PREVIOUS-YEAR QUESTIONS

IISER IAT 2022 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: B · 2: C · 3: A · 4: A · 5: D · 6: C · 7: D · 8: B · 9: A · 10: C · 11: B · 12: A · 13: D · 14: D · 15: C
Chemistry1: A · 2: B · 3: C · 4: B · 5: C · 6: C · 7: A · 8: D · 9: B · 10: D · 11: A · 12: C · 13: A · 14: D · 15: C
Mathematics1: B · 2: C · 3: C · 4: D · 5: B† · 6: A · 7: C† · 8: D · 9: A · 10: B · 11: A · 12: B · 13: A · 14: D · 15: A
Physics1: C · 2: B · 3: —† · 4: D · 5: A · 6: C · 7: A · 8: B · 9: D · 10: C · 11: D · 12: A · 13: B† · 14: B · 15: A†

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

Biology

Question 1

IAT 2022 · Biology
Match the features of a plant cell listed in Column I with their corresponding examples in Column II.
Column IColumn II
i. Cell with cytoplasm and no nucleusp. Pollen grain
ii. Cell lacking both cytoplasm and nucleusq. Synergid
iii. Cell containing more than two nucleir. Tracheid
iv. Haploid cell produced by mitosiss. Tapetum
t. Mature sieve tube
u. Phloem companion cell
Choose the CORRECT combination from the options below.
  1. i and u; ii and r; iii and s; iv and p
  2. i and t; ii and r; iii and s; iv and q
  3. i and 𝑢; ii and r; iii and q; iv and p
  4. i and r; ii and t; iii and s; iv and q

Transcription note: Retyped the damaged matching table against the 2022 paper.

Answer & worked solution
Answer B
A mature sieve-tube element retains cytoplasm but loses its nucleus, giving i–t. A mature tracheid is dead and lacks both, giving ii–r. Tapetal cells can be multinucleate, giving iii–s. Synergids develop by mitosis in the haploid female gametophyte, giving iv–q. The complete combination is therefore i–t, ii–r, iii–s, iv–q.

Question 2

IAT 2022 · Biology

    Shown below are some of the reactions that occur in the metabolic pathway leading to complete oxidation of glucose during aerobic respiration.

    i. Pyruvate Acetyl CoA

    ii. Dihydroxy acetone phosphate Glyceraldehyde-3-phosphate

    iii. Oxaloacetate Citrate

    iv. Fumarate Malate

    Choose the CORRECT sequence of reactions during the complete oxidation of glucose.

  1. i; ii; iii; iv
  2. i; iii; iv; ii
  3. ii; i; iii; iv
  4. ii; iii; i; iv

Transcription note: Removed repeated numerals in a distractor using the 2022 paper.

Answer & worked solution
Answer C
Put the reactions in their metabolic order. Conversion of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate occurs in glycolysis (ii). Pyruvate then forms acetyl-CoA (i). Acetyl-CoA enters the citric-acid cycle by combining with oxaloacetate to form citrate (iii); conversion of fumarate to malate occurs later in that cycle (iv). Thus the order is ii, i, iii, iv.

Question 3

IAT 2022 · Biology
In a given species of flowering plant, the colour of the seeds is exclusively determined by the colour of its seed coat. Seed coat colour in this species is governed by a nuclear gene with two alleles. The WHITE ( 𝑊 ) allele is dominant over brown ( 𝑤 ) allele. If a plant with brown seeds (𝑤𝑤) is crossed as a female with pollen from a white seed (𝑊𝑊) plant, what will be the seed colour and the genotype of the embryo in the resultant seeds obtained from this cross?
  1. Brown seeds with 𝑊𝑤 embryo
  2. White seeds with 𝑊𝑤 embryo
  3. Brown seeds with ww embryo
  4. White seeds with ww embryo
Answer & worked solution
Answer A
The seed coat develops from the maternal integuments, so its genotype is that of the female parent, wwww, and its colour is brown. The embryo receives ww from the mother and WW from the father, making it WwWw. Dominance in the embryo does not change the genotype of the maternal seed coat.

Question 4

IAT 2022 · Biology
During the CoVID-19 pandemic, SARS-CoV2 virus mutated multiple times giving rise to many variants. What is the genetic material of the SARS-CoV2 virus?
  1. Single stranded RNA
  2. Single stranded DNA
  3. Double stranded RNA
  4. Double stranded DNA
Answer & worked solution
Answer A
SARS-CoV-2 has a single-stranded, positive-sense RNA genome. It is therefore an RNA virus; its genome is neither double-stranded DNA nor a DNA–RNA hybrid. The positive-sense RNA can serve as a template for translation after the virus enters a suitable host cell.

Question 5

IAT 2022 · Biology
Lymph is an important body fluid. Choose the INCORRECT statement about the lymph.
  1. Fat digested in the intestine is absorbed through the lymph.
  2. It is the interstitial fluid generated by the passage of liquid between the cells of the capillary.
  3. It is colourless and has similar mineral composition as of plasma.
  4. It remains as an interstitial fluid and is never put back into circulation.
Answer & worked solution
Answer D
Lymph collects excess interstitial fluid, transports absorbed intestinal fats and carries immune cells. It ultimately drains into the venous circulation. It does not remain permanently in interstitial spaces: failure to return this fluid would cause swelling rather than normal fluid balance.

Question 6

IAT 2022 · Biology
Interneurons play an important role in the execution of spinal cord-mediated reflex action. Where are these interneurons located?
  1. Dorsal root ganglion
  2. White matter
  3. Gray matter
  4. Muscle spindle
Answer & worked solution
Answer C
Interneurons connect neurons within the central nervous system and participate in local processing. Their cell bodies are associated with grey matter. Sensory neurons carry information toward the CNS, whereas motor neurons carry commands from it to effectors; neither description replaces the interneuron's integrating role.

Question 7

IAT 2022 · Biology
Osmoreceptors are sensitive to changes in ionic concentrations and volumes of body fluids. Which among the following best describes the function of these osmoreceptors?
  1. They stimulate renin production to induce vasoconstriction and increase blood pressure
  2. They stimulate pituitary gland to facilitate the secretion of mineralocorticoids to adjust the changes in mineral composition of the body fluid.
  3. They stimulate atrial wall to release atrial natriuretic factor to induce vasodilation and reduce blood pressure.
  4. They stimulate hypothalamus to facilitate the release of the anti-diuretic hormone and increase water reabsorption.
Answer & worked solution
Answer D
An increase in blood osmolarity stimulates hypothalamic osmoreceptors. This promotes release of antidiuretic hormone, ADH, from the posterior pituitary. ADH increases water permeability in the distal nephron, especially the collecting ducts, so more water returns to the blood and urine becomes more concentrated.

