PREVIOUS-YEAR QUESTIONS

IISER IAT 2021 question paper with solutions

60 questions across Biology, Chemistry, Mathematics and Physics. Read the retyped questions, attempt them independently and open each freshly written solution when ready.

Try each question before opening its answer. Answer letters refer to the option order displayed here. A marked qualification identifies a source defect or a modelling assumption; see the explanation before scoring that item.

Open the displayed answer key
SubjectQuestion: answer
Biology1: A · 2: A · 3: A · 4: A · 5: A† · 6: A · 7: A · 8: A · 9: A · 10: A · 11: A · 12: A · 13: A · 14: A · 15: A
Chemistry1: A · 2: A · 3: A · 4: A · 5: A · 6: A · 7: A · 8: A · 9: A · 10: A · 11: A · 12: A · 13: A · 14: A · 15: A
Mathematics1: A · 2: A · 3: A · 4: A · 5: A · 6: A · 7: A · 8: —† · 9: A · 10: A · 11: A · 12: A · 13: A · 14: A · 15: A
Physics1: A · 2: A · 3: A · 4: A · 5: A · 6: A · 7: A · 8: A · 9: A · 10: A · 11: A · 12: A · 13: A · 14: A · 15: A

— = no valid listed option. † = read the qualification beside the solution.

Biology

Question 1

IAT 2021 · Biology
Which one of the following epithelial cell types is commonly found in the inner surface of the fallopian tubes?
  1. Ciliated
  2. Columnar
  3. Squamous
  4. Cuboidal
Answer & worked solution
Answer A
Cilia on the epithelial lining of the fallopian tube help move the ovum or early embryo towards the uterus. The relevant modification is therefore a ciliated epithelium. Columnar describes cell shape, but the distinctive feature asked for here is the presence of cilia.

Question 2

IAT 2021 · Biology
Which of the following is not an asexual reproductive structure?
  1. Isogametes of Cladophora
  2. Conidia of Penicillium
  3. Zoospores of Chlamydomonas
  4. Gemmules in a sponge
Answer & worked solution
Answer A
Isogametes are similar-looking gametes that fuse in sexual reproduction. Conidia, zoospores and gemmules can each produce a new individual without gamete fusion. Thus the isogametes of Cladophora are the exception.

Question 3

IAT 2021 · Biology
Which one of the following is an example of palindromic DNA sequence?
  1. 5′ GAATTC 3′
    3′ CTTAAG 5′
  2. 5′ GACTTC 3′
    3′ CTGAAG 5′
  3. 5′ GAAGTC 3′
    3′ CTTCAG 5′
  4. 5′ GACCAG 3′
    3′ CTGGTC 5′
Answer & worked solution
Answer A
A DNA palindrome must read the same in the 5′→3′ direction on the two complementary strands. For A, the upper strand reads GAATTC from left to right, and the lower strand reads GAATTC from right to left. The corresponding reverse-complement check fails for B, C and D.

Question 4

IAT 2021 · Biology
Some individuals start sneezing when the pollen content is high in the air. Primarily which Ig isotype will these individuals produce as an immune response?
  1. IgE
  2. IgA
  3. IgM
  4. IgG1
Answer & worked solution
Answer A
Pollen can act as an allergen in a type-I hypersensitivity response. Allergen-specific IgE binds receptors on mast cells; subsequent exposure can trigger mediator release and symptoms such as sneezing. Hence the immunoglobulin is IgE.

Question 5

IAT 2021 · Biology
Which of these factors is least likely to cause deviation from the Hardy-Weinberg equilibrium?
  1. Reduction in population size
  2. Mutation
  3. Gene flow
  4. Genetic drift
Answer & worked solution
Answer A · qualification below

A population bottleneck can cause genetic drift; the selected option is a qualified comparison.

Mutation introduces alleles, gene flow transfers alleles between populations, and genetic drift directly changes allele frequencies by chance. A reduction in population size does not by itself specify a change in allele frequencies, making A the least direct cause among these choices. However, a real population bottleneck can cause drift: A should not be read as saying that small populations always remain in Hardy–Weinberg equilibrium.

Question 6

IAT 2021 · Biology
Motivated by the classic experiment by Frederick Griffith and the work of Avery, Macleod and McCarty to identify the transforming principle, a scientist redesigned and performed the experiment with S-strain and R-strain of Streptococcus pneumonae as summarized in the table below.
TreatmentExperimental condition
T1S-strain (heat-killed) injected into mice
T2S-strain (heat-killed) + R-strain (live) together injected into mice
T3Nucleic acids isolated from S-strain (heat-killed) + R-strain (live), incubated with RNase, and injected into mice
T4Nucleic acids isolated from S-strain (heat-killed) + R-strain (live), incubated with DNase, and injected into mice
T5S-strain (live) injected into mice
T6R-strain (live) injected into mice
From the options below, identify the outcomes of the treatments (T1 to T6) on the viability of the mice.
  1. T1: live, T2: die, T3: die, T4: live, T5: die, T6: live
  2. T1: live, T2: live, T3: die, T4: live, T5: live, T6: live
  3. T1: live, T2: die, T3: die, T4: die, T5: live, T6: die
  4. T1: live, T2: die, T3: live, T4: die, T5: die, T6: live
Answer & worked solution
Answer A
Living S bacteria are virulent, whereas living R bacteria and heat-killed S bacteria alone do not kill the mice. Intact DNA from killed S cells can transform living R cells, so T2 and the RNase-treated T3 are lethal. DNase destroys that transforming DNA, so T4 is non-lethal. In order, the outcomes are live, die, die, live, die, live.

Question 7

IAT 2021 · Biology
Which of the following is/are produced by a plant during photosynthesis with far-red light?
  1. Only ATP
  2. ATP, NADPH and H+
  3. NADPH and H+
  4. Only NADPH
Answer & worked solution
Answer A
Far-red light preferentially excites photosystem I. With photosystem II insufficiently excited, the exam model is cyclic electron transport around PSI: the proton gradient supports ATP synthesis, but there is no net NADPH formation or water splitting. The product listed is therefore ATP alone.

Question 8

IAT 2021 · Biology
The map of a 4000 base pair (bp) plasmid DNA marking the locations of different restriction enzyme cut sites is shown in the figure below. The numbers in brackets indicate the base pair positions where the enzymes cut. This plasmid is completely digested first with the combination of restriction enzymes PstI and HindIII, and then with EcoRI and SalI. The final digested plasmid is analyzed by agarose gel electrophoresis. IAT 2021, Biology, question 8 — question diagramWhich one of the following options correctly represents the bands sizes (in bp) obtained on the gel?
  1. 220, 540, 900 and 2340
  2. 220, 540, 1200 and 2040
  3. 540, 760, 1200 and 1500
  4. 320, 540, 800 and 2340
Answer & worked solution
Answer A
Order the four cut sites around the circular plasmid: 800, 1700, 1920 and 2460 bp. Adjacent distances are 1700800=9001700-800=900, 19201700=2201920-1700=220 and 24601920=5402460-1920=540 bp. The wrap-around fragment has length 40002460+800=23404000-2460+800=2340 bp. These four fragments total 4000 bp and match A.

