advantage and disadvantage of python programming?

 Question: advantage and disadvantage of python programming?

Solution: 

Advantages of Python

1. Easy to Read, Learn and Write

Python is a high-level programming language that has English-like syntax. This makes it easier to read and understand the code.

Python is really easy to pick up and learn, that is why a lot of people recommend Python to beginners. You need less lines of code to perform the same task as compared to other major languages like C/C++ and Java.

2. Improved Productivity

Python is a very productive language. Due to the simplicity of Python, developers can focus on solving the problem. They don’t need to spend too much time in understanding the syntax or behavior of the programming language. You write less code and get more things done.

3. Interpreted Language

Python is an interpreted language which means that Python directly executes the code line by line. In case of any error, it stops further execution and reports back the error which has occurred.

Python shows only one error even if the program has multiple errors. This makes debugging easier.

4. Dynamically Typed

Python doesn’t know the type of variable until we run the code. It automatically assigns the data type during execution. The programmer doesn’t need to worry about declaring variables and their data types.

5. Free and Open-Source

Python comes under the OSI approved open-source license. This makes it free to use and distribute. You can download the source code, modify it and even distribute your version of Python. This is useful for organizations that want to modify some specific behavior and use their version for development.

6. Vast Libraries Support

The standard library of Python is huge, you can find almost all the functions needed for your task. So, you don’t have to depend on external libraries.

But even if you do, a Python package manager (pip) makes things easier to import other great packages from the Python package index (PyPi). It consists of over 200,000 packages.

7. Portability

In many languages like C/C++, you need to change your code to run the program on different platforms. That is not the same with Python. You only write once and run it anywhere.

However, you should be careful not to include any system-dependent features.

Disadvantages of Python

1. Slow Speed

We discussed above that Python is an interpreted language and dynamically-typed language. The line by line execution of code often leads to slow execution.

The dynamic nature of Python is also responsible for the slow speed of Python because it has to do the extra work while executing code. So, Python is not used for purposes where speed is an important aspect of the project.

2. Not Memory Efficient

To provide simplicity to the developer, Python has to do a little tradeoff. The Python programming language uses a large amount of memory. This can be a disadvantage while building applications when we prefer memory optimization.

3. Weak in Mobile Computing

Python is generally used in server-side programming. We don’t get to see Python on the client-side or mobile applications because of the following reasons. Python is not memory efficient and it has slow processing power as compared to other languages.

4. Database Access

Programming in Python is easy and stress-free. But when we are interacting with the database, it lacks behind.

The Python’s database access layer is primitive and underdeveloped in comparison to the popular technologies like JDBC and ODBC.

Huge enterprises need smooth interaction of complex legacy data and Python is thus rarely used in enterprises.

5. Runtime Errors

As we know Python is a dynamically typed language so the data type of a variable can change anytime. A variable containing integer number may hold a string in the future, which can lead to Runtime Errors.

Cells of which layer of anther wall are binucleate and nourishes the developing pollen grain? (A) Epidermis (B) Exothecium (C) Middle layer (D) Tapetum

Question:

Cells of which layer of anther wall are binucleate and nourishes the developing pollen grain? 

(A) Epidermis 

(B) Exothecium 

(C) Middle layer 

(D) Tapetum

Answer: option (D) "Tapetum"

Explanation:

Structure of Microsporangium: A typical microsporangium appears nearly circular in outline. It is generally surrounded by four wall layers, viz. epidermis, endothecium, middle layers and tapetum. The outer three wall layers perform the function of protection and help in dehiscence of anther to release the pollen. Tapetum is the innermost wall layer. Tapetum nourishes the developing pollen grains. Cells of the tapetum have dense cytoplasm and are generally bi-nucleate. 

Hence, option (D) "Tapetum" is the correct answer.

Function f(x) = |x| - |x-1| is monotonically increasing when (1) x < 0 (2) x > 1 (3) x < 1 (4) 0 < x < 1

Question:

Function f(x) = |x| - |x-1| is monotonically increasing when 

(1) x < 0 

(2) x > 1 

(3) x < 1 

(4) 0 < x < 1

Answer: option (4) "0<x<1"

Explanation:

Given: f(x)=|x| - |x-1| 

To find: At which condition f(x) is Monotonically increasing. 

Solution: 

f(x)=|x| - |x-1| 

for x>1 

f(x)=x-(x-1)=1 

f '(x)=0 

for 0<x<1 

f(x)=x-(1-x)=2x-1

 f(x)=2>0 

Monotonically increasing 

for x<0 

f (x)=-x-(1-x)= -1 

f '(x)=0 

0<x<1 is the interval for f(x) to be monotonically increases. 

