

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. |
The pair of atoms having the same number of neutrons is(a) 126C, 2412Mg(b) 2311Na, 199F(c) 2311Na, 2412Mg(d) 2311Na, 3919K |
Answer» Answer is : (c) 2311Na, 2412Mg Number of neutrons in the pairs given in options are as follows (a) 126C, 2412Mg No. of neutrons in C = 12 - 6 = 6 No. of neutrons in Mg = 24 - 12 = 12 (b) 2311Na, 199F No. of neutrons in Na = 23 - 11 = 12 No. of neutrons in F = 19 - 9 = 10 (c) 2311Na, 2412Mg No. of neutrons in Na = 23 - 11 = 12 No. of neutrons in Mg = 24 - 12 = 12 (d) 2311Na, 3919K No. of neutrons in Na = 23 - 11 = 12 No. of neutrons in K = 39 - 19 = 20 Thus, option (c) is correct. |
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302. |
Maximum number of electrons that can be accommodated in the subshell with azimuthal quantum number l = 4, is(a) 10(b) 8(c) 16(d) 18 |
Answer» Answer is : (d) 18 Maximum number of electrons that can be accommodated in a subshell = 2(2l+1) When l = 4 Maximum number of electrons = 2(2 x 4+1) = 18 |
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303. |
A mixture of NH4Cl and NaCl can be separated by(a) filtration(b) distillation(c) sublimation(d) decantation |
Answer» Answer is : (c) sublimation A mixture of NH4Cl and NaCl can be separated by the process of sublimation. In this process, solid directly changes to gaseous state without passing into liquid state. NH4Cl sublimes to gaseous NH3 and HCl upon heating whereas NaCl does not sublime, the reaction can be written as NH4Cl(s) → NH3(g) + HCl(g) |
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304. |
Restriction endonucleases are enzymes that are used by biotechnologists to -(A) cut DNA at specific base sequence(B) join fragments of DNA(C) digest DNA from the 3' end(D) digest DNA from the 5' end |
Answer» Correct option (A) cut DNA at specific base sequence Explanation: Restriction endonuclease enzyme breaks the phosphodiester bond on specific pallindromic sequences. |
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305. |
Excess salt inhibits bacterial growth in pickles by -(A) endosmosis(B) exosmosis(C) oxidation(D) denaturation |
Answer» Correct option (B) exosmosis Explanation: Excessive salt in pickle inhibits the bacterial growth by exosmosis because external medium become hypertonic. |
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306. |
Restriction endonucleases are enzymes that are used by biotechnologists to(a) cut DNA at specific base sequences(b) join fragments of DNA(c) digest DNA from the 3′ end(d) digest DNA from the 5′ end |
Answer» Answer is : (a) cut DNA at specific base sequences Restriction endonuclease is an enzyme that cuts dsDNA into fragments at or near specific recognition sites (palindromic sequence) within the molecule known as restriction sites. These enzymes are found in bacteria and archaea and provide a defence mechanism against invading viruses |
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307. |
Which among grass, goat, tiger and vulture in a food chain, will have the maximum concentration of harmful chemicals in its body due to contamination of pesticides in the soil?(a) Grass since it grows in the contaminated soil(b) Goat since it eats the grass(c) Tiger since it feeds on the goat which feeds on the grass(d) Vulture since it eats the tiger, which in turn eats the goat, which eats the grass |
Answer» Answer is : (d) Vulture since it eats the tiger, which in turn eats the goat, which eats the grass The increase in concentration of harmful chemical substance like pesticides in the body of living organisms at each trophic level of a food chain is called biological magnification. The organism which occurs at the highest trophic level in the food chain will have the maximum concentration of harmful chemicals in its body. Since vulture occupies the top level as it eats the tiger, which eats the goat, which eats the grass in the food chain, it will have the maximum concentration of harmful chemicals in its body. |
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308. |
