This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 25901. |
Opium is obtained from(A) Papaver somniferum(B) Erythroxylum coca(C) Cannabis Sativa(D) Atropa belladonna |
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Answer» Answer (A) Papaver somniferum |
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| 25902. |
Amplification of gene for interest may be done by(A) MMR(B) PCR(C) MRI(D) All of these |
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Answer» Answer (B) PCR |
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| 25903. |
Gametes are usually(A) Haploid(B) Diploid(C) Polyploid(D) Nulliploid |
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Answer» Answer (A) Haploid |
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| 25904. |
Amplification of gene for interest may be done by- (A) MMR (B) PCR (C) MRI (D) All of these |
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Answer» Amplification of gene for interest may be done by PCR. |
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| 25905. |
What are homologous and analogous organs? Explain by giving examples. |
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Answer» Term Homlogous was introduced by Richard Owen. The organs which have the same fundamental structure but the different in functions are called Homologous organs. These organs follows the same basic plan of organisation during their development. But in adult condition, these organs are modified to perform different functions an adaptation to different environments. The homologus organs are a result of divergent evaluation. Homology indicates common ancestry. Examples:- Fore-limbs of whole, cheetah and man represent the similar structure of bones like Humerus, Redulna etc. Analogous organs are those organs which have similar functions but are defferent in their structural details and origin. Such organs are the result of convergent evolution. Examples- Wings of insects and wings of birds. Fins of fishes and Flippers of whales. |
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| 25906. |
Draw the double helical structure of DNA as proposed by Watson and Crick. |
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Answer» DNA is a genetic material of all organism except RNA viruses. (i) DNA is made up of two polynucleotide chains. The unit of polycleotide is nucleotide. Nucleotide is made up of nitrogenoces bases, pentose sugar and phosphate. (ii) The bases are of two types- pyrines and pyrimidines. (a) Purines- Adenine (A) and Guanine (G) (b) Pyrimidines- Cytosine (C) and Thymine (T) (iii) Sugar is in the form of dexyribose (iv) Phosphate is in the form of orthophosphoric acid. Chemical linkage:- The base is attached to the number carbon of the sugar. The phosphate group is attached to the number 5-carbon of one sugar and number 3-carbon of sugar of adjacent nucleotide with loss of one molecule of water. This bond is called phosphocliester bond which forms polynucleotide. Double helix:- (i) On the basis of works by chargaff and Wilkins, Watson and crick proposed double helix model of DNA. According to this model, DNA is made up of two helical polynucleotides. Both relices are right handed and coiled around the same axis to form a double helix. (ii) The basis are on the inside of helix in pairs arranged in such a fashion that purine of one chain pairs with pyrimmidine of another chain (A -T) and (G - C). (iii) The pair between bases is faciliated by hydrogen bonding. The hydrogen bonds between A = T is two and between G ≡ C is three. That is why DNA with high G - C content has mas melting point. (iv) The distance between two nucleotides is 3.4 Å. One turn of the helix convers a distance of 34 Å, hence 10 base pairs are accommodated per turn of the helix. Since one turn of the helix involves 360° it is evident that one-base-pair at an angle of 36°. These two helices are so coiled as to generate successively a major and a minor groove on each side. (v) The diameter of double helix is about 20 Å. (vi) The two helices of DNA run anti-parallel i.e. one strand in 5'-3' direction and another in 3'-5' direction. This is necessary to present pair of bases in such position to each other as to make possible formation of H-bond. |
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| 25907. |
Define cross-pollination. Write the various agencies of cross-pollination. |
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Answer» Cross pollination- It is the transfer of pollens from one flower to the stigma of another flower belonging to genetically different plant. Cross pollination can occur only with the help of external agents called pollinating agents or agencies. The various agencies of cross pollinations are (i) Wind (ii) Water (iii) Animals |
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| 25908. |
Anemophily is pollination by(a) Birds(b) Bats(c) Snails(d) Wind |
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Answer» Answer (d) Wind |
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| 25909. |
Gametes are usually- (A) Haploid (B) Diploid (C) Polyploid (D) Nulliploid |
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Answer» Gametes are usually Haploid. |
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| 25910. |
Bio reactors provide optimal conditions to produce desired(A) Product(B) Organism(C) Medium(D) All of these |
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Answer» Answer (A) Product |
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| 25911. |
Taichung is a variety of - (A) Rice (B) Wheat (C) Maize (D) Sugarcane |
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Answer» Taichung is a variety of Rice. |
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| 25912. |
Bio reactors provided optional conditions to produce desired- (A) Product (B) Organism (C) Medium (D) All of these |
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Answer» (A) Product. |
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| 25913. |
Explain the causes and effects of deforestation. |
Answer»
Effects of deforestration:
Remedies to control the deforestration:
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| 25914. |
