 50, 2x + y = 550(d) x + y = 50, 2x − y = 550.
Answer» Let the length of plot is x m, the breadth of the plot is y m. Therefore, the area of the plot is xy m². According to the first condition, length is decreased by 50 m and breadth is increased by 50 m then the area remains same. Therefore, (x − 50)(y + 50) = xy ⇒ xy + 50x − 50y − 2500 = xy ⇒ 50x − 50y = 2500 ⇒ x − y = 50. ... (1) According to the second condition, length is decreased by 10 m and breadth is decreased by 20 m then the area is decreased by 5300 m² . Therefore, (x − 10)(y − 20) = xy − 5300 ⇒ xy − 20x − 10y + 200 = xy − 5300 ⇒ −20 x − 10y = −5500 ⇒ 2x + y = 550. ... (2) Hence, option (b) is correct.

26.

Manjit wants to donate a rectangle plot of level for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50m and breadth is increased by 50m, then its area will remain same, but if length increased by 10 m and breadth is decreased by 20m, then its area will decrease by 5300 m2.with the help of above facts, answer the following question:The relation between x and y is given by :(a) x − y = 50, 2x − y = 550(b) x − y = 50, 2x + y = 550(c) x + y = 50, 2x + y = 550(d) x + y = 50, 2x − y = 550.

Answer»

Let the length of plot is x m, the breadth of the plot is y m.

Therefore, the area of the plot is xy m².

According to the first condition, length is decreased by 50 m and breadth is increased by 50 m then the area remains same.

Therefore, (x − 50)(y + 50) = xy ⇒ xy + 50x − 50y − 2500 = xy ⇒ 50x − 50y = 2500

⇒ x − y = 50. ... (1)

According to the second condition, length is decreased by 10 m and breadth is decreased by 20 m then the area is decreased by 5300 m² .

Therefore, (x − 10)(y − 20) = xy − 5300

⇒ xy − 20x − 10y + 200 = xy − 5300

⇒ −20 x − 10y = −5500

⇒ 2x + y = 550. ... (2)

Hence, option (b) is correct.

Discussion

27.	Find all three digit natural numbers of the form (abc)10 such that (abc)10, (bca)10 and (cab)10 are in geometric progression. (Here (abc)10 is representation in base 10.)
Answer» Let us write x = (a x 10²) + (b x 10) + c; y = (b x 10²) + (c x 10) + a; z = (c x 10²) + (a x 10) + b. We are given that y² = xz. This means (b x 10²) + (c x 10) + a)² = ((a x 10²) + (b x 10) + c))((c x 10²) + (a x 10) + b)). We can solve for c and get c = (10b² - a²)/(10a - b). If a, b, c are digits leading to a solution, and if d = gcd(a; b) then d\|c. Consequently, we may assume that gcd(a; b) = 1. Now c = (999a²)/(10a - b) - (10b + 100a), showing that 10a - b divides 999a². Since a; b are relatively prime, this is possible only if 10a - b is a factor of 999. It follows that 10a - b takes the values 1,3,9,27,37. These values lead to the pairs (a, b) = (1; 9), (1; 7), (1; 1), (4; 3). We can discard the first two pairs as they lead to a value of c > 10. The third gives the trivial solution (111, 111, 111). Taking d = 2, 3, 4, 5, 6, 7, 8, 9, we get 9 solution. (abc)10 = 111, 222, 333 444, 555, 666, 777, 888, 999. The last pair gives c = 2 and hence the solution (432,324, 243). Another solution is obtained on multiplying by 2: (864, 648, 486). Thus we have (abc)₁₀ = 111, 222, 333, 444, 555, 666, 777, 888, 999, 432,864.

27.

Find all three digit natural numbers of the form (abc)10 such that (abc)10, (bca)10 and (cab)10 are in geometric progression. (Here (abc)10 is representation in base 10.)

Answer»

Let us write

x = (a x 10²) + (b x 10) + c; y = (b x 10²) + (c x 10) + a; z = (c x 10²) + (a x 10) + b.

We are given that y² = xz. This means

(b x 10²) + (c x 10) + a)² = ((a x 10²) + (b x 10) + c))((c x 10²) + (a x 10) + b)).