Question 8

IAT 2022 · Biology
A single point mutation in gene ' 𝑋 ' results in breathing difficulty, hypertension as well as partial sterility. Which among the following best explains the observed phenotypes?
  1. Incomplete dominance
  2. Pleiotropy
  3. Linkage
  4. Partial dominance
Answer & worked solution
Answer B
Pleiotropy means that one gene influences more than one phenotypic characteristic. A single altered protein may affect several tissues or pathways, producing multiple observable effects. This differs from polygenic inheritance, in which several genes contribute to one characteristic.

Question 9

IAT 2022 · Biology
In a cross between individuals of the genotypes PpQQRrSS and ppqqRrSS, what will be the expected number of progenies with the genotype ppQQRrSS in a population of 400 individuals, assuming independent assortment?
  1. 0
  2. 25
  3. 100
  4. 200
Answer & worked solution
Answer A
Every gamete from a QQQQ parent carries QQ, and every gamete from a qqqq parent carries qq. Consequently every offspring is QqQq. The dominant phenotype appears in all offspring, but the proportion with the homozygous dominant genotype QQQQ is zero.

Question 10

IAT 2022 · Biology
Organisms in which of the following phyla are triploblastic, acoelomate and have bilateral symmetry?
  1. Arthropoda
  2. Mollusca
  3. Platyhelminthes
  4. Hemichordata
Answer & worked solution
Answer C
Platyhelminthes are triploblastic, bilaterally symmetrical animals with no body cavity between the body wall and gut: they are acoelomate. Arthropods and molluscs have a true coelom, often reduced with a prominent haemocoel, and hemichordates are coelomate. Therefore only Platyhelminthes has the complete combination requested.

Question 11

IAT 2022 · Biology
Consider the following biomes: Tropical Rainforest (𝑃); Tundra (𝑄); Desert (𝑅) and Coastal zone (𝑆). The most probable order of Net Primary Productivity of these biomes is
  1. P>Q>S>R
  2. 𝑃>𝑆>𝑄>𝑅
  3. 𝑄>𝑃>𝑆>𝑅
  4. 𝑅>𝑄>𝑆>𝑃
Answer & worked solution
Answer B
Tropical rain forests (P) combine warmth, water and a long growing season, giving high net primary productivity. Productive coastal zones (S) follow in this broad comparison. Tundra (Q) is limited by a short cold growing season, and deserts (R) by severe water shortage. The expected order is therefore P>S>Q>RP>S>Q>R. Actual site measurements vary, so this is the typical biome comparison used in the question.

Question 12

IAT 2022 · Biology
Nine percent of a population cannot taste a certain food item because of a recessive allele of the gene IAT. Assuming the population is in Hardy-Weinberg equilibrium, what will be the frequency of dominant and recessive alleles, respectively?
  1. 0.7 and 0.3
  2. 0.9 and 0.1
  3. 0.3 and 0.7
  4. 0.1 and 0.9
Answer & worked solution
Answer A
Under Hardy–Weinberg equilibrium, the recessive phenotype has frequency q2q^2. Here q2=0.09q^2=0.09, so q=0.3q=0.3 and p=1q=0.7p=1-q=0.7. The dominant and recessive allele frequencies are therefore 0.70.7 and 0.30.3, respectively; it should not be confused with the dominant phenotype frequency 1q2=0.911-q^2=0.91.

Question 13

IAT 2022 · Biology
As opposed to DNA replication within the cell, discontinuous synthesis of DNA does NOT occur in a polymerase chain reaction (PCR). Why?
  1. The replication fork is formed and DNA ligase activity of the Taq polymerase joins the discontinuous fragments.
  2. The replication fork is formed and Taq polymerase extends DNA in both 3 to 5 and 5 to 3 direction.
  3. Denaturation step in PCR substitutes for the replication fork and Taq polymerase extends DNA only in the 3 to 5 direction.
  4. Denaturation step in PCR substitutes for the replication fork and Taq polymerase extends DNA only in the 5 to 3 direction.
Answer & worked solution
Answer D
PCR first separates the DNA strands by heating. After primers bind, thermostable DNA polymerase extends each primer continuously in the 535'\to3' direction along its available template. There is no advancing replication fork requiring repeated lagging-strand priming, so Okazaki fragments are not an intrinsic part of PCR amplification.

Question 14

IAT 2022 · Biology
Which of the following figures CORRECTLY represents the chemiosmotic hypothesis of ATP synthesis occurring in a mitochondrion in a cell? (Keys for the figure; IMS: Intermembrane space, IMM: Inner mitochondrial membrane, MM: Mitochondrial matrix, ETC: Electron transport chain)
  1. IAT 2022, Biology, question 14 — option A
  2. IAT 2022, Biology, question 14 — option B
  3. IAT 2022, Biology, question 14 — option C
  4. IAT 2022, Biology, question 14 — option D
Answer & worked solution
Answer D
The respiratory electron-transport chain pumps protons from the mitochondrial matrix into the intermembrane space. Protons then return to the matrix through ATP synthase; its catalytic head faces the matrix. Only the diagram in D has both the transport direction and ATP-synthase orientation consistent with this chemiosmotic arrangement.

Question 15

IAT 2022 · Biology
Let 'X' be the perpendicular distance from the centromere of each chromosome to the equatorial plane of a human cell. Which of the following stages of the cell cycle is most likely to have the HIGHEST average value of 'X'?
  1. Early anaphase
  2. Early metaphase
  3. Late anaphase
  4. Metaphase

Transcription note: Normalised quotation marks around X; this is a distance, not a derivative.

Answer & worked solution
Answer C
During metaphase, chromosomes align near the equatorial plane. In anaphase the separated sister chromatids, now daughter chromosomes, move toward opposite poles. Their centromeres are therefore farthest from the equator among the stages listed in late anaphase, making the average distance XX largest then.

Chemistry

Question 1

IAT 2022 · Chemistry
The products of the reaction between aqueous solutions of 𝐾4[Fe(CN)6] and 12H2O2 are
  1. K3[Fe(CN)6] and KOH
  2. K3[Fe(CN)6] and H2O
  3. K3[Fe(CN)6], H2O and O2
  4. K4[Fe(CN)5(OH)] and HCN

Transcription note: Restored O2 in the distractor from the 2022 paper.

Answer & worked solution
Answer A
Iron changes from +2+2 in ferrocyanide to +3+3 in ferricyanide, releasing one electron per complex. Half a mole of peroxide accepts that electron in the alkaline balance. The net equation is K4[Fe(CN)6]+12H2O2K3[Fe(CN)6]+KOH\mathrm{K_4[Fe(CN)_6]+\tfrac12H_2O_2\to K_3[Fe(CN)_6]+KOH}. Both the potassium count and the hydrogen/oxygen counts balance.