Unit of mass |
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Answer» Answer: kilogram (kg) |
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| 25915. |
Differentiate between covergent evolution and divergent evolution? |
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Answer» Convergent evolution: Different structures evolving for the same function. Eg: Analogous organs Divergent evolution: Same structure developed in different directions for different needs. Eg: Homologous organs |
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| 25916. |
Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour?1) \( Cu +2 H _{2} SO _{4} \rightarrow CuSO _{4}+ SO _{2}+2 H _{2} O \)2) \( S +2 H _{2} SO _{4} \rightarrow 3 SO _{2}+2 H _{2} O \)3) \( C +2 H _{2} SO _{4} \rightarrow CO _{2}+2 SO _{2}+2 H _{2} O \)4) \( CaF _{2}+ H _{2} SO _{4} \rightarrow CaSO _{4}+2 HF \) (NEET-II2016) |
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Answer» Answer: CaF2+H2SO4 →CaSO4 +2HF In this reaction, the oxidation number of Ca,H,S and O are not changed (oxidation number not changed means it is not a redox reaction). Hence, H2SO4 is not acting as an oxidising agent and reducing agent. |
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| 25917. |
Give two characteristics of a good contraceptive. |
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Answer» User friendly,easily available,effective |
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| 25918. |
Mention the different methods in which IUD‘s function as contraceptives. |
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Answer» (a) To delay pregency. (b) To space children. |
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| 25919. |
1. A hormone releasing IUD is2. Do mucor and rhizopus have zoospores?3. In the pressure flow hypothesis osmotic pressure at sink must be?4. The genotypes of a husband and wife are IAIB and IAi Among the blood types of their. The genotypes of a husband and wife are IAIB and IAi. Among the blood types of their children, how many different genotypes and phenotypes are possible?5. What is common to Lantana Eichhornia and African catfish?6. Which of the following techniques are used for induced breeding in fishes?7. Is buna n biodegradable?8. Is the sunflower an example of cypsela?9. Pusa A-4 is resistant to shoot and fruit borer.10. Dumb-bell shaped guard cells are found in |
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Answer» (1) LNG-20 (2) Zoospores are absent in Mucor and Rhizopus (3) In the pressure flow hypothesis, osmotic pressure at sink must be reduced (4) Genotype of Husband:- IAIB Genotype of Wife:- IAI The possible Genotypes are:- IAIA, IAI, IAIB IBI Possible Phenotypes:- A(IAIA, IAI), B(IBI), AB(IAIB) Therefore, there are 4 Genotypes and 3 Phenotypes. (5) They are exotic species (6) The common technique most used for induced breeding in fishes is Hypophysation (7) No, Buna- n is a non biodegrade substance (8) Yes, Sunflower is an example of cypsella (9) The statement is true, as Pusa A-4 is an insect-resistant variety of okra or bhindi. (10) Dumb-bell shaped guard cells are found in Sugarcane and Cereals. |
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| 25920. |
Name four reproduction related problems. |
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Answer» Reproductive related problems are following:– (i) Pregnancy (ii) Delivery (iii) Sexually transmitted disease (iv) Abortion (v) Contraception (vi) Menstrual problems (Any four) |
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| 25921. |
Discuss the characters chosen by Mendel in garden pea. Mention the advantages of selecting this plant for experiment. |
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Answer» Gregor Mendel, conducted hybridisation experiments on garden peas for seven years (1856-1863) and proposed the laws of inheritance. During Mendel’s investigations into inheritance patterns it was for the first time that statistical analysis and mathematical logic were applied to problems in biology. His experiments had a large sampling size, which have greater credibility to the data that he collected. Also, the confirmation of his inferences from experiments on successive generations of his test plants, proved that his results pointed to general rules of inheritance rather than being unsubstantiated ideas. Mendel selected only pea plant for his experiment due to presence of many types of contrasting characters, like- (i) Round and wrinkled seed (ii) Yellow and green seed (iii) Violet and white flower (iv) Inflated and constricted pod (v) Green and yellow pod (vi) Axillary and terminal flower (vii) Tall and dwarf plant Mendel investigated characters in the garden pea plant that were manifested as two opposite traits which allowed him to setup a basic frame work of rules governing inheritance. Mendel conducted such artificial pollination/cross pollination experiments using several true breeding pea lines. Advantages of selecting pea plants for experiment by Mendel:– (c) Due to short height of pea plant, it was easy to do experiment on that. (d) Pea plant had a short life cycle so that results could be had within a year. (e) Pea plant produced many seeds in one generators which helped in drawing correct conclusions. (f) Artificial cross-pollination could be easily achieved because androecium and gynoecium were completely enclosured. (g) Pea plants having each of the seven characters he selected were readily available. (h) Mendel took only one character at one time for his experiment. (i) Pea plants are self pollinated and pure. |
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| 25922. |
What is gene therapy ? Describe it by using the example of adenosine deaminase (ADA) deficiency. |