Question 9

IAT 2021 · Biology
The following plot represents the change in the rate of photosynthesis with leaf temperature for two plants, P1 and P2. IAT 2021, Biology, question 9 — question diagramWhich one of the following statements correctly describes the characteristics of P1 and P2 plants?
  1. Plant P1 is a C3 plant having temperate adaptation whereas Plant P2 is a C4 plant with tropical adaptation
  2. Plant P1 is a C4 plant having temperate adaptation whereas Plant P2 is a C3 plant with tropical adaptation
  3. Plant P1 is a C4 plant having tropical adaptation whereas Plant P2 is a C3 plant with temperate adaptation
  4. Plant P1 is a C3 plant having tropical adaptation whereas Plant P2 is a C4 plant with temperate adaptation
Answer & worked solution
Answer A
P1 peaks at the lower leaf temperature, whereas P2 maintains a higher photosynthetic rate at warmer temperatures. This is the expected comparison between a temperate C3 plant and a tropical C4 plant. The C4 carbon-concentrating mechanism reduces photorespiratory loss under warm conditions.

Question 10

IAT 2021 · Biology
The figure (i) represents two categories of animals (X and Y) with respect to their response to the external environment. The figure (ii) represents two broad categories of animals (Type a and Type b) with respect to their body surface area to volume ratio. IAT 2021, Biology, question 10 — question diagramWhich combination most suitably represents the following animals in the order: humming bird; crocodile; frog; polar bear?
  1. Xa; Yb; Ya; Xb
  2. Xb; Ya; Yb; Xa
  3. Ya; Yb; Xa; Xb
  4. Yb; Xb; Xa; Ya
Answer & worked solution
Answer A
The horizontal response X represents regulation of body temperature; the sloping response Y represents conformity with the surroundings. Type a has the greater surface-area-to-volume ratio and corresponds to smaller animals. Thus humming bird = Xa, crocodile = Yb, frog = Ya and polar bear = Xb.

Question 11

IAT 2021 · Biology
In plants, ammonium ions are produced by protonation of ammonia. Which enzyme uses these ammonium ions to convert an alpha-keto acid into an amino acid?
  1. Glutamate dehydrogenase
  2. Nitrogenase
  3. Transacetylase
  4. Lactate dehydrogenase
Answer & worked solution
Answer A
Reductive amination incorporates ammonium into an alpha-keto acid. Glutamate dehydrogenase catalyses the conversion of alpha-ketoglutarate to glutamate using ammonium and reducing power. Nitrogenase instead reduces molecular nitrogen; the other listed enzymes do not perform this amination.

Question 12

IAT 2021 · Biology
Which one of the following physiological functions is common between the small intestine and the renal tubules?
  1. Absorption of glucose
  2. Excretion of waste materials
  3. Excretion of water
  4. Absorption of proteins
Answer & worked solution
Answer A
Glucose crosses the intestinal epithelium after digestion and is reabsorbed from the filtrate in the renal tubules, mainly in the proximal tubule. Both tissues therefore absorb glucose. Whole-protein absorption and waste excretion are not the common function identified by these options.

Question 13

IAT 2021 · Biology
Match the following pairs of interating species to the corresponding names of the interactions.
Name of interacting speciesName of interaction
iHerbivores and PlantsaMutualism
iiCuckoo and CrowbPredation
iiiSea anemone and Clown fishcParasitism
ivFungus and Cyanobacteria (Lichens)dCommensalism
Pick the correct option from below.
  1. i and b; ii and c; iii and d; iv and a
  2. i and c; ii and b; iii and d; iv and a
  3. i and b; ii and d; iii and c; iv and a
  4. i and b; ii and a; iii and c; iv and d
Answer & worked solution
Answer A
Herbivores consume plant tissue, so i is predation (b). Cuckoo–crow is brood parasitism (c). In the conventional example, clown fish obtain shelter from the sea anemone without a stated effect on it, giving commensalism (d). The fungal and photosynthetic partners of a lichen form a mutualistic association (a).

Question 14

IAT 2021 · Biology
The following events are associated with meiosis:
  1. appearance of recombinant nodules
  2. formation of meiotic spindle
  3. formation of chiasmata
  4. formation of synaptonemal complex
Which of the following is the correct sequence of these events during meiosis?
  1. iv → i→ iii→ ii
  2. iii → iv→ ii→ i
  3. iv → iii→ i→ ii
  4. i → iii→ iv→ ii
Answer & worked solution
Answer A
Synapsis and synaptonemal-complex formation occur in zygotene. Recombination nodules are associated with pachytene; chiasmata become evident in diplotene. Spindle formation follows later in prophase I. The sequence is therefore iv → i → iii → ii.

Question 15

IAT 2021 · Biology
Which of the following floral formulae represents a zygomorphic flower with diadelphous androecium and superior ovary?
  1. IAT 2021, Biology, question 15 — option A
  2. IAT 2021, Biology, question 15 — option B
  3. IAT 2021, Biology, question 15 — option C
  4. IAT 2021, Biology, question 15 — option D
Answer & worked solution
Answer A
The percent-like symmetry symbol denotes a zygomorphic flower. A(9)+1A_{(9)+1} represents stamens arranged in two bundles, which is diadelphous, and the underline beneath G denotes a superior ovary. Only formula A combines all three required features; the other formulae lack the specified androecium.

Chemistry

Question 1

IAT 2021 · Chemistry
The F-P-Cl bond angles in the most stable structure of PF3Cl2 are close to
  1. 90◦ and 120◦.
  2. 90◦, 120◦, and 180◦.
  3. 90◦ only.
  4. 90◦ and 180◦.
Answer & worked solution
Answer A
The molecule has a trigonal-bipyramidal arrangement. The more electronegative fluorine atoms preferentially occupy the two axial sites; the third F and the two Cl atoms occupy equatorial sites. An axial F makes a 90° angle with an equatorial Cl, while equatorial F and Cl make 120°. There is no axial Cl to give an F–P–Cl angle of 180°.

