

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.
251. |
Which one of the following statements is correct about the tobacco mosaic virus?(a) It affects all monocotyledonous plants(b) It affects photosynthetic tissue of the infected plant(c) It does not infect other species belonging to the Solanaceae(d) It infects gymnosperms |
Answer» Answer is : (b) It affects photosynthetic tissue of the infected plant Tobacco Mosaic Virus (TMV) affects photosynthetic tissue of the infected plant. Other statements can be corrected as TMV affects all dicotyledonous plants, of which most important are tobacco and tomato. But it does not affect any monocotyledonous plant. TMV is a ssRNA virus, it infects a wide range of plants, especially tobacco and other members of the family Solanaceae. TMV does not infect gymnosperms. |
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252. |
Which one of the following statements is correct about placenta?(a) Placenta is permeable to all bacteria(b) Oxygen and carbon dioxide cannot diffuse through the placenta(c) Waste products diffuse out of placenta into maternal blood(d) Placenta does not secrete chorionic gonadotropins |
Answer» Answer is : (c) Waste products diffuse out of placenta into maternal blood Placenta allows the foetus to transfer waste products to the mother’s blood. Other statements can be corrected as Placenta gives protection against most bacteria and does not allow infections to enter the foetus. Placenta allows gaseous exchange via the mother’s blood supply, i.e. it allows diffusion of O2 and CO2. Placenta produces hormones like human Chorionic Gonadotropin (hCG), progesterone, oestrogen and human Placental Lactogen (hPL). |
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253. |
The pH values of (i) 0.1 M HCl (aq), (ii) 0.1 M KOH, (iii) tomato juice and (iv) pure water follow the order.(a) (i) < (iii) < (iv) < (ii)(b) (iii) < (i) < (iv) < (ii)(c) (i) < (ii) < (iii) < (iv)(d) (iv) < (iii) < (ii) < (i) |
Answer» Answer is : (a) (i) < (iii) < (iv) < (ii) pH stands for potenz (power) of hydrogen. The pH values for acidic solution ranges from 0 to 7, for pure water its value is 7 and for basic solution, it ranges from 7-14. As HCl is a strong acid, the pH value will be least. On the other hand, KOH is a strong base, its pH value will be maximum, tomato juice contains citric acid which is a weak acid, so its pH value will be more than HCl, but will be less than H2O and KOH. Thus, the correct order of pH values will be (i) < (iii) < (iv) < (ii) |
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254. |
The pH values of(i) 0.1 M HCl aq (ii) 0.1 M KOH(iii) tomato juice and (iv) pure waterfollow the order:(A) (i) < (iii) < (iv) < (ii) (B) (iii) < (i) < (iv) < (ii) (C) (i) < (ii) < (iii) < (iv) (D) (iv) < (iii) < (ii) < (i) |
Answer» Correct option: (A) (i) < (iii) < (iv) < (ii) |
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255. |
The period number in the long form of the periodic table is equal to (a) magnetic quantum number of any element of the period (b) atomic number of any element of the period (c) maximum principal quantum number of any element of the period (d) maximum azimuthal number of any element of the period |
Answer» Correct option is (c) maximum principal quantum number of any element of the period Since, each period starts with the filling of electrons in a new principal quantum number, therefore the period number in the long form of the periodic table refers to the maximum principal quantum number of any element in the period. Thus, period number = maximum n of any element. (where, n = principal quantum number) |
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256. |
Among Mg, Cu, Fe, Zn the metal that does not produce hydrogen gas in reaction with hydrochloric acid is(a) Cu(b) Zn(c) Mg(d) Fe |
Answer» Answer is : (a) Cu The metals that are present below hydrogen in reactivity series will not produce hydrogen gas in reaction with hydrochloric acid. Among the given metals, Cu is present below H in reactivity series, i.e. it is less reactive than H, will not produce H2 gas in reaction with HCl acid. |
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257. |
If Avogadro’s number is A0, the number of sulphur atoms present in 200 mL of 1N H2SO4 is(A) A0/5(B) A0/2(C) A0/10(D) A0 |
Answer» Correct option (C) A0/10 Explanation: MH2SO4 = 0.5 VH2SO4 = 0.2 nH2SO4 = 0.1 no of mole of ‘S’ atom = 0.1 ∴ no of ‘s’ atom = 0.1 A0 = A0/10 |
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258. |