Which, among grass, goat, tiger and vulture, in a food chain, will have the maximum concentration of harmful chemicals in its body due to contamination of pesticides in the soil ?(A) Grass since it grows in the contaminated soil(B) Goat since it eats the grass(C) Tiger since it feeds on the goat which feeds on the grass(D) Vulture since it eats the tiger, which in turn eats the goat, which eats the grass |
Answer» Correct option (D) Vulture since it eats the tiger, which in turn eats the goat, which eats the grass Explanation: Vulture will have the maximum concentration of pesticide because it feeds on tiger which in tern eat the goat which eat the grass. |
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309. |
A gas at initial temperature T undergoes sudden expansion from volume V to 2V. Then -(A) The process is adiabatic(B) The process is isothermal(C) The work done in this process is nRTlne(2) where n is the number of moles of the gas.(D) The entropy in the process does not change |
Answer» Correct option (A) The process is adiabatic Explanation: In sudden expansion gas do not get enough time for exchange of heat. ∴ Process is adiabatic. |
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310. |
The decay profiles of three radioactive species A, B and C are given below :These profiles imply that the decay constants kA, kB and kC follow the order(A) kA > kB > kC (B) kA > kC > kB (C) kB > kA > kC (D) kC > kB > kA |
Answer» Correct option(d) Explanation: Ct=C0e-kt |
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311. |
The Arrhenius plots of two reactions, I and II are shown graphically -The graph suggests that -(A) EI > EII and AI > AII(B) EII > EI and AII > AI(C) EI > EII and AII > AI(D) EII > EI and AI > AII |
Answer» Correct option (A) EI > EII and AI > AII Explanation: For plot between In k v/s 1/T y-intercept is ln A & slope is -Ea/R therefore; EII < EI and AI > AII |
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312. |
Which of the following molecules has no dipole moment ?(A)CH3Cl (B)CHCl3 (C)CH2Cl2 (D)CCl4 |
Answer» Correct option(d) Explanation: CCl4 has zero dipole moment due to its tetrahedral shape, all C-Cl bond moment cancel each other. |
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313. |
Usain Bolt, an olympic runner, at the end of a 100 metre sprint, will have more of which of the following in his muscles?(a) ATP(b) Pyruvic acid(c) Lactic acid(d) Carbon dioxide |
Answer» Answer is : (c) Lactic acid During vigorous muscular activity like running, muscles perform anaerobic respiration after a while due to scarcity of oxygen. During anaerobic respiration in muscles, lactic acid is produced as a byproduct. Thus a runner will have lactic acid in his muscles after a 100 metre sprint. |
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314. |
At which phase of the cell cycle, DNA polymerase activity is at its highest?(a) Gap 1 (G1)(b) Mitotic (M)(c) Synthetic (S)(d) Gap 2 (G2) |
Answer» Answer is : (c) Synthetic (S) DNA polymerase is an enzyme that synthesises DNA molecules from deoxyribonucleotides, the building blocks of DNA. These enzymes are essential for DNA replication. S-phase or synthetic phase is significant due to DNA synthesis. Thus, DNA polymerase activity is highest at S-phase of cell cycle. |
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315. |
When heated in air, brown copper powder turns black. This black powder would turn brown again when heated withA) CO(B) O2 (C) H2 (D) NH3 |
Answer» Correct option (C) H2 Explanation: H2 is reducing agent |
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316. |
For a 4p orbital, the number of radial and angular nodes, respectively, are(A) 3,2(B) 1,2(C) 2, 4(D) 2,1 |
Answer» Correct option (D) 2,1 Explanation: no. of radial node = n – l – 1 = 4 – 1 – 1 = 2 no. of angular node = l = 1 |
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317. |
If the initial number of template DNA molecules in a PCR reaction is 1000, the number of product DNA molecules at the end of 20 cycles will be closest to (A) 103 (B) 106 (C) 109 (D) 1012 |
Answer» Correct option (C) 109 Explanation: During PCR, the number of DNA molecules increases by 2n . Where 'n' is number of divisions |
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318. |
