InterviewSolution
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.
| 551. |
In the given distribution, what is the value of i ?(a) 8 A (b) 0(c) 2 A (d) 5 A |
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Answer» Correct Option (c) 2 A Explanation : Applying Kirchhoff's first law i + 4 + 2 – 5 – 3 = 0 i + 8 – 8 = 0 i = 2A |
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| 552. |
If a body falls in air, the resistance of air depends to a great extent on the shape of body. Three different shapes are given. Identify the combination of air resistance which truly represent the physical situation (cross sectional areas are the same)(a) 1 < 2 < 3(b) 2 < 3 < 1(c) 3 < 2 < 1(d) 3 < 1 < 2 |
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Answer» Correct Option (c) 3 < 2 < 1 Explanation : Figure 3 is stream lined, so air resistance of it will be least. For figure 1 surface area is maximum so air resistance for it is minimum :. Correct order is 3 < 2 < 1 |
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| 553. |
Relative permitivity and permeability of a material area εr and μr respectively. Which of the following values of these quantities are allowed for a diamanetic material(a) εr = 0.5 μr = 1.5 (b) εr = 1.5, μr = 0.5(c) εr = 0.5, μr = 0.5 (d) εr = 1.5, μr = 1.5 |
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Answer» Correct option (b) εr = 1.5, μr = 0.5 Explanation: For diamangnetic material 0 < μn < 1 and for any material εn > 1 so, (b) is correct |
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| 554. |
The kind of epithelium which forms the inner walls of blood vessels is :–[a] cuboidal epithelium[b] columnar epithelium[c] ciliated columnar epithelium[d] squamous epithelium |
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Answer» Correct option [d] squamous epithelium |
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| 555. |
Phylum Mollusca can be distinguished from other invertebrates by the presence of :–[a] bilateral symmetry and exoskeleton[b] a mantle and gills[c] shell and non-segmented body[d] a mantle and non-segmented body |
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Answer» Correct option [d] a mantle and non-segmented body |
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| 556. |
The tyep of epithelial cells which line the inner surface of Fallopian tubes, bronchioles and bronchi are known as :– [a] squamous epithelium [b] ciliated epithelium [c] columnar epithelium [d] squamous epithelium |
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Answer» Correct option [b] ciliated epithelium |
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| 557. |
Which of the following has a three dimensional structure :–[a] Haemoglobin [b] Elastin[c] Collagen [d] Both [a] and [c] |
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Answer» Correct option [a] Haemoglobin |
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| 558. |
In which of the following fishes the males have brood pouch, where eggs laid by the female remain till they hatch :–[a] lung fish [b] climbing perch[c] salmon [d] sea horse |
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Answer» Correct option [d] sea horse |
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| 559. |
Amniocentesis is a process is :– (a) Determine any hereditary disease in the embryo (b) Determine any disease in heart (c) Know about the disease of brain (d) All of the above |
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Answer» Correct Option (a) Determine any hereditary disease in the embryo |
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| 560. |
Excess atmospheric CO2 increases green house effect as CO2 :–(a) Precipitates dust in the atmosphere(b) Reduces atmospheric pressure(c) Is opaque to infrared rays(d) Is not opaque to infrared rays |
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Answer» Correct Option (c) Is opaque to infrared rays |
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| 561. |
Blue baby syndrome is caused by which of the following pollutants :–[a] Cobalt [b] Fluorine [c] Nitrate [d] Cadmium |
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Answer» Correct option [c] Nitrate |
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| 562. |
Total muscle in human body are :–(a) 639 (b) 629(c) 649 (d) 739 |
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Answer» Correct Option (a) 639 |
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| 563. |
Diadelphous condition is found in :–(a) Rosaceae (b) Leguminosae(c) Papilionaceae (d) Cucurbitaceae |
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Answer» Correct Option (c) Papilionaceae |
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| 564. |
Genetic engineering is ;–(a) Making artificial genes(b) Hybridisation of DNA of one organism to that of others(c) Production of alcohol by using microorganisms(d) Making artificial limbs, diagnostic instruments such as ECG, EEG etc. |