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Answer» Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child/embryo. Here genes are inserted into a person’s cells and tissues to treat a disease. Correction of a genetic defect involves delivery of a normal gene into the individual or embryo to take over the function of and compensate for the non-functional gene. The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency. This enzyme is crucial for the immune system to function. This disorder is caused due to the deletion of the gene for adnesine deaminase. In some children ADA deficiency can be cured by bone marow transplantation; in others it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. But the problem with both of these approaches that they are not completely curative. As a first step towards gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. However, as these cells are not immortal the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the gene isolate from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure. |
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| 25923. |
How does killer T-cells act? |
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Answer» T-cells attack directly and destroy antigens. In the process, these cells move to the site of invasion and produce chemicals that attract phagocytes and stimulate them so that they can feed more vigorously on antigens. They also produce substances that attract other T-cells. |
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| 25924. |
What is pollination? Give an account of the varied means of cross pollination in plants. |
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Answer» Transfer of pollengrains after sheding from the anther to the stigma of the carpel is called pollination. Pollination in flowering plants occurs by two methods- (1) Self Pollination and (2) Cross Pollination. (1) Self Pollination – Self pollination involves the transfer of pollengrains from the anther of the flower to the stigma of the same flower or another flower of the same plant. (2) Cross Pollination – Cross pollination involves the transfer of pollengrains from the flower of one plant to the stigma of the flower of another plant. Varied means of cross pollination in plants are– (a) Dicliny – Cross pollination always occurs when the plants are unisexual and dioecious. Ex-Papaya (b) Self Sterility – When the stigma of the gynoecium of a flower is not pollinated by pollengrains of the same flower. Ex-Pea, Potato etc. (c) Dichogamy – When maturity of male and female parts of a bisexual flower occurs in different times. It is of two types. (i) Protogyny (wen gynoecious matures first. Ex-Banyan) and (ii) Protandry (When androecium matures first. Ex-Maize) (d) Herkogamy – In some flowers physical barrier present between anther and stigma which avoid self pollination. Ex- Calotropis (e) Heterostyly – The flowers of some plants have different lengths of stamens and styles so that self pollination is not possible. Ex-Primula, Oxalis etc. |
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| 25925. |
Define biodiversity. Describe in brief any three important components of biodiversity. |
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Answer» Biodiversity is the variation of life at all levels of biological organization. In other words variability found in different forms of life like plants, animals and microorganisms and in their genes are called as biodiversity. It voluntarily integrates terresterial, aquatic and ecosystem diversity. There are three basic components of biodiversity:– (1) Genetic diversity, (2) Species diversity and (3) Ecological diversity. (d) Genetic diversity – Genetic diversity is related to the variations of genes within species. A single species might shows high diversity at the genetic level over its distributional range. For example, genetic variations shown by the medicinal plant Rauwolfia vomitoria growing in different Himalayan ranges might be in terms of the potency and concentration of the active chemical (reserpine) that the plant produces. India has more than 50,000 genetically different strains of rice, and 1,000 varieties of mango. (e) Species diversity – The diversity at the species level is called species diversity. For example, the Western Ghats have a greater amphibian species diversity than Eastern Ghats. Similarly, taking example of plants, brinjal (Solanum melanogena), Potato (Solanum tuberosum) and tomato (Solanum lycopersicon) belong to the same genus Solanum but they all too differ at the species level. (f) Ecological diversity – The diversity at the ecosystem level is called ecological diversity. It is related to the different types of ecosystems, e.g., terrestrial (Deserts, rain forest, grass land, etc.) and aquatic (rivers, ponds, lakes, sea, etc.) ecosystems. Ecological diversity is of three types – Alpha (α), Beta (β) and Gamma (γ) diversity (i) Alpha diversity – It refers to the diversity of those organism which belongs to same community and habitat. (ii) Beta diversity – Diversity between diferent communities of same habitat is called beta diversity. If there is more heterogeneity in the habitats in a region or more dissimilarity between communities, there will be more beta diversity. (iii) Gamma diversity – Diversity at landscape level is called gamma diversity. It envolves both types of diversity – alpha and beta. It refers to the diversity of the habitats in the whole geographical region. |
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| 25926. |
Milky water of green coconut is- (a) Liquid female gametophyte (b) Secretion of seedcoat (c) liquid endosperm (d) Degenerated nuclear tissue |
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Answer» Milky water of green coconut is liquid endosperm. |
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| 25927. |
Name the three important components of biodiversity. |
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Answer» Three important components of biodiversity are (a) Genetic Biodiversity (b) Species Biodiversity (c) Ecological Biodiversity |
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| 25928. |
Direction: Answer the following question by selecting the most appropriate option.Which is not a teacher-centered instruction?1. Individualized instruction2. Demonstration3. Modelling4. Lecture |
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Answer» Correct Answer - Option 3 : Modelling During the last few decades, there has been a vast change in the instructional process followed by teachers in the classroom.
