Datasets:

InfoBayAI
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English_QA_Dataset

Sno int64 1 50	Question stringlengths 0 618	Options stringlengths 36 2.29k	Answer stringclasses 4 values	Explanation stringlengths 4 4.08k	Subject stringclasses 4 values
1	Which one of the following aspects is an exclusive characteristic of living things ?	{'A': 'Isolated metabolic reactions occur in vitro', 'B': 'increase in mass from inside only', 'C': 'Perception of events happening in the environment and their memory', 'D': 'Increase in mass by accumulation of material both on surface as well as internally'}	C	Sol: 1. Metabolism is the process by which all living things assimilate energy and use it for various purposes such as growth, movement, locomotion etc. This mechanism of metabolism is not shown by nonliving objects. However, some of the metabolic reactions can be carried out in a cell-free system or outside the cells....	Biology
2	Which of the following statement is incorrect?	{'A': 'The Darwin’s variation are large and do not require accumulate while mutation are small and need accumulation.', 'B': 'Variation are directional while mutation are directionless and they appear in all possible direction', 'C': 'The single step large mutation which can cause speciation was named as saltation by d...	A	Sol: The Darwin’s variations are small, which means they cannot bring sudden change in the life forms and they can bring change when they accumulates. While mutation are large and they bring sudden change even in the span of single generation. Mutation do not need accumulation.	Biology
3	Assertion Assertion E.coli having pBR322 with DNA insert at BamHI site cannot grow in medium containing tetracycline. Reason Reason Recognition site for BamHI is present in tetR region of pBR322.	{'A': 'Both Assertion and Reason are correct and Reason is the correct explanation for Assertion', 'B': 'Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion', 'C': 'Assertion is correct but Reason is incorrect', 'D': 'Both Assertion and Reason are incorrect'}	A	Sol: pBR322 has recognition sites for several commonly used restriction enzymes. Recognition site for BamHI is present in tetr region i.e., region responsible for tetracycline resistance. When an insert is added at the BamHI recognition site the gene for tetracycline resistance becomes non-functional and the recombinan...	Biology
4	Polyomavirus (a DNA virus) causes tumors in "nude mice" (nude mice do not have a thymus, because of a genetic defect) but not in normal mice. The best interpretation is that	{'A': 'Macrophages are required to reject polyomavirus-induced tumors.', 'B': 'Natural killer cells can reject polyomavirus-induced tumors without help from T lymphocytes.', 'C': 'T lymphocytes play an important role in the rejection of polyomavirus-induced tumors.', 'D': 'B lymphocytes play no role in rejection of pol...	C	Sol: The thymus is the site where the maturation of T-lymphocytes takes place. In nude mice, there is no thymus due to genetic defect. The absence of T-cells is important for the rejection of the tumours caused by the introduction of the polyomavirus. The normal mice with the thymus is not affected by this virus. Thus ...	Biology
5	Which one of the following conditions correctly describes the manner of determining the sex?	{'A': 'Homozygous sex chromosomes (Z Z ) determine female sex in birds.', 'B': 'XO type of sex chromosomes determine male sex in grasshopper.', 'C': "XO condition in humans as found in Turner's syndrome, determines female sex.", 'D': 'Homozygous sex chromosomes (X X ) produce male in Drosophila'}	B	Sol:	Biology
6	Read the following four statements (A-D) about certain mistakes in two of them. (A) The first transgenic buffalo, Rosie produced milk which was human alphalactalbumin enriched. (B) Restriction enzymes are used in isolation of DNA from other macromolecules. (C) Downstream processing is one of the steps of rDNA technolog...	{'A': 'B and C', 'B': 'C and D', 'C': 'A and C', 'D': 'A and B XO type of sex chromosomes determine male sex in grasshoppers. This type of sexdetermination comes under XX-XO type. Its common examples are cockroaches, grasshoppers and bugs. The female has two homomorphic sex chromosomes XX and is homogametic. It produce...	D	Sol:	Biology
7	In which one of the following male and female gametophytes do not have free living independent existence?	{'A': 'Polytrichum', 'B': 'Cedrus', 'C': 'Pieris', 'D': 'Funaria'}	B	Sol:	Biology
8	Which one of the following animals is correctly matched with its one characteristic and the taxon? Animal Characteristic Taxon	{'A': 'Millipede Ventral nerve Arachnida cord', 'B': 'Sea anemone Triploblastic', 'C': 'Silverfish Pectoral and Chordata pelvic fins', 'D': 'Duckbilled Oviparous Mammalia platypus'}	D	Sol: (d) : In 1997, the first transgenic cow, Rosie, produced human protein enriched milk. The milk contained the human alpha-lactalbumin and was nutritionally a more balanced product for human babies than natural cow-milk. Isolation of DNA from other macromolecule is achieved by treating the bacterial cells/plant or a...	Biology
9	Which of the following structures will not be common to mitotic eell of a higher plant?	{'A': 'Centriole', 'B': 'Spindle fibre', 'C': 'Cell plate', 'D': 'Centromere'}	A	Sol:	Biology
10	Statement 1: Statement 1: A potted plant placed in a window shows bending of stem towards light. Statement 2: Statement 2: This is an example of Teleology.	{'A': 'Both Statement 1 and Statement 2 are true but Statement 2 is not the correct explanation of Statement 1', 'B': 'Both Statement 1 and Statement 2 are true and the Statement 2 is correct explanation of the Statement 1', 'C': 'The Statement 1 is true but the Statement 2 is false', 'D': 'The Statement 1 is false but...	A	Sol: Light stimulates the movement of different plant parts in specific direction. It is called photrotropism. On keeping a potted plant near an open window of a dark room, we (Lepisma) belongs to nonchordata. It is an insect. (a) : The centrioles occur in nearly all animal cells and in motile plant cells, such as zoos...	Biology
11	Statement 1: Statement 1:In fungi like black mould and algae, meiosis occurs directly after zygote formation. Statement 2: Statement 2: Because the organism is haploid and it produces haploid gametes.	{'A': 'Both Statement 1 and Statement 2 are true but Statement 2 is not the correct explanation of Statement 1', 'B': 'Both Statement 1 and Statement 2 are true and the Statement 2 is correct explanation of the Statement 1', 'C': 'The Statement 1 is true, but the Statement 2 is false', 'D': 'The Statement 1 is false bu...	B	Sol: In black moulds like Rhizopus and algae, meiosis occurs directly after zygote formation. Meiosis occurs prior to the germination of the zygospore. Each diploid zygospore undergoes meiosis to form four haploid daughter nuclei. This is a condition similar to that of Spirogyra (An algae. Two of these haploid nuclei a...	Biology
12	Statement 1 : Some bacteria have the capacity to retain Gram stain after treatment with acid alcohol. Statement 2 : They are known as Gram positive as they are attracted towards positive pole under influence of electric current.	{'A': 'Both the statement 1 and the statement 2 are true and the statement 2 is a correct explanation of the statement 1', 'B': 'Both the statement 1 and statement 2 are true but the statement 2 is not a correct explanation of the statement 1', 'C': 'Both the statement 1 and statement 2 are false', 'D': 'The statement ...	D	Sol: The cell wall of Gram-negative bacteria contains alcohol-soluble lipid, while the cell wall of Gram-positive bacteria lacks the lipids. Hence they resist decolourisation and retain the primary stain, appearing violet. Gram-negative bacteria are thus decolorized by organic solvents and therefore, take the counterst...	Biology
13	Statement 1 : Detorsion is the characteristic of mollusca. Statement 2 : Detorison is an arrested stage of torsion.	{'A': 'Both the statement 1 and the statement 2 are true and the statement 2 is a correct explanation of the statement 1', 'B': 'Both the statement 1 and statement 2 are true but the statement 2 is not a correct explanation of the statement 1', 'C': 'The statement 1 is true but the statement 2 is false', 'D': 'Both the...	B	Sol: Torsion or twisting is a process during larval development of gastropods, that rotates the viscero- pallium anti-clockwise brought 180° from its initial position, so that mantle cavity, with its pallial complex, is through in front of the body in adult. Changes occurring in torsion are to certain extent reversible...	Biology