Question 2

IAT 2022 · Chemistry
For the colourless complex [𝑀(𝐻2𝑂)6]𝑛+, where 𝑀 is a 3𝑑 transition metal, the CORRECT ground state 𝑑-electron configuration of 𝑀 is
  1. (𝑡2𝑔)3(𝑒𝑔)0
  2. (𝑡2𝑔)6(𝑒𝑔)4
  3. (𝑡2𝑔)6(𝑒𝑔)2
  4. (𝑡2𝑔)4(𝑒𝑔)2
Answer & worked solution
Answer B
A d10d^{10} ion has all five dd orbitals filled. In an octahedral field its configuration is t2g6eg4t_{2g}^{6}e_g^{4}. There is no vacant dd level available for an ordinary dddd excitation, explaining the colourless aqua complex within the options. The other listed configurations have partially filled dd shells.

Question 3

IAT 2022 · Chemistry

Choose the correct statement about the structure of C60 fullerene

  1. 6-membered rings are fused with 5-membered rings ONLY.
  2. 6-membered rings are fused with 6-membered rings ONLY.
  3. 5-membered rings are fused with 6-membered rings ONLY.
  4. 5-membered rings are fused with both 5-membered and 6-membered rings.
Answer & worked solution
Answer C
The C60C_{60} cage contains 12 pentagons and 20 hexagons. Its pentagons are isolated: each pentagon shares its edges with hexagons, not with other pentagons. Hexagons border both pentagons and hexagons. Thus the statement that five-membered rings are fused only with six-membered rings is correct.

Question 4

IAT 2022 · Chemistry

 The first to fifth ionization energies (IE) of two p-block elements 𝑋 and 𝑌 are given below. 

IE1(eV)IE2(eV)IE3(eV)IE4(eV)IE5(eV)𝐗6.018.828.4120.0153.7𝐘8.216.333.545.1166.7

 The number of valence electrons in 𝑋 and 𝑌 respectively are 

  1. 1,4
  2. 3,4
  3. 3,5
  4. 4,5
Answer & worked solution
Answer B
Look for the sudden rise after all valence electrons have been removed. For X, the jump from the third to fourth ionisation energy indicates three valence electrons. For Y, the jump from the fourth to fifth indicates four. The required pair is therefore (3,4)(3,4).

Question 5

IAT 2022 · Chemistry

Which of the following expressions represents the hydrogen atom wave function 𝜓(𝑟) shown in the figure below? ( 𝑟 is the distance of the electron from the nucleus and 𝑎0 is a constant)

IAT 2022, Chemistry, question 5 — question diagram
  1. 1𝜋(1𝑎0)3/2𝑒𝑟/2𝑎0
  2. 16𝜋(1𝑎0)3/2(𝑟𝑎0)𝑒𝑟/2𝑎0
  3. 142𝜋(1𝑎0)3/2(2𝑟𝑎0)𝑒𝑟/2𝑎0
  4. 13𝜋(1𝑎0)3/2(32𝑟𝑎0+2𝑟29𝑎02)𝑒𝑟/3𝑎0
Answer & worked solution
Answer C
The plotted wavefunction is nonzero at the nucleus, crosses zero once, then approaches zero from below. A 2s2s orbital has exactly this radial behaviour: ψ(r)(2r/a0)er/(2a0)\psi(r)\propto(2-r/a_0)e^{-r/(2a_0)}, with its node at r=2a0r=2a_0. A 1s1s function never changes sign, while a pp function vanishes at the nucleus.

Question 6

IAT 2022 · Chemistry

 Which of the following molecules are aromatic? 

IAT 2022, Chemistry, question 6 — question diagram
  1. M and O only
  2. N and P only
  3. M, N and O only
  4. M, N, O, and P
Answer & worked solution
Answer C
M is pyridine, with six ring π\pi electrons. Protonation of its external nitrogen lone pair gives N without removing the six-electron aromatic circuit. In O, pyrrole's nitrogen lone pair contributes two electrons to the four from the double bonds. Protonating that lone pair in P interrupts the aromatic circuit. Hence M, N and O are aromatic, but P is not.

Question 7

IAT 2022 · Chemistry
Which of the following are aryl bromides? IAT 2022, Chemistry, question 7 — question diagram
  1. N and P only
  2. N and O only
  3. M, O and P only
  4. N, O and P only
Answer & worked solution
Answer A
An aryl bromide has bromine bonded directly to a carbon of an aromatic ring. N and P satisfy this definition. A bromine on a side chain is an alkyl or benzylic bromide instead, and the brominated saturated carbon in M is not an aromatic ring carbon. Therefore only N and P qualify.

Question 8

IAT 2022 · Chemistry
Benzamide is treated with Br2 and NaOH(aq) to form the product 𝑋, which is then reacted with NaNO2 and HCl(aq) at 05C to form 𝑌. 𝑌 is immediately treated with ethanol to give 𝑍. What is Z?
  1. IAT 2022, Chemistry, question 8 — option A
  2. IAT 2022, Chemistry, question 8 — option B
  3. IAT 2022, Chemistry, question 8 — option C
  4. IAT 2022, Chemistry, question 8 — option D
Answer & worked solution
Answer D
Bromine and aqueous base convert benzamide to aniline by Hofmann rearrangement, losing the carbonyl carbon. Cold nitrous acid then converts aniline into a benzenediazonium salt. Ethanol reduces the diazonium group to hydrogen, releasing nitrogen. The final aromatic product is benzene.

Question 9

IAT 2022 · Chemistry

In the following reaction sequence, the major products 𝑀,𝑁, and 𝑂 respectively are 

IAT 2022, Chemistry, question 9 — question diagram
  1. IAT 2022, Chemistry, question 9 — option A
  2. IAT 2022, Chemistry, question 9 — option B
  3. IAT 2022, Chemistry, question 9 — option C
  4. IAT 2022, Chemistry, question 9 — option D
Answer & worked solution
Answer B
Base first forms phenoxide, which reacts with methyl iodide by SN2S_N2 substitution to produce anisole M. Its methoxy group activates the ring and directs Friedel–Crafts acetylation mainly to the para position, giving N. Hypochlorite then causes the haloform oxidation of the methyl ketone to the para-methoxybenzoate salt O. This sequence is shown in B.

Question 10

IAT 2022 · Chemistry

The van der Waals equation for a real gas is given by (𝑃+𝑎𝑉2)(𝑉𝑏)=𝑅𝑇. What is the dimension of (𝑎𝑏) ?