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Answer» Correct Option (b) Hybridisation of DNA of one organism to that of others |
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| 565. |
Most abundant organic compound on earth is :–[a] cellulose [b] protein[c] lipids [d] steroids |
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Answer» Correct option [a] cellulose |
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| 566. |
Body temperature is regulated by :–(a) Hypothalamus(b) Cerebrum and hypothalamus(c) Cerebellum(d) medulla oblongata |
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Answer» Correct Option (a) Hypothalamus |
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| 567. |
The fastest enzyme is :–[a] urease [b] carbonic anghydrase[c] trypsin [d] pepsin |
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Answer» Correct option [b] carbonic anghydrase |
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| 568. |
During the process of fertilization, when one nucleus fues with egg and the second with the secondary nucleus, it is called :–(a) Triple fusion(b) Single fertilization(c) Double fertilization(d) Triple fertilization |
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Answer» Correct Option (c) Double fertilization |
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| 569. |
Identify the plant belonging to the reedswamp stage in hydrach succession : –(a) Juncus (b) Sagittaria(c) Salix (d) Trapa |
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Answer» Correct Option (b) Sagittaria |
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| 570. |
Silkworm silk is the product of :–(a) salivary gland of adult(b) salivary gland of larvae(c) salivary gland of pupa(d) salivary gland of egg cells |
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Answer» Correct Option (b) salivary gland of larvae |
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| 571. |
The intermediate compounds, that connect s glycolysis of Kreb's cycle is : –(a) oxalic acid (b) citric acid(c) pyruvic acid (d) acetyl Co–A |
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Answer» Correct Option (d) acetyl Co–A |
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| 572. |
Stomata opens by the :–[a] influx of Ca2+ ions [b] influx of K+ ions[c] outflux of K+ ions [d] influx of Mg+ ions |
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Answer» Correct option [b] influx of K+ ions |
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| 573. |
The stem of submerged hydrophytes is soft and weak due to :–(a) absence of xylem(b) absence of stomatas(c) absence of phloem(d) reduced mechanical tissue and xylem |
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Answer» Correct Option (d) reduced mechanical tissue and xylem |
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| 574. |
Sickle-cell anaemia is :–[a] caused by substitution of valine by glutamic acid in the beta-globin chain of haemoglobin[b] caused by a change in a single base pair of DNA[c] characterised by elongated sickle-like nucleated RBCs[d] an autosomal dominant trait |
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Answer» Correct option [b] caused by a change in a single base pair of DNA |
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| 575. |
The main function of Henle's loop is :–(a) Formation of urine(b) Passage of urine(c) Conservation of water(d) Filtrat ion of blood |
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Answer» Correct Option (c) Conservation of water |
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| 576. |
Gas which is toxic at troposphere while useful at stratosphere is :–[a] CFC [b] CO2 [c] O3 [d] NO2 |
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Answer» Correct option [c] O3 |
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| 577. |
The main role of bacteria in the carbon cycle invovles(a) chemosynthesis(b) digestion or breakdown of organic compounds(c) photosynthesis(d) assimilation of nitrogenous compounds |
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Answer» Correct Option (b) digestion or breakdown of organic compounds |
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| 578. |
The smallest function unit of kidney is :–(a) nephron (b) collecting tube(c) glomerulus (d) Bowman's capsule |
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Answer» Correct Option (a) nephron |
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| 579. |
Vascular cambium of stem is :–(a) partly primary and partly secondary meristem(b) primary meristem(c) secondary meristem(d) intercalary meristem |
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Answer» Correct Option (c) secondary meristem |
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| 580. |
Which typical stage in cell cycle is known for DNA replication :–[a] S-phase [b] G2-phase[c] G0-phase [d] G1-phase |
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Answer» Correct option [a] S-phase |
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| 581. |