Hence, option (4) "0<x<1" is the correct answer.

Bleaching action of H2O2 is due to its : (A) Oxidizing nature (B) Reducing nature (C) Acidic nature (D) Thermal instability

Question:

Bleaching action of H2O2 is due to its : 

(A) Oxidizing nature 

(B) Reducing nature 

(C) Acidic nature 

(D) Thermal instability

Answer: option (1) " oxidizing nature"

Explanation:

It acts as bleaching agent due to the release of nascent oxygen. Thus, the bleaching action of hydrogen peroxide is permanent and it is due to oxidation. It oxidizes the coloring matter to a colorless product. 

H2O2 → H2O + [O]

Hence, option (1) " oxidizing nature" is the correct answer.

Which of the following statements regarding the class phycomycetes is correct ? (A) These are found in aquatic habitats and on decaying wood in moist and damp places or as obligate parasites on plants (B) Mycelium in these fungi is aseptate and coenocytic (C) Asexual reproduction occurs by motile z0ospores and by non-motile apanospores (D) All of these

Question:

Which of the following statements regarding the class phycomycetes is correct ? 

(A) These are found in aquatic habitats and on decaying wood in moist and damp places or as obligate parasites on plants 

(B) Mycelium in these fungi is aseptate and coenocytic 

(C) Asexual reproduction occurs by motile z0ospores and by non-motile apanospores 

(D) All of these

Answer:  option (D) "All the above"

Explanation:

Phycomycetes are characterized by the aseptate coenocytic hyphae. Asexual reproduction occurs by motile zoospores or non-motile aplanospores produced endogenously inside sporangia. Members of Phycomycetes are found in aquatic habitats and on decaying wood in moist and damp places or as obligate parasites on plants. 

Hence, option (D) "All the above" is the correct answer.

Which one of the following statements is correct regarding Sexually Transmitted Diseases (STD)? (1) The chances of a 5-year boy contacting a STD are very little. (2) A person may contact syphilis by sharing milk with the one who is already suffering from the same disease. (3) Haemophilia is one of the STD. (4) Genital herpes and sickle-cell anaemia are both STD.

Question: 

Which one of the following statements is correct regarding Sexually Transmitted Diseases (STD)? (1) The chances of a 5-year boy contacting a STD are very little. 

(2) A person may contact syphilis by sharing milk with the one who is already suffering from the same disease. 

(3) Haemophilia is one of the STD. 

(4) Genital herpes and sickle-cell anaemia are both STD.

Answer: option (1) "The chances of a 5-years boy contracting a STD are very little"

Explanation:

Syphilis is caused by bacterium Treponema pallidum. It is a sexually transmitted disease (STD) which is transmitted through sexual intercourse with infected person. Haemophilia is an X-linked genetic disorder of blood, It is not transmitted via any sexual practice. Genital herpes is an STD while sickle-cell anaemia is an autosomal hereditary disorder. The chances of a 5 years old boy contacting an STD are very rare since he is unlikely to have sex at this age. So, the correct answer is 'The chances of a 5-years boy contracting a STD are very rare'. 

Hence, option (1) "The chances of a 5-years boy contracting a STD are very little" is the correct answer.

Antimarkonikov's addition of HBr is not observed in (A) Propene (B) But-1-ene (C) but-2-ene (D) Pent-2-ene

Question:

Antimarkonikov's addition of HBr is not observed in 

(A) Propene 

(B) But-1-ene 

(C) but-2-ene 

(D) Pent-2-ene 

Answer: option (C) but-2-ene

Explanation:

The Anti-Markovnikov's addition is also called the peroxide effect or the Kharasch effect. The Anti-Markovnikov's rule states that, in the presence of peroxide like benzoyl peroxide (C6H5CO-O-O-COC6H5), the addition of HBr to unsymmetrical alkenes takes place contrary to the Markovnikov's rule. In Markovnikov's rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has a lesser number of a hydrogen atom or more substituted carbon atom. So, in Anti- Markovnikov's rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has more number of hydrogen atoms. 

(A) Propene: The formula of Propene is CH2=CH-CH3. This is an unsymmetrical alkene. So according to the Anti- Markovnikov's rule, the negative part will attach the carbon atom with 2 hydrogen atoms. 

(B) But-1-ene: The formula of But-1-ene is CH2=CH-CH2-CH3. This is an unsymmetrical alkene. So according to the Anti- Markovnikov's rule, the negative part will attach the carbon atom with 2 hydrogen atoms.

(C)- But-2-ene: The structure of 2-Butene is CH3-CH=CH-CH3. This is a symmetrical alkene and the Anti- Markovnikov's rule will not be observed in 2-Butene. 