14	Statement 1 : The digestive system is undifferentiated in lampreys. Statement 2 : All digestive glands are present in the digestive glands of lampreys.	{'A': 'Both the statement 1 and the statement 2 are true and the statement 2 is a correct explanation of the statement 1', 'B': 'Both the statement 1 and statement 2 are true but the statement 2 is not a correct explanation of the statement 1', 'C': 'The statement 1 is true but the statement 2 is false', 'D': 'Both the...	B	Sol: Lampreys have undifferentiated gut from intestine to anus.They have no stomach, and oesophagus leads straight into intestine and waste products passes out through cloaca. Digestive glands are well developed which has a bilobed organ, liver surrounds the anterior part of intestine. Gall bladder and bile duct occur ...	Biology
15	Whereas the number of chromosomes is reduced to half in first reduction division of meiosis, then what is the need for second mitotic division?	{'A': 'For the segregation of replicated chromosomes', 'B': 'For equal distribution of haploid chromosomes', 'C': 'Fo rthe formation of four gametes', 'D': 'For the equal distribution of genes on chromosomes'}	A	Sol: The meiosis has two stages of division, meiosis I & meiosis II which are reductional & equational divisions respectively. The first meiotic division leads to the reduction of the chromosome number to half i.e haploid condition, the second meiotic division results in the formation of four different daughter cells w...	Biology
16	An adult animal that possesses bilateral symmetry is most certainly also –	{'A': 'triploblastic.', 'B': 'a deuterostome.', 'C': 'eucoelomate', 'D': 'the product of metamorphosis.'}	A	Sol: The correct answer is triploblastic. triploblastic. Explanation: Explanation: • Bilateral symmetry is most commonly associated with triploblastic animals, which have three primary germ layers: ectoderm, mesoderm, and endoderm. • Not all bilaterally symmetrical animals are deuterostomes, eucoelomates, or the produc...	Biology
17	You are trying to identify an organism. It is an animal, but it does not have nerve or muscle tissue. It is neither diploblastic nor triploblastic. It is probably a	{'A': 'flatworm', 'B': 'jelly', 'C': 'comb jelly', 'D': 'sponge'}	D	Sol: The correct answer is Sponge. Sponge. Explanation: Explanation: • Sponges Sponges are simple animals that do not have true tissues, nervous, or muscle systems. They lack organized tissues and complexity. • Sponges belong to the phylum Porifer Porifera • They are asymmetric, that is, no plane that passes through th...	Biology
18	Select the incorrect statement.	{'A': 'Male fruit fly is heterogametic', 'B': 'In male grasshoppers 50% of sperms have no sex-chromosome', 'C': 'In domesticated fowls, sex of progeny depends on the type of sperm rather than egg', 'D': 'Human males have one of their sex chromosomes much shorter than the other'}	C	Sol: Corr Correct Answer: Option 3 ect Answer: Option 3 Concept: Concept: • The sex chromosome contains the genes genes that determine the sex of an individual whether male or female. • There are two types of sex chromosomes that are generally denoted as the X chromosome and the Y chromosome. • Sex determination is bas...	Biology
19	In which one of the following, both autogamy and geitonogamy are prevented?	{'A': 'Castor', 'B': 'Maize', 'C': 'Wheat', 'D': 'Papaya'}	D	Sol: The correct answer is option 4. option 4. Concept: Concept: • The transfer of pollen grains from anther to the stigma of the same flo same flower or wer or another flo another flower wer is called pollination. • There are three types of pollination: autogamy, geitonogamy, and x autogamy, geitonogamy, and xenogamy....	Biology
20	You are given an unknown plant to study in the laboratory. You find that it has chlorophyll, no xylem. Its multicellullar sex organs are enclosed in a layer of jacket cells. Its gametophyte stage is free living. The plant probabily belongs to	{'A': 'chlorophyceae', 'B': 'bryophyte', 'C': 'pteridophyte', 'D': 'gymnosperm'}	B	Sol: According to the features described, the unknown plant belongs to bryophytes. Bryophytes refer to a group of plants comprising the mosses, liverworts, and hornworts. They do not have a true vascular system and are unable to pull water and nutrients up from the ground at any significant distance. This distinguishes...	Biology
21	What causes a green plant exposed to the light on only one side, to bend toward the source of light as it grows?	{'A': 'Green plants need light to perform photosynthesis.', 'B': 'Green plants seek light because they are phototropic', 'C': 'Light stimulates plant cells on the lighted side to grow faster.', 'D': 'Auxin accumulates on the shaded side, stimulating greater cell elongation there'}	D	Sol: Auxin accumulates on the shaded side due to which it stimulates cell elongation and enlargement. Its movement is polar and passes from shoot tip to region of elongation. It helps in the elongation of both root and shoots. Green plants need light to perform photosynthesis. Green plants seek light because they are p...	Biology
22	Which of the following structures in different group of animals have similar function?	{'A': 'Typhlosole in Earthworm, intestinal villi in Rat, and contractile vacuole in Amoeba.', 'B': 'Nephridia in Earthworm, Malpighian tubules in Cockroach and uriniferous tubules in Rat.', 'C': 'Antennae in Cockroach, tympanum in Frog and Clitellum in Earthworm.', 'D': 'Incisors in Rat, gizzard [proventriculus] in Coc...	B	Sol: Nephridia in Earthworm, Malpighian tubules in Cockroach and uriniferous tubules in Rat are involved in excretion. Typhlosole in Earthworm, intestinal villi in Rat increase the surface area for digestion and absorption of food. Contractile vacuole in Amoeba helps in removing extra water from cell in fresh water. An...	Biology
23	I am a fish like creature and possess neither parapods nor muscular foot. I have neither eyes nor jaws, nor ventral nerve cord. But I have a lot of gill-slits and a notocord. I am a burrowing animal of marine water. I show ciliary mode of feeding. I have flame cells for excretion. On the basis of above features, guess ...	{'A': 'Herdmania', 'B': 'Amphioxus', 'C': 'Balanoglosus', 'D': 'Petromyzon'}	B	Sol: Amphioxus is a marine, burrowing animal. It lives in shallow waters. In the day ... From the pharynx, water passes into the atrium through the gill slits.	Biology
24	Which one of the following NOT correct about PCT?	{'A': 'Lined with simple cuboidal brush border epithelium.', 'B': 'All essential nutrient and 70-80% electrolyte and water are reabsorbed here.', 'C': 'It helps in PH maintenance of body fluid by selection H+ ion and by absorption of HCO3-', 'D': 'It does not help in maintenance of ionic balance of body fluid.'}	D	Sol: PCT also helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate and by absorption of HCO3 – from it.	Biology
25	Which of the following statements is true for Tapetum?	{'A': 'Parietal in origin usually and is the inner most layer of anther wall', 'B': 'Modified endothecium of anther wall', 'C': 'Outer most layer of endothecium', 'D': 'Parietal in origin and is the inner most layer of ovule wall'}	A	Sol: Anther differentiation starts with the appearance of archesporial cells, which divide to generate primary parietal and primary sporogenous cells. The primary parietal cells form two secondary parietal layers. The anther wall comprises an epidermis, endothecium, middle layer and the secretory-type tapetum.	Biology
26	Incorrect statement among the following is	{'A': 'Transcriptionally active chromatin is loosely packed and stains light.', 'B': 'The negatively charged DNA of nucleoid is organized in large loops held by positively charged proteins.', 'C': 'Non histone chromosomal proteins are required for the packaging of chromatins at higher level.', 'D': 'Reversal of central...	D	Sol: Transcriptionally active chromatin that is loosely packed and stains light is known as euchromatin. Reverse central dogma is not universal. It is observed only in Virus. During the reverse flow of information in the central dogma of molecular biology, the information flows from RNA to DNA which then forms RNA agai...	Biology
27	Which of the following differences are incorrect between leading and lagging strands of DNA? Leading Strand Lagging strand (i) It does not require DNA ligase for its growth. DNA ligase is required for joining Okazaki fragments. (ii) Formation of leading strand is slower. Formation of lagging strand is quite rapid. (iii...	{'A': '(ii) and (iv) only', 'B': '(ii), (iii) and (iv) only', 'C': '(ii) and (iii) only', 'D': '(i), (ii) and (iii) only'}	C	Sol: One new strand, which runs 5' to 3' towards the replication fork, is the easy one. This strand is made continuously, because the DNA polymerase is moving in the same direction as the replication fork. This continuously synthesized strand is called the leading strand. The other new strand, which runs 5' to 3' away ...	Biology
