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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.
| 151. |
A saample of radioactive elements contains `4xx10^(10)` active nuclei. If half-life of element is `10` days, then the number of decayed nuclei after `30` days isA. `0.5xx10^16`B. `2xx10^16`C. `3.5xx10^16`D. `1xx10^16` |
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Answer» Correct Answer - C `t=nT, X=X_0/2^n, n=t/T=30/10=3` Active nuclei X =`(4xx10^16)/"(2)"^3` and decayed nuclie X =`(X_0-X)=3.5xx10^16` |
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| 152. |
Two bar magnets having same geometry with magnetic moments M and 2M, are firstly placed in such a way what their similar poles are same side then its time period of oscillation is `T_1`. Now the polarity of one of the magnet is reversed then time period of oscillation will be:-A. `T_1 lt T_2`B. `T_1=T_2`C. `T_1 gt T_2`D. `T_2=oo` |
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Answer» Correct Answer - A `T=2pisqrt(1/"MB")rArr T prop 1/sqrtM` case I:`M_1=2M+M` Case II: `M_2=2M-M` `T_1/T_2 =sqrt(M/(3M))=1/sqrt3rArr T_2=sqrt3T_1` |
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| 153. |
The activity of a radioactive sample is measures as `N_0` counts per minute at `t = 0` and `N_0//e` counts per minute at `t = 5 min`. The time (in minute) at which the activity reduces to half its value is.A. `log _(e) 2//5`B. `(5)/(log_(e)2)`C. `5 log _(10)2`D. `5 log _(e) 2` |
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Answer» Correct Answer - D `N=N_(0)e^(-lamdat) implies (N_(0))/(e )=N_(0)e^(-lamda(5))implies lamda =(1)/(5)` `Now (N_(0))/(2)=N_(0)e^(-lamda(t)) implies t=(1)/(lamda)ln2=5ln2` |
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| 154. |
A sample of radioactive element has a mass of `10 g` at an instant `t=0`. The approximate mass of this element in the sample after two mean lives isA. 1.35 gmB. 2.50 gmC. 3.70 gmD. 6.30 gm |
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Answer» Correct Answer - A `N=N_0e^(-lambdat) rArr m=m_0e^(-lambdat)=m_0e^(-lambda(2//lambda))` `10/e^2`=1.35 gm |
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| 155. |
In the circuits shown below, the readings of the voltmeters and the ammeters will be: A. `V_(2)gtV_(1)` and `i_(1)gti_(2)`B. `V_(2)gtV_(1)` and `i_(1)=i_(2)`C. `V_(1)=V_(2)` and `i_(1)gti_(2)`D. `V_(1)=V_(2)` and `i_(1)=i_(2)` |
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Answer» Correct Answer - D Resistance for ideal voltemeter `=infty` Resistance for ideal ammeter =0 `(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))` `(1)/(R_(eq))=(1)/(10)+(1)/(infty)` `(1)/(R_(eq))=(1)/(10)+0` `R_(eq)=10 Omega` `i_(1)=(V)/(R)=(10)/(10)=1A` `v_(1)=10V` In `II^(nd)` circuit `(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))` `(1)/(R_(eq))=(1)/(10)+(1)/(10+infty)` `(1)/(R_(eq))=(1)/(10)+0` `R_(eq)=10 Omega` `i_(2)=(1-0)/(10)=1A` `v_(2)=10V` |
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| 156. |
the term' Antibiotic " was coined byA. Edward JennerB. Louis PasteurC. Selman waksmanD. Alexander Fleming |
| Answer» Correct Answer - C | |
| 157. |
Select the incorrect statement.A. Inbreeding helps in accumulation of superior genes and elimination of undesirable genesB. Inbreeding increases homozygosity.C. Inbreeding is essential to evolve purelines, in any animal.D. Inbreeding selects harmful recessive gene that reduce fertility and productivity. |
| Answer» Correct Answer - D | |
| 158. |