(D)- pent-2-ene The structure of 2-Pentene is CH3-CH=CH-CH2-CH3. This is an unsymmetrical alkene. So according to the Anti- Markovnikov's rule, the negative part will attach the carbon atom with the methyl group because it is less substituted. 

Hence, option (C) "But-2-ene" is the correct answer.

The property which regularly increases down the group in the periodic table is : (a) lionization enthalpy (b) Electronegativity (c) Reducing character (d) Electron gain enthalpy.

Question: 

The property which regularly increases down the group in the periodic table is : 

(a) lionization enthalpy 

(b) Electronegativity 

(c) Reducing character 

(d) Electron gain enthalpy.

Answer: option (c) " Reducing character"

Explanation:

The metallic character is used to define the chemical property that metallic elements present. Generally, metals tend to lose electrons to form cations. Nonmetals tend to gain electrons to form anions. They also have a high oxidation potential therefore they are easily oxidized and are strong reducing agents. Thus the property which regularly increases down the group in the periodic table is Reducing nature. 

Hence, option (c) " Reducing character" is the correct answer.

In which one of the following would you expect to find Glyoxysomes (A) Endosperm of wheat (B) Endosperm of castor (C) Palisode cells in leaf (D) Root hairs

Question: 

In which one of the following would you expect to find Glyoxysomes 

(A) Endosperm of wheat 

(B) Endosperm of castor 

(C) Palisode cells in leaf 

(D) Root hairs

Answer: option (B) "Endosperm of castor"

Explanation:

Glyoxysomes are the special type of peroxisomes which are found in plant cells. It is mainly present in cells where fat conversion takes place. Glyoxysomes are the microbodies that contain enzymes for the beta-oxidation of fatty acids and glyoxylate pathways. It is quite abundant in germinating oilseeds. They are basically responsible for the glyoxylate cycle in which fat is converted into carbohydrates. It can be found in seeds containing fats and oils like corn, soybean, sunflower, peanut, castor, and pumpkin. So, Glyoxysomes are found in the endosperm of castor because it contains oil. In castor seeds, endosperm persists in the mature seed and can be used during seed germination. 

Hence, option (B) "Endosperm of castor" is the correct answer.

Law of independent assortment is applicable to (A) Closely linked genes (B) Distantly located genes on same chromosome (C) Two genes present on different chromosomes (D) Both 2) & 3)

Question:

 Law of independent assortment is applicable to 

(A) Closely linked genes 

(B) Distantly located genes on same chromosome 

(C) Two genes present on different chromosomes 

(D) Both (B) & (C)

Answer: option (C) "Two genes present on different chromosomes"

Explanation:

Law of independent assortment tells about segregation and distribution of factors governing two different traits. Tendency of closely placed genes on a chromosome to stay together during inheritance i.e. linkage and therefore no independent assortment make the linked traits exception to law of independent assortment. To exhibit law of independent assortment, two genes should be present distantly on one chromosome or should be present on two different chromosomes only. 

Hence, option (C) "Two genes present on different chromosomes" is the correct answer.

What does the filiform apparatus do at the entrance into ovule ? (1) It brings about opening of the pollen tube (2) It guides pollen tube from a synergid to egg (3) It helps in the entry of pollen tube into a synergid (4) It prevents entry of more than one pollen tube into the embryo sac

Question:

What does the filiform apparatus do at the entrance into ovule ? 

(1) It brings about opening of the pollen tube 

(2) It guides pollen tube from a synergid to egg 

(3) It helps in the entry of pollen tube into a synergid 

(4) It prevents entry of more than one pollen tube into the embryo sac

Answer: option (3) "It helps in the entry of pollen tube into a synergid "

Explanation:

Filiform system assists in entering a pollen tube into an ovule synergy. The filiform device is a finger-like projection consisting of an enclosed core of microfibrils in a sheath. The filiform apparatus resembles the transfer cells meaning the movement of metabolites at short distances. The filiform system which extracts food from the nucleus. 

Hence, option (3) "It helps in the entry of pollen tube into a synergid " is the correct answer.

Which substance is synthesized by Granular endoplasmic reticulum (RER)?

 Question:

Which substance is synthesized by Granular endoplasmic reticulum (RER)?

Answer: Protein

Explanation:

Rough (Granular) endoplasmic reticulum (RER), series of connected flattened sacs, part of a continuous membrane organelle within the cytoplasm of eukaryotic cells, that plays a central role in the synthesis of proteins. The rough endoplasmic reticulum (RER) is so named for the appearance of its outer surface, which is studded with protein-synthesizing particles known as ribosomes. This feature distinguishes it superficially and functionally from the other major type of endoplasmic reticulum (ER), the smooth endoplasmic reticulum (SER), which lacks ribosomes and is involved in the synthesis and storage of lipids. RER occurs in both animal and plant cells. 