  1. 𝑀𝐿2𝑇1
  2. 𝑀𝐿1𝑇2
  3. 𝑀𝐿𝑇2
  4. 𝑀𝐿2𝑇2
Answer & worked solution
Answer D
The correction a/V2a/V^2 must have the dimensions of pressure, so [a]=[P][V]2[a]=[P][V]^2. Also [b]=[V][b]=[V]. Hence [a/b]=[P][V]=ML2T2[a/b]=[P][V]=ML^2T^{-2}, the dimensions of energy. This uses dimensional consistency of the equation, without assigning numerical values to either constant.

Question 11

IAT 2022 · Chemistry

 The major product 𝑃 of the following reaction is 

IAT 2022, Chemistry, question 11 — question diagram
  1. IAT 2022, Chemistry, question 11 — option A
  2. IAT 2022, Chemistry, question 11 — option B
  3. IAT 2022, Chemistry, question 11 — option C
  4. IAT 2022, Chemistry, question 11 — option D
Answer & worked solution
Answer A
Without a peroxide initiator, HBr adds by the ionic pathway. Proton addition that leaves a benzylic carbocation is favoured because the positive charge is resonance-stabilised by the phenyl ring. Bromide captures that carbocation, placing Br on the carbon directly attached to the ring, as shown in A.

Question 12

IAT 2022 · Chemistry

An aqueous solution contains 1.0 M of X2+ and 0.001 M of Y2+ ions at 25C.X2+ and Y2+ ions do not interact with each other. This solution is put in an electrolytic cell and the voltage is gradually increased till a current begins to flow through the cell. The voltage is maintained at this point and a deposit is observed on the cathode. What is the composition of the material deposited on the cathode?

(Given: Atomic weight of 𝑋 is 63 and 𝑌 is 200.)

𝑋2++2𝑒𝑋,𝐸0=0.35𝑌2++2𝑒𝑌,𝐸0=0.40 V

( 𝐸0 is the standard reduction potential.)

  1. Only Y
  2. 24%𝑋 and 76%𝑌 by weight
  3. Only X
  4. 98%𝑋 and 2%𝑌 by weight
Answer & worked solution
Answer C
Use the actual concentrations in the Nernst equation, not just the standard potentials. For X, EX=0.35E_X=0.35 V because [X2+]=1[X^{2+}]=1 M. For Y, EY=0.40+(0.0591/2)log10(0.001)=0.31135E_Y=0.40+(0.0591/2)\log_{10}(0.001)=0.31135 V. X has the higher reduction potential under these conditions and begins depositing first. At the threshold specified, the deposit is only X.

Question 13

IAT 2022 · Chemistry

2 g of naphthoic acid (molecular weight =172 g mol1 ) dissolved in 20 mL of benzene shows a freezing point depression of 2 K . For benzene, the freezing point depression constant, 𝐾𝑓=5 Kkgmol1 and the density is 0.88 g mL1. What is the magnitude of the van't Hoff factor?

  1. 0.605
  2. 605.0
  3. 0.688
  4. 688.0
Answer & worked solution
Answer A
The solvent mass is 20×0.88=17.620\times0.88=17.6 g, or 0.01760.0176 kg. The solute amount is 2/1722/172 mol, so m=(2/172)/0.0176m=(2/172)/0.0176. From ΔTf=iKfm\Delta T_f=iK_fm, i=2/[5(2/172)/0.0176]=0.605440.605i=2/[5(2/172)/0.0176]=0.60544\approx0.605. The value below one is consistent with association of acid molecules in benzene.

Question 14

IAT 2022 · Chemistry
For the reaction involving ideal gases, 𝐴(𝑔)+2𝐵(𝑔)=2𝐶(𝑔)+3𝐷(𝑔), which of the following plots is qualitatively correct? ( 𝐾𝑝 and 𝐾𝑐 are the equilibrium constants in terms of pressure and concentration respectively. 𝑇 is the absolute temperature.)
  1. IAT 2022, Chemistry, question 14 — option A
  2. IAT 2022, Chemistry, question 14 — option B
  3. IAT 2022, Chemistry, question 14 — option C
  4. IAT 2022, Chemistry, question 14 — option D
Answer & worked solution
Answer D
For A+2B2C+3DA+2B\rightleftharpoons2C+3D, the change in gaseous mole number is Δn=(2+3)(1+2)=2\Delta n=(2+3)-(1+2)=2. Therefore Kp/Kc=(RT)2K_p/K_c=(RT)^2. As a function of absolute temperature this is an upward-curving quadratic through the origin, matching D. This conclusion concerns the ratio, even though each equilibrium constant may have a more complicated temperature dependence.

Question 15

IAT 2022 · Chemistry
Identify the correct order of the molecules with respect to the magnitude of their dipole moment:
  1. IAT 2022, Chemistry, question 15 — option A
  2. IAT 2022, Chemistry, question 15 — option B
  3. IAT 2022, Chemistry, question 15 — option C
  4. IAT 2022, Chemistry, question 15 — option D

Transcription note: Restored the para-difluoro structure from the 2022 paper; the secondary diagram showed different para substituents.

Answer & worked solution
Answer C
The para-difluoro compound has zero dipole moment by symmetry. Meta-dichlorobenzene has a smaller resultant than its ortho isomer because its substituent dipoles are farther apart in angle. The ortho-difluoro compound is slightly more polar than ortho-dichlorobenzene in the ordering used here, giving para-difluoro < meta-dichloro < ortho-dichloro < ortho-difluoro. The first structure in C has been restored to para-difluorobenzene from the paper; a secondary diagram incorrectly showed unlike para substituents. Electronegativity alone is not a reliable numerical rule for comparing molecular dipole moments.

Mathematics

Question 1

IAT 2022 · Mathematics

A randomly chosen card from a deck of 52 cards is given to be a black card (Spade or Club). What is the probability that it is either a face card (King, Queen or Jack) or a Spade?

  1. 7/13
  2. 8/13
  3. 9/13
  4. 19/26
Answer & worked solution
Answer B
Condition on the 26 black cards. There are 13 spades and six black face cards, but three cards belong to both groups. The favourable count is 13+63=1613+6-3=16. Thus the conditional probability is 16/26=8/1316/26=8/13.

Question 2

IAT 2022 · Mathematics

    Let 𝜔 be a complex root of the quadratic polynomial 𝑥2+𝑥+1. The value of

    (𝜔+1𝜔)(𝜔2+1𝜔2)(𝜔100+1𝜔100)

    is

  1. 233
  2. 231
  3. 233
  4. 231
Answer & worked solution
Answer C
Since ω3=1\omega^3=1 and 1+ω+ω2=01+\omega+\omega^2=0, each factor is 22 when its exponent is divisible by 3 and 1-1 otherwise. Among 1 through 100 there are 33 multiples of 3 and 67 other integers. The product is 233(1)67=2332^{33}(-1)^{67}=-2^{33}.