Question 2

IAT 2021 · Chemistry
The correct statement about the bond angles and bond lengths in Al2Cl6 is
(Clt = terminal Cl; Clb = bridging Cl)
  1. CltAlClt>ClbAlClb\angle\mathrm{Cl_t-Al-Cl_t}>\angle\mathrm{Cl_b-Al-Cl_b} and AlClb>AlClt\mathrm{Al-Cl_b}>\mathrm{Al-Cl_t}.
  2. CltAlClt>ClbAlClb\angle\mathrm{Cl_t-Al-Cl_t}>\angle\mathrm{Cl_b-Al-Cl_b} and AlClt>AlClb\mathrm{Al-Cl_t}>\mathrm{Al-Cl_b}.
  3. CltAlClt=ClbAlClb\angle\mathrm{Cl_t-Al-Cl_t}=\angle\mathrm{Cl_b-Al-Cl_b} and AlClt>AlClb\mathrm{Al-Cl_t}>\mathrm{Al-Cl_b}.
  4. ClbAlClb>CltAlClt\angle\mathrm{Cl_b-Al-Cl_b}>\angle\mathrm{Cl_t-Al-Cl_t} and AlClb>AlClt\mathrm{Al-Cl_b}>\mathrm{Al-Cl_t}.
Answer & worked solution
Answer A
Each Al centre has two terminal chlorides and two bridging chlorides. The bridging Al–Cl contacts are longer and weaker than the terminal bonds. Bridging also compresses the Cl_b–Al–Cl_b angle relative to the terminal Cl_t–Al–Cl_t angle. Both comparisons are those in A.

Question 3

IAT 2021 · Chemistry
Among CH3SiCl3, (CH3)2SiCl2 and (CH3)3SiCl, which one is used to synthesize straight chain (linear) and which one is used to prepare branched chain (cross-linked) silicone polymer, respectively?
  1. (CH3)2SiCl2 and (CH3)SiCl3
  2. (CH3)SiCl3 and (CH3)2SiCl2
  3. (CH3)2SiCl2 and (CH3)3SiCl
  4. (CH3)3SiCl and (CH3)2SiCl2
Answer & worked solution
Answer A
Hydrolysis of (CH3)2SiCl2(\mathrm{CH_3})_2\mathrm{SiCl_2} gives a silicon unit with two linking sites, allowing a linear siloxane chain. CH3SiCl3\mathrm{CH_3SiCl_3} provides three linking sites and can form a cross-linked network. The monofunctional (CH3)3SiCl(\mathrm{CH_3})_3\mathrm{SiCl} tends to terminate chains rather than build the requested network.

Question 4

IAT 2021 · Chemistry
Which one of the following statements is INCORRECT about a complex of a divalent ion with atomic number 25? (BM = Bohr Magneton)
  1. Complex with weak field ligands in tetrahedral geometry has a magnetic moment of 1.73 BM.
  2. Complex with weak field ligands in tetrahedral geometry has a magnetic moment of 5.92 BM.
  3. Complex with strong field ligands in octahedral geometry has a magnetic moment of 1.73 BM.
  4. Complex with weak field ligands in octahedral geometry has a magnetic moment of 5.92 BM.
Answer & worked solution
Answer A
Atomic number 25 identifies manganese, so the divalent ion is Mn2+\mathrm{Mn^{2+}}, a d5d^5 ion. In a weak-field tetrahedral complex all five d electrons are unpaired, giving μ=5(5+2)=355.92\mu=\sqrt{5(5+2)}=\sqrt{35}\approx5.92 BM. A moment of 1.73 BM corresponds to one unpaired electron and is not correct for the stated tetrahedral case.

Question 5

IAT 2021 · Chemistry
Ni(CO)4 is diamagnetic because
  1. Ni has completely filled 3d orbitals.
  2. it is a square planar complex.
  3. CO is a strong field ligand.
  4. it has synergic bonding.
Answer & worked solution
Answer A
In Ni(CO)4\mathrm{Ni(CO)_4}, nickel is in oxidation state zero and the occupied metal d shell is d10d^{10}. All electrons are paired, so the complex is diamagnetic. It is tetrahedral rather than square planar; merely naming strong-field or synergic bonding does not identify the electron-count reason.

Question 6

IAT 2021 · Chemistry
The correct order for the relative energies of the following sawhorse conformations isIAT 2021, Chemistry, question 6 — question diagram
  1. IV > III = II > I.
  2. IV > III > II > I.
  3. II≈ III > IV > I.
  4. I > II = III > IV.
Answer & worked solution
Answer A
Conformation I is the anti staggered arrangement and has the lowest steric and torsional energy. II and III are equivalent eclipsed arrangements involving methyl–hydrogen eclipsing. IV places the two methyl groups in the most unfavourable eclipsed relationship. Thus EIV>EIII=EII>EIE_{\mathrm{IV}}>E_{\mathrm{III}}=E_{\mathrm{II}}>E_{\mathrm I}.

Question 7

IAT 2021 · Chemistry
Among the haloarenes (I–IV) shown below, the correct order of reactivity for the substitution reaction with NaOH isIAT 2021, Chemistry, question 7 — question diagram
  1. II > III≈ I > IV.
  2. I > IV > III > II.
  3. II > III > IV > I.
  4. IV > I≈ III > II.
Answer & worked solution
Answer A
A para nitro group stabilises the anionic intermediate in nucleophilic aromatic substitution by resonance, strongly activating II. A meta nitro group cannot provide the same resonance stabilisation, so III is much less activated and is grouped approximately with I in the options. Para methoxy donates electron density and opposes the substitution, placing IV last.

Question 8

IAT 2021 · Chemistry
Consider the following reaction sequence,IAT 2021, Chemistry, question 8 — question diagramand identify the correct statements:
  1. The products P and R can form addition product with HCN.
  2. The product Q has no chiral center.
  3. The product R can undergo Cannizzaro reaction.
  4. The product S is a saturated hydrocarbon.
  1. (i) and (iv)
  2. (iii) and (iv)
  3. (i) and (ii)
  4. (ii) and (iii)
Answer & worked solution
Answer A
Rosenmund reduction changes propanoyl chloride into propanal P. Addition of MeMgBr followed by acid gives 2-butanol Q, which has a chiral carbon. Oxidation produces butan-2-one R; reduction of its carbonyl gives butane S. P and R both add HCN, R does not undergo the Cannizzaro reaction, and S is saturated. Therefore only statements (i) and (iv) are correct.

Question 9

IAT 2021 · Chemistry
The major product of the reaction of salicylaldehyde with one equivalent of MeMgBr, followed by acid neutralization isIAT 2021, Chemistry, question 9 — question diagram
  1. II.
  2. I.
  3. IV.
  4. III.
Answer & worked solution
Answer A
One equivalent of MeMgBr is consumed by the acidic phenolic hydrogen: the Grignard reagent acts as a base and produces methane. There is then no remaining equivalent available to add to the aldehyde. Acid work-up reprotonates the phenoxide, returning salicylaldehyde, structure II.