Suppose we have to cover the xy-plane with identical tiles such that no two tiles overlap and no gap is left between the tiles. Suppose that we can choose tiles of the following shapes; equilateral triangle, square, regular pentagon, regular hexagon. Then the tiling can be done with tiles of - (A) all four shapes(B) exactly three of the four shapes(C) exactly two of the four shapes(D) exactly one of the four shapes |
Answer» Correct option (B) exactly three of the four shapes Explanation: We can cover the plane using squares definitely using equilateral triangle, we can also cover the plane. Also regular hexagon is made of equilateral triangles. But pentagon cannot cover the plane because of its shape. |
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259. |
The gall bladder is involved in(a) synthesising bile(b) storing and secreting bile(c) degrading bile(d) producing insulin |
Answer» Answer is : (b) storing and secreting bile The gall bladder is involved in storing and secreting bile. The gall bladder is a pear-shaped, hollow structure located under the liver and on the right side of the abdomen. Its primary function is to store and concentrate bile, a yellow brown digestive enzyme produced by the liver. |
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260. |
A person with blood group AB has (A) antigen A and B on RBCs and both anti-A and anti-B antibodies in plasma(B) antigen A and B on RBCs, but neither anti-A nor anti-B antibodies in plasma(C) no antigen on RBCs but both anti-A and anti-B antibodies in plasma(D) antigen A on RBCs and anti-B antibodies in plasma |
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Answer» Correct option (B) antigen A and B on RBCs, but neither anti-A nor anti-B antibodies in plasma Explanation:
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261. |
The process of transfer of electrons from glucose to molecular oxygen in bacteria and mitochondira is known as - (A) TCA cycle(B) Oxidative phosphorylation(C) Fermentation(D) Glycolysis |
Answer» Correct option (B) Oxidative phosphorylation Explanation: The process of electron from glucose to molecular oxygen in bacteria and mitochondrion is occur by electron transport system which leads to oxidative phosphorylation. |
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262. |
Deficiency of which one of the following vitamins can cause impaired blood clotting?(a) Vitamin-B(b) Vitamin-C(c) Vitamin-D(d) Vitamin-K |
Answer» Answer is : (d) Vitamin-K Vitamin-K is a cofactor for the enzyme responsible for chemical reactions that maintains blood clotting factors : prothrombin; factor VII, IX, X; and proteins. Thus vitamin-K plays a key role in helping the blood clot thereby preventing excessive bleeding. |
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263. |
In which one of the following phyla is the body segmented ? (A) Porifera (B) Platyhelminthes (C) Annelida(D) Echinodermata |
Answer» Correct option (C) Annelida Explanation: Metameric segmentation is present in (1) Annelida (2) Arthropoda (3) Chordata |
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264. |
Water soluble pigments found in plant cell vacuoles are (a) chlorophylls (b) carotenoids (c) anthocyanins (d) xanthophylls |
Answer» Correct option is (c) anthocyanins Anthocyanins are water soluble vacuolar pigments that may appear red, purple or blue depending on pH . It is impermeable to cell membranes of plants and can leak out only when membrane is damaged or dead. |
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265. |
A convex lens is used to form an image of an object on a screen. If the upper half of the lens is blackened, so that it becomes opaque, then(a) only half of the image will be visible(b) the image position shifts towards the lens(c) the image position shifts away from the lens(d) the brightness of the image reduces |
Answer» Answer is : (d) the brightness of the image reduces When a lens is cut into half or its half part is blackened, image is formed at same place but its intensity is reduced. |
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266. |
Melting temperature for double stranded DNA is the temperature at which 50% of the double stranded molecules are converted into single stranded molecules. Which one of the following DNA will have the highest melting temperature? (A) DNA with 15% guanine(B) DNA with 30% cytosine(C) DNA with 40% thymine(D) DNA with 50% adenine |
Answer» Correct option (B) DNA with 30% cytosine Explanation: Melting temperature of DNA ∝ GC content |
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267. |
Which one of the following statements about mitosis is correct?(a) One nucleus gives rise to 4 nuclei(b) Homologous chromosomes synapse during anaphase(c) The centromeres separate at the onset of anaphase(d) Non-sister chromatids recombine |