Question 3

IAT 2022 · Mathematics

Let 𝑓(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛1𝑥𝑛1++𝑎1𝑥+𝑎0 be a polynomial. Suppose that 𝑓(0)=0,

𝑑𝑓𝑑𝑥]𝑥=0=1,𝑑2𝑓𝑑𝑥2]𝑥=0=4

and

𝑑3𝑓𝑑𝑥3=𝑑5𝑓𝑑𝑥5

Then 𝑓(5)=

  1. 25
  2. 35
  3. 55
  4. 105
Answer & worked solution
Answer C
If a polynomial has degree at least three, its third and fifth derivatives cannot be identical nonzero polynomials because their degrees differ. The identity therefore forces its degree to be at most two. Write f(x)=ax2+bx+cf(x)=ax^2+bx+c. The initial data give c=0c=0, b=1b=1 and 2a=42a=4, so f(5)=2(25)+5=55f(5)=2(25)+5=55.

Question 4

IAT 2022 · Mathematics
Let 𝑆 be the set of all unit vectors in the 𝑋𝑌-plane. Then the set 𝑆 has
  1. 8 elements
  2. 2 elements
  3. 4 elements
  4. Infinitely many elements
Answer & worked solution
Answer D
Every angle θ\theta defines a unit vector (cosθ,sinθ)(\cos\theta,\sin\theta) in the plane. As θ\theta ranges continuously over [0,2π)[0,2\pi), these give infinitely many distinct vectors. Restricting to the coordinate-axis directions would omit almost all of the unit circle.

Question 5

IAT 2022 · Mathematics
A lab reports that the global average temperature in the year 2020 was 14.9C and predicts that the global average temperature will increase at the rate of 1% per year. What will be the global average temperature in the year 2035?
  1. 15.049C
  2. 17.298C
  3. 17.135C
  4. 17.471C
Answer & worked solution
Answer B · qualification below

This is the stipulated numerical growth model, not a physical climate model.

Annual growth by 1% multiplies the reported numerical temperature by 1.011.01 each year. There are 15 years from 2020 to 2035, so the stipulated model gives 14.9(1.01)15=17.2984C14.9(1.01)^{15}=17.2984\ldots{}^\circ\mathrm C, closest to B. This is the mathematical model stated in the question; percentage change of Celsius temperature is scale-dependent and is not a physical climate model.

Question 6

IAT 2022 · Mathematics
Let 𝑋 be the set of all 2×2 matrices with real entries and 𝑅𝑋×𝑋 be the relation 𝑅= {(𝐴,𝐵):𝐴𝐵=𝐵𝐴}. Which of the following statements is true?
  1. 𝑅 is reflexive and symmetric but not transitive
  2. 𝑅 is reflexive and transitive but not symmetric
  3. 𝑅 is symmetric and transitive but not reflexive
  4. 𝑅 is an equivalance relation
Answer & worked solution
Answer A
Every matrix commutes with itself, so the relation is reflexive. If AB=BAAB=BA, the same equation proves symmetry. Transitivity fails: let A=(1000)A=\begin{pmatrix}1&0\\0&0\end{pmatrix}, B=IB=I, and C=(0100)C=\begin{pmatrix}0&1\\0&0\end{pmatrix}. Both A and C commute with B, but AC=CAC=C and CA=0CA=0.

Question 7

IAT 2022 · Mathematics
For a natural number nn, let CnC_n be the curve in the XY-plane given by y=xny=x^n, where 0x10\le x\le1. Let AnA_n denote the area of the region bounded between CnC_n and Cn+1C_{n+1}. Then the largest value of AnA_n is
  1. 1/2
  2. 1/3
  3. 1/6
  4. 1/12

Transcription note: Restored the subscript n+1 from the paper.

Answer & worked solution
Answer C · qualification below

Natural numbers are taken to begin at 1, as intended by the paper.

On 0x10\le x\le1, xnxn+1x^n\ge x^{n+1}. Hence An=01(xnxn+1)dx=1/(n+1)1/(n+2)=1/[(n+1)(n+2)]A_n=\int_0^1(x^n-x^{n+1})\,dx=1/(n+1)-1/(n+2)=1/[(n+1)(n+2)]. For natural numbers starting at 1, this decreases with n and is largest at n=1n=1, giving 1/61/6. Including zero in the convention for natural numbers would change the answer to 1/21/2; the paper uses the positive-integer convention.

Question 8

IAT 2022 · Mathematics
Let 𝑓 be a continuous function on [0,1] and 𝐹 be its antiderivative. If 𝐹(0)=1 and 01𝑓(𝑥)𝑑𝑥=1, then 𝐹(1) is
  1. 0
  2. 1/2
  3. 1
  4. 2
Answer & worked solution
Answer D
The fundamental theorem of calculus gives F(1)F(0)=01f(x)dx=1F(1)-F(0)=\int_0^1 f(x)\,dx=1. Since F(0)=1F(0)=1, it follows that F(1)=2F(1)=2. The additive constant in an antiderivative is fixed by the initial value.

Question 9

IAT 2022 · Mathematics
Let 𝑎 be a nonzero real number and 𝑓:𝐑𝐑 be a continuous function such that 𝑓(𝑥)>0 for all 𝑥𝑅. Consider 𝑔(𝑥)=𝑓(2𝑎2𝑥𝑎𝑥2). Then 𝑔 has
  1. Local maxima at 𝑥=𝑎 if 𝑎>0
  2. Local maxima at 𝑥=𝑎 if 𝑎<0
  3. Local minima at 𝑥=𝑎 if 𝑎>0
  4. A point of inflection at 𝑥=𝑎
Answer & worked solution
Answer A
The condition f>0f'>0 makes f strictly increasing, so it preserves the location and type of extrema of its inner argument. Write 2a2xax2=a3a(xa)22a^2x-ax^2=a^3-a(x-a)^2. If a>0a>0, this has a strict maximum at x=ax=a; if a<0a<0, it has a minimum there. Thus A is the true statement.

Question 10

IAT 2022 · Mathematics
Let 𝐴 be the matrix [cos𝜃0sin𝜃111sin𝜃0cos𝜃]. For any natural number 𝑘, the determinant of 𝐴𝑘 is
  1. 0
  2. 1
  3. -1
  4. (1)𝑘
Answer & worked solution
Answer B
Expand the determinant along the second column: detA=cos2θ+sin2θ=1\det A=\cos^2\theta+\sin^2\theta=1. Determinants multiply under matrix products, so det(Ak)=(detA)k=1\det(A^k)=(\det A)^k=1 for every positive integer k.