Select the correct sequence of organs in the alimentary canal of cockroach starting from mouth:A. Pharynx `to` Oesophagus `to` Ileum `to` Crop `to` Gizzard `to` Colon `to` RectumB. Pharynx `to` Oesophagus `to` Crop `to` Gizzard `to` Ileum `to` Colon `to` RectumC. Pharynx `to` Oesophagus `to` Gizzard `to` Crop `to` Ileum `to` Colon `to` RectumD. Pharynx`to` Oesophagus `to` Gizzard `to` Ileum `to` Crop `to` Colon `to` Rectum |
| Answer» Correct Answer - B | |
| 159. |
Plant group with largest ovule, largest tree, and largest gamets:A. GymnospermB. AngiospermC. BryoBIOtaD. PteriodoBIOta |
| Answer» Correct Answer - A | |
| 160. |
Which one of the following hydrolyses internal phosphodiester, bonds in a polynucleotide chainA. LipaseB. proteaseC. ExonucleaseD. Endonuclease |
| Answer» Correct Answer - D | |
| 161. |
What is true for a cyclic processA. W=0B. `DeltaE=0`C. `DeltaH=0`D. `DeltaEne0` |
| Answer» Correct Answer - B::C | |
| 162. |
Increasing order of bond length isA. `NO^(-)ltNOltNO^+ltO_2^-`B. `O_2^(-) ltNOltNO^(-)ltO^+`C. `O_2^(-)ltNO^(-)ltNOltNO^+`D. `NO^+ltNOltNO^(-)ltO_2^(-)` |
| Answer» Correct Answer - D | |
| 163. |
Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology ?A. Genetic code is specificB. Genetic code is not ambiguousC. Genetic code is redundantD. Genetic code is nearly universal |
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Answer» Correct Answer - D As genetic code is nearly universal means almost all organism will have amino acids coded by same kind of codons as given in checkerboard. So this property is utilised to produce human insulin using bacteria. |
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| 164. |
A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the populationA. 0.16 (M), 0.36 (Aa), 0.48 (aa)B. 0.36 (M), 0.48 (Aa), 0.16 (aa)C. 0.16 (M), 0.24 (Aa), 0.36 (aa)D. 0.16 (M), 0.48 (Aa), 0.36 (aa) |
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Answer» Correct Answer - D Frequency of dominant allele (p) = .4 Appling Hardy Weinberg equilibrium - p + q = 1 q = 1 - .4 = .6 `p^2 + q^2 + 2pq = 1` Frequency of homozygous dominant genotype `(p^2 // A A) = (.4)^2 = .16` Frequency of Heterozygote `(2pq // Aa) = 2 xx .4 xx .6 = .48` Frequency of homozygous recessive genotype `(q^2 // a a) = (.6)^2 = .36` |
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| 165. |
Vegetative propagation in Pistia occurs byA. offsetB. RunnerC. SuckerD. Stolon |
| Answer» Correct Answer - A | |
| 166. |
Two plants can be conclusively said to belong to the same species if theyA. Have same number of chromosomesB. Can reproduce freely with each other and form seedsC. Have more than 90 percent similar genesD. Look similar and possess identical secondary metabolites. |
| Answer» Correct Answer - B | |
| 167. |
Which group of vertebrates comprises the highest number of endangered speciesA. MammalsB. FishesC. ReptilesD. Birds |
| Answer» Correct Answer - C | |
| 168. |
Which one is the principal cation in the plasma of blood ?A. `K^+`B. `Mg^(+2)`C. `Ca^(+2)`D. `Na^+` |
| Answer» Correct Answer - D | |
| 169. |
Activation energy `(E_(a))` and rate constants (`k_(1)` and `k_(2)`) of a chemical reaction at two different temperatures (`T_(1)` and `T_(2)`) are related byA. `ln.(k_(2))/(k_(1))=-(E_(a))/(R)((1)/(T_(2))-(1)/(T_(1)))`B. `ln.(k_(2))/(k_(1))=-(E_(a))/(R)((1)/(T_(2))+(1)/(T_(1)))`C. `ln.(k_(2))/(k_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`D. `ln.(k_(2))/(k_(2))=-(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))` |
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Answer» Correct Answer - A::C Ace. To Arrheneis equation `ln.(K_(2))/(K_(1))=(E_(a))/(R)[(1)/(T_(1))-(1)/(T_(2))]` option (1) and (3) are same |
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| 170. |
Ergot (of Rye ) is obtained fromA. ClavicepsB. PhytophthoraC. UncinulaD. Ustilago |
| Answer» Correct Answer - A | |
| 171. |