Hence, protein substance is synthesized by granular endoplasmic reticulum (RER).

Why should magnesium ribbon be cleaned before burning it?

Question:

Why should magnesium ribbon be cleaned before burning it?

Explanation:

Magnesium ribbon should be cleaned with sandpaper before burning in the air. It should be cleaned before burning due to the following reasons: 

1. Magnesium very reactive element which rapidly reacts with oxygen in the air to form a white layer of magnesium oxide and this layer will not burn. 

2. To remove the Magnesium oxide layer from the ribbon which may prevent or slow down the burning of magnesium ribbon. 

3. Unwanted impurities deposited on the magnesium ribbon can be removed and only pure magnesium can be used for the reaction. Hence, magnesium ribbon be cleaned before burning in air.

All the tissue exterior to vascular cambium refers to (A) Periderm 2 (B) Bark (C) Phellogen (D) Phelloderm

Question:

 All the tissue exterior to vascular cambium refers to 

(A) Periderm 2 

(B) Bark 

(C) Phellogen 

(D) Phelloderm

Answer: option (B) Bark

Explanation:

Bark, in woody plants, tissues external to the vascular cambium (the growth layer of the vascular cylinder); the term bark is also employed more popularly to refer to all tissues outside the wood. The inner soft bark, or bast, is produced by the vascular cambium; it consists of secondary phloem tissue whose innermost layer conveys food from the leaves to the rest of the plant. The outer bark, which is mostly dead tissue, is the product of the cork cambium (phellogen). Layered outer bark, containing cork and old, dead phloem, is known as rhytidome. The dead cork cells are lined with suberin, a fatty substance that makes them highly impermeable to gases and water. Gas exchange between the inner tissues of bark-covered roots and stems and their surroundings takes place through spongy areas (lenticels) in the cork. Bark is usually thinner than the woody part of the stem or root. Both inner bark (secondary phloem) and wood (secondary xylem) are generated by the vascular cambium layer of cells: bark toward the outside where the oldest layers may slough off, and wood toward the inside where it accumulates as dead tissue. 

Hence, the correct answer is option (B) "Bark". 

The ploidy levels of the cells of nucellus, megaspore mother cell, the functional megaspore and female gametophyte are, respectively :- (A) 2n, 2n, n, n (B) 2n, 2n, 2n, n (C) 2n, 2n, n, 2n (D) n, 2n, n, n

Question:

The ploidy levels of the cells of nucellus, megaspore mother cell, the functional megaspore and female gametophyte are, respectively :- 

(A) 2n, 2n, n, n 

(B) 2n, 2n, 2n, n

(C) 2n, 2n, n, 2n 

(D) n, 2n, n, n  

Explanation:

Considering if the ploidy of plant is 2n, 1)Nucellus(located within the integumentary) have abundance reserve food materials. Its ploidy is 2n. MMC (produces haploid gametophytes by meiotic division) Since, it undergoes meiotic division its ploidy is 2n. Female gametophytes also called as embryo sac (produced by meiotic division from Megaspore mother cell) has ploidy n. Similarly, the functional megaspore is haploid in nature. 

Hence, the correct answer is option (A) "2n, 2n, n, n".

In some plants where flowering occurs more than once what would you will call the inter flowering period: (A) Juvenile (B) Mature (C) Senescent (D) Aging

Question:

In some plants where flowering occurs more than once what would you will call the inter flowering period: 

(A) Juvenile 

(B) Mature 

(C) Senescent 

(D) Aging

Answer: option (A) Juvenile

Explanation:

Option A: The plants which flower more than once, their inter - flowering period represents the juvenile phase. Although they pass the mature phase of flowering for the first time. But for flowering for the next time, they require further development. Therefore this is the correct option. 

Option B : Mature phase of a plant is when it has the ability to flower and reproduce. In plants where flowering occurs more than once, the plant regresses back to the juvenile phase before it flowers again. Therefore this is an incorrect option. 

Option C: Senescent phase marks the end of the reproductive phase of the plant. Senescent phase signifies that after this phase the plant cannot reproduce. Therefore this is an incorrect option.

Option D: Plant senescence is the process of aging in plants. Plants have both stress-induced and age-related developmental aging. Chlorophyll degradation during leaf senescence reveals the carotenoids, such as anthocyanin and xanthophyll's and is the cause of autumn leaf color in deciduous trees. Leaf senescence has the important function of recycling nutrients, mostly nitrogen, to growing and storage organs of the plant. Therefore this is an incorrect option. 

Hence, the correct answer is option (A) "Juvenile".