Question 11

IAT 2022 · Mathematics
Consider the vectors 𝑎=ıˆ+𝑥ȷˆ+2𝑘ˆ,𝑏=ıˆ+2ȷˆ+𝑥𝑘ˆ,𝑐=2ıˆ+ȷˆ+3𝑘ˆ. The values of 𝑥 for Which there is at least one nonzero vector perpendicular to the vectors 𝑎,𝑏 and 𝑐 are
  1. 0,2
  2. 2,2
  3. 7/2,0
  4. 4,2
Answer & worked solution
Answer A
A nonzero vector perpendicular to all three exists precisely when the three vectors do not span R3\mathbb R^3. Set their scalar triple product to zero: det(1x212x213)=2x(x2)=0\det\begin{pmatrix}1&x&2\\1&2&x\\2&1&3\end{pmatrix}=2x(x-2)=0. Therefore x=0x=0 or x=2x=2.

Question 12

IAT 2022 · Mathematics
Consider the tangent lines to the circle 𝑥2+𝑦2=1 at points 𝑃=(1,0) and 𝑄=(12,12). If 𝑅 is the point of intersection of these two tangent lines, then 𝑃𝑅𝑄 is:
  1. 𝜋4
  2. 3𝜋4
  3. 5𝜋6
  4. 𝜋6
Answer & worked solution
Answer B
The radii OP and OQ subtend π/4\pi/4 at the centre. Each radius is perpendicular to its tangent, so in quadrilateral OPRQ the two angles at P and Q are right angles. Hence PRQ=2πππ/4=3π/4\angle PRQ=2\pi-\pi-\pi/4=3\pi/4.

Question 13

IAT 2022 · Mathematics
The function given by 𝑓(𝑥)=2𝑥315𝑥2+36𝑥5 is
  1. Increasing on the interval (0,2)
  2. Decreasing on the interval (3,0)
  3. Increasing on the interval (2,3)
  4. Decreasing on the interval (3,)
Answer & worked solution
Answer A
Differentiate: f(x)=6x230x+36=6(x2)(x3)f'(x)=6x^2-30x+36=6(x-2)(x-3). It is positive for x<2x<2 and x>3x>3, and negative for 2<x<32<x<3. Thus the function is increasing throughout (0,2)(0,2), while each of the other descriptions has the wrong sign.

Question 14

IAT 2022 · Mathematics

The value of the integral

1100[𝑥]𝑥𝑑𝑥

where [𝑥] is the greatest integer less than or equal to 𝑥 for any real number 𝑥, is

  1. log(1009898!)
  2. log(1009998!)
  3. log(1009899!)
  4. log(1009999!)
Answer & worked solution
Answer D
On [k,k+1)[k,k+1), the floor function is k. Thus the integral equals k=199klog[(k+1)/k]\sum_{k=1}^{99}k\log[(k+1)/k]. Collecting the logarithms gives 99log100k=299logk=log(10099/99!)99\log100-\sum_{k=2}^{99}\log k=\log(100^{99}/99!). Values at isolated integer endpoints do not change the integral.

Question 15

IAT 2022 · Mathematics
For arbitrary constants 𝛼,𝛽, the differential equation representing the family of curves 𝑦= (𝛼𝑥+𝛽)𝑒𝑥 is
  1. 𝑦2𝑦+𝑦=0
  2. 𝑦𝑦+𝑦=0
  3. 𝑦2𝑦𝑦=0
  4. 𝑦𝑦𝑦=0
Answer & worked solution
Answer A
Multiply the family by exe^{-x} to get exy=αx+βe^{-x}y=\alpha x+\beta. Its second derivative is zero. Differentiating twice gives ex(y2y+y)=0e^{-x}(y''-2y'+y)=0, so the required differential equation is y2y+y=0y''-2y'+y=0.

Physics

Question 1

IAT 2022 · Physics
A particle experiences an acceleration 𝑎=𝛼𝑣, where 𝑣 is the velocity of the particle and 𝛼 is a constant. If the distances traveled by the particle in the time intervals 𝑡2𝑡1 and 𝑡3𝑡1 are 𝑆12 and 𝑆13, respectively, which of the following relations is true?
  1. 𝑆13𝑆12=log[𝛼(𝑡3𝑡1)]log[𝛼(𝑡2𝑡1)]
  2. 𝑆13𝑆12=exp[𝛼(𝑡3𝑡1)]exp[𝛼(𝑡2𝑡1)]
  3. 𝑆13𝑆12=exp[𝛼(𝑡3𝑡1)]1exp[𝛼(𝑡2𝑡1)]1
  4. 𝑆13𝑆12=log[𝛼(𝑡3𝑡1)]1log[𝛼(𝑡2𝑡1)]1

Answer & worked solution
Answer C
Solving dv/dt=αvdv/dt=\alpha v from time t1t_1 gives v(t)=v1eα(tt1)v(t)=v_1e^{\alpha(t-t_1)}. The direction stays fixed, so integrating the speed gives S1j=v1[eα(tjt1)1]/αS_{1j}=|v_1|[e^{\alpha(t_j-t_1)}-1]/\alpha. Taking the ratio cancels the prefactor and yields C. For α=0\alpha=0, interpret the expression by its constant-speed limit.

Question 2

IAT 2022 · Physics
A point mass 𝑚 attached to a massless string is undergoing circular motion in a vertical plane. The length of the string is 𝑅 and the acceleration due to gravity is 𝑔. If the minimum value of the tension in the string is 2 mg , the maximum speed of this circular motion of the point mass is
  1. 6𝑔𝑅
  2. 7𝑔𝑅
  3. (7/2)𝑔𝑅
  4. 4𝑔𝑅
Answer & worked solution
Answer B
Tension is smallest at the top. There, Ttop+mg=mvtop2/RT_{\rm top}+mg=mv_{\rm top}^2/R, so vtop2=3gRv_{\rm top}^2=3gR. Falling through height 2R2R to the bottom adds 4gR4gR to the squared speed by energy conservation. Hence the maximum speed is 7gR\sqrt{7gR}.

Question 3

IAT 2022 · Physics
An object is released from rest from the inner edge of a hemispherical bowl, and it falls under gravity. The coefficient of kinetic friction between the object and the bowl is 𝜇. If the object covers an angular displacement 𝜃 with respect to the center of the hemisphere when it stops for the first time, which of the following expressions is correct?
  1. 𝜇=cot(𝜃)
  2. 𝜇=cot(𝜃2)
  3. 𝜇=tan(𝜃)
  4. 𝜇=tan(𝜃/2)
Answer & worked solution
No valid listed option

None of the options gives the exact sliding-bowl result with the full normal reaction.