There are various teacher-centered instructional methods: Lecture Method:
Demonstration Method:
Individualized instruction:
Learner-centred Method: Modelling:
Hence, modelling is not teacher-centred instruction. |
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| 25929. |
Choose the appropriate intonation.He has passed with distinction in English, however ...1. rising2. falling3. rising-falling4. falling-rising |
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Answer» Correct Answer - Option 4 : falling-rising The correct answer is 'falling-rising'.
Hence, we can conclude that the appropriate intonation for the sentence- He has passed with distinction in English, however, is falling-rising
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| 25930. |
In a listening-speaking assessment activity, indicate how you will assess the following exchange:Teacher: Have you watched any English film lately? Tell your friend about it. Student 1: Of course. Especially animations.Student 2: Me too. I like them a lot.1.Student 1: Incorrect responseStudent 1: Correct response2.Student 1: Correct responseStudent 1: Incorrect response3.Student 1: Incorrect responseStudent 2: Incorrect response4.Student 1: Correct responseStudent 1: Correct response |
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Answer» Correct Answer - Option 3 : Student 1: Incorrect response Student 2: Incorrect response Intonation is the melodic pattern and primarily a matter of variation in the pitch level of the voice, but in the English language, stress, and rhythm are also involved. They convey differences of expressive meaning (e.g., surprise, anger, wariness).
Hence, we can conclude that student 1: incorrect response and student 2: incorrect response. |
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| 25931. |
When we say "survival of the fittest", does it mean thata. Those which are fit only survive, orb. Those that survive are called fitComment. |
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Answer» Those individuals which survive and reproduce in their respective environment are called fit. |
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| 25932. |
When we say "survival of the fittest", does it mean thata. Those which are fit only survive, or b. Those that survive are called fit Comment |
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Answer» Those individuals which survive and reproduce in their respective environment are called fit. |
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| 25933. |
The solubility of Ag2Cro4 at 85°c is 8.0 ×10.5 m. Find the solubility product of Ag2Cro4. |
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Answer» Ag2Cro4 → 2 Ag+ + cro4-2 using law at mass action K = \(\frac {[Ag^+]^2[Cro_4^{-2}}{Ag_2Cro_4}\) = K [Ag2Cro4] = [Ag+]2 [Crp4-2] = Ksp = [Ag+]2 [Cro4-2] (Ksp = K [Ag2Cro4]) Let say solubility of Ag2Cro4 = S M = 8.0 x 10-5m ∴ Ksp = [2s]2 [s] Ksp = 4s3 Ksp = 4 x (8.0 x 10-5)3 Ksp = 2048 x 10-15 Ksp = 2.048 x 10-12 Hence the solubility product at Ag2Cro4 will be 2.048 x 10-12 |
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| 25934. |
Choose the correct answer out of the four available choices given under each question.A glass tube made of pyrex is called aHeating tubeBoiling tubeMelting tubeCracking tubeFrost is the ___________ state of water.SolidLiquidGasVapourFill in the blanks and rewrite the sentences:A molecule of nitrogen consists of __________atoms of element nitrogen only.A compound is formed by a combination of at least _____elements.In solids, the molecules are__________.retain their shape.Water is_____________ |