Schottky defect is observed in crystals when.....

Question:

Schottky defect is observed in crystals when.....

Answer: when equal numbers of cations and anions are missing from the lattice.

Explanation:

Schottky defect in crystals is observed when equal numbers of cations and anions are missing from the lattice. It is important that an equal number of cations and anions are missing, otherwise the electrical neutrality of the crystal will get affected.

Hence, Schottky defect is observed in crystals when equal numbers of cations and anions are missing from the lattice.

Number of water molecules consumed by one of P4O10 to undergo complete hydrolysis in order to produce ortho phosphoric acid is....

   Question: 

   Number of water molecules consumed by one of P4O10 to             undergo complete hydrolysis in order to produce ortho                 phosphoric acid is....

   Answer: 6

Water vascular system present in....A... and water canal system present in....B.... (1) Sponges and Echinoderms (2) Echinoderms and Sponges (3) Sponges and Cnidarians (4) Cnidarians and Ctenophores

Question:

 Water vascular system present in....A... and water canal system present in....B.... 

(1) Sponges and Echinoderms 

(2) Echinoderms and Sponges 

(3) Sponges and Cnidarians 

(4) Cnidarians and Ctenophores 

Answer: option (2) Echinoderms and Sponges

Explanation:

Solution: The water vascular system is a hydraulic system used by Echinoderms, such as sea stars and sea urchins, for locomotion, food and waste transportation and respiration. The system is composed of canals connecting numerous tube feet. They move by alternately contracting muscles that force water into the tube feet, causing them to extend and push against the ground, then relaxing to allow the feet to retract. 

Phylum Porifera members also known as sponged have a water transport or canal system. Water enters through the minute pores of sponges called ostia in the body wall into a central cavity called spongocoel, from where it goes out through the osculum. 

Hence, the correct answer is option (2) That is A (Echinoderms) and B (Sponges).

valance electron of sodium and magnesium are (a) 1 and 2 (b) 1 and 3 (c) 1 and 4 (d) 1 and 5

Question:

valance electron of sodium and magnesium are 

(a) 1 and 2 

(b) 1 and 3 

(c) 1 and 4 

(d) 1 and 5 

Answer: option (a) 1 and 2

Explanation:

The electrons in the outermost shell are the Valence electrons. These are the electrons in An atom that can participate in a chemical Reaction. 

The electron configuration for sodium (Na) is: 

1s²2s²2p⁶3s¹ 

The outer energy level for this atom is n = 3, And it has one electron in this energy level. Therefore, Sodium has one valence Electrons. 

The electron configuration for magnesium is: 

1s²2s²2p⁶3s²

The outer energy level for this atom is n = 3, And it has two electrons in this energy level. Therefore, magnesium has two valence Electrons. So, Valance electron of sodium and magnesium Are 1 and 2 respectively. Hence, the correct answer is option (a) 1 and 2.

Hence the answer is (A) 1 and 2

if the sum of two roots of equation x^3 - 2px^2 + 3qx - 4r =0 then r=?

 

Find the fourth term from the end in the expansion of (((3/2)x^2) - (1/3x))^9

 

The most accepted theory of origin of life is ___________________.

Question:

The most accepted theory of origin of life is ___________________.

Explanation:

The most accepted theory of origin of life is Oparin -Haldane​ theory. This theory explained the origin of life through aggregation of non-living inorganic matter that gave rise to complex substances called the coecervates in a reducing environment.

Hence, the correct answer is 'Oparin - Haldane theory'

 

The minimum value of charge on any charged body may be (a) 1.6 × 10^-19 (b) 1 coulomb (c) 1μC (d) 4.8 × 10^-12 coulomb

Question:

The minimum value of charge on any charged body may be 

(a) 1.6 × 10^-19 

(b) 1 coulomb 

(c) 1μC 

(d) 4.8 × 10^-12 coulomb

Answer:

Correct option: (a) 1.6 × 10^-19

Explanation:

The charge of any charged body in this universe is always a multiple of the charge of an electron. So the minimum charge of a body is ±e ( where e≈ 1.602×10^-19 Coulomb). If charge of a body is Q then Q= ±ne, where n is an integer.

Hence, the correct answer is option (a) "1.6×10^-19".

Double circulation plays important role in maintenance of Homeostasis in human beings. Validate the above point giving the suitable explanation.

Question: 

Double circulation plays important role in maintenance of Homeostasis in human beings. Validate the above point giving the suitable explanation.

Explanation:

A circulatory path is a path taken by the blood , wherein it travels throughout the different organs of the body through arteries and veins. In humans, it is a closed circulatory system that exists, as blood flows in closed blood vessels. The circulatory system is responsible for the transport of gases, nutrients, waste products etc. 