Plants having little or no secondary growth are(1) Grasses(2) Deciduous angiosperms(3) Conifers(4) Cycads |
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Answer» The correct option is(1) Grasses Explanation: Secondary growth occurs due to the presence of vascular cambium. Grasses are monocot and lack vascular cambium. Therefore, they do not show secondary growth. Deciduous angiosperms are usually woody dicot plants and show secondary growth. Conifers and cycads are gymnosperms and usually show anomalous secondary growth. |
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| 582. |
Use of bioresources by multinational companies and organisations without authorisation from the concerned country and its people is called(1) Bio-infringement(2) Biopiracy(3) Biodegradation(4) Bioexploitation |
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Answer» Correct Option(2) Biopiracy Explanation: Biopiracy is referred to the use of bioresources bl multinational companies and other organisations without proper authorisation front the countries and people concerned without compensatory payment. Bio-infringement is the commission of a prohibited act with respect to a patented invention without permission from the patent holder. Bio-exploitation means taking advantage of biological resources of other countries without permission. Biodegradation is a biological breakdown of organic material by bacteria, fungi, etc. |
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| 583. |
Sweet potato is a modified(1) Stem(2) Adventitious root(3) Tap root (4) Rhizome |
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Answer» The correct option is(2) Adventitious root Explanation: Sweet potato is a modified adventitious root which is meant for storage of food. It does not assume a definite shape and occurs singly. Tap roots develop from the radicle of the embryo. They gradually become narrow towards the tip. Stem is usually the above-ground erect ascending part of the plant body. It bears leaves and flowers. Rhizome is modified underground stem, e.g, Zingiber Officinale. |
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| 584. |
For the preparation of nitrosyl chloride from NO and Cl2 a mechanism involving steps has been proposed.Step I : NO(g) + Cl2 (s) → NOCl2(g)Step II : 2NOCl (g) + NO(g) → 2NOCl(g)Rate law for the reaction is Rate = k[NO][Cl2]The rate limiting step for this reaction is :–(a) Step I (b) Step II(c) Both (a) & (b) (d) either of the spets I or II |
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Answer» Correct option (a) Step I Explanation: Rate law suggest that rate determining step is ste (I) |
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| 585. |
The structural formula of monomer of polymethymethacrylate (PMMA) is :–(a) CH2 = CHCOOCH3 (c) CH3COOCH = CH2 |
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Answer» Correct Option (b) |
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| 586. |
Which part of male reproductive system is the site of sperm maturation and storage ? (a) Vas deferens (b) Testes (c) Epididymis (d) Seminal vesicle |
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Answer» Correct Option (c) Epididymis |
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| 587. |
An ideal vector must have which among the followingI. An origin of replicationII. Single cloning siteIII. Have a gene encoding transcriptionIV. Have a gene encoding resistance to antibioticChoose the correct combination(a) I and IV (b) III and IV(c) II and IV (d) I and II |
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Answer» Correct Option (a) I and IV |
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| 588. |
Assertion (A) : In human eye, cornea absorb UV-B radiation and a high dose of UV-B causes inflammation of Cornea.Reason (R) : Such exposure may permanently damage the cornea.(a) Both A and R are true and R is the correct explanation of A(b) Both A and R are true and R is not the correct explanation of A(c) A is true but R is false(d) Both A and R is false |
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Answer» Correct Option (a) Both A and R are true and R is the correct explanation of A |
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| 589. |
Haemophilic man marries a normal woman. Their off springs will be all(a) haemophilic (b) boys haemophilic(c) girls (d) Normal |
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Answer» Correct Option (d) Normal |
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| 590. |
With respect to eukaryotic gene regulation, which of the following statements is not true for enhancers?(a) They function in tissue specific manner(b) They function in any orientation(c) They function as promoters(d) They function even when are at a distance from the gene |
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Answer» Answer is : (c) They function as promoters Enhancers are DNA sequences that facilitate the expression of a given gene. They may be located few hundred base pairs away from the gene. Whereas, promoter is a DNA sequence at which RNA polymerase may bind, leading to the initiation of transcription. |