We can solve for c and get

c = (10b² - a²)/(10a - b).

If a, b, c are digits leading to a solution, and if d = gcd(a; b) then d|c. Consequently, we may assume that gcd(a; b) = 1. Now

c = (999a²)/(10a - b) - (10b + 100a),

showing that 10a - b divides 999a². Since a; b are relatively prime, this is possible only if 10a - b is a factor of 999. It follows that 10a - b takes the values 1,3,9,27,37. These values lead to the pairs

(a, b) = (1; 9), (1; 7), (1; 1), (4; 3).

We can discard the first two pairs as they lead to a value of c > 10. The third gives the trivial solution (111, 111, 111). Taking d = 2, 3, 4, 5, 6, 7, 8, 9, we get 9 solution. (abc)10 = 111, 222, 333 444, 555, 666, 777, 888, 999.

The last pair gives c = 2 and hence the solution (432,324, 243). Another solution is obtained on multiplying by 2: (864, 648, 486).

Thus we have

(abc)₁₀ = 111, 222, 333, 444, 555, 666, 777, 888, 999, 432,864.

Discussion

28.	The larger of the two supplementary angles exceeds the smaller by 18°. Find them.
Answer» We know that two angles are supplementary angles if their sum is 180°. Let smaller angle is x°. Given that difference between supplementary angles are 18° Therefore, the other angle is x° + 18°. Hence, x° & x° + 18° are supplementary angles. Therefore, x° + x° + 18° = 180°. (By definition of supplementary angles) ⇒ 2 x° + 18° = 180°⇒ 2x° = 180° – 18° = 162° ⇒ x° = \(\frac{162°}{2}\) = 81°. And x° + 18° = 81°+18° = 99°. Hence, the angles are 81°, 99°

28.

The larger of the two supplementary angles exceeds the smaller by 18°. Find them.

Answer»

We know that two angles are supplementary angles if their sum is 180°.

Let smaller angle is x°.

Given that difference between supplementary angles are 18°

Therefore, the other angle is x° + 18°.

Hence, x° & x° + 18° are supplementary angles.

Therefore, x° + x° + 18° = 180°. (By definition of supplementary angles)

⇒ 2 x° + 18° = 180°⇒ 2x° = 180° – 18° = 162°

⇒ x° = \(\frac{162°}{2}\) = 81°.

And x° + 18° = 81°+18° = 99°.

Hence, the angles are 81°, 99°

Discussion

29.	Describe Bhandhani of Gujrat. What are the different textiles from Gujrat?
Answer» The tie-dye from Gujarat called Bandhani is regarded for its fine resist dots and intricate designs. Traditionally the tie-dye is done on silk, cotton and wool. The motifs created by outlining with tiny dots are animal and human figures, flowers, plants and trees. The products range varies from odhanis, saris, shawls to stitched garments like kurta and skirts The major centres of bandhani in gujarat are Jamnagar, Bhavnagar, rajkot and Porbandar. Gharcholu: A popular bandhani textile produced in gujarat is called gharchola or gharcholu, a traditional odhani for Hindu brides, which is nowadays available as a sari worn on auspicious occasions. The tie-dyed textile in cotton or silk is red in colour and the layout is a checkerboard created with woven gold threads. Each square within the check contains a different tie-dyed motif like dancing lady, parrot, elephant, peacock, flowering shrub and geometric forms. Chandrokhani: The traditional odhani for a Muslim bride in red and black colour is called chandrokhani. It is a tie-dyed textile with a big medallion in the centre surrounded by four smaller medallions and wide borders. Motifs created with small tie dye dots are small paisleys, zig zag lines, sunflowers etc.

29.

Describe Bhandhani of Gujrat. What are the different textiles from Gujrat?

Answer»

The tie-dye from Gujarat called Bandhani is regarded for its fine resist dots and intricate designs. Traditionally the tie-dye is done on silk, cotton and wool. The motifs created by outlining with tiny dots are animal and human figures, flowers, plants and trees. The products range varies from odhanis, saris, shawls to stitched garments like kurta and skirts The major centres of bandhani in gujarat are Jamnagar, Bhavnagar, rajkot and Porbandar.