28	A patient with chronic pneumonia presents with a challenging case to the physician. The physician decides to check for the pathogen so he could administer the precise therapeutic for the patient’s recovery. Early tests had revealed the presence of Streptococcus pneumoniae. After prolonged trials, the physician discover...	{'A': 'Adaptive radiation', 'B': 'Convergent evolution', 'C': 'Artificial selection', 'D': 'Natural selection by anthropogenic action'}	D	Sol: The selective survival of Haemophilus influenzae and not Streptococcus pneumoniae in presence of antibiotic serves as evidence of Natural Selection by anthropogenic action.	Biology
29	Which of the following statement is NOT incorrect?	{'A': 'Clones are morphologically similar but genetically dissimilar.', 'B': 'The product of binary fission in bacteria are clones.', 'C': 'The two daughter cells produced as a result of mitosis are not clones.', 'D': 'The two plasmid DNA produced as a result of replication from a parent plasmid DNA cannot be termed as...	B	Sol: Asexual reproduction results into production of morphologically and genetically similar organisms, called as clones. Binary fission is a method of asexual reproduction. The morphologically and genetically alike individuals are called clones. Mitosis produces new cells, and replaces cells that are old, lost, or dam...	Biology
30	In rainy season, the population of insects exhibit explosive increase, but in the end they disappear. Which of the following results can be deduced from this statement?	{'A': 'The plants mature and die at the end of the rainy season.', 'B': 'J-type population growth curve is shown by insects.', 'C': 'Population of insect predators shows explosive increase in the end of season.', 'D': 'Insects show S-shaped or sigmoid growth curve.'}	B	Sol: J-shaped curve is observed when the organism grows in a predator-free environment. In this curve, the organism shows explosive increase and then exhibit a sudden crash due to overexploitation of resources and scarcity of food. This curve is shown by the insects in the rainy season. The plants does not mature and d...	Biology
31	Under which of the following conditions will there be no change in there reading frame of following mRNA? 5’AACAGCGGUGCUAUU3’	{'A': 'Insertion of A and G at 4th and 5th positions respectively.', 'B': 'Deletion of GGU from 7th, 8th and 9th positions.', 'C': 'Insertion of G at 5th position.', 'D': 'Deletion of G from 5th position.'}	B	Sol: Deletion of GGU from 7th, 8th and 9th positions will not change the reading frame. It just leads to the formation of one less amino acid. Insertion of A and G at 4th and 5th positions respectively, will change the reading frame as every codon is a sequence of three amino acid. Insertion of G at 5th position will c...	Biology
32	A plant variety has 20 pg DNA and 8 chromosomes in its egg cell, then what was the amount of DNA and its corresponding number of chromosomes in the megaspore mother cell at G1-phase of the same plant variety?	{'A': '80 pg and 8 chromosomes', 'B': '20 pg and 32 chromosomes', 'C': '40 pg and 16 chromosomes', 'D': '80 pg and 16 chromosomes'}	C	Sol: The amount of DNA and corresponding number of chromosomes in a haploid gamete cell is denoted by c and n, respectively, while the amount of DNA and corresponding number of chromosomes of a meiocyte at its G1 stage of cell division is 2c and 2n. So for a plant variety if c value is 20 pg and n is 8 chromosomes, the...	Biology
33	Drosophila flies with XXY genotype are females, but human beings with such genotype are abnormal males. It shows that	{'A': 'Y-chromosome is essential for sex determination in Drosophila', 'B': 'Y-chromosome is female determining in Drosophila', 'C': 'Y-chromosome is male determining in human beings', 'D': 'Y-chromosome has no role in sex detemination either in Drosophila or in human beings'}	C	Sol: 'Genic Balance Theory' was proposed by C.B. Bridges in 1922 for sex determination in Drosophila. It states that ratio of number of X-chromosome to that of complete set of autosome determines the sex of Drosophila. According to this theory of sex determination, if the ratio of X chromosome to total number of sets o...	Biology
34	A normal woman whose father was colour blind, is married to a normal man. The sons would be	{'A': '75% colour blind', 'B': '50% colour blind', 'C': 'all normal', 'D': 'all colour blind'}	B	Sol: If the father of a normal women is colour blind then she will be a carrier of the disease, since she will receive one X chromosome from both her parents and the X chromosome from the father will be diseased. If she now marries a normal man then there will be 50% chances of her sons being colourblind.	Biology
35	Which one of the following groups of structures/organs have similar function?	{'A': 'Nephridia in earthworm, Malpighian tubules in cockroach and urinary tubules in rat.', 'B': 'Typhlosole in earthworm, intestinal villi in rat and contractile vacuole in Amoeba.', 'C': 'Incisors of rat, gizzard (proventriculus) of cockroach and tube feet of starfish.', 'D': 'Antennae of cockroach, tympanum of frog...	A	Sol: Nephridial organs, or nephridia, are excretory structures that evolved in many invertebrates, including flatworms, nemerteans, rotifers, annelids, mollusks, and lancelets. Each nephridial organ consists of simple or branching tubes that typically open to the outside of the body through excretory pores, called neph...	Biology
36	A cell at the telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase stage. There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result ...	{'A': 'Aneuploidy', 'B': 'Polyploidy', 'C': 'Somaclonal variation', 'D': 'Polyteny'}	B	Sol: Polyploid cells have a chromosome number that is more than double the haploid number. Polyploidy is common in nature and provides a major mechanism for adaptation and speciation. The cells of these plants do not form cell plate at the end of cell division thus the cells have an extra number of chromosomes as compa...	Biology
37	Which of the following statements is true?	{'A': 'The direction of osmosis depends upon pressure gradient, but the rate of osmosis is independent of the pressure gradient.', 'B': 'The external pressure applied to help in the diffusion of water is osmotic pressure.', 'C': 'Numerically, the osmotic pressure is greater than osmotic potential.', 'D': 'Osmotic press...	D	Sol: The symbol π is called osmotic pressure and is the negative of Ψs. That is, π has positive values, and Ψs has negative values. "Osmotic pressure" is the term that physical chemists, zoologists, and many others use to denote the effect of dissolved solutes on the free energy of water.	Biology
38	How does pruning of plants promote branching?	{'A': 'The activity of abscisic acid increases.', 'B': 'Axillary buds get synthesized to gibberellin.', 'C': 'The activity of ethylene decreases.', 'D': 'Axillary buds get synthesized to cytokinin.'}	D	Sol: Pruning is cutting off the apical portion of branches to induce growth which in turn lowers the amount of auxin synthesized as shoot apex. The lowered auxin synthesis allows translocation of cytokinins to lateral bud stimulating cell division and growth of lateral branches.	Biology
39	Which one of the following statement is incorrect about menstruation?	{'A': 'The menstrual fluid can easily clot.', 'B': 'The beginning of the cycle of menstruation is called menarche.', 'C': 'During normal menstruation, about 40 ml of blood is lost.', 'D': 'At menopause in females, there is an especially abrupt increase in gonadotropic hormones.'}	A	Sol: Menstrual blood is loaded with anticoagulants that break down the thick blood clots and prevents blood from clotting. This helps in maintaining a free flow consistency. In menstrual blood, plasminogen activators are also present which causes fibrinolysis, which dissolves the clot. Thus, the correct option is 'The ...	Biology
40	Which of the following does not characterise an ecosystem?	{'A': 'During stress conditions, the count of rare species gets reduced in an ecosystem.', 'B': 'An ecosystem is considered to be unstable if its structure and functions remain more or less the same throughout the year.', 'C': 'An ecosystem having high population diversity and low dominance is regarded as less producti...	B	Sol: An ecosystem is an integrated unit or zone of variable size, comprising vegetation, fauna, microbes and the environment. Most ecosystems characteristically possess a well- defined soil, climate, flora and fauna (or communities) and have their own potential for adaptation, change and tolerance. During stress condit...	Biology
41	Which of the following types of disorder is transferred from a phenotypically normal, but carrier female to only some of the male progeny?	{'A': 'Autosomal recessive', 'B': 'Sex-limited dominant', 'C': 'Sex-linked dominant', 'D': 'Sex-linked recessive'}	D	Sol: Sex-linked recessive inheritance is a mode of inheritance in which a mutation in a gene on the X chromosome causes the phenotype to be always expressed in males (who are necessarily homozygous for the gene mutation because they have one X and one Y chromosome) and in females who are homozygous for the gene mutatio...	Biology