The Circulatory System delivers oxygen and nutrients in the blood to the surrounding cells to maintain Homeostasis. The blood transports nutrients, oxygen, and water to other body cells throughout the body. is necessary for a human being to separate oxygenated and de - oxygenated blood because this makes their circulatory system more efficient and helps in maintaining constant body temperature.

The human circulatory system is a double circulatory system. It has two separate circuits and blood passes through the heart twice. the pulmonary circuit is between the heart and lungs.

Hence, double circulation plays an important role in maintenance of Homeostasis.

Parathyroid hormone is involved in (A) Demineralization (B) Bone resorption (C) Both A and B (D) Antibodies production

Question:

Parathyroid hormone is involved in 

(A) Demineralization 

(B) Bone resorption 

(C) Both A and B 

(D) Antibodies production 

Answer:

Correct option: (C) Both A and B

Explanation:

Calcium and phosphorus homeostasis are maintained by the influence of parathyroid hormone on bone tissue and on the kidneys. An overabundance of the hormone causes demineralization of the bone tissue and the depletion of calcium and phosphorus from the body. Excess phosphorus is excreted by the kidneys. 

Parathyroid hormone regulates serum calcium through its effects on bone, kidney, and the intestine: In bone, PTH enhances the release of calcium from the large reservoir contained in the bones. Bone resorption is the normal destruction of bone by osteoclasts, which are indirectly stimulated by PTH.

PTH stimulates bone remodeling overall, bone resorption predominates when continuous exposure to high levels of PTH ensues, whereas administration of low, intermittent doses of PTH leads to a net increase in bone mass.

Hence, Option (C) Both A and B is the correct answer. 

Ethoxy ethane and methoxy propane are (A) Geometrical isomers (B) Optical isomers (C) Functional group isomers (D) Metamers

Question: 

Ethoxy ethane and methoxy propane are 

(A) Geometrical isomers 

(B) Optical isomers 

(C) Functional group isomers 

(D) Metamers

Answer:

Correct option: (D) Metamers

Explanation:

Structure of Ethoxyethane:

CH₃-CH₂-O-CH₂-CH₃

Structure of Methoxy propane:

CH₃-O-CH₂-CH₂-CH₃

Ethoxy ethane and Methoxy propane are metamers. Because, both are isomeric ethers but oxygen is attached to different alkyl groups in two compounds. Ethoxyethane and Methoxy propane are metamers. Because, both are isomeric ethers but oxygen is attached to different alkyl groups in two compounds.

Hence, option (D) "Metamers" is the correct answer.

If two liquids A and B have P_A^o:P_B^o=1:2 and have mole fraction in solution 1:2, then mole fraction of Ain vapour is (1) 0.33 (2) 0.25 (3) 0.52 (4) 0.2

 

Rhizoids of Funaria are (1) Unicellular and simple (2) Unicellular and tuberculatated (3)Multicellular and branched (4) Multicellular and unbranched

Question:

 Rhizoids of Funaria are...

(1) Unicellular and simple

(2) Unicellular and tuberculatated

(3)Multicellular and branched

(4) Multicellular and unbranched

Answer:

Correct option: (3) Multicellular and branched

Explanation:

In Funaria root like structures called Rhizoids are present. these Rhizoids in Funaria are branched , slender and multicellular. They are stout, brown, almost cable like and form the main anchoring strands. The Rhizoids are colorless in the young stage, but become red or brown at maturity. The rhizoidal branches have oblique cross walls and grow down into soil. They help in anchorage and absorption.

Hence, option (3) "Multicellular and branched" is the correct answer.

Antimarkonikov's addition of HBr is not observed is (a) Propene (b) But-1-ene (c) But-2-ene (d) Pent-2-ene

Question: 

Antimarkonikov's addition of HBr is not observed is 

(a) Propene 

(b) But-1-ene 

(c) But-2-ene 

(d) Pent-2-ene

Answer:

Correct option: (c) But-2-ene

Explanation:

The Anti-Markovnikov's addition is also called the peroxide effect or the Kharasch effect. The Anti-Markovnikov's rule states that, in the presence of peroxide like benzoyl peroxide ( C₆H₅CO–O–O-COC₆H₅), the addition of HBr to unsymmetrical alkenes takes place contrary to the Markovnikov's rule. In Markovnikov's rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has a lesser number of a hydrogen atom or more substituted carbon atom. So, in Anti- Markovnikov's rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has more number of hydrogen atoms.

(A)- Propene: 

The formula of Propene is CH₂=CH–CH₃. This is an unsymmetrical alkene. So according to the Anti- Markovnikov's rule, the negative part will attach the carbon atom with 2 hydrogen atoms.