Question 10

IAT 2021 · Chemistry
The total number of stereoisomers possible for the following structure is CH3CH(OH)CH(OH)HC=CHCH3\mathrm{CH_3CH(OH)CH(OH)HC{=}CHCH_3}
  1. 8 .
  2. 4 .
  3. 2 .
  4. 6 .
Answer & worked solution
Answer A
The two CH(OH) carbons are stereogenic and contribute 22=42^2=4 configurations. The alkene has two different substituents on each double-bond carbon, so it also has E and Z forms. The unsymmetrical ends prevent a meso reduction. Hence the total is 4×2=84\times2=8.

Question 11

IAT 2021 · Chemistry
For a homogeneous reaction involving ideal gases, equilibrium constants at 27C27^\circ\mathrm C are Kp=9.98×1027K_p=9.98\times10^{27} and Kc=4.0×1024K_c=4.0\times10^{24}. If the enthalpy change of the reaction is 18.4 kJmol1-18.4\ \mathrm{kJ\,mol^{-1}}, the internal energy change is
[Use R=8.314 JK1mol1R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}]
  1. 20.89 kJmol1-20.89\ \mathrm{kJ\,mol^{-1}}.
  2. 15.91 kJmol1-15.91\ \mathrm{kJ\,mol^{-1}}.
  3. 13.41 kJmol1-13.41\ \mathrm{kJ\,mol^{-1}}.
  4. 23.39 kJmol1-23.39\ \mathrm{kJ\,mol^{-1}}.
Answer & worked solution
Answer A
Using the stated numerical convention, Kp/Kc=2495RTK_p/K_c=2495\approx RT at 300 K, so Δng=1\Delta n_g=1 in Kp=Kc(RT)ΔngK_p=K_c(RT)^{\Delta n_g}. For ideal gases, ΔH=ΔU+ΔngRT\Delta H=\Delta U+\Delta n_gRT. Therefore ΔU=18.4(8.314×300/1000)=20.8942 kJmol1\Delta U=-18.4-(8.314\times300/1000)=-20.8942\ \mathrm{kJ\,mol^{-1}}, which rounds to A.

Question 12

IAT 2021 · Chemistry
The probability density of an electron in 1s orbital for an H atom is maximum
  1. at the nucleus.
  2. at the Bohr radius.
  3. at twice the Bohr radius.
  4. at infinite distance.
Answer & worked solution
Answer A
The 1s probability density is proportional to e2r/a0e^{-2r/a_0} and decreases continuously with distance from the nucleus. It is therefore greatest at r=0r=0. Do not confuse this density with the radial probability distribution 4πr2ψ24\pi r^2|\psi|^2, whose maximum is at the Bohr radius.

Question 13

IAT 2021 · Chemistry
Which of the following statements is INCORRECT?
  1. Half-life of a zero order reaction is proportional to the rate constant.
  2. Half-life of a zero order reaction is proportional to the initial concentration of reactant.
  3. Half-life of a first order reaction is independent of the initial concentration of reactant.
  4. Half-life of a first order reaction is inversely proportional to the rate constant.
Answer & worked solution
Answer A
For a zero-order reaction, [A]=[A]0kt[A]=[A]_0-kt, giving t1/2=[A]0/(2k)t_{1/2}=[A]_0/(2k). Its half-life is proportional to initial concentration but inversely proportional to the rate constant. For first order, t1/2=ln2/kt_{1/2}=\ln2/k. Thus A, which says proportional to the rate constant for zero order, is the incorrect statement.

Question 14

IAT 2021 · Chemistry
The time required for reducing 1 mole of MnO4(aq.)\mathrm{MnO_4^-}(\mathrm{aq.}) to Mn2+(aq.)\mathrm{Mn^{2+}}(\mathrm{aq.}) when a current of 2.5 A2.5\ \mathrm A is passed during electrolysis is
  1. 53.6 hours.
  2. 12.9 microeconds.
  3. 10.7 hours.
  4. 64.8 microseconds.
Answer & worked solution
Answer A
Manganese changes from +7 in permanganate to +2, requiring five electrons per ion. Reducing one mole therefore requires Q=5F482425Q=5F\approx482425 C. With I=2.5I=2.5 A, t=Q/I192970t=Q/I\approx192970 s, or 192970/360053.6192970/3600\approx53.6 h.

Question 15

IAT 2021 · Chemistry
Based on the relative stabilities and barriers, which among the following schematic energy diagrams would best correspond to the reaction scheme shown below? [kk and kk\prime are similar in magnitude]IAT 2021, Chemistry, question 15 — question diagram
  1. IAT 2021, Chemistry, question 15 — option A
  2. IAT 2021, Chemistry, question 15 — option B
  3. IAT 2021, Chemistry, question 15 — option C
  4. IAT 2021, Chemistry, question 15 — option D
Answer & worked solution
Answer A
For X ⇌ Y, the equilibrium ratio is 4k/(2k)=2>14k/(2k)=2>1, so Y lies below X in free energy. For Y ⇌ Z the ratio is 8k/(3k)=8/3>18k'/(3k')=8/3>1, placing Z below Y. The second forward step is faster for comparable k and k′, consistent with a smaller forward barrier in the intended schematic. A is the only diagram with the required descending free-energy order and barrier pattern.

Mathematics

Question 1

IAT 2021 · Mathematics
What is the maximum value of cos4(x)+sin2(x)+cos(x)\cos^4(x)+\sin^2(x)+\cos(x), when x0x\geq0?
  1. 2
  2. 1
  3. 2\sqrt2
  4. 222\sqrt2
Answer & worked solution
Answer A
Put u=cosxu=\cos x, so 1u1-1\leq u\leq1 and the expression is u4+1u2+uu^4+1-u^2+u. Since u4u2u^4\leq u^2 and u1u\leq1, this is at most 2. Equality holds at u=1u=1, for example x=0x=0, so the maximum is attained and equals 2.

Question 2

IAT 2021 · Mathematics
Out of a pack of ten cards numbered 1 to 10, a boy draws a card at random and keeps it back. Then a girl draws a card at random from the same pack. If the boy’s card reads m, and the girl’s card reads n, then what is the probability that m>n , given that m is even?
  1. 12\frac12
  2. 13\frac13
  3. 14\frac14
  4. 15\frac15
Answer & worked solution
Answer A
Given that m is even, m is uniformly distributed over 2, 4, 6, 8 and 10. For these five values, respectively 1, 3, 5, 7 and 9 of the ten possible n values satisfy n < m. Hence the conditional probability is (1+3+5+7+9)/(5×10)=25/50=1/2(1+3+5+7+9)/(5\times10)=25/50=1/2.