Gharcholu: A popular bandhani textile produced in gujarat is called gharchola or gharcholu, a traditional odhani for Hindu brides, which is nowadays available as a sari worn on auspicious occasions. The tie-dyed textile in cotton or silk is red in colour and the layout is a checkerboard created with woven gold threads. Each square within the check contains a different tie-dyed motif like dancing lady, parrot, elephant, peacock, flowering shrub and geometric forms.

Chandrokhani: The traditional odhani for a Muslim bride in red and black colour is called chandrokhani. It is a tie-dyed textile with a big medallion in the centre surrounded by four smaller medallions and wide borders. Motifs created with small tie dye dots are small paisleys, zig zag lines, sunflowers etc.

Discussion

30.	What are the basic steps to create bhandhani textile? Describe it elaborately.
Answer» The basic steps of creating a bandhani textile are as follows: Pre-preparation of fabric: The fabric generally used for tie and dye is finer variety of cotton and silk, so that dye can penetrate deep into the layers of tied fabrics. It is soaked in water overnight and washed thoroughly to remove the starch in order to improve its dye uptake. The fabric is bleached by drying it in the sun. Tracing of design: The fabric is folded into four or more layers for convenience of tying as well as to achieve symmetry in design. The design layout is marked on the folded fabric with wooden blocks, dipped in washable colours like neel or geru. Tying of fabric: as per the design, the folded fabric is raised with a pointed metal nail worn over the finger. A cotton thread coated with wax is wrapped tightly around the raised area to create a simple fine dot: bundi or bindi, which is the basic motif of the design. Dyeing of fabric in the lightest colour: after tying, the fabric is dyed in the lightest colour first from the selected colour scheme. After dyeing, fabric is washed, rinsed and dried. Renewal of tying and dyeing in next-darker colour: Parts of the fabric to be retained in the lighter colour are covered with tying and then the fabric is dyed in the next darker colour. The Process of re-tying and dyeing is continued till the darkest colour in the scheme is applied. Washing: Following the final dyeing, the textile is washed to remove excess dye and starched. Opening the ties: The ties of the tie-dyed fabric are kept tied till purchased by a consumer in order to differentiate between a bandhani textile and a printed imitation. Only a portion of the bandhani textile is opened to display the colour scheme to the customer. To unravel the ties, the bandhani textile is stretched crosswise to open all ties at the same time.

30.

What are the basic steps to create bhandhani textile? Describe it elaborately.

Answer»

The basic steps of creating a bandhani textile are as follows:

Pre-preparation of fabric: The fabric generally used for tie and dye is finer variety of cotton and silk, so that dye can penetrate deep into the layers of tied fabrics. It is soaked in water overnight and washed thoroughly to remove the starch in order to improve its dye uptake. The fabric is bleached by drying it in the sun.
Tracing of design: The fabric is folded into four or more layers for convenience of tying as well as to achieve symmetry in design. The design layout is marked on the folded fabric with wooden blocks, dipped in washable colours like neel or geru.
Tying of fabric: as per the design, the folded fabric is raised with a pointed metal nail worn over the finger. A cotton thread coated with wax is wrapped tightly around the raised area to create a simple fine dot: bundi or bindi, which is the basic motif of the design.
Dyeing of fabric in the lightest colour: after tying, the fabric is dyed in the lightest colour first from the selected colour scheme. After dyeing, fabric is washed, rinsed and dried.
Renewal of tying and dyeing in next-darker colour: Parts of the fabric to be retained in the lighter colour are covered with tying and then the fabric is dyed in the next darker colour. The Process of re-tying and dyeing is continued till the darkest colour in the scheme is applied.
Washing: Following the final dyeing, the textile is washed to remove excess dye and starched.
Opening the ties: The ties of the tie-dyed fabric are kept tied till purchased by a consumer in order to differentiate between a bandhani textile and a printed imitation. Only a portion of the bandhani textile is opened to display the colour scheme to the customer. To unravel the ties, the bandhani textile is stretched crosswise to open all ties at the same time.