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| 591. |
Oxidative phosphorylation means :–(a) Photo-oxidation during photosynthesis(b) Photolysis of water(c) ATP production in respiration(d) Anerobic respiration |
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Answer» Correct Option (c) ATP production in respiration |
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| 592. |
Which of these belongs to same family :–(a) Cotton, hemp, china rose, lady's finger(b) Cotton, hemp, china rose, brinjal(c) Cotton, hemp, china rose, tobacco(d) Cooton, hemp, china rose, rose |
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Answer» Correct Option (a) Photo-oxidation during photosynthesis |
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| 593. |
Chemonasty in Mimosa pudica is due to :–[a] chloroform [b] florigen[c] auxin [d] gibberelline |
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Answer» Correct option [a] chloroform |
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| 594. |
Identify the cells whose secretion protects the lining of gastro-intestinal tract from various enzymes.(1) Chief Cells(2) Goblet Cells(3) Oxyntic Cells(4) Duodenal Cells |
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Answer» The correct option is (2) Goblet Cells. Explanation: Goblet Cells - The intestinal mucosal epithelium has goblet cells which secrete mucus. - wherein The secretions of the brush border cells of the mucosa along with the secretions of the goblet cells constitute the intestinal juice or succus entericus Secretion of globet cells i.e. mucus protects the GI tract from various enzyme. |
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| 595. |
The human biochemical disorder Tay-Sachs disease is an example of(a) incomplete dominance(b) codominance(c) epistasis(d) multiple alleles |
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Answer» Answer is : (a) incomplete dominance Tay-Sachs disease is an example of incomplete dominance. In this disease, mutations in the gene coding for an enzyme, hexosaminidase that causes neurological dysfunction. This enzyme breaks down lipid byproducts gangliosides in the cell’s lysosomes. In Tay-Sachs disease, homozygous recessive individuals are severely affected with a fatal lipid-storage disorder, whereas, heterozygous individuals have only 50% of enzyme activity found in normal individuals. |
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| 596. |
If λ1 and λ2 are the wavelengths of the first members of the Lyman and Paschen series respectively, then λ1 : λ2 is(a) 1 : 3(b) 1 : 30(c) 7 : 50(d) 7 : 108 |
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Answer» Answer is : (d) 7 : 108 For first line of Lyman series, n1 = 1 and n2 = 2 \(\therefore\frac{1}{\lambda_1}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)\) \(=R\left(1-\frac{1}{4}\right)=\frac{3R}{4}\) For first line of Paschen series, n1 = 3 and n2 = 4 \(\therefore\frac{1}{\lambda_2}=R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)\) \(=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{144}\) \(\frac{\lambda_1}{\lambda_2}=\frac{7R}{144}\times\frac{4}{3R}=\frac{7}{108}\) |
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| 597. |
Match the following columns :Column–IColumn–IIA. Azoic 1. Era of ancient lifeB. Proterozoic 2. Era of no lifeC. Paleozoic3. Era of medieval lifeD. Mesozoic4. Era of early lifeABCD(a)1342(b)2413(c)1234(d)4321 |
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Answer» Correct Option (b) |
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| 598. |
Hassal's bodies which area potent souce potent source of the cytokine TS2P, are found in one of the following organ : (a) liver (b) thymus (c) adranal (d) thryoid |
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Answer» Correct Option (b) thymus |
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| 599. |
If your Renal Plasma Flow is 650 mL/min and Glomerular Filtratino Rate is 125 mL/min. Calculate your Filtration Fraction : (a) Given informatons are not complete (b) 0.19 (c) 0.5 (d) 5.5 |
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Answer» Correct Option (b) 0.19 |
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| 600. |
Match the corresponding entries of Column I with Column II and choose the correct option from the codes given belowColumn IColumn IIA.Astigmatism1.Convex lensB.Hypermetropia2.Concave lensC.Myopia3.Cylindrical lensCodesABC(a)321(b)312(c)123(d)213 |
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Answer» Answer is : (b)
(A) Astigmatism-It can be corrected with the use of cylindrical lens. (B) Hypermetropia-To overcome from this problem a convex lens is used. (C) Myopia-It is also know by short sightedness and can be corrected with the use of concave lens. |
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