Question 3

IAT 2021 · Mathematics
Suppose a,bRa,b\in\mathbb R are such that the points (a,b)(a,b), (a2,b2)(a^2,b^2), and (a3,b3)(a^3,b^3) in the coordinate plane are distinct, collinear, and the line passing through these points is not parallel to the yy-axis. Then, for all such choices of (a,b)(a,b), the slope of the line passing through these three points can take
  1. exactly two values.
  2. infinitely many values.
  3. exactly three values.
  4. exactly one value.
Answer & worked solution
Answer A
The collinearity determinant factors as ab(a1)(b1)(ba)ab(a-1)(b-1)(b-a). Values a=0a=0 or a=1a=1 force a vertical line or coincident points and are excluded. Thus the remaining possibilities are b=0b=0, b=1b=1, or b=ab=a. The first two give horizontal lines of slope 0, and the last gives slope 1. Both slopes occur for valid choices, for example (a,b)=(2,0)(a,b)=(2,0) and (2,2)(2,2).

Question 4

IAT 2021 · Mathematics
Define f:[2,22]Rf:[2,22]\to\mathbb R by f(x)=max{n(1x(2n+1)):n=1,2,,10}f(x)=\max\{n(1-|x-(2n+1)|):n=1,2,\ldots,10\}. The area of the region {(x,y):0yf(x),x[2,22]}\{(x,y):0\leq y\leq f(x),x\in[2,22]\} is
  1. 55.
  2. 60.
  3. 5.
  4. 10.
Answer & worked solution
Answer A
On [2n,2n+2][2n,2n+2], the nth expression is a non-negative triangular tent of base 2 and height n. Every other expression is non-positive in its interior, so the maximum equals this tent. The ten intervals tile [2,22]. Their total area is n=11012(2)n=1++10=55\sum_{n=1}^{10}\frac12(2)n=1+\cdots+10=55.

Question 5

IAT 2021 · Mathematics
The value of the limit limx0+(sinx)x(ex+x)1/x\displaystyle\lim_{x\to0^+}(\sin x)^{\sqrt x}(e^x+x)^{1/x} is
  1. e2e^2.
  2. 1.
  3. ee.
  4. 0.
Answer & worked solution
Answer A
Take the logarithm of the positive expression. It becomes xln(sinx)+ln(ex+x)/x\sqrt{x}\ln(\sin x)+\ln(e^x+x)/x. The first term tends to zero because sinxx\sin x\sim x and xlnx0\sqrt{x}\ln x\to0. Since ex+x=1+2x+O(x2)e^x+x=1+2x+O(x^2), the second term tends to 2. Exponentiating gives e2e^2.

Question 6

IAT 2021 · Mathematics
Let f:RRf:\mathbb R\to\mathbb R be a continuous function satisfying f(x)=ex2/2+0xtf(t)dtfor all x.f(x)=e^{x^2/2}+\int_0^x tf(t)\,dt\quad\text{for all }x. Then which of the following is correct?
  1. 5<f(2)<65<f(\sqrt2)<6
  2. 2<f(2)<32<f(\sqrt2)<3
  3. 3<f(2)<43<f(\sqrt2)<4
  4. 4<f(2)<54<f(\sqrt2)<5
Answer & worked solution
Answer A
The fundamental theorem of calculus gives f(x)=xex2/2+xf(x)f'(x)=xe^{x^2/2}+xf(x) and f(0)=1f(0)=1. Multiply the differential equation by ex2/2e^{-x^2/2}: ddx(fex2/2)=x\frac{d}{dx}(fe^{-x^2/2})=x. Integration gives f(x)=ex2/2(1+x2/2)f(x)=e^{x^2/2}(1+x^2/2). At x=2x=\sqrt2 this is 2e5.4362e\approx5.436, between 5 and 6.

Question 7

IAT 2021 · Mathematics
For each aRa\in\mathbb R, define pa(z)=z2+2eaeaz+eaea.p_a(z)=z^2+2e^{a-e^a}z+e^{a-e^a}. Then, which one of the followings is true?
  1. pap_a has only non-real complex roots for all aRa\in\mathbb R.
  2. pap_a has a real root for all aRa\in\mathbb R.
  3. pap_a has a real root if and only if a1a\geq1.
  4. pap_a has a real root if and only if a1a\leq-1.
Answer & worked solution
Answer A
Let t=eaeat=e^{a-e^a}. The inequality ea1+ae^a\geq1+a gives 0<te1<10<t\leq e^{-1}<1. The quadratic is z2+2tz+tz^2+2tz+t, with discriminant 4t24t=4t(t1)<04t^2-4t=4t(t-1)<0. It therefore has two non-real conjugate roots for every real a.

Question 8

IAT 2021 · Mathematics
The number of functions f:{1,2,3,4,5}{1,2,3,4,5}f:\{1,2,3,4,5\}\longrightarrow\{1,2,3,4,5\} such that f(f(n))=nf(f(n))=n for all n{1,2,3,4,5}n\in\{1,2,3,4,5\}, is
  1. 41.
  2. 25.
  3. 31.
  4. 120.
Answer & worked solution
No valid listed option

Correct count 26 is absent from all four printed options.

The condition ff=idf\circ f=\mathrm{id} makes f a permutation whose cycles have lengths 1 or 2. There is 1 identity permutation, (52)=10\binom52=10 choices with one transposition, and (54)×3=15\binom54\times3=15 choices with two disjoint transpositions. The total is 1+10+15=261+10+15=26. This is absent from the printed options 41, 25, 31 and 120. The question and options are retained exactly; none of A–D is mathematically correct.

Question 9

IAT 2021 · Mathematics
Let P1:x+y+z=1P_1:x+y+z=1, P2:2x+y+z=3P_2:2x+y+z=3 be two planes, and let LL denote the line of intersection of P1P_1 and P2P_2. Let PP be the plane passing through the point (1,2,1)(1,2,1), and normal to LL. Which of the following equations represents PP?
  1. yz=1y-z=1
  2. x+z=2x+z=2
  3. x+2y+z=6x+2y+z=6
  4. x+y+2z=5x+y+2z=5
Answer & worked solution
Answer A
Normals to the two given planes are (1,1,1)(1,1,1) and (2,1,1)(2,1,1). Their cross product, (0,1,1)(0,1,-1), is a direction vector for their intersection L. A plane normal to L has this vector as its normal. Through (1,2,1), its equation is (y2)(z1)=0(y-2)-(z-1)=0, or yz=1y-z=1.