42	Which among the following fruits is a simple fleshy, false fruit developing from a syncarpous inferior ovary having axile placentation?	{'A': 'Drupe', 'B': 'Berry', 'C': 'Pepo', 'D': 'Pome'}	D	Sol: A pome is a fleshy fruit with a cartilaginous endocarp derived from an inferior ovary, with the bulk of the fleshy tissue from the outer, adnate hypanthial tissue, as in Malus (apple) and Pyrus (pear). A pome is an accessory fruit composed of one or more carpels surrounded by accessory tissue. The accessory tissue...	Biology
43	Which is of the following statements is incorrect?	{'A': 'Morels and truffles are edible delicacies', 'B': 'Claviceps is a source of many alkaloids and LSD', 'C': 'Conidia are produced exogenously and ascospores endogenously', 'D': 'Yeasts have filamentous bodies with long thread-like hyphae'}	D	Sol: The explanation that said Yeasts have filamentous bodies with long strings like hyphae is bogus. Yeast is a sac parasite. Yeast doesn't have a filamentous structure or hyphae. Yeast is a solitary cell or unicellular living being, known as Saccharomyces cerevisiae. It requires dampness, warmth, and food to flourish...	Biology
44	Choose the correct statement about the transmission of impulses:	{'A': 'At electrical synapses, the membranes of pre-and post-synaptic neurons are situated far from each other.', 'B': 'The synaptic vessels fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft.', 'C': 'The released neurotransmitters bind to their specific receptors, present on the pr...	B	Sol: At electrical synapses, the membranes of pre-and post-synaptic neurons are in very close proximity. Electrical current can flow directly from one neuron into the other across these synapses. Impulse transmission across an electrical synapse is always faster than that across a chemical synapse. The axon terminals c...	Biology
45	Read the following statement and find out the incorrect statement:	{'A': 'The altered understanding of the plant kingdom describes that Fungi and members of the Monera and Protista having cell walls have now been excluded from Plantae', 'B': "Cyanobacteria that are also referred to as blue-green algae are not 'algae' any more", 'C': 'Numerical taxonomy is based on chromosome number st...	C	Sol: The branch of science that deals with the naming and classification of living organisms on the basis of similarities and dissimilarities. is called taxonomy. Numerical taxonomy based on old observable characteristics. It is done by using computers. Numbers and codes are assigned to those characteristics and then p...	Biology
46	Which one of the following pairs of plant structures has haploid number of chromosomes?	{'A': 'Nucellus and antipodal cells', 'B': 'Egg nucleus and secondary nucleus', 'C': 'Megaspore mother cell and antipodal cells', 'D': 'Egg cell and antipodal cells'}	D	Sol: Megaspore is the initial cell or beginning of the female gametophyte or embryo sac. The nucleus of the megaspore undergoes divisions and gives rise to the embryo sac or female gametophyte, which is called megagametogenesis. During development, the single nucleus of the functional megaspore (of the chalazal end) un...	Biology
47	Which of the following is not true of RNA processing?	{'A': 'Exons are excised and hydrolysed before mRNA moves out of the nucleus', 'B': 'An initial RNA transcript is much longer than the final RNA molecule that may leave the nucleus', 'C': 'RNA splicing may be catalyzed by spliceosomes', 'D': 'Existence of exons and introns may facilitate crossing over between regions o...	A	Sol: In most eukaryotic genes, coding regions (exons) are interrupted by noncoding regions (introns). During transcription, the entire gene is copied into a pre-mRNA, which includes exons and introns. During the process of RNA splicing, introns are excised and hydrolysed before mRNA moves out of the nucleus and exons a...	Biology
48	Which among the following is incorrect about tissues in a plant?	{'A': 'A cluster of cells that have a common origin and work together to achieve a specific function', 'B': 'A plant tissue varies from that of an animal because they vary in their functions and characters', 'C': 'In plant there are two types of tissues, namely, Meristematic and permanent tissues', 'D': 'Secondary Meri...	D	Sol: A cluster of cells that have a common origin and work together to achieve a specific function. A plant tissue varies from that of an animal because they vary in their functions and characters. In plant there are two types of tissues, namely, Meristematic and permanent tissues. Simple permanent tissues comprise of ...	Biology
49	Which of the following is a wrong statement about myelin sheath?	{'A': 'The gaps between two adjacent myelin sheaths are called nodes of Ranvier.', 'B': 'Schwann cells are lipid-rich cells enveloping the axon.', 'C': 'It is mainly secreted by the Schwann cells in the cranial and spinal nerves.', 'D': 'Myelin sheath is seen only as an outermost layer of a whole Schwann cell.'}	D	Sol: Neurilemma (also known as neurolemma, sheath of Schwann, or Schwann's sheath) is the outermost nucleated cytoplasmic layer of Schwann cells (also called neurilemmocytes) that surrounds the axon of the neuron.In the central nervous system, axons are myelinated by oligodendrocytes, thus lack neurilemma. Schwann cell...	Biology
50	Which of the following statements regarding heart sounds is incorr incorrect ect?	{'A': 'Heart sounds are a result of a beating heart and resultant blood flow.', 'B': 'Heart murmurs are pathological noises that results from abnormal blood flow in the heart or blood vessels.', 'C': 'Turbulent blood flow is a result of stenotic (narrowed) valves or blood vessels.', 'D': 'Third heart sound is pathologi...	D	Sol: S3 is a dull, low-pitched sound best heard with the bell placed over the cardiac apex with the patient lying in the left lateral decubitus position. This heart sound when present in a child or young adult implies the presence of a supple ventricle that can undergo rapid filling. A third heart sound occurs early in...	Biology
1	138 g of N2O4 is placed in a 8.2 L container at 300 K . The equilibrium vapor density of the gas mixture is found to be 30.67 . Calculate the total pressure of the gas mixture at equilibrium.	{'A': '4.32 atm', 'B': '5.67 atm', 'C': '6.80 atm', 'D': '7.12 atm'}	C	Sol: The vapor density of the mixture is given as 30.67 . The vapor density is defined as the molar mass of the substance divided by the molar mass of hydrogen gas ( H2 = 2 g/mol ). Therefore, the molar mass of the gas mixture is: Molar mass of the mixture = 30.67 × 2 = 61.34 g/mol The molar mass of N2O4 is: Molar mass...	Chemistry
2	Ammonia prepared by treating ammonium chloride with calcium hydroxide is completely utilized by CoCl2 ⋅6H2O to form a stable coordination compound. Assume that both reactions proceed to 100% completion. If 2670g of ammonium chloride and 1420 g of CoCl2 ⋅6H2O are used, calculate the combined weight (in grams) of calcium...	{'A': '4290', 'B': '4360', 'C': '4280', 'D': '4380'}	D	Sol: 1. Ammonia generation: 2NH4Cl + Ca(OH)2 →2NH3 + CaCl2 + 2H2O 2. Coordination compound formation: CoCl2 ⋅6H2O + 6NH3 →[Co(NH3)6]Cl2 + 6H2O NH4Cl : 53.5 g/mol Ca(OH)2 : 74 g/mol CoCl2 ⋅6H2O : 237 g/mol CaCl2 : 111 g/mol [Co(NH3)6]Cl2 : 267.5 g/mol Mass of NH4Cl : 2670 g Moles of NH4Cl : 2670 53.5 = 50 mol From the r...	Chemistry
3	A compound (X) containing C, H, and O is unreactive towards sodium. It also does not react with Schiff's reagent. On refluxing with an excess of hydroiodic acid. (X) yields only one organic product (Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorous and iod...	{'A': 'Methanol', 'B': 'Ethanol', 'C': 'Propanol', 'D': 'butanol'}	B	Sol: Explanation: Explanation: Given, Overall Reaction is: Comparing both, Compound Z is Ethanol (C Comparing both, Compound Z is Ethanol (C2H5OH) OH)	Chemistry
4	Co-ordination number of Cr in CrCl3.6H2O is six. The volume of 0.1 M AgNO3 needed to precipitate the chlorine in ionisation sphere in 200 ml of 0.01 M solution of the complex cannot be:	{'A': '60 ml', 'B': '40 ml', 'C': '80 ml', 'D': '20 ml'}	D	Sol: (1) [Cr(H2O)6]Cl3 + AgNO3 → 3AgCl 200 × 0.01 × 3 = 0.1 × V. V = 60 ml. (2) [Cr(H2O)5Cl]Cl2.H2O + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl] (NO3)2 number of mole of complex = 200 × 0.01 = 2 required m mole of AgNO3 = 4 m mole = m × Vml ⇒ 4 = 0.1 × Vml or Vml = 40 ml (3) [Cr(H2O)4Cl2]Cl.2H2O + AgNO3 → AgCl + [Cr(H2O)5Cl2] (NO3)...	Chemistry