The advertised relation μ=cot(θ/2)\mu=\cot(\theta/2), option B, results from omitting the centripetal contribution to the normal force. It is not the exact sliding-bowl result. Let ϕ\phi be the angle travelled from the rim and u=v2u=v^2. The normal reaction is N=m(gsinϕ+u/R)N=m(g\sin\phi+u/R), giving u+2μu=2gR(cosϕμsinϕ)u'+2\mu u=2gR(\cos\phi-\mu\sin\phi). With u(0)=0u(0)=0, u=2gR1+4μ2[3μcosϕ+(12μ2)sinϕ3μe2μϕ]u=\frac{2gR}{1+4\mu^2}[3\mu\cos\phi+(1-2\mu^2)\sin\phi-3\mu e^{-2\mu\phi}]. The first positive zero determines the stopping angle. No listed expression gives this relation in general; B is the source's intended simplified answer, not an exact consequence of the stated mechanics.

Question 4

IAT 2022 · Physics
As shown in the figure a block is resting on a frictionless floor and is attached to the free end of a spring. The right edge of the block in equilibrium is at a distance 𝑑 from the wall. When the spring is compressed by a distance 𝑑/2 and released, the time-period of the motion is 𝑇. Similarly, when compressed by a distance 2𝑑 and released, the time-period is 𝑇0. Considering elastic collision between the block and the wall, what is the value of the quantity 𝑇0/𝑇 ? IAT 2022, Physics, question 4 — question diagram
  1. 1/4
  2. 1
  3. 3/4
  4. 2/3
Answer & worked solution
Answer D
With the smaller amplitude, the block never reaches the wall and T=2π/ωT=2\pi/\omega. For amplitude 2d2d, set x=2dcosωtx=-2d\cos\omega t, measured from equilibrium. It first hits the wall at x=dx=d, so ωt=2π/3\omega t=2\pi/3. The elastic rebound retraces the motion to x=2dx=-2d in the same time. Thus T0=4π/(3ω)T_0=4\pi/(3\omega) and T0/T=2/3T_0/T=2/3.

Question 5

IAT 2022 · Physics
A liquid of density 𝜌 in a container weighs 𝑊. A cubic block of side 𝐿 and density 𝜌𝑏<𝜌 is pushed by a stick to completely submerge the block in the liquid without touching the bottom. If 𝑔 is the acceleration due to the gravity and liquid displaced by the stick is negligible, what is the new weight of the container as registered by the weighing machine below? IAT 2022, Physics, question 5 — question diagram
  1. 𝑊+𝜌𝑔𝐿3
  2. 𝑊+𝜌𝑏𝑔𝐿3
  3. 𝑊+(𝜌𝜌𝑏)𝑔𝐿3
  4. 𝑊+(𝜌𝑏𝜌)𝑔𝐿3
Answer & worked solution
Answer A
A fully submerged block displaces volume L3L^3, so the liquid exerts buoyancy B=ρgL3B=\rho gL^3 upward on it. The block exerts an equal downward force on the liquid. Therefore the scale reading increases by ρgL3\rho gL^3, giving W+ρgL3W+\rho gL^3. The externally held stick supplies the additional force needed to keep the light block submerged.

Question 6

IAT 2022 · Physics
A monoatomic ideal gas at pressure 𝑃 and volume 𝑉 is first adiabatically compressed to volume 𝑉/8 and then is allowed to expand isothermally back to its original volume. What is the final pressure of the gas?
  1. 𝑃
  2. 2𝑃
  3. 4𝑃
  4. 𝑃/2
Answer & worked solution
Answer C
For a monatomic ideal gas, γ=5/3\gamma=5/3. Adiabatic compression gives T2/T1=(V/(V/8))γ1=82/3=4T_2/T_1=(V/(V/8))^{\gamma-1}=8^{2/3}=4. The subsequent expansion is isothermal at 4T14T_1. Once the volume returns to V, the ideal-gas law gives the final pressure 4P4P.

Question 7

IAT 2022 · Physics
A uniform taut string with two point charges 𝑞 and 𝑞 attached to its ends passes over two massless pullies kept 𝐿 distance apart as shown in the fgure. If 𝑓 is the fundamental frequency of the part of the string over pullies, which of the following statements is correct? IAT 2022, Physics, question 7 — question diagram
  1. 𝑓𝐿2
  2. 𝑓𝐿1
  3. 𝑓𝐿
  4. 𝑓𝐿2
Answer & worked solution
Answer A
The charge separation shown is proportional to L, so Coulomb attraction, and therefore the string tension, scales as L2L^{-2}. For constant linear mass density, wave speed scales as TL1\sqrt{T}\propto L^{-1}. The fundamental frequency of the segment is f=v/(2L)f=v/(2L), giving fL2f\propto L^{-2}.

Question 8

IAT 2022 · Physics
Two uniformly charged concentric thin spherical shells of radii 𝑟1 and 𝑟2 have charges +𝑄 and 𝑄, respectively. How does the electrostatic potential vary with distance from the center of the shells? IAT 2022, Physics, question 8 — question diagram
  1. IAT 2022, Physics, question 8 — option A
  2. IAT 2022, Physics, question 8 — option B
  3. IAT 2022, Physics, question 8 — option C
  4. IAT 2022, Physics, question 8 — option D
Answer & worked solution
Answer B
Choose zero potential at infinity. Outside the outer shell the total enclosed charge is zero and V=0V=0. Between the shells, V=kQ(1/r1/r2)V=kQ(1/r-1/r_2), which decreases continuously to zero. Inside the inner shell, V=kQ(1/r11/r2)V=kQ(1/r_1-1/r_2) is constant. The potential cannot jump at either thin shell, so B gives the appropriate qualitative graph.

Question 9

IAT 2022 · Physics

Consider a very long cylinder of radius 𝑎 having a uniform positive charge density 𝜌. A sphere of radius 𝑎 has been carved out (see figure), leaving no charge in that region. The distance radially outward to the cylinder, as measured from the center of the sphere is 𝑥. At what value of 𝑥 will the electric field be maximum?

IAT 2022, Physics, question 9 — question diagram
  1. 𝑎
  2. (2/3)𝑎
  3. (3/2)𝑎
  4. (4/3)𝑎
Answer & worked solution
Answer D
Superpose the full charged cylinder and a negatively charged sphere representing the cavity. Inside the cavity, E=ρx/(2ϵ0)ρx/(3ϵ0)=ρx/(6ϵ0)E=\rho x/(2\epsilon_0)-\rho x/(3\epsilon_0)=\rho x/(6\epsilon_0), which increases. For xax\ge a, E=ρϵ0[a2/(2x)a3/(3x2)]E=\frac{\rho}{\epsilon_0}[a^2/(2x)-a^3/(3x^2)]. Setting its derivative to zero gives a2/(2x2)+2a3/(3x3)=0-a^2/(2x^2)+2a^3/(3x^3)=0, hence x=4a/3x=4a/3. The derivative changes from positive to negative there.