Question 10

IAT 2021 · Mathematics
Let f(x)=ln(1+x)f(x)=\ln(1+x) for x0x\geq0. The value of 0π/2f(cosθ3)f(sinθ3)+f(cosθ3)dθ\int_0^{\pi/2}\frac{f(\sqrt[3]{\cos\theta})}{f(\sqrt[3]{\sin\theta})+f(\sqrt[3]{\cos\theta})}\,d\theta is
  1. π4\frac\pi4.
  2. π6\frac\pi6.
  3. π3\frac\pi3.
  4. π2\frac\pi2.
Answer & worked solution
Answer A
Call the integral I. Under θπ/2θ\theta\mapsto\pi/2-\theta, sine and cosine exchange, so the transformed numerator is the other term in the unchanged denominator. Adding the two equal integrals gives 2I=0π/21dθ=π/22I=\int_0^{\pi/2}1\,d\theta=\pi/2. Hence I=π/4I=\pi/4.

Question 11

IAT 2021 · Mathematics
The area enclosed by the curves y=1+sinxy=1+|\sin x|, y=sinxy=-|\sin x| and the lines x=0x=0, x=2πx=2\pi is
  1. 8+2π8+2\pi.
  2. 8+4π8+4\pi.
  3. 8+6π8+6\pi.
  4. 8+8π8+8\pi.
Answer & worked solution
Answer A
The upper curve minus the lower curve is 1+2sinx1+2|\sin x|. On one full cycle, 02πsinxdx=4\int_0^{2\pi}|\sin x|\,dx=4. Therefore the enclosed area is 2π+2(4)=2π+82\pi+2(4)=2\pi+8.

Question 12

IAT 2021 · Mathematics
Consider the function f:RRf:\mathbb R\to\mathbb R defined by f(x)=x2x.f(x)=x^2|x|. Then, which of the following statements is correct?
  1. ff' is differentiable but ff'' is not differentiable.
  2. ff is continuous but not differentiable.
  3. ff is differentiable but ff' is not differentiable.
  4. ff'' is differentiable.
Answer & worked solution
Answer A
Write f(x)=x3f(x)=|x|^3. Its first derivative is 3xx3x|x| and its second derivative is 6x6|x|, including value zero at the origin. Thus f′ is differentiable everywhere. The left and right slopes of f″ at zero are −6 and +6, so f″ is not differentiable there.

Question 13

IAT 2021 · Mathematics
Let AA be a 2×22\times2 matrix such that A2+A+(0110)=(0000).A^2+A+\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}. Let II denote the 2×22\times2 identity matrix. Which of the following statements is correct?
  1. Both A and A +I are invertible.
  2. A is invertible but A +I may not be invertible.
  3. A +I is invertible but A may not be invertible.
  4. Neither A +I nor A may be invertible.
Answer & worked solution
Answer A
Rearrange the equation as A(A+I)=JA(A+I)=-J, where J=(0110)J=\begin{pmatrix}0&1\\1&0\end{pmatrix}. Since det(J)=10\det(-J)=-1\ne0, the product has nonzero determinant. The identity det(A(A+I))=detAdet(A+I)\det(A(A+I))=\det A\det(A+I) then forces both factors to have nonzero determinants, so both matrices are invertible.

Question 14

IAT 2021 · Mathematics
If three real numbers a,b,ca,b,c are in arithmetic progression, the value of the determinant x2+3x2+4x2+5x2+4x2+5x2+6x2+ax2+bx2+c\begin{vmatrix}x^2+3&x^2+4&x^2+5\\x^2+4&x^2+5&x^2+6\\x^2+a&x^2+b&x^2+c\end{vmatrix} is
  1. 0.
  2. 2a2a.
  3. a+cba+c-b.
  4. x2+2bx^2+2b.
Answer & worked solution
Answer A
Let the columns be C1, C2 and C3. In each of the first two rows, first entry minus twice the second plus the third is zero. The third row gives a2b+c=0a-2b+c=0 because a, b and c are in arithmetic progression. Hence C12C2+C3=0C_1-2C_2+C_3=0 and the determinant is zero.

Question 15

IAT 2021 · Mathematics
Consider the two data sets S1={1,2,4,8,9,11,15,20,27,29,33},S_1=\{1,2,4,8,9,11,15,20,27,29,33\}, S2={51,52,54,58,59,61,65,70,77,79,83}.S_2=\{51,52,54,58,59,61,65,70,77,79,83\}. Let m1,m2m_1,m_2 and v1,v2v_1,v_2 be the means and the variances of S1S_1 and S2S_2, respectively. Then, which of the following relations is correct?
  1. m2=m1+50, v2=v1m_2=m_1+50,\ v_2=v_1.
  2. m2=m1+50, v2=v1+50m_2=m_1+50,\ v_2=v_1+50.
  3. m2=m1, v2=v1m_2=m_1,\ v_2=v_1.
  4. m2=m1+50, v2<v1m_2=m_1+50,\ v_2<v_1.
Answer & worked solution
Answer A
Every observation in S2 is the corresponding observation in S1 plus 50. The mean therefore increases by 50. Deviations from the mean are unchanged: (x+50)(m1+50)=xm1(x+50)-(m_1+50)=x-m_1. Their squared average, the variance, stays the same.

Physics

Question 1

IAT 2021 · Physics
A door of mass MM and width LL is hinged at one end and rotates about a vertical axis without friction. A bullet of mass mm (mM)(m\ll M) fired perpendicularly to the door at a speed vv gets embedded in it at a distance xx from its axis of rotation. Assuming the door was stationary initially, how does the resultant angular speed ω\omega of the door vary as a function of xx?
  1. IAT 2021, Physics, question 1 — option A
  2. IAT 2021, Physics, question 1 — option B
  3. IAT 2021, Physics, question 1 — option C
  4. IAT 2021, Physics, question 1 — option D
Answer & worked solution
Answer A
Conserve angular momentum about the hinge during impact: mvx=(ML2/3+mx2)ωmvx=(ML^2/3+mx^2)\omega. Because mMm\ll M and 0xL0\leq x\leq L, the bullet's moment of inertia is negligible, so ω3mvx/(ML2)\omega\approx3mvx/(ML^2), a straight-line increase. For x beyond the door width the bullet misses it, explaining the zero branch beyond L. This is the approximation represented by A.