Discussion

31.	What are the End Uses of Chikankari?
Answer» Traditionally the embroidery was done mainly for male garments such as kurta, bandi, choga etc. for summer wear. Presently chikankari is being explored for apparel as well as home products on different fabrics like crepe silks, chiffons, georgettes and cotton polyester blends. Ladies suits, saares, scarves, duppatas are also made from chikankari. Bed spreads, cushion Covers, curtains are also made from chikankari.

Discussion

32.	Write Dry Brush Technique in relevance to material required, steps, precautions and results.
Answer» The dry brush effect adds a three dimensional feel to the print. An interesting stencil can be made, kept over a sheet of paper and a dry brush can be brushed cover the stencil creating interesting shapes with textures. Material required: Drawing/Cartridge sheets(Different textures) Acrylic colours/poster paints mixed with fevicol Paintbrush Different sizes Water container Colour palette/Mixing bowl Step: Take a drawing sheet. Put the dry paint brush in acrylic paints. Make any design of your choice on the sheet Use different colour and sizes of brush for making of different designs Precautions: Wash the brush properly before dipping and using another colour Dry the brush before putting it in acrylic colours. Results: Different textured sheet after application of dry brush will look little more embossed and the entire design gets a three dimension feel.

32.

Write Dry Brush Technique in relevance to material required, steps, precautions and results.

Answer»

The dry brush effect adds a three dimensional feel to the print. An interesting stencil can be made, kept over a sheet of paper and a dry brush can be brushed cover the stencil creating interesting shapes with textures.

Material required:

Drawing/Cartridge sheets(Different textures)
Acrylic colours/poster paints mixed with fevicol
Paintbrush Different sizes
Water container
Colour palette/Mixing bowl

Step:

Take a drawing sheet.
Put the dry paint brush in acrylic paints.
Make any design of your choice on the sheet
Use different colour and sizes of brush for making of different designs

Precautions:

Wash the brush properly before dipping and using another colour
Dry the brush before putting it in acrylic colours.

Results: Different textured sheet after application of dry brush will look little more embossed and the entire design gets a three dimension feel.

Discussion

33.	How do you identify the Blotch prints?
Answer» Identification of Blotch Prints:- The blotch print background color is lighter on backside of the fabric. Possibilities of large background color areas of the print are not covered with full depth of colors. Precise control is necessary. If pigments are used in blotch prints, then fabrics very often result in objectionable stiff hand.

33.

How do you identify the Blotch prints?

Answer»

Identification of Blotch Prints:-

The blotch print background color is lighter on backside of the fabric.
Possibilities of large background color areas of the print are not covered with full depth of colors.
Precise control is necessary.
If pigments are used in blotch prints, then fabrics very often result in objectionable stiff hand.

Discussion

34.	Special effects to a design can be given to create a new appearance or to enhance an existing design. Some of these techniques are Wax Resist Technique, Fevicol Resist Technique, Etching Technique. These techniques bring originality, add texture and give a three dimensional look to the print design being developed.” What do you understand by Etching? List down any two precautions to be carried out while doing etching technique.
Answer» The etching technique consists of layering two or more colors over each other and then etching out a design from the top layer with a blade/ scraper to bring out color of the lower layer. Precautions to be kept in mind while preparing it: 1. While coloring the first layer makes sure no white spaces are left. 2. For the second layer, make sure the previous lower layer is not visible. 3. While using poster colour use thick paint and minimum amount of water. 4. While using a blade or sharp object, do not uses too much pressure or you may tear the paper.

34.

Special effects to a design can be given to create a new appearance or to enhance an existing design. Some of these techniques are Wax Resist Technique, Fevicol Resist Technique, Etching Technique. These techniques bring originality, add texture and give a three dimensional look to the print design being developed.” What do you understand by Etching? List down any two precautions to be carried out while doing etching technique.

Answer»

The etching technique consists of layering two or more colors over each other and then etching out a design from the top layer with a blade/ scraper to bring out color of the lower layer.