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Answer» A glass tube made of pyrex is called a heating tube Frost is the solid state of water. A molecule of nitrogen consists of two atoms of element nitrogen only. A compound is formed by a combination of at least two elements. In solids, the molecules are closely packed and retain their shape. Water is fluid. |
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| 25935. |
Name the major nutrients in our food.Name the following:(a) The nutrients which mainly give energy to our body.(b) The nutrients that are needed for the growth and maintenance of our body.(c) A vitamin required for maintaining good eyesight.(d) A mineral that is required for keeping our bones healthy.Name two foods each rich in:(a) Fats(b) Starch(c) Dietary fiber(d) Protein |
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Answer» 1. The major nutrients in our food are named carbohydrates, proteins, fats, vitamins and minerals. 2 a) Carbohydrates b) Proteins c) vitamin A d) Calcium 3) a) Fats:- Cheese, Ghee b) Starch:- Rice, bread c) Dietary fibre:- Spinach, cabbage d) Protein:- Meat, Paneer |
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| 25936. |
If the molar conductivity of a weak monobasic acid at 0.1 M and zero concentration are 15.8 and 226. 4 S `cm^(2) mol^(-1)` respectively calculate the degree of dissociation and the dissociation constant for acid |
| Answer» `z= 0.06988 , K_(a) = 4.872 xx 10^(-4) ` | |
| 25937. |
If the molar ionic conductivities of` K^(+)` and `I^(-) ` at zero concentration are `73.4 ` and `76.3 Omega^(-1) cm^(2) mol^(-1)` respectively calculate the molar conductivity of K1 at zero concentration |
| Answer» ` 149. 7 Omega^(-1) cm^(2) mol^(-1)` | |
| 25938. |
A conductivity cell filled with 0.01 M KC1 gives at `25^(@) C` a resistance of 484 ohms. The conductivity of 0.01 M KC1 at `25^(@) C` is 0.00141 `Omega^(-1) cm^(-1) 0.001 M ` of `NaC1` solution when filled in the same cell gives a resistance of 5490 ohms at `25^(@) C` Calculate the molar conductivity of NaC1 solution . |
| Answer» `^^_(m) =124 .3 Omega^(-1) cm^(2) mol^(-1)` | |
| 25939. |
`1 g` charcoal is placed in `100 mL` of `0.5 M CH_(3)COOH` to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of `CH_(3)COOH` reduces to `0.49`. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal `=3.01xx10^(2)m^(2)//g`.A. `5 xx 10^(-19) m^(2)`B. `5 xx 10^(20) m^(2)`C. `5 xx 10^(-21) m^(2)`D. `5 xx 10^(-23) m^(2)` |
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Answer» Correct Answer - A Initial moles of acetic acid `= (0.5 xx 100)/(1000) = 0.05` Moles of acetic acid left `= (0.49 xx 100)/(1000) = 0.049` Moles of acetic acid absorbed `0.05 - 0.049 = 0.01` mol in a litre Molecules `6.022 xx 10^(23) xx 0.001 = 6.022 xx 10^(2)` in `100 ml` Area occupied by single molecule of acetic acid `= ("Total area")/("no. of molecules") = (3.01 xx 10^(2))/(6.022 xx 10^(20)) = 5 xx 10^(-19) m^(2)` |
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| 25940. |
The energy produced by the de`-`excitation of `(1)/(1000)` moles of `H -` atoms from `1^(st)` excited to ground state is supplied to `2` moles of an ideal gas. `10` litres of that gas at `27^(@)C` are allow to expand reversibly & isothermally. The final volume fo the gas will beA. `15l`B. `- 20l`C. `- 25l`D. `- 30l` |