In an experiment represented in the schematic below, a plant species was grown in different day and night cycles and its photoperiodic flowering behaviour was noted. This species is a(A) short day plant and actually measures day length to flower(B) short day plant and actually measures night length to flower (C) long day plant and actually measures night length to flower(D) long day plant and actually measures day length to flower |
Answer» Correct option (B) short day plant and actually measures night length to flower Explanation: SDP (Short day plants) or LNP (Long Night Plants) flowers only when photoperiod is below critical day length (Critical photoperiod) or they are responsible to night length and flower when night length is above critical dark period. In this experiment plant flowers when dark period is above 8 hrs. So, it is SDP and actually measures night length to flower. |
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319. |
A box is falling freely inside the box, a particle is projected with some velocity v with respect to the box at angle θ.For an observer sitting in the box, path of particle is |
Answer» Answer is : (b) With respect to observer, there is no acceleration in the vertical velocity component. So, path of particle is a straight line as in option (b). |
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320. |
The figure of a centimeter scale below shows a particular position of the vernier calipers. In this position the value of x shown in the figure is ( figure is not to scale) (A) 0.02 cm(B) 3.65 cm(C) 4.15 cm(D) 0.03 cm |
Answer» Correct option (D) 0.03 cm Explanation: x = 0.3 cm – 3 x scale division of vernier calipers = 0.3 cm -3 x 9/100 = 30 - 27/100 = 30 - 27/100 = 3/100 = 0.03 cm |
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321. |
Mendeleev’s periodic law states that the properties of elements are a periodic function of their(a) reactivity of elements(b) atomic size(c) atomic mass(d) electronic configuration |
Answer» Answer is : (c) atomic mass According to Mendeleev’s periodic law, the physical and chemical properties of the elements are a periodic function of their atomic mass. |
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322. |
Mendeleev’s periodic law states that the properties of elements are a periodic function of their(A) reactivity of elements(B) atomic size(C) atomic mass(D) electronic configuration |
Answer» Correct option (C) atomic mass Explanation: Mendeleev’s periodic table state that the property of elements are a periodic function of their atomic mass |
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323. |
For any real number r, let Ar = {eiπrn : n is a natural number} be a set of complex numbers. Then(A) A1, A1/π, A0.3 are all infinite sets(B) A1 is a finite set and A1/π, A0.3 are infinite sets(C) A1, A1/π , A0.3 are all finite sets(D) A1, A0.3 are finite sets and A1/π is an infinite sets |
Answer» Correct option (D) A1, A0.3 are finite sets and A1/π is an infinite sets Explanation: eiπrn is always a finite set when r is a rational & is infinite when r = 1/π . |
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324. |
Let a, b, c, d, e, be real numbers such that a + b < c + d, b + c < d + e, c + d < e + a, d + e < a + b. Then(A) The largest is a and the smallest is b(B) The largest is a and the smallest is c(C) The largest is c and the smallest is e(D) The largest is c and the smallest is b |
Answer» Correct option (A) The largest is a and the smallest is b Explanation: (i) a + b < c + d (ii) b + c < d + e (iii) c + d < e + a (iv) d + e < a + b from (i) & (iii) a + b < e + a ⇒ b < e from (ii) & (iv) b + c < a + b c < a (i) – (ii) a – c < c – e ⇒c > e (i) – (iv) (a – e) + (b – d) < (c – a) + (d – b) from thus d > b (i) + (iii) – (ii c > d overall a is greatest, b is least |
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325. |
In an in vitro translation experiment, poly (UC) RNA template produced poly (Ser-Leu), while poly (AG) RNA template produced poly (Arg-Glu) polypeptide. Which one of the following options represents correct interpretations of the codons assignments for Ser, Leu, Arg and Glu?(a) Ser-UCU, Leu-CUC, Arg-AGA, Glu-GAG(b) Ser-CUC, Leu-GAG, Arg-UCU, Glu-AGA(c) Ser-AGA, Leu-UCU, Arg-GAG, Glu-CUC(d) Ser-GAG, Leu-AGA, Arg-CUC, Glu-UCU |