(B)- But-1-ene: 

The formula of But-1-ene is CH₂=CH–CH₂–CH₃. This is an unsymmetrical alkene. So according to the Anti- Markovnikov's rule, the negative part will attach the carbon atom with 2 hydrogen atoms.

(C)- But-2-ene:

The structure of 2-Butene is CH₃-CH=CH–CH₃. This is a symmetrical alkene and the Anti- Markovnikov's rule will not be observed in 2-Butene.

(D)- pent-2-ene: 

The structure of 2-Pentene is CH₃–CH=CH–CH₂–CH₃. This is an unsymmetrical alkene. So according to the Anti- Markovnikov's rule, the negative part will attach the carbon atom with the methyl group because it is less substituted.

Hence, option (c) "But-2-ene" is the correct answer.

What are pseudo first order reactions? Give one example and explain why it is pseudo first order.

Question: 

What are pseudo first order reactions? Give one example and explain why it is pseudo first order.

Explanation:

Pseudo first-order reaction: 

Reaction which appears to be a second-order reaction, but actually is the first-order reaction is called pseudo-first-order reaction. This condition occurs in a chemical reaction between two substances when one reactant is present in a large amount. The concentration of reactant present in excess does not get altered much during the course of the reaction. Due to this reaction behaves as the first-order reaction.

Example:

Hydrolysis of esters

CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH

Rate law for this reaction is;

Rate K[CH₃COOC₂H₅][H₂O]

But the concentration of water does not change during the course of the reaction. 

So; [H₂O] is constant.

Therefore rate=K₁[CH₃COOC₂H₅]

Where, K=K'[H₂O]

So, the reaction behave as a first order reaction.

Which of the following are bone forming cells? a) Osteocytes b) Osteoblasts c) Osteoclasts d) None of these

Question: 

Which of the following are bone forming cells? 

a) Osteocytes 

b) Osteoblasts 

c) Osteoclasts 

d) None of these

Answer:

Correct option: b) Osteoblasts

Explanation:

Osteoblasts, osteocytes and osteoclasts are the three cell types involved in the development, growth and remodeling of bones. Osteoblasts are bone-forming cells, osteocytes are mature bone cells and osteoclasts break down and reabsorb bone. There are two types of ossification: intramembranous and endochondral.

Hence, the correct answer is option (b) "Osteoblasts".

Identify the wrong match. (a) MO - Nitrogen metabolism (b) B - Carbohydrate translocation (c) Zn - Auxin synthesis (d) Fe - Splitting of water

Question: Identify the wrong match. 

(a) MO - Nitrogen metabolism 

(b) B - Carbohydrate translocation 

(c) Zn - Auxin synthesis 

(d) Fe - Splitting of water

Answer:

wrong match option: (d) Fe - Splitting of water

Explanation:

Option (a):

MO - nitrogen metabolism:

The molybdenum nitrogenase is responsible for most biological nitrogen fixation, a prokaryotic metabolic process that determines the global biogeochemical cycles of nitrogen and carbon. 

Option (b):

B - Carbohydrate translocation:

Boron is necessary for the translocation of sugar (carbohydrates) in plants. 

Option (c):

Zn - Auxin synthesis:

Auxin is a growth hormone which is synthesized by tryptophan as a precursor by tryptophan-dependent auxin biosynthesis pathway. Zinc helps in the biosynthesis of auxin and its oxidative attack by free radical. It suppresses the production of oxygen during the synthesis of auxin and also acts as a coenzyme and activates the enzyme for auxin formation.

Option (d):

Fe - Splitting of water:

The enzyme has four Mn ions. Light energized  Mn and removes electrons from OH component of water forming oxygen. So, the essential element required for water splitting in photosynthesis leading to release of oxygen is Mn.

Hence, the wrong match is (d) "Fe - Splitting of water".

Number of accessory gland in male Reproductive system. (a). 2 (b). 3 (c). 4 (d). 5

Question: 

Number of accessory gland in male Reproductive system. 

(a). 2 

(b). 3 

(c). 4 

(d). 5

Answer:

Correct option: (b). 3

Explanation:

accessory glands are specialized structures found in males that produce fluids essential for the motility, nourishment and protection of sperm.

Males have three of these glands, and each one contributes to the production of semen. They are:

(1) the seminal vesicles

(2) the prostate gland

(3) the bulbourethral glands

Hence, the correct answer is option (b) "3".

Equation of the plane in intercept form is....

      Question: Equation of the plane in intercept form is...

      Answer: 

      (x/a) + (y/b) + (z/c) = 1.

