PREVIOUS-YEAR QUESTIONS

IISER IAT 2026 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: A · 6: C · 7: A · 8: C · 9: A · 10: A · 11: B · 12: B · 13: A · 14: A · 15: A
Chemistry1: A · 2: C · 3: C · 4: D · 5: B · 6: A · 7: D · 8: A · 9: A · 10: D · 11: A · 12: B · 13: D · 14: C · 15: C
Mathematics1: B · 2: D · 3: D · 4: C · 5: D · 6: D · 7: B · 8: A · 9: D · 10: D · 11: A · 12: A · 13: D · 14: C · 15: A
Physics1: C · 2: C · 3: D · 4: B · 5: A · 6: B · 7: A† · 8: D · 9: D · 10: A · 11: B† · 12: C · 13: B† · 14: D · 15: C

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

Biology

Question 1

IAT 2026 · Biology
If cell wall was used as the only criterion for classifying organisms, then Mycoplasma would have belonged to which one of the following groups?
  1. Plants
  2. Animals
  3. Protists
  4. Fungi
Answer & worked solution
Answer B
Mycoplasma cells lack a cell wall. If that single feature were used, the absence of a wall would place them with animals among the listed choices. Their actual classification remains bacterial; the question deliberately replaces normal classification with a one-feature rule.

Question 2

IAT 2026 · Biology
Which one of the following choices matches the organs in column I with their vascular system arrangement described in column II?

Column IColumn II
P.Dicot rooti.Radial, open, diarch to tetrarch
Q.Dicot stemii.Ring, conjoint, open
R.Monocot rootiii.Radial, closed, polyarch
S.Monocot stemiv.Scattered, conjoint, closed
  1. P - (ii); Q - (i); R - (iv); S - (iii)
  2. P - (iii); Q - (iv); R - (i); S - (ii)
  3. P - (i); Q - (ii); R - (iii); S - (iv)
  4. P - (iv); Q - (iii); R - (ii); S - (i)
Answer & worked solution
Answer C
Dicot roots have radial vascular bundles with a small number of xylem groups, matching i. Dicot stems have conjoint open bundles arranged in a ring, matching ii. Monocot roots are polyarch and match iii; monocot stems have scattered conjoint closed bundles and match iv. The sequence is P–i, Q–ii, R–iii, S–iv.

Question 3

IAT 2026 · Biology
For which one of the following enzyme activity plots, will the 𝐾𝑚 of the enzyme for substrate 'S' be 20? [S] indicates substrate concentration.
  1. IAT 2026, Biology, question 3 — option A
  2. IAT 2026, Biology, question 3 — option B
  3. IAT 2026, Biology, question 3 — option C
  4. IAT 2026, Biology, question 3 — option D
Answer & worked solution
Answer A
For Michaelis–Menten kinetics, v=Vmax[S]/(Km+[S])v=V_{\mathrm{max}}[S]/(K_m+[S]). Thus v=Vmax/2v=V_{\mathrm{max}}/2 when [S]=Km[S]=K_m. Read the plateau at approximately 16 units and locate half of it, 8 units: graph A reaches that rate at the stated substrate concentration of 20. The other plots do not satisfy both the plateau and half-maximum condition.

Question 4

IAT 2026 · Biology
Which of the following processes result in development of a proton gradient across the thylakoid membrane during photosynthesis?

i. Release of protons into the lumen of the thylakoid by plastoquinone
ii. Consumption of protons in the stroma during the reduction of NADP+
iii. Release of protons into the lumen of the thylakoid by ATP synthase
iv. Release of protons into the lumen of the thylakoid by water splitting reaction
  1. i, ii and iv
  2. i, ii and iii
  3. ii, iii and iv
  4. i, iii and iv
Answer & worked solution
Answer A
A proton gradient increases when protons are released into the thylakoid lumen, moved from stroma to lumen during electron transport, or removed from the stroma during NADPH formation. These are the effects described in i, ii and iv. ATP synthase instead lets protons flow back to the stroma, dissipating the gradient while making ATP; it is not a gradient-building step.

Question 5

IAT 2026 · Biology
Which one of the following correctly describes the mode of action of Follicle Stimulating Hormone (FSH) and estrogen?
  1. FSH interacts with membrane-bound receptor and generates cyclic AMP, while estrogen interacts with intracellular receptor and regulates gene expression
  2. Both FSH and estrogen regulate gene expression via intracellular receptors
  3. FSH interacts with intracellular receptor and generates cyclic AMP, while estrogen interacts with membrane-bound receptor and regulates cellular metabolism
  4. Both FSH and estrogen regulate cellular metabolism via membrane-bound receptors
Answer & worked solution
Answer A
FSH is a peptide hormone and acts through a receptor at the cell membrane, with cAMP as a second messenger. Estrogen is a steroid hormone that can bind an intracellular receptor; the hormone–receptor complex regulates transcription. Only A assigns these mechanisms to the correct hormones.