Common cold differs from pneumonia in, thatA. Pneumonia is caused by a virus while the common cold is caused by the bacterium haemophilus influenzaeB. Pneumonia pathogen infects alveoli wheras the common cold affects nose and respiratory passage but not the lungsC. Pneumonia is a communicable disease whereas the common cold is a nutritional deficiency diseaseD. Pneumonia can be prevented by a live attenuated bacterial vaccine whereas the common cold has no effective vaccine |
| Answer» Correct Answer - B | |
| 172. |
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What be the percentage purity of magnsesium carbonate in the sample?A. 60B. 84C. 75D. 96 |
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Answer» Correct Answer - B `MgCO_(3) overset(Delta)toMgO+CO_(2)` 40 g `MgO` obtained from 84g `MgCO_(3)` then 8 g obtained from `(84)/(40)xx8` =16.8g `MgCO_(3)` `therefore%`purity`=(16.8)/(20)xx100=84%` |
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| 173. |
Three moles of an ideal gas expanded spontaneously into vacuum. The work done will beA. 3 JoulesB. 9 JoulesC. ZeroD. Infinite |
| Answer» Correct Answer - C | |
| 174. |
A `0.66 kg` ball is moving wih a speed of `100 m//s`. The associated wavelength will be.A. `6.6xx10^(-34)m`B. `1.0xx10^(-35)m`C. `1.0xx10^(-32)m`D. `6.6xx10^(-32)m` |
| Answer» Correct Answer - B | |
| 175. |
The test-tube baby programme employs which one of the following techniquesA. Gamete intra fallopian transfer (GIFT)B. Zygote intra fallopian transfer (ZIFT)C. Intra cytoplasmic sperm injection (ICSI)D. Intra uterine insemination (IUI) |
| Answer» Correct Answer - B | |
| 176. |
In which of the following interaction both partners are adversely affected ?A. MutualismB. CompetitionC. predationD. Parasitism |
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Answer» Correct Answer - B Competition is the interaction where fitness of both interacting species decreases (-/-). |
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| 177. |
Which of the following are not membrane bound ?A. MesosomesB. VacuolesC. RibosomesD. Lysosomes |
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Answer» Correct Answer - C Ribosomes are membrane less cell organelle. |
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| 178. |
Choose the wrong statementA. yeast is unicellular and useful in fermentationB. Penicillium is multicellular and produces antibioticsC. neurosphora is used in the study of biochemical geneticsD. morels and truffles are poisonous mushrooms |
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Answer» Correct Answer - D Morels and truffles are edible member of ascomycetes. |
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| 179. |
Which one of the following animals has two separate circulatory pathwaysA. SharkB. FrogC. LizardD. Whale |
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Answer» Correct Answer - D Whale have double circulation which consist of (a) Pulmonary circulation (b) Systemic circulation. |
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| 180. |
Axile placentation is present inA. ArgemoneB. DianthusC. LemonD. Pea |
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Answer» Correct Answer - C Argemone:Parietal Dianthus : Free central Lemon: Axile Pea: Marginal |
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| 181. |
An association of individuals of different species living in the same habitat and having functional interactions isA. populationB. Ecological nicheC. Biotic communityD. Ecosystem |
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Answer» Correct Answer - C Community is group of interacting populations of different species occupying same habitat. |