Answer» Answer is : (a) Ser-UCU, Leu-CUC, Arg-AGA, Glu-GAG Serine is coded by UCU, UCC, UCA, UCG, AGU, AGC Leucine is coded by CUU, CUC, CUA, CUG, UUA, UUG Arginine is coded by AGA, AGG, CGU, CGC, CGA, CGG Glutamic acid is coded by GAA, GAG Therefore, option (a) is the correct interpretation of the assigned amino acids. Ser-UCU, Leu-CUC, Arg-AGA, Glu-GAG |
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326. |
Among the elements Li, N, C and Be, one with the largest atomic radius is(a) Li(b) N(c) C(d) Be |
Answer» Answer is : (a) Li As the given elements Li, N, C and Be belong to same period i.e. 2nd period, so on moving from left to right in a period the atomic radius decreases because the effective nuclear charge increases. Thus, Li has the largest atomic radius among them all. |
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327. |
Ascorbic acid is a/an(a) strong inorganic acid(b) hormone(c) vitamin(d) enzyme |
Answer» Answer is : (c) vitamin Ascorbic acid is also known as vitamin-C, a vitamin found in citrus fruits and is an essential nutrient involved in the repair of connective tissue and the enzymatic production of certain neurotransmitters. In the body, it acts as an antioxidant, helping to protect cells from the damage caused by free radicals. |
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328. |
Ascorbic acid is a/an :(A) Strong inorganic acid (B)Hormone (C) Vitamin (D) Enzyme |
Answer» Ascorbic acid is a Vitamin. |
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329. |
If birds are moved from 30°C - 10°C, their body temperature(a) changes from 30°C - 10°C(b) increases by 10°C(c) does not change at all(d) decreases by 10°C |
Answer» Answer is : (c) does not change at all Birds and mammals are endothermic animals, i.e. their core body temperature is kept nearly constant through thermal homeostasis. Therefore, if birds are moved from 30°-10°C their body temperature does not change at all. |
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330. |
If birds are moved from 30°C to 10°C, their body temperature :(A) changes from 30°C to 10°C (B) increases by 10°C(C) does not changes at all (D) decreases by 10°C |
Answer» If birds are moved from 30°C to 10°C, their body temperature does not changes at all. |
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331. |
Which one of the following is a modified leaf?(a) Sweet potato(b) Ginger(c) Onion(d) Carrot |
Answer» Answer is : (c) Onion Onion is a bulb, i.e. it is a modified leaf. A bulb is an underground pyriform-spherical structure that possesses a reduced convex or slightly conical disc-shaped stem and several fleshy scales enclosing a terminal bud. In Onion, the fleshy scales represent leaf bases in the outer part and scale leaves in the central region. |
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332. |
Which one of the following is the smallest in size ?(A) Bacteria (B) Mitochondrion (C) Mammalian cell (D) Virus |
Answer» Virus is the smallest in size. |
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333. |
Which one of the following is a modified leaf ?(A) Sweet potato(B) Ginger(C) Onion(D) Carrot |
Answer» Correct option (C) Onion Explanation: In onion modified leaves are present for food storage |
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334. |
Which of the following pairs are both polysaccharides ?(A) Cellulose and glycogen(B) Starch and glucose(C) Cellulose and fructose(D) Ribose and sucrose |
Answer» Correct option (A) Cellulose and glycogen Explanation: Cellulose → Homopolysaccharide of β glucose Glycogen → Homopolysaccharide of α glucose |
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335. |
Considering the average molecular mass of a base to be 500 Da, what is the molecular mass of a double stranded DNA of 10 base pairs ?(A) 500 Da(B) 5 kDa(C) 10 kDa(D) 1 kDa |
Answer» Correct option (C) 10 kDa Explanation: Molecular Mass of a base = 500 da Total No. of bases = 10 bp = 10 x 2 = 20 bases Molecular mass of 20 bases = 20 x 500 da = 10000 dalton = 10 kda |
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336. |
Considering the average molecular mass of a base to be 500 Da, what is the molecular mass of a double-stranded DNA of 10 base pairs?(a) 500 Da(b) 5 kDa(c) 10 kDa(d) 1 kDa |