Precautions to be kept in mind while preparing it:

1. While coloring the first layer makes sure no white spaces are left.

2. For the second layer, make sure the previous lower layer is not visible.

3. While using poster colour use thick paint and minimum amount of water.

4. While using a blade or sharp object, do not uses too much pressure or you may tear the paper.

Discussion

35.	What are the precautions to be taken while doing Etching Technique?
Answer» The precautions to be taken while doing Etching Technique are:- While colouring the first layer make sure no white spaces are left. For the second layer, make sure the previous lower layer is not visible. While using poster colour use thick paint and minimum amount of water.

35.

What are the precautions to be taken while doing Etching Technique?

Answer»

The precautions to be taken while doing Etching Technique are:-

While colouring the first layer make sure no white spaces are left.
For the second layer, make sure the previous lower layer is not visible.
While using poster colour use thick paint and minimum amount of water.

Discussion

36.	On acid hydrolysis, propanenitrile gives _______. (A) propanal (B) acetic acid (C) propionamide (D) propanoic acid
Answer» (D) propanoic acid CH₃ - CH₂ - C (Propanenitrile) ≡ N + 2H₂O + HCl \(\overset{Δ}\longrightarrow\) CH₃ - CH₂ - COOH (Propanoic acid) + NH₄Cl

36.

On acid hydrolysis, propanenitrile gives _______. (A) propanal (B) acetic acid (C) propionamide (D) propanoic acid

Answer»

(D) propanoic acid

CH₃ - CH₂ - C (Propanenitrile) ≡ N + 2H₂O + HCl \(\overset{Δ}\longrightarrow\) CH₃ - CH₂ - COOH (Propanoic acid) + NH₄Cl

Discussion

37.	Select the compound whose 0.1 M solution is basic : (A) ammonium chloride (B) ammonium acetate (C) ammonium sulphate (D) sodium acetate
Answer» Correct option: (D) sodium acetate Explanation: Since sodium acetate is salt of (WA + SB) so its pH > 7.

37.

Select the compound whose 0.1 M solution is basic : (A) ammonium chloride (B) ammonium acetate (C) ammonium sulphate (D) sodium acetate

Answer»

Correct option: (D) sodium acetate

Explanation:

Since sodium acetate is salt of (WA + SB) so its pH > 7.

Discussion

38.	Metals react with acids and produce .........gas. a) H2 b) O2 c) N2 d) Cl2
Answer» Metals react with acids and produce H₂ gas.

Discussion

39.	100% pure gold is expressed as .......carat gold. a) 24 b) 26 c) 18 d) 21
Answer» 100% pure gold is expressed as 24 carat gold.

Discussion

40.	What would happen to iron railings on the road side if they are not painted? Why does it happen so?
Answer» If the iron railing on the road side is not painted, a brown rust would form on its surface because the moist air of the atmosphere reacts with iron to form brown flaky substance on its surface. The rust is hydrated iron (III) oxide, Fe₂O₃.xH₂O.

Discussion

41.	Which of the following is correct for repulsice process in real gas ?A. `Z gt 1`, repulsive forces are dominatingB. `Z lt 1`, repulsive forces are dominatingC. short range interactionD. long range interactin
Answer» Correct Answer - A Higher the value oof z, difficult is to compress a gas.

Discussion

42.	What is 24-carat gold? How will you convert it into 18-carat gold?
Answer» 24-carat gold is pure gold. Pure gold is very soft and not suitable for making jewellery. Therefore, to increase its hardness, it is allowed either with copper or silver. 18-carat gold is prepared by alloying 18 parts pure gold with 6 parts of either copper or silver.

Discussion

43.	The solubility of CH3COOAg in a buffer solution with ph=4 whose ksp=10-12 and ka=(10-4)/3 is Answer is 2x10-6 how
Answer» When CH3COOAg(s) is dissolved in water an equilibrium is established with the solid salt at a given temperature forming ions as follows. CH₃COOAg(s) ⇋ CH₃COO^- + Ag⁺ ...[1] If the salt is dissolved in buffer solution of pH=4 ie [H+]=10^-4 then acetate ions in the solution goes to combine with H+ ions as follows and proceeds towards another equilibrium state CH₃COO^- + H⁺ ⇋ CH₃COOH.....[2] Both the reactions establish a overall equilibrium state. Let at the final equilibrium state the concentration of [Ag⁺] be x(M). This will represent the solubility of the salt in buffer solution. If a(M) be the concentration of acetic formed at equilibrium then concentration of acetate ion will be (x-a) (M) Concentration of H⁺ ion remains unaltered due to buffer action. So for equation(1) K_sp=(x-a)x. =>(x-a)x=10^-12......(3) And for equation (2) 1/k_a=a/((x-a)10^-4) =>3/10^-4=a/((x-a)×10^-4) =>3=a/(x-a) =>3x-3a=a =>a=(3/4)x...(4) Combining (3) and (4) we get (x-3x/4)x=10^-12 =>x²=4×10^-12 =>x=2×10^-6 M