Answer» Answer is : (c) The centromeres separate at the onset of anaphase In anaphase, sister chromatids separate from centromeres so, number of chromosome becomes double. Other statements about mitosis can be corrected as Mitosis is a single nuclear division that results in two nuclei. Synapsis takes place during prophase-I of meiosis not during mitosis. Non-sister chromatids recombine during prophase-I of meiosis. During mitosis, each sister chromatid separates and moves to opposite pole of the cell at anaphase. |
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268. |
A 100 mark examination was administered to a class of 50 students. Despite only integer marks being given, the average score of the class was 47.5. Then, the maximum number of students who could get marks more than the class average is(a) 25(b) 35(c) 45(d) 49 |
Answer» Answer is : (d) 49 Total number of students = 50 Average marks of student = 47.5 ∴Total marks of students = 50x 47.5 = 2375 Now, the student get integer marks Hence, the maximum number of students we will divide total mark by 48. ∴ 2375/48 = 49 |
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269. |
The rate constant of a chemical reaction at a very high temperature will approach(A) Arrhenius frequency factor divided by the ideal gas constant(B) activation energy(C) Arrhenius frequency factor(D) activation energy divided by the ideal gas constant |
Answer» Correct option (C) Arrhenius frequency factor Explanation: K = Ae–Ea/RT T →∞ k = A |
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270. |
The standard reduction potentials (in V) of a few metal ion/metal electrodes are given below. Cr3+/Cr = –0.74; Cu2+/Cu = +0.34; Pb2+/Pb = –0.13; Ag+ /Ag = +0.8. The reducing strength of the metals follows the order(A) Ag > Cu > Pb > Cr(B) Cr > Pb > Cu > Ag(C) Pb > Cr > Ag > Cu(D) Cr > Ag > Cu > Pb |
Answer» Correct option (B) Cr > Pb > Cu > Ag Explanation : SRP ↓ Reducing power↑ |
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271. |
Which of the following graphs best describes the oxygen dissociation curve where pO2 is the partial pressure of oxygen ? |
Answer» Correct option (D) Explanation: Oxyhaemoglobin dissociation curve is sigmoid shaped. |
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272. |
Study the following graph of metabolic rate of various terrestrial mammals as a function of their body mass and choose the correct option below.(A) Animals are distributed throughout the curve with the smaller animals towards the left and progressively bigger animals towards the right (B) The smaller animals below a certain critical mass cluster at the left end of the curve and the larger animals above the critical mass cluster on the right end(C) Animals are distributed throughout the curve with the larger animals towards the left and progressively smaller animals towards the right(D) The larger animals above a certain critical mass cluster at the left end of the curve and the smaller animals below the critical mass cluster on the right end |
Answer» Correct option (A) Animals are distributed throughout the curve with the smaller animals towards the left and progressively bigger animals towards the right. Explanation: The metabolic theory of Ecology (MTE) is an extension of Kleiber's law and states that the metabolic rate of organism is the fundamental biological rate that governs most observed patterns in ecology |
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273. |
Match the human disorders shown in Group I with the biochemical processes in Group II. Choose the correct combinationGroup IGroup IIP. Phenylketonuriai. Melanin synthesisQ. Albinismii. Conversion of Phenylalanine to Tyrosine R. Homocystinuriaiii. Tyrosine degradationS. Argininemiaiv. Methionine metabolismv. Urea Synthesis (A) P-ii, Q-i, R-iv, S-v (B) P-i, Q-iv, R-ii, S-v (C) P-ii, Q-i, R-v, S-iii (D) P-v, Q-iii, R-i, S-ii |
Answer» Correct option (A) P-ii, Q-i, R-iv, S-v Explanation: Phenylketonuria is due to non-conversion of phenylalanine into tyrosine. Albinism is non-synthesis of melanin pigment. Homocystinuria is associated with methionine metabolism. Argininemia is associated with urea synthesis. |
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274. |
Most common type of phospholipids in the cell membrane of nerve cell is (a) phosphatidylcholine (b) phosphatidylinositol (c) phosphatidylserine (d) sphingomyelin |
Answer» Correct option is (a) phosphatidylcholine Phosphatidylcholine is a class of phospholipids that are a major component of biological membranes (i.e. nerve cell membrane). It functions in the production of brain chemical called acetylcholine used for nerve impulse transmission at the synapse. |
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275. |