Answer» Answer is : (c) 10 kDa Molecular mass of a base = 500 Da Number of base in a dsDNA = 10 BP or 20 bases Thus, molecular mass of a dsDNA with 20 bases = 20 x 500 = 10 kDa |
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337. |
A 25,000 Da protein contains a single binding site for a molecule (ligand), whose molecular weight is 2,500 Da. Assuming high affinity and physiologically irreversible binding, the amount of the ligand required to occupy all the binding sites in 10 mg protein will be(a) 0.1 mg(b) 1 mg(c) 10 mg(d) 100 mg |
Answer» Answer is : (b) 1 mg Assuming x as the amount of ligand to occupy all the binding sites in 10 mg protein. x mg/Protein in grams = Ligand molecular weight/Protein molecular weight x = 2500/25000 x 10 = 1 mg |
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338. |
A baby is born with the normal number and distribution of rods, but no cones in his eyes. We would expect that the baby would be(a) colourblind(b) nightblind(c) blind with both eyes(d) blind with one eye |
Answer» Answer is : (a) colourblind Absence of cone cells in eyes is known as total colour blindness or monochromacy. This person views everything as if it were in a black and white television. Monochromacy occurs when 2 or all 3 of cone pigments are missing and colour and light vision is reduced to one dimension. |
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339. |
Albinism in humans is controlled by a recessive allele. How many copies of this allele will be found at one of the poles of a cell at telophase-I of meiosis in an albino person? (a) 23 (b) 4 (c) 2 (d) 1 |
Answer» Correct option is (c) 2 Since the allele is recessive, both homologous chromosomes in a somatic cell of an albino person would have the allele. After meiosis-I, each end would have a homologous chromosome with the allele. As the chromosome is existing as a each chromosome and hence each end would have two copies of the allele. |
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340. |
The stalk of a plant leaf is derived from which one of the following types of plant tissue?(a) Sclerenchyma(b) Parenchyma(c) Chlorenchyma(d) Collenchyma |
Answer» Answer is : (d) Collenchyma The stalk of plant leaf (petiole) is derived from collenchyma. Collenchyma cells are elongated cells with irregular thick cell walls that provide structural support, particularly in growing shoots and leaves. Their thick cell walls are composed of the compounds cellulose and pectin. |
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341. |
A spring balance A reads 2 kg when a block of mass m suspended from it. Another balance B reads 3 kg when a beaker with a liquid is put on its pan. The two balances are now so arranged that the hanging mass m is fully immersed inside the liquid in to be beaker as shown in the figure. In this situation.(A) the balance A will read 2 kg and B will read 5 kg.(B) the balance A will read 2 kg and B will read 3 kg.(C) the balance A will read less than 2 kg and B will read between 3 kg and 5 kg.(D) the balance A will read less than 2 kg and B will read 3 kg. |
Answer» (C) the balance A will read less than 2 kg and B will read between 3 kg and 5 kg. |
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342. |
A solid square plate is spun around different axes with the same angular speed. In which of the following choice of axis of rotation will the kinetic energy of the plate be the largest ?(A) through the central normal to the plate. (B) along one of the diagonals of the plate.(C) along one of the edges of the plate. (D) through one corner normal to the plate. |
Answer» Correct option: (D) through one corner normal to the plate. |
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343. |
A metallic wire is loaded at ends with two masses is placed over a slab of ice.This wire passes through slab without splitting it into two pieces. This is due to(a) depression of melting point(b) elevation of melting point(c) high conductivity of metal wire(d) high specific heat of ice |
Answer» Answer is : (a) depression of melting point Ice melts at lower temperature due to increase in pressure. As wire passes, the water formed is again freezes and hence wire passes without cutting ice. |
