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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.
| 301. |
Vasa efferentia are the ductules leading from:A. Testicular lobules to rete testisB. Rete testis to vas deferensC. Vas deferens to epididymisD. Epididymis to urethra |
| Answer» Correct Answer - B | |
| 302. |
An atom has electronic configuration `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3P^(6) 3d^(3) 4s^(2)` . In which group wouold it be placed ?A. FifthB. FifteenthC. SecondD. Third |
| Answer» Correct Answer - A | |
| 303. |
Volume occupied by one molecule of water (density = 1 g `cm^(-3)`)A. `3.0xx10^(-23)cm^(3)`B. `5.5xx10^(-23)cm^(3)`C. `9.0xx10^(-23)cm^(3)`D. `6.023xx10^(-23)cm^(3)` |
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Answer» Correct Answer - A Weight of `6.023xx10^(23)` molecules of water = 18 g As volume occupied by `6.023xx10^(23)` molecules of water (dneisty `= "1 g cm"^(-3)`) will be `=(18g)/(1gcm^(-3))` `="18 cm"^(3)"or mL"` So volume occupied by one molecule of water `=(18)/(6.023xx10^(23))=2.988xx10^(-23)` `=3.0xx10^(-23)cm^(3)` |
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| 304. |
Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` isA. `-93" kJ mol"^(-1)`B. `245" kJ mol"^(-1)`C. `93" kJ mol"^(-1)`D. `-245" kJ mol"^(-1)` |
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Answer» Correct Answer - A `H_(2)+Cl_(2)rarr 2HCl` ltBrgt `DeltaH_("reaction")=Sigma(BE)_("reactant")-Sigma(BE)_("product")` `=[(BE)_(H-H)+(BE)_(Cl-Cl)]-[2(BE)_(H-Cl)]` `=434+242-(431)xx2` `=-186kJ` As `DeltaH_("reaction")=-186kJ` So enthalpy of formation of HCl `=(-186kJ)/(2)` `=-93" kJ mol"^(-1)` |
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| 305. |
Which one of the following does not follow the central dogma of molecular biologyA. PeaB. MucorC. ChlamydomonasD. HIV |
| Answer» Correct Answer - D | |
| 306. |
The second maturation division of the mammalian ovum occursA. Shortly after ovulation before the ovum makes entry into the Fallopian tubeB. Until after the ovum has been penetrated by a spermC. Until the nucleus of the sperm has fused with that of the ovumD. in the Grafian follicle following the first maturation division |
| Answer» Correct Answer - B | |
| 307. |
heart wood differs from sapwood inA. Presence of rays and fibresB. Abscence of vessels and parenchymaC. Having dead and non-conducting elementsD. Being susceptible to pests and pathogens |
| Answer» Correct Answer - C | |
| 308. |
Purines found both in DNA and RNA areA. Cytosine and thymineB. Adenine and thymineC. Adenine and guanineD. Guanine and cytosin |
| Answer» Correct Answer - C | |
| 309. |
Satellite DNA is useful tool inA. Organ transplantationB. Sex determinationC. Forensic scienceD. Genetic engineering |
| Answer» Correct Answer - C | |
| 310. |
The one aspect which is not a salient feature of genetic code, is its beingA. DegenerateB. AmbiguousC. UniversalD. Specific |
| Answer» Correct Answer - B | |
| 311. |
When we move from dark to light, we fail to see for some time but soon the visibility become normal. It isA. AccomodationB. AdaptationC. MutationD. Photoperiodism |
| Answer» Correct Answer - B | |
| 312. |
Biochemical Oxygen Demand (BOD) may not be a good index for pollution for water bodies receiving effluents fromA. domestic sewageB. dairy idustryC. petroleum industryD. sugar industry |
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Answer» Correct Answer - C Petrolium industry release inorganic chemical pollutants which can not be oxidized by microbes bionomically. So not be oxidized by microbes biochemical. So not indicates BOD. |
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| 313. |
What is the intensity of sound during normal conversation?A. 10 – 20 decibalB. 30 – 60 decibalC. 70 – 90 decibalD. 120 – 150 decibal |
| Answer» Correct Answer - B | |
| 314. |
Which one of the following statements is not true with respect to viability of mammalian sperm?A. Viability of sperm is determined by its motilityB. Sperms must be concentrated in a thick suspensionC. Sperm is viable for only up to 24 hoursD. Survival of sperm depends on the pH of the medium and is more active in alkaline medium |
| Answer» Correct Answer - C | |
| 315. |
Seminal plasma in human males is rich in –A. fructose and calciumB. glucose and calciumC. DNA and testosteroneD. ribose and potassium |
| Answer» Correct Answer - A | |
| 316. |
Which one of the following correctly described the location of some body parts in the earthworm Pheretima ?A. Ball and socket jointB. Pivot jointC. Hinge jointD. Gliding joint |
| Answer» Correct Answer - 2 | |
| 317. |