5	Cr2O2− 7 + X acidic → medium Cr3+ + H2O + Oxidised product of X X in the above reaction cannot be:	{'A': 'Fe2+', 'B': 'SO32-', 'C': 'SO42-', 'D': 'C2O42-'}	C	Sol: Given redox reaction is: Dichromate ion Cr2O72- is a strong oxidising agent as it oxidation number decreases from +6 to +3. Thus, it can oxidise X. A. Fe2+ Cr2O72- oxidises ferrous (Fe2+) to ferric (Fe3+) ion as shown below: B. SO32- Cr2O72- oxidises sulphite (SO32-) to sulphate (SO42-) as shown below: C. It oxidi...	Chemistry
6	A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted to CO on the addition of graphite. Calculate the equilibrium constant at a total pressure of 0.8 atm?	{'A': '1.8 atm', 'B': '3 atm', 'C': '0.3 atm', 'D': '0.18 atm'}	A	Sol: Given: Given: Pressure of CO2(g) = 0.5 atm Temperature = 1000 K Total pressure = 0.8 atm We have to find the value of equilibrium constant. Equilibrium between CO2(g) and CO(g) is given as: where, α = Change in pressure Total pressure at equilibrium is given as: P = Pco2(g) + Pco(g) ........ (i) On substituting th...	Chemistry
7	5.1g NH4SH is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327°C is (R = 0.082 L atm mol-1 K-1, molar mass of S = 32g mol-1, molar mass of N = 14g mol-1)	{'A': '1 x 10-4 atm2', 'B': '4.9 x 10-3 atm2', 'C': '0.242 atm2', 'D': '0.24 x 10-4 atm2.'}	C	Sol: Given: Mass of NH4SH = 5.1 g Volume of flask = 3.0 L Temperature = 327C° + 273 = 600 K % of NH4SH decomposed = 30% R = 0.082 L atm mol-1 K-1 Molar mass of S = 32 g mol-1 Molar mass of N = 14 g mol-1 We have to find the value of Kp of the reaction. Molar mass of NH4SH = 14 + 4 x 1 + 32 + 1 = 51 g mol-1 No. of moles...	Chemistry
8	Sodium nitrate on reduction with Zn in presence of NaOH solution produces NH3. Mass of sodium nitrate absorbing 1 mole of electrons will be:	{'A': '7.750', 'B': '10.625', 'C': '8.000', 'D': '9875'}	B	Sol: Sodium nitrate NaNO3 is reduced to ammonia NH3 by the action of Zn/ NaOH as shown below: NaNO3 Zn/NaOH → NH3 Atomic mass of Na = 23 g Atomic mass of N = 14 g Atomic mass of O = 16 g Molecular mass of NaNO3 = 23 + 14 + 3 × 16 = 85 g Reduction of nitrate to ammonia can be represented as: Zn + 2 − OH →ZnO2− 2 + 2H + ...	Chemistry
9	If 10-5% reactant molecules are crossing over the barrier in transition state at 298 K, the activation energy is:	{'A': '39.94 kJ', 'B': '49.94 kJ', 'C': '79.94 kJ', 'D': '97.97 kJ'}	A	Sol: Given: Percentage of reactant molecules that are crossing over the energy barrier = 10-5% Temperature = 298 K We have to find the activation energy. Let it be Ea. Arrhenius equation is given as: k = Ae−Ea /RT ...(i) where k = Rate constant A = Frequency factor Ea = Activation energy R = Gas constant = 8.3145 JK-1 ...	Chemistry
10	The equivalent conductance of M 32 solution of a weak monobasic acid is 8.0 mho cm2 and at infinite dilution is 400 mho cm2. The dissociation constant of this acid is:	{'A': '1.25 × 10−5', 'B': '1.25 × 10−6', 'C': '6.25 × 10−4', 'D': '1.25 × 10−4'}	A	Sol: Given: Equivalent conductance of solution of a weak acid ( monobasic acid ) Λc = 8.0 mho cm2 The concentration of weak monobasic acid = M 32 Equivalent conductance at infinite dilution, Λ = 400 mho cm2 We have to find the dissociation constant of the acid. Now, the degree of dissociation ∝ is given by: ∝ = λc λ .....	Chemistry
11	Shikhar was to perform a neutralization reaction in laboratory. In order to neutralize 500 ml of 2 M NaOH solution he was supposed to add certain volume of 1 M H3PO3 solution, but by mistake, he added the same volume of water instead. What is the molality of the resulting solution – Assume density of the original solut...	{'A': '0.67', 'B': '1', 'C': '1.2', 'D': '2'}	B	Sol: Wt. of original solution = Density x volume = 1.08 g/cc× 500 ml=540 g Wt. of NaOH + = No. of moles x molecular Wt. = 500×2 1000 × 40 g =40 g Wt. of solvent i.e. water in original solution = 540-40 = 500 g if V2 be the volume of H3PO3 required for neutralizing the solution then- M1V1=M2V2 ⇒ V2 =M1V1/M2= 2×500 1×2 m...	Chemistry
12	Fixed mass of an ideal gas contained in a 24.63 L sealed rigid vessel at 1 atm is heated from −73∘C to 27∘C . Calculate change in Gibbs energy if entropy of gas is a function of temperature as S = 2 + 10−2T(J/K) : (Use 1 atm L = 0.1 kJ)	{'A': '1231.5 J', 'B': '1281.5 J', 'C': '781.5 J', 'D': '0'}	C	Sol: Ideal gas equation PV = nRT At constant volume, P1 T1 = P2 T2 ⇒P2 = 1 × 300 200 = 3 2 and V1 = 24.63L for single phase ∵ dG = Vdp – S dT ΔG = V ⋅ΔP −∫(2 + 10−2T) ⋅dT = 1231.5 −200 −10−2×50,000 2 = 781.5 J	Chemistry
13	To completely neutralize 0.135 g H2AO4 , 2.75×10–3 g equivalents of NaOH is required. What will be the number of neutrons in A (Consider that number of neutrons and protons are equal in A) ?	{'A': '16', 'B': '12', 'C': '9', 'D': '32'}	A	Sol: Let Molar weight of H2AO4 be ′M ′ g equivalents of H2AO4= g equivalents of NaOH (0.135×2) M = 2.75 × 10−3 Molar mass of H2AO4 (M)= 98 g/mol 2 (+1)+M + 4(16)= 98 g/mol M = 32 g/mol ∴Number of protons = Number of neutrons So, Number of neutrons = 32 2 = 16	Chemistry
14	When 100 mL solution of NaOH and Na2CO3 was first titrated with HCl in presence of HPh, 17.5 mL were used till end point is obtained. After this end point MeOH was added and 2.5 mL of same HCl were required to attain new end point. The amount of NaOH in mixture is:	{'A': '0.06 g per 100 mL', 'B': '0.06 g per 200 mL', 'C': '0.05 g per 100 mL', 'D': '0.012 g per 200 mL'}	A	Sol: NaOH xg + Na2CO3 yg In presence of HPh Eq. of NaOH + 1 2 × eq. ofNa2CO3 = Eq. of HCl x 40 × 1 + 1 2 × y 106 × 2 = 17.5 1000 × 1 10 ........ (1) After this MeOH is added 1 2 × eq. ofNa2CO3 = eq. of HCl 1 2 × y 106 × 2 = 2.5 × 1 10 × 1 1000 ...... (2) Placing the value of Eqn. …. (1) x 40 + 2.5 10000 = 17.5 10000 x ...	Chemistry
15	When a 20 mL of 0.08 M weak base BOH is titrated with 0.08 M HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 mL of 0.04 M NaOH is added to the resulting solution? [given log 2 = 0.30 and log3 = 0.48]	{'A': '5.40', 'B': '5.88', 'C': '4.92', 'D': 'None of these'}	B	Sol: BOH + HCl →BCl + H2O at equilibrium point N1V1 = N2V2; V2= 20mL [BCl] = 20×0.04 20+20 = 0.04 pH = 1 2 [pKw−pKb−log(C)]pKb= 5.4 B+ + OH −→BOH; Basic Buffer is formed Initial milli −moles 1.6 0.4 Final mili −moles 1.2 0.4 pOH = pKb+log [B+] [BOH] = 5.4 + log ( 1.2 0.4 )= 5.4 + 0.48 pOH = 5.88	Chemistry
16	The longest wavelength of He atom in Balmer series is ‘x’ What is the longest wave length in the paschen series of Li2+?	{'A': '80x/63', 'B': '63/80x', 'C': '40x/23', 'D': '23/40x'}	A	Sol: For Balmer series, longest wavelength n1 = 2 ⇒n1 = 3 For He+ 1 λ1 = RH × (2)2 × [ 1 22 − 1 32 ] = RH × 4 × 5 36 λ1 = 9 5 RH = x For Paschen series, longest wavelength n1 = 3 ⇒n1 = 4 For Li2+ 1 λ2 = RH × (3)2 × [ 1 32 − 1 42 ] 1 λ2 = RH × 9 × 7 9 × 16 λ2 = 16 7RH = 16 7 × 5 9x λ2 = 80 63 x	Chemistry
17	The shortest wavelength of H atom in the Lyman series is λ1. The longest wavelength in the Balmer series of He+ is -	{'A': '36λ1 5', 'B': '5λ1 9', 'C': '9λ1 5', 'D': '27λ1 5'}	C	Sol: As we know that, ΔE = hc λ Hence, λ will be, λ = hc ΔE For, λ = minimum, i.e. shortest, ΔE = maximum For Lyman series, n = 1 and for ΔEmax transition must be form n = ∞to n = 1 So, 1 λ = RHZ 2 ( 1 n2 1 − 1 n2 2 ) (∵n1 = 1 and n2 = ∞) 1 λ = RH ( 1 12 − 1 ∞2 ) × (1)2 1 λ = RH(1 −0) ⇒1 λ = R × (1)2 ⇒λ1 = 1 R For long...	Chemistry
18	A solution contains Na2CO3 and NaHCO3 . 10 ml of the solution required 2.5 mL of 0.1 M H2SO4 for neutralization using phenolphthalein as indicator . Methyl orange is then added when a further 2.5 mL of 0.2 M H2SO4 was added . The amount of Na2CO3 and NaHCO3 in 1 litre of the solution is :	{'A': '5.3 g and 4.2 g', 'B': '3.3 g and 6.2 g', 'C': '4.2 g and 5.3 g', 'D': '6.2 g and 3.3 g'}	A	Sol: For phenolphthalein : 1 2 Meq. of Na2CO3 = 2.5 × 0.1 × 2 = 0.5 For methyl orange : 1 2 Meq. of Na2CO3 + Meq. of NaHCO3 = 2.5 × 0.2 × 2 = 1.0 On solving the 2 equations , we get the mili gm equivalents and hence the weight can be calculated by the formula ( no. of gm eq. = w/ E ) . Meq of Na2CO3 = 1 Meq of NaHCO3 =...	Chemistry
19	A mixture of CO and CO2 having a volume of 20 ml is mixed with x ml of oxygen and electrically sparked. The volume after explosion is (16 +x) ml under the same conditions. What would be the residual volume if 30 ml of the original mixture is treated with aqueous NaOH?	{'A': '12 ml', 'B': '10 ml', 'C': '9 ml', 'D': '8 ml'}	A	Sol: CO + 1 2 O2 →CO2 CO2 + O2 → No reaction Let a ml CO & b ml CO2 are present in the mixture so, a + b = 20 ) ×2 . . . . . . (1) After the explosion a mL CO2 is formed so, a + b+ x 2 = 16 + x or 2a + 2b −x = 32 . . . . (2) from eq (1) & (2) x = 8 ml Therefore volume of CO in mixture = 8 mL volume of CO2 = 20 −8 = 12 ...	Chemistry
20	A mixture of 200 ml of CO, CH4 and N2 was burnt in excess of O2 resulting in reduction of 13 ml of volume. The residual gas was then treated with KOH solution to show a contraction of 14 ml in volume. Calculate volume of CO, CH4 and N2 in mixture. All measurements are made at constant pressure and temperature.	{'A': '10 ml, 4 ml, 6 ml', 'B': '6 ml, 8 ml, 6ml,', 'C': '6 ml, 9 ml, 5 ml', 'D': '4 ml, 8 ml, 8 ml'}	A	Sol: Let a ml CO, b ml CH4, c ml N2 be present in mixture a + b + c = 20 CO+ 1 2 O2 →CO2 . . . . . . . . (1) V olume of CO = a ∴volume of CO2 = a CH4 + 2O2 →CO2 + 2H2O (l) V olume of CH4 = b ∴V olume of CO2 = b N2 + O2 →No reaction V olume of CO2 formed = V ol absorbed by KOH a + b = 14 mL . . . . . . . (2) Now initial...	Chemistry