Question 2

IAT 2021 · Physics
A body of mass mm executes simple harmonic motion along a line with time period TT and energy EE. What is the magnitude of the maximum acceleration of the body?
  1. 22πTEm\frac{2\sqrt2\pi}{T}\sqrt{\frac Em}
  2. 2πTEm\frac{2\pi}{T}\sqrt{\frac Em}
  3. 2TEm\frac{\sqrt2}{T}\sqrt{\frac Em}
  4. πTEm\frac{\pi}{T}\sqrt{\frac Em}
Answer & worked solution
Answer A
For SHM, E=12mω2A2E=\frac12m\omega^2A^2, so ωA=2E/m\omega A=\sqrt{2E/m}. The maximum acceleration is ω2A=ω2E/m\omega^2A=\omega\sqrt{2E/m}. Using ω=2π/T\omega=2\pi/T gives amax=(22π/T)E/ma_{\mathrm{max}}=(2\sqrt2\pi/T)\sqrt{E/m}.

Question 3

IAT 2021 · Physics
An athlete runs on a straight track. She starts from rest and runs with a constant acceleration for the first 2 seconds, reaching a speed of 9 ms19\ \mathrm{m\,s^{-1}}. She then continues at this constant speed for some time before slowing down to a halt at a constant deceleration. The total time taken, from start to finish, is 12 seconds. If the magnitude of her acceleration is twice the magnitude of deceleration, then what is the total distance covered by her?
  1. 81 m
  2. 108 m
  3. 90 m
  4. 72 m
Answer & worked solution
Answer A
The initial acceleration is 9/2=4.5 ms29/2=4.5\ \mathrm{m\,s^{-2}}, so the deceleration magnitude is 2.25 and stopping takes 9/2.25=49/2.25=4 s. Constant-speed running lasts 1224=612-2-4=6 s. The three distances are 12(2)(9)=9\frac12(2)(9)=9 m, 6(9)=546(9)=54 m and 12(4)(9)=18\frac12(4)(9)=18 m, totalling 81 m.

Question 4

IAT 2021 · Physics
A capacitor of capacitance CC consists of two large parallel metal plates. The coefficient of linear expansion of the metal is α\alpha. What is the change in capacitance if the temperature of the plates rises by ΔT\Delta T, while the gap between the plates is kept fixed?
  1. 2αΔTC2\alpha\Delta T C
  2. αΔTC\alpha\Delta T C
  3. αΔTC-\alpha\Delta T C
  4. 2αΔTC-2\alpha\Delta T C
Answer & worked solution
Answer A
With fixed plate spacing, capacitance is proportional to plate area: C=εA/dC=\varepsilon A/d. Each linear dimension expands by the factor 1+αΔT1+\alpha\Delta T, so area becomes A(1+αΔT)2A(1+2αΔT)A(1+\alpha\Delta T)^2\approx A(1+2\alpha\Delta T). To first order in thermal expansion, ΔC=2αΔTC\Delta C=2\alpha\Delta T C.

Question 5

IAT 2021 · Physics
What are the charges stored in the 2μF2\,\mu\mathrm F and 4μF4\,\mu\mathrm F capacitors in the given circuit a long time after the key K is closed?IAT 2021, Physics, question 5 — question diagram
  1. 8μC8\,\mu\mathrm C and 8μC8\,\mu\mathrm C respectively
  2. 18μC18\,\mu\mathrm C and 18μC18\,\mu\mathrm C respectively
  3. 163μC\frac{16}{3}\,\mu\mathrm C and 83μC\frac83\,\mu\mathrm C respectively
  4. 83μC\frac83\,\mu\mathrm C and 163μC\frac{16}{3}\,\mu\mathrm C respectively
Answer & worked solution
Answer A
At steady state the capacitor branch carries no current. The upper and lower resistor combinations are each 6 kΩ, so the 12 V supply produces a 6 V difference between the two side nodes. The series capacitors have Ceq=2×4/(2+4)=4/3μFC_{\rm eq}=2\times4/(2+4)=4/3\,\mu\mathrm F. Both carry equal charge Q=CeqV=8μCQ=C_{\rm eq}V=8\,\mu\mathrm C.

Question 6

IAT 2021 · Physics
Consider a point charge +q+q moving with a constant velocity v=vk^\vec v=v\hat k in vacuum in the presence of an electric field E=Exi^+Eyj^\vec E=E_x\hat i+E_y\hat j and a magnetic field B=Bxi^+Byj^\vec B=B_x\hat i+B_y\hat j. Unit vectors i^\hat i, j^\hat j, and k^\hat k are in the directions of xx, yy, and zz axes, respectively. Which of the following relations is correct?
  1. Ex=vBy, Ey=vBxE_x=vB_y,\ E_y=-vB_x
  2. Ex=vBx, Ey=vByE_x=-vB_x,\ E_y=-vB_y
  3. Ex=vBy, Ey=vBxE_x=-vB_y,\ E_y=vB_x
  4. Ex=vBx, Ey=vByE_x=vB_x,\ E_y=vB_y
Answer & worked solution
Answer A
Constant velocity requires zero net Lorentz force, so E=v×B\vec E=-\vec v\times\vec B. With v=vk^\vec v=v\hat k, the cross product is vBxj^vByi^vB_x\hat j-vB_y\hat i. Therefore Ex=vByE_x=vB_y and Ey=vBxE_y=-vB_x.

Question 7

IAT 2021 · Physics
Four point charges q,2q,3qq,-2q,-3q and 4q4q are placed at the four vertices of a regular tetrahedron of side LL, while a charge 5q5q is placed at its center. What is the total electrostatic energy of the system?
(Vacuum permittivity is denoted by ε0\varepsilon_0.)
  1. 15q24πε0L-\frac{15q^2}{4\pi\varepsilon_0 L}
  2. 15q24πε0L\frac{15q^2}{4\pi\varepsilon_0 L}
  3. 30q24πε0L-\frac{30q^2}{4\pi\varepsilon_0 L}
  4. 30q24πε0L\frac{30q^2}{4\pi\varepsilon_0 L}
Answer & worked solution
Answer A
The four vertex charges sum to zero, and each is the same distance from the centre. Their total interaction with the central charge therefore cancels. For vertex pairs, i<jqiqj=12[(qi)2qi2]=(1+4+9+16)q2/2=15q2\sum_{i<j}q_iq_j=\frac12[(\sum q_i)^2-\sum q_i^2]=-(1+4+9+16)q^2/2=-15q^2. Dividing by the common separation factor gives U=15q2/(4πε0L)U=-15q^2/(4\pi\varepsilon_0L).

Question 8

IAT 2021 · Physics
The half life of a radioactive element is 2000 hours. Approximately how much time is required for the decay of 2/3 of its nuclei?
  1. 3170 hours
  2. 3000 hours
  3. 1170 hours
  4. 2830 hours
Answer & worked solution
Answer A
If two-thirds have decayed, one-third remains. From N/N0=2t/2000=1/3N/N_0=2^{-t/2000}=1/3, obtain t=2000ln3/ln23169.9t=2000\ln3/\ln2\approx3169.9 h. The closest listed value is 3170 hours.