Which one of the following statements is true regarding digestion and absorption of food in humans?A. Two pairs of accessory glands in 16–18 segmentsB. Two pairs of testes in 10th and 11th segments.C. Four pairs of spermathecae in 4–7 segmentsD. One pair of ovaries attached at intersegmental septum of 14th and 15 th segments. |
| Answer» Correct Answer - 3 | |
| 318. |
Which one of the following statements about morula humans is correct?A. It has almost equal quantity of cytoplasm as an uncleaved zygote but much more DNAB. It has far less cytoplasm as well as less DNA than in an uncleaved zygoteC. It has more or less equal quantity of cytoplasm and DNA as in uncleaved zygoteD. It has more cytoplasm and more DNA than an uncleaved zygote |
| Answer» Correct Answer - A | |
| 319. |
In a regularly cycling human female, which can be the root cause of menstrual failure?A. Retention of well-developed corpus luteumB. Fertilization of the ovumC. Maintenance of the hypertrophical endometrial liningD. Maintenance of high concentration of sex- hormones in the blood stream |
| Answer» Correct Answer - 2 | |
| 320. |
A fruit developed from hypanthodium inflorescence is calledA. CaryopsisB. HesperidiumC. SorosisD. Syconus |
| Answer» Correct Answer - 4 | |
| 321. |
Foetal ejection reflex in human female is induced by:A. Differentiation of mammary glandsB. Pressure exerted by amniotic fluidC. Release of oxytocin from pituitaryD. Fully developed foetus and placenta |
| Answer» Correct Answer - 4 | |
| 322. |
Which of the following is not correctly matched for the organism and its cell wall degrading enzyme.A. Bacteria-LysozymeB. Plant cells- CellulaseC. Algae-MethylaseD. Fungi - Chitinase |
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Answer» Correct Answer - C Methylase enzyme is used for methylation. |
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| 323. |
Which of the following are likely to be present in deep sea water?A. ArchaebacteriaB. EubacteriaC. Blue-green algaeD. Saprophytic fungi |
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Answer» Correct Answer - A In deep sea water no light is reach the archaebacteria like sulphur bacteria methanomonas are present and their nutritional category is chemoautotrophs or chemosynthetic bacteria. |
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| 324. |
The colonies of recombinant bacteria appear white in contrast to blue colonies of non-recombinant bacteria because ofA. Non-recombinant bacteria containing betagalactosidaseB. Insertional inactivation of alphagalactosidase in non-recombinant bacteriaC. Insertional inactivation of alphagalactosidase in recombinant bacteriaD. Inactivation of glycosidase enzyme in recombinant bacteria |
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Answer» Correct Answer - A Lac operon Z-gene produce beta galactosidase that convert X-gal chromogen into blue colour in non recombinant bacteria if this gene is affected by insertion inactivation this reaction does not take place and bacteria appears white. But this inactivation take place in beta-galactosidase gene not in `alpha`-galactosidase gene in recombinant bacteria so right answer is non recombinant bacteria containing beta galactosidase enzyme that cause blue colour colonies of bacteria. |
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| 325. |
Which one of the following is a vascular cryptogam?A. CedrusB. EquisetumC. GinkgoD. Marchantia |
| Answer» Correct Answer - 2 | |
| 326. |
Selaginella and Salvinia are considered to represent a significant step toward evolution of seed habit becauseA. Embryo develops in female gametophyte which is retained on parent sporophyte.B. Female gametophyte is free and gets dispersed like seeds.C. Female gametophyte lacks archegonia.D. Megaspore possess endosperm and embryo surrounded by seed coat. |
| Answer» Correct Answer - A | |
| 327. |
Which of the following coordination compounds would exhibit optical isomerism?A. Diamminedichloroplatinum (II)B. Trans-dicyanobis (ethylenediamine) chromium (III) chlorideC. Tris – (ethylenediamine) cobalt (III) bromideD. Pentaamminenitrocobalt (III) iodide |
| Answer» Correct Answer - C | |
| 328. |
The rapid change of `pH` near the stoichiometric point of an acid-base titration is the basic of indicator detection. `pH` of the solution is related to the ratio of the concentration of conjugate acid `(Hin)` and base `(In^(-))` forms of the indicator by the expressionA. `"log""[Hln]"/([In^-])=pK_"In"-pH`B. `"log""[Hln]"/([In^-])=pH-pK_"In"`C. `"log" ([In^-])/"[Hln]"=pH-pK_"In"`D. `"log" ([In^-])/"[Hln]"=pK_"In"-pH` |
| Answer» Correct Answer - A::C | |
| 329. |
Which is wrong for cytochrome P–450A. It contains FeB. It concern with oxidationC. It is a pigmentD. It is a coloured cell |
| Answer» Correct Answer - D | |
| 330. |