Three uniformly watered plants i, ii and iii were kept in 45% relative humidity, 45% relative humidity with blowing wind and 95% relative humidity, respectively. Arrange, these plants in the order (fastest to slowest) in which they will dry up.(a) i → ii → iii(b) ii → i → iii(c) iii → ii → i(d) iii → i → ii |
Answer» Answer is : (b) ii → i → iii The plants in the order (fastest to slowest) in which they will dry up is ii → i → iii. Relative humidity is the amount of water vapour present in air expressed as a percentage of the amount needed for saturation at the same temperature. As plants transpire, the humidity around saturates leaves with water vapour. When relative humidity levels are too high or there is a lack of air circulation, a plant cannot make water evaporate by transpiration or draw nutrients from the soil. Therefore 95% relative humidity will dry up the slowest, and the 45% relative humidity with blowing wind will dry up the fastest. |
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276. |
Three uniformly watered plants i, ii and iii were kept in 45% relative humidity, 45% relative humidity with blowing wind and 95% relative humidity, respectively.Arrange these plants in the order (faster to slowest) in which they will dry up. (A) i = ii, iii (B) ii, i, iii (C) iii, ii, i (D) iii, i = ii |
Answer» Correct option(B) ii, i, iii |
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277. |
Sister chromatids of a chromosome have (A) different genes at the same locus (B) different alleles of the same gene at the same locus (C) same alleles of the same gene at the same locus (D) same alleles at different loci |
Answer» Correct option:(C) same alleles of the same gene at the same locus Explanation: Sister chromatids are identical DNA molecules (In somatic cells) |
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278. |
A regular check on the unborn baby of a lady towards the end of her pregnancy showed a heart rate of 80 beats per minute. What would the doctor infer about the baby's heart condition from this? (A) Normal heart rate (B) Faster heart rate (C) Slower heart rate (D) Defective brain function |
Answer» Correct option(C) Slower heart rate Explanation: Infants have higher heart rate |
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279. |
A diabetic individual becomes unconscious after self-administering insulin. What should be done immediately to revive the individual? (A) Provide him sugar (B) Give him high dose of insulin (C) Provide him salt solution (D) Provide him lots of water |
Answer» Correct option(A) Provide him sugar Explanation: Insulin lowers blood sugar level and in this case brain is getting inadequate sugar/ glucose |
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280. |
Considering air resistance, if t1 = time for a thrown ball in upward journey and t2 = time taken for downward journey, then(a) t1 = t2 (b) t1 > t2 (c) t2 > t1 (d) 3t2 = 2t1 |
Answer» Correct option is (c) t2 > t1 Time for downward journey is higher as ball can be thrown with any velocity but its downward velocity is always less than or equal to terminal velocity. |
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281. |
What is the maximum number of orbitals that can be identified with the following quantum numbers?n = 3, l = 1 and ml = 0(a) 1(b) 2(c) 3(d) 4 |
Answer» Answer is : (a) 1 The given value of n = 3 suggests that the shell is 3. For n = 1, l has 3 values, i.e. +1, 0 and −1 hence there occur 3 orbitals in p-subshell namely px, py and pz. Thus, the given values for n = 3, l = 1 and ml = 0 suggests that the orbital is 3py. Hence, the maximum number of orbitals that can be identified with given quantum number is only 1. |
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282. |
Potential energy between two molecules as a function of their separation is as shown below.Force between particles is zero at (a) x = 0.4 Å (b) x = 0.6 Å(c) x = 1.2 Å (d) x = 1.8 Å |
Answer» Correct option is (a) x = 0.4 Å Force is zero when potential energy is minimum. |
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283. |
Fermi energy level for an electron is (a) a possible energy value that an electron can have in free state (b) an unfilled energy level that can be occupied by two electrons of opposite spins (c) lowest energy value possible for a bound electron (d) highest occupied energy level at absolute zero kelvin upto which every possible energy levels are filled |
Answer» Correct option is (d) highest occupied energy level at absolute zero kelvin upto which every possible energy levels are filled Fermi energy level is last filled energy level at zero kelvin. |
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284. |
Solar cookers are not very popular because (a) they are bulky (b) they are not put into kitchen (c) they cook food in large time (d) sun changes its position rapidly |