Question 10

IAT 2022 · Physics
Consider that in a fission of a single Plutonium ( Pu239 ) atom 207 MeV energy is released. Assuming all atoms of Plutonium undergo fission, which of the following options is the closest estimate of the amount of Plutonium required for a 20,000 units of TNT explosion? (1 unit of TNT =4.184×109 J,1eV=1.6×1019 J )
  1. 60.0 g
  2. 0.5 kg
  3. 1 kg
  4. 10 kg
Answer & worked solution
Answer C
The total energy in the stated comparison is 20000(4.184×109)=8.368×101320000(4.184\times10^9)=8.368\times10^{13} J. Divide by the stipulated energy per fission, 207×106(1.6×1019)207\times10^6(1.6\times10^{-19}) J, to obtain 2.53×10242.53\times10^{24} atoms. Using 239 g per mole gives about 1.00×1031.00\times10^3 g. Thus 1 kg is the closest estimate under the question's assumption that every atom fissions.

Question 11

IAT 2022 · Physics
In the given circuit, the box 𝐵 either contains a capacitor, or an inductor or a resistor. The current 𝐼 versus time 𝑡 plots for three cases (𝑎1,𝑎2,𝑎3) are shown in figures. The switch 𝑆 is closed at time 𝑡=0. Which of the following is correct? IAT 2022, Physics, question 11 — question diagram IAT 2022, Physics, question 11 — question diagram
  1. 𝑎1 corresponds to an inductor, 𝑎2 corresponds to a capacitor, 𝑎3 correspond to a resistor.
  2. 𝑎1 corresponds to a resistor, 𝑎2 corresponds to an inductor, 𝑎3 corresponds to capacitor
  3. 𝑎1 corresponds to an inductor, 𝑎2 corresponds to a resistor, 𝑎3 corresponds to capacitor.
  4. 𝑎1 corresponds to a resistor, 𝑎2 corresponds to a capacitor, 𝑎3 corresponds to inductor.
Answer & worked solution
Answer D
A resistor produces an immediate constant current, matching a1a_1. A capacitor initially conducts while charging, after which the current decays to zero, matching a2a_2. An inductor initially opposes the change in current, which then rises toward its steady value, matching a3a_3. Hence D gives the correct correspondence.

Question 12

IAT 2022 · Physics
In the given circuit each of the resistors is of 1kΩ resistance and each of the capacitors has 4𝜇 F capacitance. What is the charge in the capacitor between the points B and F ? IAT 2022, Physics, question 12 — question diagram
  1. 10𝜇𝐶
  2. 15𝜇𝐶
  3. (40/7)𝜇𝐶
  4. (60/7)𝜇𝐶
Answer & worked solution
Answer A
At DC steady state, all capacitors are open circuits. The conducting path from A to D contains four equal resistors: AB, BC, CG and HD, with G and H joined by wire. Taking VA=0V_A=0 and VD=5V_D=5 V gives VB=1.25V_B=1.25 V and VH=3.75V_H=3.75 V. The unpowered resistor branch through E and F carries no current, so VF=VHV_F=V_H. Thus QBF=4μF×2.5V=10μCQ_{BF}=4\,\mu\mathrm F\times2.5\,\mathrm V=10\,\mu\mathrm C.

Question 13

IAT 2022 · Physics
A thin lens made of material of refractive index nln_l forms the image at the position I of a point object held at O on the central axis of the lens, as is shown in the ray diagram below. Consider that the refractive index of the medium is nmn_m and nl>nmn_l>n_m for the figure given. Where does the image form when nl<nmn_l<n_m?IAT 2022, Physics, question 13 — question diagram
  1. At point A
  2. At point B
  3. At point 𝐶
  4. At point D

Transcription note: Restored the missing ray diagram and damaged inequality from the 2022 paper. This is Physics question 12 in that PDF; numbering here follows the supplied year-wise collection.

Answer & worked solution
Answer B · qualification below

The image lies between object and lens; exact distance requires additional refractive-index data.

The lens-maker factor is nl/nm1n_l/n_m-1. When the surrounding medium becomes optically denser than the lens, this factor changes sign, so the biconvex lens becomes diverging. A real object then produces an upright virtual image on the object side, between the object and the lens. The marked point in that region is B. The exact position depends on the new refractive indices, which are not numerically specified.

Question 14

IAT 2022 · Physics
A conducting sphere of radius 𝑎 initially contains a uniform volume charge density 𝜌. What is the electric flux Φ through a spherical surface of radius 2𝑎 centered at a point on the surface of the charged sphere (see figure) at a later time 𝑡 ? ( 𝜖0 is the permittivity of free space.) IAT 2022, Physics, question 14 — question diagram
  1. Φ=16𝜖0𝜋𝑎2𝜌
  2. Φ=43𝜖0𝜋𝑎3𝜌
  3. Φ=323𝜖0𝜋𝑎3𝜌
  4. Φ=4𝜖0𝜋𝑎2𝜌
Answer & worked solution
Answer B
The displaced Gaussian sphere encloses the entire charged conducting sphere, including its surface after charge redistribution. The conserved enclosed charge is Q=ρ(4πa3/3)Q=\rho(4\pi a^3/3). Gauss's law therefore gives Φ=Q/ϵ0=4πa3ρ/(3ϵ0)\Phi=Q/\epsilon_0=4\pi a^3\rho/(3\epsilon_0). Changing the internal charge distribution does not change the total flux through a surface enclosing all the charge.

Question 15

IAT 2022 · Physics
The figures in the options show the photocurrent 𝐽 of a photoelectric material versus the frequency 𝑓 of an incident light beam. The light beam can have three different intensities 𝐼1,𝐼2,𝐼3, with 𝐼3>𝐼2> 𝐼1. Which of the options is correct?
  1. IAT 2022, Physics, question 15 — option A
  2. IAT 2022, Physics, question 15 — option B
  3. IAT 2022, Physics, question 15 — option C
  4. IAT 2022, Physics, question 15 — option D
Answer & worked solution
Answer A · qualification below

The horizontal branches are schematic; current need not be frequency-independent at fixed power.

The threshold frequency is fixed by the material's work function and is the same for all intensities. Above threshold, a greater intensity gives a greater photocurrent under otherwise identical collection conditions. A is the only sketch with both features. Its perfectly flat branches are a simplified qualitative convention: at fixed optical power and quantum efficiency the photon rate is I/(hf)I/(hf), so actual current need not be exactly independent of frequency.

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