Anaphase promoting complex (APC) is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in a human cell, which of the following is expected to occurA. Chromosomes will not condenseB. Chromosomes will be fragmentedC. Chromosomes will not segregateD. Recombination of chromosome arms will occur |
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Answer» Correct Answer - C Anaphase Promoting Complex (APC) is a protein necessary for separation of daughter chromosomes during anaphase. If APC is defective then the chromosomes will fail to segregate during anaphase. |
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| 331. |
MALT constitutes about__________ percent of the lymphoid tissue in human bodyA. `50%`B. `20%`C. `70%`D. `10%` |
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Answer» Correct Answer - A MALT is Mucosa Associated Lymphoid Tissue and it constitutes about 50 percent of the lymphoid tissue in human body. |
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| 332. |
Hypersecretion of Growth Hormone in adults does not cause further increase in height, becauseA. Growth Hormone becomes inactive in adultsB. Epiphyseal plates close after adolescenceC. Bones loose their sensitivity to Growth Hormone in adultsD. Muscle fibres do not grow in size after birth |
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Answer» Correct Answer - B Epiphyseal plate is responsible for the growth of bone which close after adolescence so hypersecretion of growth hormone in adults does not cause further increase in height. |
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| 333. |
Which of the following is correctly matched for the product produced by themA. Acetobacter aceti : AntibioticsB. Methanobacterium : Lactic acidC. Penicillium notatum : Acetic acidD. Saccharomyces cerevisiae : Ethanol |
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Answer» Correct Answer - D Saccharomyces cerevisiae is commonly called Brewer’s yeast. It causes fermentation of carbohydrates producing ethanol. |
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| 334. |
Which of the following RNAs should be most abundant in animal cell ?A. r-RNAB. t-RNAC. m-RNAD. mi-RNA |
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Answer» Correct Answer - A rRNA is most abundant in animal cell. It constitutes 80% of total RNA of the cell. |
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| 335. |
In the genetic code dictionary, how many codons are used to code for all the 20 essential amino acids ?A. 20B. 64C. 61D. 60 |
| Answer» Correct Answer - C | |
| 336. |
There exists a close associatin between the algae and the fungus within a lichen. The fungusA. Provides food for the algaB. Provides protection, anchorage and absorption for the algaC. Fixes the atomospheric nitrogen for the algaD. releases oxygen for the alga |
| Answer» Correct Answer - B | |
| 337. |
The input resistance of a silicon transistor is `100 Omega`. Base current is changed by `40 muA` which results in a change in collector current by `2 mA`. This transistor is used as a common-emitter amplifier with a load resistance of `4k Omega`. The voltage gain of the amplifier isA. 3000B. 4000C. 1000D. 2000 |
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Answer» Correct Answer - D `"Current gain "(beta)=(DeltaI_(C))/(DeltaI_(B))=(2xx10^(-3))/(40xx10^(-6))` `beta=50` `"Voltage gain "=beta((R_("out"))/(R_("in")))` `=50((4xx10^(3))/(100))` `=2000` |
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| 338. |
A rectangular block of mass m and area of cross-section A floats in a liquid of density `rho`. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period T. ThenA. ` T prop sqrt(rho)`B. `T prop 1/sqrt(A)`C. `T prop 1/prop`D. `T prop 1/sqrt(m)` |
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Answer» Correct Answer - B Restoring force=`Axrhog=kximpliesk=arhog implies T=2pi sqrt(m)/(Arhog)` |
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| 339. |
The dimensions of `1/2 in_(0) E^(2)` (`in_(0)`: permittivity of free space, E: electric field) is-A. `ML^(2)T^(-2)`B. `ML^(-1)T^(-2)`C. `ML^(2)T^(-1)`D. `MLT^(-1)` |
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Answer» Correct Answer - B `[(1)/(2)in_(0)E^(2)]=["Energy Density "]` `=(ML^(2)T^(-2))/(L^(3))=Ml^(-1)T^(-2)` |
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| 340. |
A man of `50 kg` mass is standing in a gravity free space at a height of `10m` above the floor. He throws a stone of `0.5 kg` mass downwards with a speed `2m//s`. When the stone reaches the floor, the distance of the man above the floor will beA. 9.9 mB. 10.1 mC. 10 mD. 20 m |