      Explanation:

Which taxonomic category will have similar taxon in the classification of Mango and wheat? (a) Class (b) Division (c) Order (d) Family

Question: 

Which taxonomic category will have similar taxon in the classification of Mango and wheat? 

(a) Class 

(b) Division 

(c) Order 

(d) Family

Answer:

correct option: (b) Division

Explanation:

Taxonomic categories of Mango:

Class: Dicotyledonae

Division: Angiospermae 

Order: Sapindales 

Family: Anacardiaceae 

Taxonomic categories of wheat:

Class: Monocotyledonae

Division: Angiospermae

Order: Poales

Family: Poaceae 

So, by above conclusion it is clear that Division (Taxonomic category) have similar taxon in classification of mango and wheat.

Hence, the correct answer is option second "Division".

Fusion dissimilar gamete is (1) Allogamy (2) Fertilization (3) Autogamy (4) Dichogamy

Question: 

Fusion dissimilar gamete is.....

(1) Allogamy 

(2) Fertilization 

(3) Autogamy 

(4) Dichogamy 

Answer:

Correct option: (2) Fertilization

Explanation:

Cross-pollination in a crop plant is known as allogamy. Allogamy is defined as fertilization of ovum of a flower by the pollen of another flower on the different plant. The pollen is transferred from the anther of a male flower to the stigma of a female flower. Allogamy takes place because of the male and female part of a particular flower mature at a different time. 

Dichogamy is a condition when the stamens and pistils of a flower mature at different times to prevent self-fertilization.

Autogamy is the phenomena of fusion of gametes produced from the same individual. It involves the transfer of pollen from the anther to the stigma of the same flower. It is avoided when the number of seed is less, or the produced pollen is not able to fuse and germinate with the ovule of the same flower or the produced pollen are not healthy. 

Fusion of dissimilar gametes is fertilization. It takes place during sexual reproduction where the male and female gametes fuse to form a zygote. 

Hence, the correct answer is option (ii) "Fertilization".

Find the integrating factor of differential equation dy=cosx(2-ycosecx)dx

         Question: Find the integrating factor of differential equation        dy=cosx(2-ycosecx)dx

      Answer:

      Correct answer is sinx

      Explanation:

How many type of stomata are found in plants (1).1 (2).2 (3).4 (4).5

Question: 

 How many type of stomata are found in plants 

(1).1 

(2).2 

(3).4 

(4).5

 Answer:

correct option (4).5

Explanation:

Stomata:

Stomata are the tiny openings present on the epidermis of leaves. We can see stomata under the light microscope. In some of the plants, stomata are present on stems and other parts of plants.  

Types of Stomata:

There are five types of stomata and they are mainly classified based on their number and characteristics of the surrounding subsidiary cells. Listed below are the different types of stomata.

Anomocytic Stomata:

They are surrounded by epidermal cells, which have a fixed shape and size.  The stomata appear to be embedded in epidermal cells. There is no definite number and arrangement of cells surrounding the stomata.

Anisocytic Stomata:

Stomata are surrounded by three subsidiary cells having unequal sizes, one is smaller compared to the other two.

Diacytic Stomata :

The stomata are surrounded by a pair of subsidiary cells that are perpendicular to the guard cell.

Paracytic Stomata:

The stomata are continuously surrounded by two subsidiaries, which are arranged parallel to the stomatal pore and the guard cells.

Gramineous Stomata:

Each stoma possesses two guard cells, which are shaped like dumbbells. The subsidiary cells are parallel to the guard cells. The guard cells are found narrow in the middle and wider at the ends.

Hence, the correct answer is option (4) "5".

 

Write a program of "simple calculator" in python language

 # Program make a simple calculator

# This function adds two numbers

def add(x, y):

    return x + y

# This function subtracts two numbers

def subtract(x, y):

    return x - y

# This function multiplies two numbers

def multiply(x, y):

    return x * y

# This function divides two numbers

def divide(x, y):

    return x / y

print("Select operation.")

print("1.Add")

print("2.Subtract")

print("3.Multiply")

print("4.Divide")

while True:

    # Take input from the user

    choice = input("Enter choice(1/2/3/4): ")

    # Check if choice is one of the four options

    if choice in ('1', '2', '3', '4'):

        num1 = float(input("Enter first number: "))

        num2 = float(input("Enter second number: "))

        if choice == '1':

            print(num1, "+", num2, "=", add(num1, num2))

        elif choice == '2':

            print(num1, "-", num2, "=", subtract(num1, num2))

        elif choice == '3':

            print(num1, "*", num2, "=", multiply(num1, num2))

        elif choice == '4':

            print(num1, "/", num2, "=", divide(num1, num2))

        break

    else:

        print("Invalid Input")