43.

The solubility of CH3COOAg in a buffer solution with ph=4 whose ksp=10-12 and ka=(10-4)/3 is Answer is 2x10-6 how

Answer»

When CH3COOAg(s) is dissolved in water an equilibrium is established with the solid salt at a given temperature forming ions as follows.

CH₃COOAg(s) ⇋ CH₃COO^- + Ag⁺ ...[1]

If the salt is dissolved in buffer solution of pH=4 ie [H+]=10^-4 then acetate ions in the solution goes to combine with H+ ions as follows and proceeds towards another equilibrium state

CH₃COO^- + H⁺ ⇋ CH₃COOH.....[2]

Both the reactions establish a overall equilibrium state.

Let at the final equilibrium state the concentration of [Ag⁺] be x(M). This will represent the solubility of the salt in buffer solution. If a(M) be the concentration of acetic formed at equilibrium then concentration of acetate ion will be (x-a) (M)

Concentration of H⁺ ion remains unaltered due to buffer action.

So for equation(1)

K_sp=(x-a)*x.

=>(x-a)*x=10^-12......(3)

And for equation (2)

1/k_a=a/((x-a)*10^-4)

=>3/10^-4=a/((x-a)×10^-4)

=>3=a/(x-a)

=>3x-3a=a

=>a=(3/4)x...(4)

Combining (3) and (4) we get

(x-3x/4)*x=10^-12

=>x²=4×10^-12

=>x=2×10^-6 M

Discussion

44.	In solid phase, `XeF_(6)` consists ofA. `XeF_(5)^(+) + F^(-)`B. `XeF_(4) + F_(2)`C. `XeF_(5)^(+) + XeF_(2)^(-)`D. `XeF_(2) + 2F_(2)`
Answer» Correct Answer - A `XeF_(6) rarr XeF_(5)^(+) + F^(-)`

Discussion

45.	In electrorefining, the impure metal is made _____.A. CathodeB. AnodeC. ElectrolyteD. Both `(a)` and `(c )`
Answer» Correct Answer - B Impure metal in anode is converted into its cation , and is deposited on cathode.

Discussion

46.	The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions.\(CaCO_3\rightleftharpoons CaO(s)+CO_ 2(g)\)For this equilibrium, the correct statement(s) is(are) (A) ΔH is dependent on T (B) K is independent of the initial amount of CaCO3 (C) K is dependent on the pressure of CO2 at a given T (D) ΔH is independent of the catalyst, if any
Answer» (A) ΔH is dependent on T (B) K is independent of the initial amount of CaCO₃ (D) ΔH is independent of the catalyst, if any For the equilibrium CaCO₃ (s) \(\rightleftharpoons\) CaO (S) + CO₂ (g). The equilibrium constant (K) is independent of initial amount of CaCO₃ where as at a given temperature is independent of pressure of CO₂. ∆H is independent of catalyst and it depends on temperature.

46.

The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions.\(CaCO_3\rightleftharpoons CaO(s)+CO_ 2(g)\)For this equilibrium, the correct statement(s) is(are) (A) ΔH is dependent on T (B) K is independent of the initial amount of CaCO3 (C) K is dependent on the pressure of CO2 at a given T (D) ΔH is independent of the catalyst, if any

Answer»

(A) ΔH is dependent on T

(B) K is independent of the initial amount of CaCO₃

(D) ΔH is independent of the catalyst, if any

For the equilibrium CaCO₃ (s) \(\rightleftharpoons\) CaO (S) + CO₂ (g). The equilibrium constant (K) is independent of initial amount of CaCO₃ where as at a given temperature is independent of pressure of CO₂. ∆H is independent of catalyst and it depends on temperature.