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| 182. |
The oxygen evolved during photosynthesis comes from water molecules . Which one of the following pairs of elemnets is involved in this reaction ?A. Magnesium and chlorineB. manganese and chlorineC. manganese and potassiumD. magnesium and molybdenum |
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Answer» Correct Answer - B Photolysis of water helped by minerals of oxygen evolving complex (OEC) which involves mangnese and chlorine. |
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| 183. |
In which can change in entropy is negativeA. `2H(g)toH_(2)(g)`B. Evaporation of waterC. Expansion of a gas at constant temperatureD. sublimation of solid to gas. |
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Answer» Correct Answer - A Entropy=measurement of disorderness if `trianglen_(g)lt0` then `triangleSlt0` |
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| 184. |
Which one of the alkali metals forms only the normal oxide, `M_(2)O`, on heating in air ?A. LiB. NaC. RbD. K |
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Answer» Correct Answer - A `4Li+O_(2) overset(Delta)to 2Li_(2)O` |
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| 185. |
A bamboo plant is growing in a forest. What will be its trophic level?A. First trophic level `(T_1)`B. Second trophic level `(T_2)`C. Third trophic level `(T_3)`D. Fourth trophic level `(T_4)` |
| Answer» Correct Answer - A | |
| 186. |
The dimensions of universal gravitational constant are :-A. `ML^2 T^(-1)`B. `M^(-2)L^3 T^(-2)`C. `M^(-2) L^2 T^(-1)`D. `M^(-1) L^3 T^(-2)` |
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Answer» Correct Answer - D `because F=(Gm_1m_2)/r^2` `therefore [G] =[(Fr^2)/m^2]=(MLT^(-2)L^2)/M^2` `=M^(-1)L^3T^(-2)` |
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| 187. |
A nucleus represented by the symbol `._Z^AX` has.A. Z protons and A –Z neutronsB. Z protons and A neutronsC. A protons and Z –A neutronsD. Z neutrons and A –Z protons |
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Answer» Correct Answer - A Z = number of protons In `._Z^AX` A-Z= number of neutrons |
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| 188. |
If radius of earth shrinks by 1% then for acceleration due to gravity :A. No change at polesB. No change at equatorC. Max. change at equatorD. Equal change at all locations |
| Answer» Correct Answer - D | |
| 189. |
A radio-active elements emits one `alpha` and `beta` particles then mass no. of daughter element is :A. Decreased by 4B. Increased by 4C. Decreased by 2D. Increased by 2 |
| Answer» Correct Answer - A | |
| 190. |
A light of amplitude A and wavelength `lamda` is incident on a metallic surface, then saturation current flows is proportional to (assume cut off wave length = `lamda_0`) -A. `A^2, if lamdalgtlamda_0`B. `A^2, if lamdalltlamda_0`C. `A, if lamdagtlamda_0`D. `A, if lamdaltlamda_0` |
| Answer» Correct Answer - B | |
| 191. |
The current conduction in a discharged tube is due toA. Electrons onlyB. `+Ve` ions and `-Ve` ionsC. `-Ve` ions and electronsD. `+Ve` ions, and electrons |
| Answer» Correct Answer - D | |
| 192. |
What is terminal potential differnce of a cell? Can its value be greater than the emf of a cell? Explian.A. A battery of less emf is connected in its seriesB. A battery of higher emf is connected in its seriesC. A battery of higher emf is connected in its parallelD. A battery of less emf is connected in its parallel |
| Answer» Correct Answer - C | |
| 193. |