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Answer» Correct Answer - B `DE = -2.18 xx 10^(-18) [1 - (1)/(4)]` `= -2.18 xx 10^(-18) xx (3)/(4)` J/ atom `= -2.18 xx 10^(-18) xx (3)/(4) xx 6.022 xx 10^(20)` Jouls for `(1)/(100)` moles `= -2.18 xx (3)/(4) xx 6.022 xx 10^(2)J` `= -3940 J` Now for isothermal expansion `DeltaU = 0` `:. dq = -dw` so `w = -2.303 nRT log. (v_(2))/(v_(1))` `- 39400 = -2.303 xx 2 xx 8.314 xx 300 xx "log" (v_(2))/(v_(1))` `"log"(v_(2))/(v_(1)) = (39400)/(11488.3)` `"log" (v_(2))/(10) = 0.3` or `(v_(2))/(10) = 2` `v_(2) = 20` litre |
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| 25941. |
The composition of sample is `F_(0.93)O_(1.00)`. Find the correct code for correct statements. (i) the `%` of `Fe^(+)` by mass is `11.5%` (ii) The ratio of `Fe^(+3) , Fe^(+2)` ion is `0.17` (iii) `Fe^(+3)` is `7.84 gm`.A. I, II & IIIB. I & IIC. I & IIID. II & III |
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Answer» Correct Answer - D Let no. of `Fe^(2+) = x` no. of `Fe^(2+) = 0.93 - x` then `2x + 3 (0.93 - x) = 2` `x = 0.79` |
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| 25942. |
A conductivity cell filled with 0.02 M `AgNO_(3)` Gives at `25^(@) C` resistance of 947 ohms . If the cell constant is 2.3 `cm^(-1) ` what is the molar conductivity of 0.02 M `AgNO_(4) ` at `25^(@) C` ? |
| Answer» Molar conductivity `= ^^_(m) = 121 .5 Omega^(-1) cm^(2) mol^ (-1)` | |
| 25943. |
Calculate the heat of reaction of the following reaction `CH_(3) COOH_((1))+ 2O_(2(g)) to 2CO_(2(g)) + 2H_(2) O_((1))` Given that heats formation of `CO_(2(g)) H_(2) O_((1))` and `CH_(3) COOH_((1))` are `-395 kJ mol^(-1) - 285 .77 kJ mol^(-1) ` and `-490 .85 kJ mol^(-1)` respectively |
| Answer» `871 .69 kJ mol^(-1)` | |
| 25944. |
Calculate `K_(p)` for the reaction `C_(2)H_(4(g)) + H_(2(g)) hArr C_(2) H_(6(g))` `Delta G^(@) =- 100 kJ mol^(-1) " at " 25^(@) C` |
| Answer» `K_(p)` for the reaction `=3.39 xx 10^(17)` | |
| 25945. |
Calculate the heat of combustion of glucose the from the following data : (i) `C_("graphite") + O_(2(g)) to CO_(2(g)) , Delta_(C) H =- 395.00 kJ mol^(-1)` (ii) `H_(2(g)) + .(1)/(2) O_(2(g)) to H_(2) O_((1)) , Delta _(f) H =- 269.4 kJ "mol"^(-1)` (iii) `6C_("graphite ") + 6H _(2(g)) + 3O_(2(g)) to C_(6) H_(12) O_(6(g))` , `Delta H = - 1169. kJ "mol"^(-1)` |
| Answer» `-2816 .6 kJ mol^(-1)` | |
| 25946. |
Calculate the internal energy at 298 K for the formation of one mole of ammonia if the enthalpy change at constant pressure is - 42.0 `kJ mol^(-1)` (Given `: R = 8.314 JK^(-1) mol^(-1)) ` |
| Answer» `Delta U = - 39. 523 KJ` | |
| 25947. |
Calculate the standard N- H bond enthalpy from the following data : `N_(2(g)) + 2H_(2(g)) to 2NH_(3(g)) Delta H^(@) = - 83 kJ` `Delta H^(@) (N= N) = 946 kJ mol^(-1) , Delta H^(@) (H-H) 436 kJ mol^(-1)` |
| Answer» `Delta H_(N- H)^(@) = 389 kJ mol^(-1)` | |
| 25948. |
Determine whether the following reaction is spontaneous under standard conditions. `2H_(2)O_((1)) + O_(2(g)) to 2H_(2) O_(2(1)) Delta H^(@) = + 196 kJ , Delta S^(@) =- 126 JK^(-1)` Does is have a cross - over termperature ? |
| Answer» The reaction is non- spontaneous . There is no cross- over the temperature for the reaction | |
| 25949. |
Calculate `Delta H^(@) ` of the reaction `CH_(4(g)) + O_(2(g)) to CH_(2) O_(g) + H_(2)O_(g)` From the following data : |
| Answer» `Delta H^(@) = - 346 kJ` | |
| 25950. |
`K_(p)` for the reaction `2SO_(2(g)) + O_(2(g)) to 2SO_(3(g)) " is " 7.1 xx 10^(24) ` at `25^(@)C `. Calculate `Delta G^(@)` for the reactions . `(R = 8. 314 KJ^(-1) mol^(-1))` |
| Answer» `Delta G^(@) =- 141.8 kJ mol^(-1)` | |