21	How many gram moles of HCl will be required to prepare one litre of a buffer solution (containing NaCN and HCN ) of pH 8.5 using 0.10 formula mass of NaCN ? Ka for HCN = 4.1 × 10−10.	{'A': '0.0089 moles', 'B': '0.001 moles', 'C': '0.0050 moles', 'D': '0.002 moles'}	A	Sol: Let a mole of HCl be added. It will combine with NaCN to form HCN NaCN + HCl →NaCl + HCN [NaCN] = (0.01 −a); [HCN] = a Applying the equation 8.5 = log o.01−a a −log 4.1 × 10−10 So, a = 0.01 1.1296 = 0.0089 mole	Chemistry
22	A mixed solution of potassium hydroxide and sodium carbonate 15 mL of an N/20 HCl solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid. The amount of KOH present in the solution is	{'A': '0.014 g', 'B': '0.14 g', 'C': '0.028 g', 'D': '1.4 g'}	A	Sol: Given that a solution contains a mixture of KOH and Na2CO3. During titration with phenolphthalein indicator, complete neutralisation of KOH and half neutralisation of sodium carbonate occurs. During titration with methyl orange indicator, complete neutralisation of KOH and complete neutralisation of sodium carbona...	Chemistry
23	100 mL of a gas at NTP was heated with tin. Tin converted into stannous sulphide and hydrogen was left. This hydrogen when passed over hot CuO, produced 0.081 g of water. If the vapour density of the gas is 17, find its formula.	{'A': 'H2S4', 'B': 'H2S', 'C': 'CH6O', 'D': 'H2O2'}	B	Sol: In this problem it is clear that the gas contains H and S only. Let the formula of the gas be HxSy. Thus HxSy+Sn →SnS+H3 (H2 passed over hot CuO) → H2O Since all the H of the gas converted into H2O, Applying POAC for H atom, we have x × moles of HxSy = 2 × moles of H2O x × 100 22400 = 2 × 0.081 18 ; x = 2. Hence, ...	Chemistry
24	7.5 ml of a gaseous hydrocarbon was exploded with 36 mL of oxygen. The volume of gases on cooling was found to be 28.5 mL, 15 mL of which was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol. If all volumes are measured under the same conditions, deduce the formula of the hydrocarbon .	{'A': 'C2H6', 'B': 'C2H4', 'C': 'C3H6', 'D': 'CH4'}	B	Sol: Volume of CO2 produced = 15 mL (absorbed by KOH) Volume of unused O2 = (28.5 – 15) mL = 13.5 mL. (absorbed by pyrogallol) ∴ volume of O2 reacted with 7.5 mL of hydrocarbon = (36.0 - 13.5) mL = 22.5 mL Thus, CxHy (say) + O2 → CO2 + H2O 7.5 mL 22.5 mL 15 mL or 7.5 moles 22.5 moles 15 moles Applying POAC for C, H and...	Chemistry
25	138 g of N2O4 is placed in a 8.2 L container at 300 K . The equilibrium vapor density of the gas mixture is found to be 30.67 . Calculate the total pressure of the gas mixture at equilibrium.	{'A': '4.32 atm', 'B': '5.67 atm', 'C': '6.80 atm', 'D': '7.12 atm'}	C	Sol: The vapor density of the mixture is given as 30.67 . The vapor density is defined as the molar mass of the substance divided by the molar mass of hydrogen gas ( H2 = 2 g/mol ). Therefore, the molar mass of the gas mixture is: Molar mass of the mixture = 30.67 × 2 = 61.34 g/mol The molar mass of N2O4 is: Molar mass...	Chemistry
26	A saturated solution of silver benzoate (AgOCOC6H5) has pH of 8.6. Ka for benzoic acid is 5.0 × 10−5. The value of Ksp for silver benzoate is (log 2 = 0.3)	{'A': '1.4 × 10−2', 'B': '6.4 × 10−3', 'C': '6.4 × 10−4', 'D': '0.282'}	A	Sol: pH = 8.63, pOH = 5.37, [ ⊖ OH] = 4.3 × 10−6 ∴ [C6H5COO−] = (4.3×10−6)(4.3×10−6) Kh … . . . . . . . . (i) Kh = Kw Ka = 10−14 6.5×10−5 Substituting the value of Kh in equaiton (i) Ksp = [Ag⊕] [C6H5COOΘ] = (0.12)2 = 1.4 × 10−2 Second method: Use formula for the pH of salt of SB/WA C6H5COO⊖+ H2O → C6H5COOH+ ⊖ O H Init...	Chemistry
27	In what ratio should you mix. 0.2 M NaNO3 and 0.1 M Ca(NO3)2 solution so that in resulting solution, the concentration of negative ion in 50% greater than the concentration of positive ions	{'A': 'v1 v2 = 2', 'B': 'v1 v2 = 1 2', 'C': 'v1 v2 = 3', 'D': 'v1 v2 = 1 4'}	B	Sol: Let V1mL of NaNO3 is mixed with V2mL of Ca(NO3)2. mmoles of NaNO3 mixed = 0.2 × V1 mmoles of Ca(NO3)2 mixed = 0.1 × V2 pH = 1 2 (pKw + pKa + log C) 8.63 = 1 2 (14 + 4.187 + log C) ∴ C = 0.12 ⸪ Mol ratio of Ca2+ : NO⊖ 3 in Ca(NO3)2 is 1: 2 ∴ Molarity of NO⊖ 3 in the mixture = [NO⊖ 3 ] of NaNO3 + [NO⊖ 3 ] of Ca(NO3)...	Chemistry
28		{'A': '3.39 kg mol-1', 'B': '33.9 kgmol-1 The freezing point of a solution of acetic acid (mole fraction is 0.02) in benzene is 277.4K. Acetic acid exists partly as a dimer 2A ⇌A2. Determine the equilibrium constant for dimerisation. Freezing point of benzene is 278.4K and (Kf for benzene is 5)', 'C': '6.78 kg mol-1', ...	A	Sol: 2A ⇌A2 c 0 c −c ∝ c ∝/2 Kc = c∝ 2 (c(1−∝))2 = ∝ 2c(1−∝)2 i = 1+ ∝( 1 n −1) (n = 2) i = 1 −∝ 2 m = Xsolute × 100 Msolvent(1−Xsolute) m = 0.02×1000 78(1−0.02) = 0.262m ΔTf = i × kf × m (278.4 −277.4) = (1 −∝ 2 ) × 5 × 0.262 ∝= 0.48 Kc = ∝ 2c(1−∝)2 = 0.48 2×0.268 (1 −0.48)2 Kc = 3.39	Chemistry
29	Two element A and B form compounds having molecular formula AB2 and AB4. When dissolved in 20.0 g of benzene (C6H6), 1.0 g AB2 lowers the freezing point by 2.3°C whereas 1.0 g of AB4 lowers the freezing point by 1.3°C. The molal depression constant for benzene is 5.1 K kg mol–1. The atomic masses of A and B respectivel...	{'A': '74.26, 36.45', 'B': '42.64, 25.57', 'C': '25.00, 75.00', 'D': '25.57, 42.64'}	D	Sol: For AB2, ΔTf = Kf × w2×1000 M2×w1 or, 2.3 = 5.1 × 1×1000 M2×20 ∴M2 = 50×5.1 2.3 = 110.86 Similarly for AB4, 1.3 = 5.1 × 1×1000 M2×20 ∴ M2 = 50×5.1 1.3 = 196.15 Now, molecular weight of AB2 = 110.86, molecular weight of AB4 = 196.15 AB4 = A + 4B = 196.15 … (i) AB2 = A + 2B = 110.86 … (ii) (i) −(ii) gives 2B = 85.29...	Chemistry
30	A mixture which contains 0.550 g of camphor and 0.045 g of an organic solute freezes at 157°C. The solute contains 93.46% of C and 6.54% of H by weight . What is the molecular formula of the compound ? (Freezing point of camphor = 178.4°C and Kf = 37.70)	{'A': 'C10H2O', 'B': 'C12H10', 'C': 'C12H22', 'D': 'C10H18'}	B	Sol: Molality = ΔTf kf = 178.4−157 37.70 = 21.4 37.70 Molecular weight of solute = M Molality = Moles of solute weight of solvent in gm × 1000 = 0.045/M 0.550 × 1000 = 4500 55M Thus, 4500 55M = 21.4 37.70 ; M = 144.14 Now, from the given weight % of C and H we get Moles of C = 93.46 12 = 7.79 Moles of H = 6.54 1 = 6.54...	Chemistry
31	If 0.1M H2SO4(aq.) solution shows freezing point of –0.3906oC then what is the Ka2 for H2SO4? (Assume m = M and Kf(H2O) = 1.86 K kg mol−1 )	{'A': '0.122', 'B': '0.0122', 'C': '1.11 × 10–3', 'D': 'None of these'}	B	Sol: ΔT f = Kf. m. i ⇒0.3906 = i × 1.86 × 0.1, i = 2.1 first step of H2SO4 is strong so H2SO4 0 →H + C + HSO− 4 C(1 −α) HSO− 4 C(1 −α) ⇌H + Cα + SO2− 4 Cα i = C(1−α)+Cα+Cα+C C ⇒2.1; α = 0.1 Now, HSO− 4 ⇌H + + SO2− 4 C(1 −α)(Cα + C)Cα Ka2 = C(1+α)×Cα C(1−α) ⇒ 1.1×0.1×0.1 0.9 = 0.0122	Chemistry
32	A definite amount of gaseous hydrocarbon having (carbon atoms less than 5) was burnt with sufficient amount of O2. The volume of all reactants was 600 mL, after the explosion the volume of the products [(CO2(g) and H2O (g)] was found to be 700 ml under the similar conditions. The molecular formula of the compound is?	{'A': 'C3H8', 'B': 'C3H6', 'C': 'C3H4', 'D': 'C4H10'}	A	Sol: CxHy + (x+ y 4 )O2 → xCO2 + y 2 H2O a (x+ y 4 )a ax ay 2 a + (x+ y 4 )a = 600 ax + a y 2 = 700 6x + 3y = 7 + 7x+ 7y 4 7 +x = 5y 4 x < 5 Put the value if x = 3 10 = 5y 4 ⇒y = 8 Ans. is C3H8	Chemistry
33	0.5 g mixture of K2Cr2O7 and KMnO4 was treated with excess of KI in acidic medium. Iodine liberated required 150cm3 of 0.10 N solution of thiosulphate solution for titration. Find the percentage of K2Cr2O7 in the mixture	{'A': '14.64', 'B': '34.2', 'C': '65.69', 'D': '50'}	A	Sol: Solution: Solution: Reactions of K2Cr2O7 and KMnO4 with KI may be given as: K2Cr2O7 + 7H2SO4 + 6KI →4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2 2KMnO4 + 8H2SO4 + 10KI →6K2SO4 + Cr2(SO4)3 + 7H2O + 3I2 2KMnO4 + 8H2SO4 + 10KI →6K2SO4 + 2MnSO4 + 5I2 Thus, equivalent wt. of K2Cr2O7 = 294 6 = 49 Equivalent wt. of KMnO4 = 158 5 = 31...	Chemistry
34	An ideal gaseous mixture of ethane and ethene (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O mole fraction at C2H6 in the mixture is:	{'A': '0.6', 'B': '0.4', 'C': '0.5', 'D': '0.8'}	B	Sol: C2H6 + 3.5O2 →2CO2 + 3H2O C2H4 + 3O2 →2CO2 + 2H2O Let volume of ethane is x litre, 22.4 × 4 = 3.5x + 3(28 −x) ⟹ x = 11.2 litre At constant T and P, V ∝ n: ∴ Mole fraction of C2H6 in mixture 11.2 28 = 0.4 [ xB = nB/nT ]	Chemistry