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Answer» Correct Answer - B Let distance of man from the floor be (10+x) m. As centre of mass of system remains at 10 m above the floor. So `50 (x ) =0.5 (10)implies x=0.1m` `implies `Distance of the man above the floor =10+0.1 |
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| 341. |
In the circuit shown if a conducting wire is connected between points A and B, the current in this wire will :- A. Flow from A to BB. Flow in the direction which will be decided by th value of VC. Be zeroD. Flow from B to A |
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Answer» Correct Answer - D `V_(A)-V_(B)=[V-((V)/(8)xx4)]-[V-((V)/(4)xx1)]=-(V)/(2)+(V)/(4)=-(V)/(4) implies V_(B)gtV_(A)` |
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| 342. |
In producing chrlorine through electorlysis, 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated ? Electrochemical equivalent for chlorine `=0.367xx10^(-6) kg C^(-1)`A. `1.76xx10^(-3)kg`B. `9.67xx10^(-3) kg`C. `17.6xx10^(-3) kg`D. `3.67xx10^(-3)kg` |
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Answer» Correct Answer - C `m=Zlt=Z((P)/(V))t` `=(0.367xx10^(-6))((100xx10^(3))/(125))(60)` `=17.61xx10^(-3)kg` |
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| 343. |
In producing chrlorine through electorlysis, 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated ? Electrochemical equivalent for chlorine `=0.367xx10^(-6) kg C^(-1)`A. 17.6mgB. 21.3mgC. 24.3mgD. 13.6mg |
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Answer» Correct Answer - A m = ZIt = `(Z) ((P)/(V))(t)=(0.367xx 10^(-6)) ((100)/125)(60) ` =`1.76xx10^(-5)kg=17.6mg` |
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| 344. |
A steady current of `1.5 A` flows through a copper voltameter for 10 min. If the electrochemical equivalent of copper is `30 xx 10^(-5) gC^(-1)`, the mass of copper deposited on the electrode will beA. 0.27 gmB. 0.40 gmC. 0.50 gmD. 0.67 gm |
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Answer» Correct Answer - A `m=ZIt=(30xx10^(-3))(1.5x)10xx60)=0.27`gm |
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| 345. |
A cell can be balanced against `110 cm` and `100 cm` of potentiometer wire, respectively with and without being short circuited through a resistance of `10 Omega`. Its internal resistance isA. 2.0 ohmB. zeroC. 1.0 ohmD. 0.5 ohm |
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Answer» Correct Answer - C `E prop I_(1) and E-(E)/(R+r),r prop l_(2)` `rArr (r+R)/(R)=(l_(1))/(l_(2))` `rArr r=((l_(1))/(l_(2))-1).R` `=(110-100)/(100)xx10=1Omega` |
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| 346. |
A stationary partical explodes into two partical of a masses `m_(1) and m_(2)` which move in opposite direction with velocities `v_(1) and v_(2)`. The ratio of their kinetic energies `E_(1)//E_(2)` isA. `m_2//m_1`B. `m_1//m_2`C. 1D. `m_1v_2//m_2v_1` |
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Answer» Correct Answer - A `m_1v_1=m_2v_2(P_1=P_2)` `E_1/E_2=(1/2m_1v_1^2)/(1/2m_2V_2^2)=((P_1^2)/(2m_1))/((P_2^2)/(2m_2))=m_2/m_1` |
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| 347. |
Two coils of self-inductance `2mH` and `8 mH` are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coil isA. 10 mHB. 6mHC. 4 mHD. 16 mH |
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Answer» Correct Answer - C By using M=K`sqrt(L_(1)L_(2))` here K=1,`L_(1)=2mH` `L_(2)8mHimpliesM=sqrt16=4mH` |
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| 348. |
A monkey of mass `20kg` is holding a vertical rope. The rope will not break when a mass of `25kg` is suspended from it but will break it the mass exeeds `25kg` . What is the maximum acceleration with which the monkey can climb up along the rope? `(g=10m//s^(2))` .A. `5 m//s^2`B. `10 m//s^2`C. `25 m//s^2`D. `2.5 m//s^2` |
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Answer» Correct Answer - D `T_"max"`=25 g , ma=`T_"max"`-mg `rArr a=g/4=10/4=2.5 m//s^2` |
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| 349. |
In Ornithine cycle which one pair of the following wastes as removed from the blood?A. `CO_(2)` and ammoniaB. Ammonia and ureaC. `CO_(2)` and ureaD. Urea and urine |
| Answer» Correct Answer - A | |
| 350. |
In a common emitter transistor transistor amplifier, the audio signal voltage across the collector is `3kOmega`. If current gain is 100 and the base resistance is `2kOmega`, the voltage and power gain of the amplifier areA. 200 and 1000B. 15 and 200C. 150 and 15000D. 20 and 2000 |
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Answer» Correct Answer - C Current gains `(beta)=100` Voltage gain `(A_(V))=beta(R_(C))/(R_(b))` =`100((3)/(2))` `=150` Power gain `=A_(V) beta ` `=150(100)` =15000 |
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