Answer» Correct option is (d) sun changes its position rapidly As sun changes its position rapidly, so reflector of solar cooker is to be adjusted nearly in every 1/2 hour duration. |
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285. |
Biological diversity increases with the productivity of an ecosystem. In which of the following habitats do we see the greatest diversity of species?(a) Tropical dry grasslands(b) Temperate deciduous forests(c) Alpine grasslands(d) Tropical evergreen forests |
Answer» Answer is : (d) Tropical evergreen forests Diversity of species is highest in the tropical evergreen forests primarily because there are fewer ecological obstacles for biodiversity. Like the climate is wet and warm, plants and animals have the greatest access to consistent energy, water and carbon, etc. This reduces the selection for traits that emphasise the ability to withstand environmental stresses such as cold and drought, etc., and promotes higher rates of speciation. |
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286. |
Biological diversity increases with the productivity of an ecosystem. In which of the following habitats do we see the greatest diversity of species?(A) Tropical dry grasslends(B) Temperate deciduous forests(C)Alpine grasslends(D) Tropical evergreen forests |
Answer» (D) Tropical evergreen forests | |
287. |
Stratification is more common in which of the following ? (A) Deciduous forest(B) Tropical rain forest(C) Temperate forest(D) Tropical savannah |
Answer» Correct option (B) Tropical rain forest Explanation: Tropical rain forest shows vertical zonation i.e. stratification. |
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288. |
Where is the third ventricle of the brain located ?(A) Cerebrum(B) Cerebellum (C) Pons varoli(D) Diencephalon |
Answer» Correct option (D) Diencephalon Explanation: Cavity of Diancephelon is called as diocoel or third ventricle. |
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289. |
Which one of the following metabolic conversion requires oxygen?(a) Glucose to pyruvate(b) Glucose to CO2 and ethanol(c) Glucose to lactate(d) Glucose to CO2 and H2O |
Answer» Answer is : (d) Glucose to CO2 and H2O Conversion of glucose to CO2 and H2O requires oxygen. In aerobic respiration glucose reacts with oxygen forming ATP, carbon dioxide and water are released as byproducts. C6H12O6 + 6O2 → 6CO2 + 6H2O + ATP |
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290. |
Which of the following are not polymeric? (a) Nucleic acid (b) Proteins (c) Polysaccharides (d) Lipids |
Answer» Correct option is (d) Lipids Among the given options, except lipids all are polymers. These are formed by the polymerisation of monomers. The basic unit of lipid are fatty acids and glycerol molecules that do not form repetitive chains. Instead they form triglycerides from 3 fatty acids and one glycerol molecules. Protein monomers are amino acids and they bond together in repetitive chains just as carbohydrate monomers are monosaccharides. |
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291. |
Life cycle of Ectocarpus and Fucus respectively are (a) haplontic, diplontic (b) diplontic, haplodiplontic (c) haplodiplontic, diplontic (d) haplodiplontic, haplontic |
Answer» Correct option is (c) haplodiplontic, diplontic Ectocarpus and Fucus respectively show haplodiplontic and diplontic life cycle. In Ectocarpus, sporic meiosis occurs and haploid biflagellate meiozoospores are formed. They germinate to produce gametophytic thalli. The gametophytes liberate gametes which fuse to form diploid zygote which gives rise to a diploid plant. In Fucus, there is a single somatic phase. It is diploid and produces haploid gametes. They fuse during fertilisation to give rise to diploid individual. |
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292. |
Zygotic meiosis is a characteristic of (a) Marchantia (b) Fucus (c) Funaria (d) Chlamydomonas |
Answer» Correct option is (d) Chlamydomonas Zygotic meiosis is represented in the haplontic life cycle of many algae including Chlamydomonas. In such a life cycle, all cells are haploid except zygote. This is because meiosis occurs in the zygote itself resulting into four haploid cells that give rise to haploid plants. |
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293. |
The ornithine cycle removes two waste products from the blood in liver. These products are (a) CO2 and urea (b) ammonia and urea (c) CO2 and ammonia (d) ammonia and uric acid |
Answer» Correct option is (b) ammonia and urea Ornithine cycle removes both ammonia and urea from the blood. It converts ammonia into urea (in liver) and transports it to kidneys by the blood. Hence, it plays a key role in detoxification of our blood. This cycle occurs in the liver. |