35	If 250 mL of N2 over water at 30°C and a total pressure of 740 torr is mixed with 300 mL of Ne over water at 25°C and a total pressure of 780 torr, what will be the total pressure if the mixture is in a 500 mL vessel over water at 35°C. (Given: Vapour pressure (Aqueous tension) of H2O at 25°C, 30°C and 35°C are 23.8, 3...	{'A': '760 torr', 'B': '828.4 torr', 'C': '807.6 torr', 'D': '870.6 torr'}	D	Sol: Ideal gas equation PV=nRT nN2 = ( 708.2 760 ×0.25) 0.0821×303 = 9.36 × 10−3 nO2 = ( 756.2 760 )×0.3 (0.0821)×298 = 0.0122 ntotal moles = 0.02156 Pressure in final vessel = P (ntotal )RT V = 0.02156×00821×308 0.5 P = 1.09 atm or 828.4 torr Ptotal = P(O2+N2) + V . pr. of H2O = 828.4 + 42.2 = 870.6 torr	Chemistry
36	Two mole of an ideal gas is heated at constant pressure of one atmosphere from27°C to 127°C. If Cv,m = 20 + 10−2T JK−1 ⋅mol−1 then q and ∆U for the process are respectively:	{'A': '6362.8 J, 4700 J', 'B': '3037.2 J, 4700 J', 'C': '7062.8 J, 5400 J', 'D': '3181.4 J, 2350 J'}	A	Sol: w = −nRΔT = −2 × 8.314 × 100 = −1662.8J The internal energy at cons tan t volume = ΔU = n ∫Cv,mdT = 2 × ∫(20 + 10−2T)dT = 2 × 20 × (T2 −T1) + 2 × 10−2 × (T 2 2 −T 2 1 ) 2 = 4700 J 4700 = q – 1662.8 ∴ q = 6362.8 J	Chemistry
37	1 mole of a gaseous PQ3 is present in 10L container at pressure of 2.5 atm and at 273 K temperature. On increasing the temperature to 546K , PQ3 dissociates into PQ2(g) and Q2 (g). If the degree of dissociation of PQ3 is 80%, then final pressure at 546 K is:	{'A': '6.26 atm', 'B': '1.25 atm', 'C': '10 atm', 'D': '5 atm First law of thermodynamicsdU = dq + dW'}	A	Sol: Initial concentration of PQ3 = 1 10 mol/L = 0.1 mol/L 2PQ3(g) ⟺2PQ2(g) + Q2(g) Conc. at t = 0 0.1 M 0 0 Conc. at teq (0.1 −0.08)M 0.08M 0.04M Total number of moles per litre at equilibrium = 0.02 + 0.08 + 0.04 = 0.14 P = n v RT = (0.14 × 0.082 × 546)atm = 6.26 atm.	Chemistry
38	Which of the following expression for % ionization of a monoacidic base (BOH) in aqueous solution, at an appreciable concentration is not correct ?	{'A': '100 × √Kb c', 'B': '1 1+10(pKb−pOH)', 'C': 'Kw[H +] Kb+Kw', 'D': 'Kb Kb+[OH−]'}	C	Sol: BOH (aq. ) ⇌B+ (aq. ) + OH−(aq. ) An equilibrium : c(1 – α) cα cα Kb = [B+][OH−] [BOH] ⇒ cα2 (1−α) ; Neglect α w.r.t. 1 as concentration is Appreciable % α = 100 × √Kb c Total dissolved base present in solution as BOH and B+ So α = [B+] [B+]+[BOH] ⇒ 1 1+ [BOH] [B+] = 1 1+ [OH−] Kb ⇒ Kb Kb+[OH−] ⇒ Kb.[H +] Kb[H +]+...	Chemistry
39	At a certain temperature and 2 atm pressure equilibrium constant (Kp) is 25 for the reaction SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g) Initially if we take 2 moles of each of the four gases and 2 moles of inert gas, what would be the equilibrium partial pressure of NO2?	{'A': '1.33 atm', 'B': '0.1665 atm', 'C': '0.133 atm', 'D': 'None of these'}	C	Sol: SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) Initial moles 2 2 2 2 at eqm 2-x 2-x 2+x 2+x (∵ Qp < Kp) Total no. of moles of gases at equilibrium = 2 −x + 2 −x + 2 + x + 2 + x + 2 = 10 [ 2 moles of inert gas also taken] = 8 + 2 = 10 Kp = PSO3.PNO PSO2.PNO2 ⇒ 25 = ( 2+x 10 ×P) 2 ( 2−x 10 ×P) 2 ⇒ 5 = 2+x 2−x ⇒ x = 1.33 = 0.666 1...	Chemistry
40	N2 + 3H2 ⇌ 2NH3 The equilibrium constant for the given reaction is 1.64 × 10–4 atm at 400ºC. The value of equilibrium constant at 500ºC, if the heat of reaction in this temperature range is –105185.8 J, will be	{'A': '0.144 × 10–4 atm', 'B': '0.144 × 103 atm', 'C': '0.144 × 10-2 atm', 'D': '0.144 atm'}	A	Sol: For the reaction, N2 + 3H2 ⇌ 2NH3 Partial pressure of NO2 = xNO2 × Ptotal = 2 −x 10 × Ptotal KP(400ºC) = 1.64 × 10–4 atm, ΔHreaction = -105185.8 J T1= 400 + 273 = 673K T2= 500 + 273 = 773K We know that, R = 8.314J/K/mole On applying the van′t hoff equation logKp2 −logKp1 = ΔH 2.303R ( T2−T1 T1T2 ) logKP1− log (1.6...	Chemistry
41	A reaction system in equilibrium according to reaction 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) in one litre vessel at a given temperature was found to be 0.12 mole each of SO2 and SO3 and 5 mole of O2 . In another vessel of one litre contains 32 g of SO2 at the same temperature. What mass of O2 must be added to this vessel in ord...	{'A': '0.4125 g', 'B': '11.6 g', 'C': '1.6 g', 'D': 'None of these'}	B	Sol: 2SO2(g) + O2(g) ⇌ 2SO3(g) Kc = [SO3]2 [ SO2]2×[O2] Kc = (0.12)2 (0.12)2×5 = 0.2 Another vessel 2SO2(g) + O2(g) ⇌ 2SO3(g) moles at eqm 0.5 - 2x y-x 2x as per given 2x = 20 100 × 0.5 = 0.1 Kc = [SO3]2 [ SO2]2×[O2] Kc = (0.1)2 (0.4)2(y−0.05) = 0.20 y = 0.3625 mole ∴ mass of O2 added =11.6 g	Chemistry
42	Ka for the reaction; Fe3+ (aq. ) + H2O(l)Fe(OH)2+ (aq. ) + H3O+ (aq. ) is 6.5 × 10−3. What is the max. pH value which could be used so that at least 80% of the total iron (III) in a dilute solution exists as Fe3+?	{'A': '2', 'B': '2.41', 'C': '2.79', 'D': '1.59'}	D	Sol: Fe3+ (aq. ) + H2O(l) ⇌ Fe(OH)2+ (aq. ) + H3O+ (aq. ) Ka = [Fe(OH)2+][H3O+] [Fe3+] ⇒6.5 × 10−3 = 0.20 0.80 × [H3O+] ⇒6.5 × 10−3 = 0.20 0.80 × [H3O+] [H3O+] = 26 × 10−3 P H = −log [H +] P H = −log [26 × 10−3] P H = 1.59	Chemistry
43	A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted to CO on the addition of graphite. Calculate the equilibrium constant at a total pressure of 0.8 atm?	{'A': '1.8 atm', 'B': '3 atm', 'C': '0.3 atm', 'D': '0.18 atm'}	A	Sol: Given: Given: Pressure of CO2(g) = 0.5 atm Temperature = 1000 K Total pressure = 0.8 atm We have to find the value of equilibrium constant. Equilibrium between CO2(g) and CO(g) is given as: where, α = Change in pressure Total pressure at equilibrium is given as: P = Pco2(g) + Pco(g) ........ (i) On substituting th...	Chemistry
44	Two liquids X and Y form are ideal solution. At 300 K, vapour pressure of the solution containing 1 mole of X and 3 mole of Y is 550 mm Hg. At the same temperature, if 1 mole of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Find the Vapour pressure (in mm Hg) of X and Y in ...	{'A': '200 and 300', 'B': '300 and 400', 'C': '500 and 600', 'D': '400 and 600'}	D	Sol: Initially Pm = P ∘ X. XX + P ∘ Y . XY 550 = P ∘ X ( 1 1+3 ) + P ∘ Y ( 3 1+3 ) or P ∘ X + 3P ∘ Y = 2200.. . (i) When 1 mole of Y is further added to it Pm = P ∘ X. XX + P ∘ Y . XY 560 = P ∘ X ( 1 1+4 ) + P ∘ Y ( 4 1+4 ) ∴P ∘ X + 4P ∘ Y = 2800.. . (ii) Byeqs. (i)and (ii) P ∘ X= 400 mm and P ∘ Y = 600mm.	Chemistry
45	A mixture of two volatile liquids A and B for 1 and 3 moles respectivley has a V.P. of 300 mm at 27∘C. If one mole of A is further added to this solution, the vapour pressure becomes 290 mm at 27∘C. The vapour pressure of pure A is -	{'A': '250 mm', 'B': '316 mm', 'C': '220 mm', 'D': '270 mm'}	A	Sol: P = xAP0 A + xBP0 A 300 = 1 1+3 P0 A + 3 1+3 P0 R . . . (1) Now, I mole of A is more added. 290 = 2 2+3 P0 A + 3 2+3 P0 B . . . (2) 300 = 0.25PA0 + 0.75P0 B 290 = 0.4P0 A + 0.6P0 B solving, we get P0 A = 250 mm nA = 1; nB = 3; V ⋅P = 300 mm nA = 1; nB = 3; V ⋅P = 300 mm nA = 2; nB = 3, V ⋅P = 290 mm	Chemistry
46		{'A': '120 torr', 'B': '140 torr', 'C': '260 torr', 'D': '20 torr'}	C	Sol: XA ⟶1 ⇒lim XA→1 PA XA = P 0 A P = 130XA +130 tends to P 0 A when XA tends to 1 ⇒P 0 A = 130 + 130 = 260 Hence, choice (c) is correct while (a), (b) and (d) are incorrect.	Chemistry
47	2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms dimer in solution?	{'A': '50%', 'B': '70%', 'C': '99%', 'D': '35%'}	C	Sol: At 300 k, the vapour pressure in torr of mercury of A −B solution is represented by the equation : P = 130 XA + 130 where XA is the mole fraction of A. Then, the value of lim XA→1 PA XA is We know, ΔT = i × Kf × w2×1000 M2×w1 Given : ΔT = 1.62K; Kf = 4.9 kg mol−−1; w2 = 2 g; M2 = 122 g mol−−1; w1 = 25g Substitutin...	Chemistry
48	In the given gaseous reaction initial pressure of A is P0. Assume that this is a first order reaction A(g) → 2B(g) + C(g) + 3D(g). After time t, the pressure of gaseous mixture is found to be Pt then the Rate Constant will be	{'A': 'K = 1 t ln 7P0 5P0−Pt', 'B': 'K = 1 t ln 5P0 6P0−Pt', 'C': 'K = 1 t ln 2P0 3P0−Pt', 'D': 'None of these'}	B	Sol: A(g) → 2B(g) + C(g) + 3D(g) P0	Chemistry
49	The half-life of a first-order reaction is 10 min . In what time, the rate of reaction will decrease from 6.0 × 1021 molecules lit−1 s−1 to 4.5 × 1025 molecules lit−1 min−1 (NA = 6.0 × 1023)	{'A': '10 min', 'B': '20 min', 'C': '40 min', 'D': '30 min'}	B	Sol: For 1st Order reaction t1/2 = 0.693 k k = 0.693 10 = 0.0693 s−1 We know, Rate = K [Reactant] P0 −x 2x x 3x Pt ⇒P0 −x + 2x + x + 3x = P0 + 5x x = (Pt−P0) 5 K = 1 t ln 5P0 5P0−(Pt−P0) K = 1 t ln 5P0 6P0−Pt R1 = 6 × 1021 = 0.0693 [A0] R2 = 4.5 × 1025 = 0.0693[ A] ⇒[A] = 0.64 × 1023 Integrated Equation, log [ [A0] [A]...	Chemistry
50	For cell : Zn\|Zn2+ (0.01M) \|\| HCl \| H2\| Pt. What will be the minimum weight of NaOH required to be added to R.H.S to consume all the H+ present in R.H.S of cell of EMF 0.701 V at 25°C before its use [E 0 Zn2+/Zn = −0.76V ]	{'A': '0.2g', 'B': '0.8g ; ⇒[A0] = 0.86 × 1023', 'C': '1.264g', 'D': '0.4g'}	D	Sol: Ecell = E 0 cell − 0.059 2 log [Zn2+] [H +]2 0.701 = 0.76 −0.059 2 log 0.01 [H +]2 2 = log 0.01 – 2 log [H+] 2 = – 2 – 2 log[H+] ⇒ –2 = log [H+] ⇒ [H+] = 10–2M Equivalent of NaOH = equivalent of H+ WNaOH = 0.01 × 40 = 0.4g	Chemistry