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344. |
Let S = {x ∈ R : cos(x) + cos ( √2 x) < 2}. Then (A) S = ϕ (B) S is a non-empty finite set (C) S is an infinite proper subset of R\{0} (D) S = R\{0} |
Answer» Correct option (D) S = R\{0} Explanation: Cos x + cos( √2 x) will always be less than 2 except when both cos x = 1 & cos (√2 x) = 1 cos x = 1 ⇒ x = 2nπ cos ( √2 x) = 1 ⇒ x = √2 mπ both can simultaneously be 1 only when x = 0 ⇒ S = R – {0} |
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345. |
A positive integer K is said to be good if there exists a partition of {1, 2, 3,....,20} K in to disjoint proper subsets such that the sum of the numbers in each subset of the partition is K. Then good number are there (a) 5 (b) 6 (c) 7 (d) 4 |
Answer» Correct option is (b) 6 Let us partition in to n parts and each part has sum = K, then nK = 1 + 2 + 3 .... + 20 nK = 210 ∴K divides 210. Also, K must be ≥ 20 Now, 210 = 2 x 3 x 5 x 7 So, proper divisor of 210 are {1, 2, 3, 5, 6,7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210} ⇒ K can be 21, 30, 35, 42, 70,105 For K = 21, we have (1, 20) (2,19) .....(10, 11) ⇒ 21 is a good number. For K = 42, join two-two pairs For K = 105, join five-five pairs ⇒ 42 and 105 are also good numbers For K = 30, we have {20,10}, {19,11}, {18,12}, {17,13}, {16,14}, {15, 9, 6}, {1, 2, 3, 4, 5, 6, 7, 8} ⇒ K = 30 is also good number Similarly, 35 and 70 also good numbers. ∴There are total 6 good numbers. |
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346. |
Let a = cos 1° and b = sin 1°. We say that a real number is algebraic if it is a root of a polynomial with integer coefficients. Then (A) a is algebraic but b is not algebraic (B) b is algebraic but a is not algebraic (C) both a and b are algebraic (D) neither a nor b is algebraic |
Answer» Correct option (C) both a and b are algebraic Explanation: If cos 1° = p +√q (p, q ∈ Q) then it will be root of a quadratic equation whose other root is p – √q ⇒ cos 1° is algebraic ⇒ sin 1° is algebraic |
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347. |
Let R be a relation on the set of all natural numbers given by a R b ⇔ a divides b2 . Which of the following properties does R satisfy ?I. ReflexivityII. Symmetry III. Transitivity(A) I only(B) III only (C) I and III only (D) I and II only |
Answer» Correct option (A) I only Explanation: (I) This relation is reflexive relation because every natural no. divides square of itself a R a ⇔ a divides a2 (II) not symmetric eg. 5 R 10 ⇒ 5 Divide 100 But 10 R 5 ≠ 10 Divide 25 (III) Not transitivity for example if 8 R 4 & 4 R 2 ≠ 8 R 2 only (I) Option |
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348. |
Suppose p, q, r are real numbers such that q = p (4 – p), r = q (4 – q), p = r (4 – r). The maximum possible value of p + q + r is (A) 0(B) 3(C) 9(D) 27 |
Answer» Correct option (C) 9 Explanation: Add all these p + q + r = p3 + p2 + r2 for maximum value p = 3, q = 3, r = 3 Answer is 9. |
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349. |
Interferons combat viral infection by(A) inhibiting viral packaging directly.(B) increasing the binding of antibodies to viruses.(C) binding to the virus and agglutinating them.(D) restricting viral spread to the neighboring cells. |
Answer» Correct option (D) restricting viral spread to the neighboring cells. Explanation: In a typical scenario, a virus defected cell will release interferons causing near by cells to heighten their anti-viral defense |
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350. |
Which one of the following statements is true about trypsinogen?(a) It is activated by enterokinase(b) It is activated by renin(c) It is activated by pepsin(d) It does not need activation |
Answer» Answer is : (a) It is activated by enterokinase Trypsinogen is an inactive substance secreted by the pancreas, from which the digestive enzyme trypsin is formed in the duodenum. Trypsinogen is converted into its active form trypsin by an enzyme enterokinase. This results in the subsequent activation of pancreatic digestive enzymes. |
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