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294. |
A box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show(a) the water coming down forming a parabolic stream(b) the water going up forming a parabolic stream(c) the water coming out in a straight line(d) no water coming out |
Answer» Answer is : (d) no water coming out When box with hole is in free fall, both water and box cover equal distance downwards in equal time. Hence, no water comes out of hole in free fall of box. |
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295. |
In Griffith's experiments mice died when injected with - (A) heat killed S-strain(B) heat killed S-strain combined with R-strain(C) heat killed R-strain(D) live R-strain |
Answer» Correct option (B) heat killed S-strain combined with R-strain Explanation: In Griffith experiment, mice died when injected with combination heat killed s-strain + Live R strain which resulted in the transformation of R II into S III form. |
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296. |
A desert lizard (an ectotherm) and a mouse (an endotherm) are placed inside a chamber at 15ºC and their body temperature [T(L) for the lizard and T(M) for the mouse] and metabolic rates [M(L) for the lizard and M(M) for the mouse] are monitored. Which one of the following is correct -(A) T(L) and M(L) will fall while T(M) and M(M) will increase(B) T(L) and M(L) will increase while T(M) and M(M) will fall(C) T(L) and M(L) will fall, T(M) will remain same and M(M) will increase(D) T(L) and M(L) will remain same and T(M) and M(M) will decrease |
Answer» Correct option (C) T(L) and M(L) will fall, T(M) will remain same and M(M) will increase Explanation: Desert lizard is an ectotherm (poikilotherm) whose body temperature varies according to environmental temperature. So at 15ºC T(L) will fall due to fall in T(M) While mouse is an endotherm (Homeotherm) whose body temperature remains constant always due to variation is metabolic rate In homeotherms metabolic rate is inversely proportional to environmental temperature. So at 15ºC T(M) remain same and M((M) will increased. |
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297. |
You have a measuring cup with capacity 25 ml and another with capacity 110 ml, the cups have no markings showing intermediate volumes. Using large container and as much tap water as you wish. What is the smallest amount of water you can measure accurately?(a) 1 ml(b) 5 ml(c) 10 ml(d) 25 ml |
Answer» Answer is : (b) 5 ml Put x time of water of 110 ml to container and take y time of water of 25 ml from container. Then, container contains 110x - 25y = 5(22x - 5y) ∴ Container contains multiple of 5. ∴ Smallest amount of water be measure accurately 5 ml. |
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298. |
Three friends Ajay, Vijay and Sanjay move along a circular path of length 1.2 km with speeds of 6 km/h, 8 km/h and 9 km/h respectively. Ajay and Vijay move in the same direction but Sanjay move in opposite direction, if they all start at the same time and from same place. How many time will Ajay and Sanjay meets anywhere on the path by the time Ajay and Vijay for the first time anywhere on the path?(a) 6 times(b) 7 times(c) 8 times(d) 9 times |
Answer» Answer is : (b) 7 times Time taken by Ajay and Vijay to meet first time anywhere on the path = Distance/Relative speed = 1.2/8.6 = 0.6 h Time taken by Ajay and Sanjay to meet anywhere = Distance/Relative speed \(=\frac{1.2}{9+6}\) = 0.8 h The number of times Ajay and Sanjay meets anywhere on the path by the time Ajay and Vijay meets each other for the first time = 36/4.8 = 7 1/2, i.e. 7 times. |
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299. |
Enzyme X extracted from the digestive system hydrolyses peptide bonds. Which of the following is probable candidate to be enzyme X?(a) Amylase(b) Lipase(c) Trypsin(d) Maltase |
Answer» Answer is : (c) Trypsin In duodenum, trypsin enzyme catalyses the hydrolysis of peptide bonds, breaking down proteins into smaller peptides. Amylase hydrolyses starch into maltose inside the mouth. Lipase breaks down dietary fats into fatty acids and glycerol. Maltase hydrolyses maltose into simple sugar glucose. |
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300. |
Ratio of nuclear density of nuclei \(\frac{142}{53 }\)I and \(\frac{139}{56}\)Ba is(a) 142 : 139 (b) 53 : 56 (c) 139 : 142 (d) None of these |
Answer» Correct option is (d) None of these Nuclear density is a constant (ρ = 2.38 × 1017 kg m−3). It is independent of nuclear size and number of nucleons. So, ratio is 1:1. |
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