From the following bond energies `H-H` bond energy `431.37 kJ mol^(-1)` `C=C` bond energy `606.10 kJ mol^(-1)` `C-C` bond energy `336.49 kJ mol^(-1)` `C-H` bond energy `410.5 kJ mol^(-1)` Enthalpy for the reaction `overset(H)overset(|)underset(H)underset(|)(C )=overset(H)overset(|)underset(H)underset(|)(C )+H-HtoH-overset(H)overset(|)underset(H)underset(|)(C )-overset(H)overset(|)underset(H)underset(|)(C )-H` will beA. 553.0kJ `mol^(-1)`B. 1523.6 kJ `mol^(-1) `C. `-243.6 " kJ mol"^(-1)`D. `-120.0 "kJ mol"^(-1)` |
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Answer» Correct Answer - D `DeltaH` = dissociation energy of reactant -Bond dissociation of energy of product . `DeltaH=(606.10+4xx410.5+431.37)-(6xx410.50+336.49)` ` =-120.0 kJ//"mol"` |
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| 194. |
The coagulation value in millimoles per litre of the electrolyes used for the coagulation of `As_(2)S_(3)` are given below: I. `(NaCl)=52` , II. `(BaCl_(2))=0.69` III. `(MgSO_(4))=0.22` The correct order of their coagulating power isA. IgtIIgtIIIB. IIgtIgtIIIC. IIIgtIIgtID. IIIgtIgtII |
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Answer» Correct Answer - C Coagulation power`prop(1)/("coagulation of no. or value")` `I lt II lt III` |
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| 195. |
Which of the following molecules represents the order of hybridisation `sp^(2),sp^(2),sp,sp` from left to right atoms ?A. `CH_(2)=CH-CH=CH_(2)`B. `CH_(2)=CH-C-=CH`C. `HC-=C-C-=CH`D. `CH_(3)-CH=CH-CH_(3)` |
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Answer» Correct Answer - B `overset(sp^(2))(CH_(2))=overset(sp^(2))(CH)-overset(sp)(C)-=overset(sp)(CH)` Number of orbital require in hybridization =Number of `sigma` -bonds around each carbon atom. |
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| 196. |
A potentiometer is an accurate and versatile device to make electrical measurements of `E.M.F.` because the method involvesA. CellsB. Potential gradientsC. A condition of no current flow through the galvanometerD. A combination of cells , galvanometer and resistances |
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Answer» Correct Answer - C Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit. |
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| 197. |
A gas mixture consists of 2 moles of oxygen and 4 of Argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system isA. 4RTB. 15RTC. 9RTD. 11RT |
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Answer» Correct Answer - D `U=n_(1)(f_(1))/(2)RT+n_(2)(f_(2))/(2)RT` =`2 xx(5)/(2)RT+4xx(3)/(2)RT` `=5RT+6RT` `U=11RT` |
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| 198. |
A transistor is operated in common emitter configuration at `V_(c)=2 V` such that a change in the base current from `100 muA` to `200 muA` produces a change in the collector current from `5 mA` to `10 mA`. The current gain isA. 50B. 75C. 100D. 150 |
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Answer» Correct Answer - A `DeltaI_(B)=100 muA` `DeltaI_(C)=5mA ` `beta=(DeltaI_(C))/(DeltaI_(B))=(5xx10^(-3))/(100xx10^(-6))=50` |
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| 199. |
Which of the following gastric cells indirectly help in erythropoiesis ?A. Goblet cellsB. Mucous cellsC. Chief cellsD. Parietal cells |
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Answer» Correct Answer - D Parietal or oxyntic cell is a source of HCl and intrinsic factor. HCl converts iron present in diet from ferric to ferrous form so that it can be absorbed easily and used during erythropoiesis. Intrinsic factor is essential for the absorption of vitamin `B_(12)` and its deficiency causes pernicious anaemia. |
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| 200. |
By photoelectric effect, Einstein, provedA. E = hVB. `KE = (1)/(2) mv^(2)`C. `E = mc^(2)`D. `E = (Rhc^(2))//(n^(2))` |
| Answer» Correct Answer - A | |