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Multimodal Educational Question Answering Dataset

Overview

The Multimodal Educational Question Answering Dataset is a large-scale educational corpus designed for Vision Language Models (VLMs), Large Language Models (LLMs), Retrieval-Augmented Generation (RAG), document intelligence systems, OCR research, and multimodal reasoning applications.

The dataset combines text, images, mathematical equations, diagrams, charts, tables, and detailed explanations, enabling AI systems to reason across both textual and visual information.

Unlike conventional QA datasets that contain only text, this dataset preserves educational illustrations, scientific figures, mathematical expressions, graphs, and document layouts, making it ideal for next-generation multimodal AI.

Dataset Highlights

Property	Value
Dataset Type	Multimodal Educational QA
Modalities	Text + Images + Equations + Tables + Charts
Languages	12
Formats	JSON & PDF
Image Support	✅
Mathematical Equations	LaTeX & MathML
Multiple Choice Questions	✅
Detailed Explanations	✅
AI Ready	✅

Modalities Included

The dataset combines multiple information sources within a single sample.

Text
Educational Images
Scientific Diagrams
Flowcharts
Tables
Graphs
Mathematical Formulae
Chemical Structures
Physics Diagrams
Biology Illustrations
OCR-readable Documents
Structured Metadata

Supported Languages

English
Hindi
Telugu
Marathi
Bengali
Arabic
Tamil
Gujarati
Kannada
Punjabi
Odia
Malayalam

Subject Coverage

STEM

Mathematics
Physics
Chemistry
Biology
Engineering
Computer Science
Information Technology
Medical Sciences
Environmental Science
Agriculture
General Science

Non-STEM

History
Geography
Economics
Business Studies
Commerce
Law
Political Science
Sociology
Teaching
Communication
General Knowledge
Languages

Sample Dataset Record

{
  "question": "Identify the labeled part of the human heart shown in the diagram.",
  "image": "heart_diagram.png",
  "options": [
    "Left Atrium",
    "Right Atrium",
    "Left Ventricle",
    "Aorta"
  ],
  "answer": "Aorta",
  "explanation": "The labeled structure represents the main artery carrying oxygenated blood from the left ventricle."
}

Dataset Features

Image-grounded question answering
Multiple-choice questions
Rich educational explanations
Embedded diagrams
Scientific illustrations
Mathematical equations
Charts and graphs
OCR-compatible documents
Structured JSON annotations
Human-readable PDF references
Vision-language learning ready
Retrieval-friendly format

AI Tasks Supported

Visual Question Answering (VQA)
Multimodal Question Answering
Document Understanding
OCR
Image Captioning
Chart Understanding
Table Understanding
Diagram Reasoning
Mathematical Reasoning
Scientific Reasoning
Retrieval-Augmented Generation (RAG)
Instruction Tuning
Supervised Fine-Tuning (SFT)
Vision-Language Model Training

Potential Applications

Vision Language Models (VLMs)
Educational AI Tutors
AI Teaching Assistants
Intelligent OCR Systems
Document AI
Academic Search Engines
Digital Libraries
Interactive Learning Platforms
Educational Chatbots
AI Examination Systems
Scientific Document Analysis
Knowledge Retrieval Systems

Industries

Artificial Intelligence
Education Technology (EdTech)
Higher Education
Schools
Research Organizations
Digital Publishing
Healthcare Education
Government Education
Scientific Computing
Enterprise Knowledge Management

Dataset Structure

Multimodal_Dataset/
│
├── PDFs/
│   ├── Mathematics.pdf
│   ├── Biology.pdf
│   ├── Physics.pdf
│   └── ...
│
├── JSONs/
│   ├── Mathematics.json
│   ├── Biology.json
│   ├── Physics.json
│   └── ...
│
├── Images/
│   ├── diagrams/
│   ├── charts/
│   ├── tables/
│   ├── figures/
│   └── illustrations/

Advantages

Large-scale multimodal educational corpus
Supports both text and vision models
Rich reasoning annotations
Multiple educational domains
Mathematical notation preserved
High-quality structured metadata
Ready for enterprise AI workflows
Compatible with Hugging Face, ModelScope, and PyTorch ecosystems

Recommended Models

This dataset is suitable for training or fine-tuning:

Vision Language Models (VLMs)
GPT-style Multimodal Models
LLaVA
Qwen-VL
InternVL
Florence
BLIP-2
IDEFICS
Kosmos
MiniCPM-V
OCR + LLM Pipelines

License

This dataset is released under the Creative Commons Attribution 4.0 International (CC BY 4.0) License.

Users are free to use, modify, distribute, and build upon the dataset with proper attribution.

Citation

@dataset{multimodal_educational_qa,
  title={Multimodal Educational Question Answering Dataset},
  year={2026},
  license={CC BY 4.0}
}

Conclusion

The Multimodal Educational Question Answering Dataset provides a comprehensive resource for developing intelligent AI systems capable of understanding both textual and visual educational content. By combining questions, answers, explanations, images, diagrams, tables, equations, and multilingual content, it supports a wide range of multimodal learning tasks, making it an ideal dataset for training state-of-the-art Vision Language Models, document AI systems, educational assistants, and retrieval-augmented generation pipelines.

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