Discussion

47.	A spherical balloon of volume 5 litre is to be filled up with `H_2` at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at `0^@C`.The no of balloons that can be filled up is :
Answer» Correct Answer - 6 Volume of gas at NTP is , `P_1V_1=P_2V_2` `6xx 6 =1xxV_2` `V_2`=36 litres. Outcoming gas = 36-6 =30 litres No of balloon =`30/5=6`

Discussion

48.	What is the minimum pH required to prevent the precipitation of ZnS in a solution that is `0.01 M ZnCl_2` and saturated with `0.10 M H_2S`? Given `K_(sp)` of `ZnS=10^(-21)` , for `H_2S K_(a_1)xK_(a_2)=10^(-20)`
Answer» Correct Answer - 1 `K_(sp)=[Zn^(2+)][S^(-2)]` :. [5^2]=10^(-21)/10.01=10^(-19)` `K_(a_1)xxK_(a_2)=([H^+]^(2)[10^(-19)])/0.1implies 10^(-20)=([H^+]^(2)[10^(-19)])/0.1` `[H^+]=10^(-1)` pH=1

Discussion

49.	A compound formed by elements A and B crystallises in cubic structure in which A atoms are at the corners of the cube while B atoms are at the centre of cubic. Formula of the compound isA. `A_(2)B`B. `AB_(2)`C. `A_(4)B`D. `A_(8)B_(3)`
Answer» Correct Answer - D

Discussion

50.

According to Molecular Orbital Theory (A) C\(^{-2}_2\) is expected to be diamagnetic (B) O\(^{2+}_2\) expected to have a longer bond length than O2 (C) N\(^{+}_2\) and N2 - have the same bond order (D) He\(^{+}_2\) has the same energy as two isolated He atoms

Answer»

(A) C\(^{-2}_2\) is expected to be diamagnetic

(C) N\(^{+}_2\) and N2 - have the same bond order

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

A stone of mass ‘m’ is to be thrown to a height h1. What is the acceleration of the stone?2. With what minimum velocity should it be thrown.3. At what height does the KE and PE become equal?4. Find the velocity at that height

Given p(A) = 3/5 and p(B) = 1/5. Find P(A or B), if A &amp; B are mutually exclusive events.

In a group of 600 students in a school, 150 students were found to be taking tea, 225 and taking coffee and 100 were taking both tea and coffee. How may students were taking neither tea nor coffee.

Name the parameter which is a measure of degree of elasticity of a body.

A function f is defined by f(x) = 2x – 5.Write down the values of(i) f(0),(ii) f(7),(iii) f(–3)

Find the domain and range of the following real function: (i) f(x) = –|x| (ii) () = √(9 − 2)

A = {x: x is a natural number less than 8}1. Write in roster form.2. Write a subset of A containing all even numbers in A.3. Which of the following could not be the number of elements of power set of a set [2, 8, 10, 16]?

Let A and B be two sets such that n( A) = 20, n(A ∪ B) = 42, n(A ∩ B) = 4. Find1. n(B)2. n(B – A)3. n(A – B)

Write the following in set builder form.1. [0, 10]2. [-2, 7)3. (3, 4)

If cos∅ = 1/2 then the value of cosec∅ is –(A) 2(B) 2/√3(C) √3/2(D) 1/√3

the mode of the data 21, 26, 22, 29, 23, 29, 26, 29, 22, 23 is?

In an online test, sami scores 18 points for correct answers and -3 points for incorrect answers. Find his score.

Consider the set Q of rational numbers. Let * be the operation on Q defined by a * b = a + b - ab. The identity element under * is

what is diamond made up of like water is made of H2O

What are the two types of ecosystem?