Question 9

IAT 2021 · Physics
A 2-input exclusive OR (XOR) gate with inputs XX and YY produces the output XˉY+XYˉ\bar X Y+X\bar Y. In the Boolean circuit shown below, which values of the inputs PP and QQ will produce the output 0?IAT 2021, Physics, question 9 — question diagram
  1. P=0,Q=1P=0,Q=1
  2. P=0,Q=0P=0,Q=0
  3. P=1,Q=0P=1,Q=0
  4. P=1,Q=1P=1,Q=1
Answer & worked solution
Answer A
The upper gate produces 1P=Pˉ1\oplus P=\bar P; the lower gate produces 1Q=Q1\cdot Q=Q. A NAND gate has output zero only when both its inputs are one. Thus Pˉ=1\bar P=1 and Q=1Q=1, requiring P=0,Q=1P=0,Q=1.

Question 10

IAT 2021 · Physics
In an experiment on the photoelectric effect, the de Broglie wavelength of the emitted electron is λB\lambda_B. The energy of the photon incident on the metal is five times the work function. If hh is Planck’s constant and mem_e is the electron mass, then what is the work function?
  1. h28meλB2\frac{h^2}{8m_e\lambda_B^2}
  2. h210meλB2\frac{h^2}{10m_e\lambda_B^2}
  3. h212meλB2\frac{h^2}{12m_e\lambda_B^2}
  4. h24meλB2\frac{h^2}{4m_e\lambda_B^2}
Answer & worked solution
Answer A
Let the work function be W. The photon supplies 5W, leaving maximum kinetic energy K=5WW=4WK=5W-W=4W. With de Broglie wavelength λB\lambda_B, K=h2/(2meλB2)K=h^2/(2m_e\lambda_B^2) in the nonrelativistic exam model. Therefore W=h2/(8meλB2)W=h^2/(8m_e\lambda_B^2).

Question 11

IAT 2021 · Physics
An ambulance traveling at a speed 20 ms120\ \mathrm{m\,s^{-1}} emits a sound of frequency 540 Hz from its siren. Sunanda is driving a car which approaches the ambulance from the opposite direction at a speed of 20 ms120\ \mathrm{m\,s^{-1}}. What will be the change in detected frequency by Sunanda, as she crosses the ambulance? (Given, the speed of sound in air is 340 ms1340\ \mathrm{m\,s^{-1}}.)
  1. 127.5 Hz
  2. 128.8 Hz
  3. 135.5 Hz
  4. 72.00 Hz
Answer & worked solution
Answer A
Before crossing, both motions increase the detected frequency: fapp=540(340+20)/(34020)=607.5f_{\rm app}=540(340+20)/(340-20)=607.5 Hz. Afterwards both motions decrease it: frec=540(34020)/(340+20)=480f_{\rm rec}=540(340-20)/(340+20)=480 Hz. The frequency falls by 607.5480=127.5607.5-480=127.5 Hz.

Question 12

IAT 2021 · Physics
Consider a Young’s double slit experiment with monochromatic light of wavelength 600 nm. The intensity of the light is I0I_0 at a point on the screen where the path difference is 600 nm. What would be the intensity of light at a point on the screen where the path difference is 100 nm?
  1. 34I0\frac34 I_0
  2. 14I0\frac14 I_0
  3. 12I0\frac12 I_0
  4. 32I0\frac{\sqrt3}{2}I_0
Answer & worked solution
Answer A
A path difference of one wavelength gives a bright maximum, so I0=ImaxI_0=I_{\mathrm{max}}. At 100 nm, the phase difference is 2π(100/600)=π/32\pi(100/600)=\pi/3. For the usual equal-intensity slits, I=Imaxcos2(δ/2)=I0cos2(π/6)=3I0/4I=I_{\mathrm{max}}\cos^2(\delta/2)=I_0\cos^2(\pi/6)=3I_0/4.

Question 13

IAT 2021 · Physics
Consider a mixture of O2\mathrm O_2 and N2\mathrm N_2 gases at temperature 27C27^\circ\mathrm C. Which of the following relations is correct?
  1. RMS speed of O2\mathrm O_2 molecules < RMS speed of N2\mathrm N_2 molecules
  2. Average kinetic energy of O2\mathrm O_2 molecules < Average kinetic energy of N2\mathrm N_2 molecules
  3. Average kinetic energy of O2\mathrm O_2 molecules > Average kinetic energy of N2\mathrm N_2 molecules
  4. RMS speed of O2\mathrm O_2 molecules > RMS speed of N2\mathrm N_2 molecules
Answer & worked solution
Answer A
At the same temperature, both ideal gases have the same mean translational kinetic energy, 3kBT/23k_BT/2. Their RMS speeds depend on molecular mass as vrms=3RT/Mv_{\rm rms}=\sqrt{3RT/M}. Since oxygen has molar mass 32 and nitrogen 28, oxygen has the smaller RMS speed.

Question 14

IAT 2021 · Physics
Two identical objects A and B are at initial temperatures TAT_A and TBT_B (TA>TBT_A>T_B), respectively. The specific heat capacity of the material of these objects increases with temperature. If these two objects are brought in contact then their final equilibrium temperature is TT. Assuming that there is no heat exchange with the surroundings, then
  1. T>TA+TB2T>\frac{T_A+T_B}{2}.
  2. T>TAT>T_A.
  3. T=TA+TB2T=\frac{T_A+T_B}{2}.
  4. T<TA+TB2T<\frac{T_A+T_B}{2}.
Answer & worked solution
Answer A
Energy conservation requires TTAc(u)du=TBTc(u)du\int_T^{T_A}c(u)\,du=\int_{T_B}^T c(u)\,du. At the arithmetic midpoint the two temperature intervals are equal, but the hotter interval has larger c and therefore contains more energy. Raising T reduces the hot-side loss and increases the cold-side gain until they balance. Consequently T is above the arithmetic mean, while still between the initial temperatures.

Question 15

IAT 2021 · Physics
Which one of the following expressions has the dimension of electrical resistance where ee is the charge of an electron and hh is Planck’s constant?
  1. he2\frac h{e^2}
  2. e2h\frac{e^2}h
  3. eh\frac eh
  4. he\frac he
Answer & worked solution
Answer A
Planck's constant has units J·s, and charge squared has units C². Thus h/e2h/e^2 has units Js/C2=(J/C)/(C/s)=V/A=Ω\mathrm{J\,s/C^2}=(\mathrm{J/C})/(\mathrm{C/s})=\mathrm{V/A}=\Omega. It has the dimensions of resistance.

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