Terrestrial Ecosystems

Forest Ecosystem

Grassland Ecosystem

Tundra Ecosystem

Desert Ecosystem

Aquatic Ecosystem

Freshwater Ecosystem

Marine Ecosystem

Structure of the Ecosystem

Biotic Components

Abiotic Components

Functions of Ecosystem

Important Ecological Concepts

How hard wood differ form soft wood

Let \( y=e^{3 x} \) and \( f(x)=-4 e^{-3 x}+e^{6 x} \)i) Find \( f^{\prime}(x) \) and \( f^{\prime \prime}(x) \)ii) Show that \( f^{\prime \prime}(x)=36\left[y^{2}-\frac{1}{y}\right] \)

H.C.F of \( \left(x^{2}-3 x+2\right) \) and \( \left(x^{2}-4 x+3\right) \) is

Describe the structure of chloroplast.

Name the building blocks of proteins and explain the structural aspects of proteins.

Name the building blocks of protein explain the structural aspects of protein.

Which of the following is the distance of a point P(x, y) from the origin?(a) x2 + y2 (b) x2 - y2(c) √(x2+y2)(d) √(x2-y2)

Which of the following is the first positive term of the given A.P. –11, –8, –5, .......?(a) 1 (b) –2 (c) 2 (d) 3

The symbol for circumcenter is

Find all three digit natural numbers of the form (abc)10 such that (abc)10, (bca)10 and (cab)10 are in geometric progression. (Here (abc)10 is representation in base 10.)

The larger of the two supplementary angles exceeds the smaller by 18°. Find them.

Describe Bhandhani of Gujrat. What are the different textiles from Gujrat?

What are the basic steps to create bhandhani textile? Describe it elaborately.

What are the End Uses of Chikankari?

Write Dry Brush Technique in relevance to material required, steps, precautions and results.

How do you identify the Blotch prints?

What are the precautions to be taken while doing Etching Technique?

On acid hydrolysis, propanenitrile gives _______. (A) propanal (B) acetic acid (C) propionamide (D) propanoic acid

Select the compound whose 0.1 M solution is basic : (A) ammonium chloride (B) ammonium acetate (C) ammonium sulphate (D) sodium acetate

Metals react with acids and produce .........gas. a) H2 b) O2 c) N2 d) Cl2

100% pure gold is expressed as .......carat gold. a) 24 b) 26 c) 18 d) 21

What would happen to iron railings on the road side if they are not painted? Why does it happen so?

Which of the following is correct for repulsice process in real gas ?A. `Z gt 1`, repulsive forces are dominatingB. `Z lt 1`, repulsive forces are dominatingC. short range interactionD. long range interactin

What is 24-carat gold? How will you convert it into 18-carat gold?

The solubility of CH3COOAg in a buffer solution with ph=4 whose ksp=10-12 and ka=(10-4)/3 is Answer is 2x10-6 how

In solid phase, `XeF_(6)` consists ofA. `XeF_(5)^(+) + F^(-)`B. `XeF_(4) + F_(2)`C. `XeF_(5)^(+) + XeF_(2)^(-)`D. `XeF_(2) + 2F_(2)`

In electrorefining, the impure metal is made _____.A. CathodeB. AnodeC. ElectrolyteD. Both `(a)` and `(c )`

A spherical balloon of volume 5 litre is to be filled up with `H_2` at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at `0^@C`.The no of balloons that can be filled up is :

What is the minimum pH required to prevent the precipitation of ZnS in a solution that is `0.01 M ZnCl_2` and saturated with `0.10 M H_2S`? Given `K_(sp)` of `ZnS=10^(-21)` , for `H_2S K_(a_1)xK_(a_2)=10^(-20)`

A compound formed by elements A and B crystallises in cubic structure in which A atoms are at the corners of the cube while B atoms are at the centre of cubic. Formula of the compound isA. `A_(2)B`B. `AB_(2)`C. `A_(4)B`D. `A_(8)B_(3)`

According to Molecular Orbital Theory (A) C\(^{-2}_2\) is expected to be diamagnetic (B) O\(^{2+}_2\) expected to have a longer bond length than O2 (C) N\(^{+}_2\) and N2 - have the same bond order (D) He\(^{+}_2\) has the same energy as two isolated He atoms

Given p(A) = 3/5 and p(B) = 1/5. Find P(A or B), if A & B are mutually exclusive events.