Question 6

IAT 2026 · Biology
For which one of the following parents, their children will NOT have the same blood group phenotype as either of the parents?
  1. Father: AB; Mother: A
  2. Father: A; Mother: O
  3. Father: AB; Mother: O
  4. Father: O; Mother: B
Answer & worked solution
Answer C
An AB parent has genotype IAIBI^AI^B and an O parent has genotype ii. The first parent contributes either IAI^A or IBI^B, while the second always contributes i. Children therefore have A or B blood groups, with neither AB nor O produced by this cross. Match these offspring groups to C.

Question 7

IAT 2026 · Biology
In which one of the following cases, will an anti-Rh antibody treatment prevent erythroblastosis foetalis?
  1. Anti-Rh antibody to the Rh-negative mother after delivery of the first Rh-positive child
  2. Anti-Rh antibody to the Rh-negative mother after delivery of the first Rh-negative child
  3. Anti-Rh antibody to the Rh-positive mother after delivery of the first Rh-negative child
  4. Anti-Rh antibody to the Rh-positive mother after delivery of the first Rh-positive child
Answer & worked solution
Answer A
In the standard Rh-incompatibility scenario, an Rh-negative mother can become sensitised after exposure to Rh-positive fetal blood. A later Rh-positive fetus may then be affected by maternal anti-Rh antibodies. Giving anti-Rh immunoglobulin after the first Rh-positive delivery removes fetal Rh-positive cells from the maternal circulation before active sensitisation develops. That preventive combination is A; an Rh-positive mother does not create this incompatibility.

Question 8

IAT 2026 · Biology
Which one of the following correctly describes the sequence of events leading to muscle contraction following acetylcholine release at the neuromuscular junction?
  1. Decrease of Ca++ in sarcoplasm, masking of active sites for myosin, 'Z' lines pulled inwards
  2. Increase of Ca++ in sarcoplasm, masking of active sites for myosin, 'Z' lines pulled inwards
  3. Increase of Ca++ in sarcoplasm, unmasking of active sites for myosin, 'Z' lines pulled inwards
  4. Increase of Ca++ in sarcoplasm, unmasking of active sites for myosin, 'Z' lines pulled outwards
Answer & worked solution
Answer C
Acetylcholine initiates muscle excitation, leading to calcium release into the sarcoplasm. Calcium binding moves the troponin–tropomyosin complex so that myosin-binding sites on actin become accessible. Filament sliding shortens the sarcomere and draws the Z lines inward. The correct sequence is increased calcium, unmasking, inward Z-line movement.

Question 9

IAT 2026 · Biology
A female child is born with all the primary oocytes required during her lifetime. At which one of the following stages of cell division are these oocytes found at birth?
  1. Prophase I
  2. Metaphase I
  3. Telophase I
  4. Anaphase I
Answer & worked solution
Answer A
Primary oocytes enter meiosis during fetal development and then arrest in prophase I, specifically the diplotene stage. They do not complete meiosis I before birth. Therefore the stage asked for is prophase I, option A.

Question 10

IAT 2026 · Biology
Which of the following schematics correctly depict a lac operon that can be negatively regulated?
IAT 2026, Biology, question 10 — question diagram
  1. i and iii
  2. ii and iv
  3. i and ii
  4. iii and iv
Answer & worked solution
Answer A
For inducible control of the structural genes, the operator must be positioned so that repressor binding can block transcription from the promoter serving those genes. In the supplied arrangements, i and iii place the controlling operator with that transcription unit. The placements in ii and iv do not provide the stated control of the complete structural-gene set. Hence the matching choice is A.

Question 11

IAT 2026 · Biology
Which one of the following is the correct sequence of the coding strand of the given gene?
IAT 2026, Biology, question 11 — question diagram
  1. 5' GTGAAGCCGTTACAGCAC 3'
  2. 5' GTGCTGTAACGGCTTCAC 3'
  3. 5' CACTTCGGCAATGTCGTG 3'
  4. 5' CACGACATTGCCGAAGTG 3'
Answer & worked solution
Answer B
Identify the coding strand as the strand complementary and antiparallel to the template. The lower strand in the diagram is therefore read from right to left to obtain its 5′→3′ sequence. That gives 5′-GTGCTGTAACGGCTTCAC-3′, which is B. Reading the same strand left to right would reverse the required orientation.

Question 12

IAT 2026 · Biology
The following pedigree diagram shows the inheritance of a rare genetic disorder in a family (filled shapes depict affected individuals). Which one of the following is the most likely pattern of inheritance of the disorder?
IAT 2026, Biology, question 12 — question diagram
  1. Mitochondrial
  2. X-linked dominant
  3. Y-linked
  4. X-linked recessive
Answer & worked solution
Answer B
The pedigree shows affected fathers transmitting the trait to daughters but not to sons. That rules out Y-linkage and is the characteristic paternal transmission pattern of an X-linked dominant trait. Transmission through affected females is also compatible with this model, giving B.

Question 13

IAT 2026 · Biology
Which one of the following molecules can be used for RNA interference?
  1. IAT 2026, Biology, question 13 — option A
  2. IAT 2026, Biology, question 13 — option B
  3. IAT 2026, Biology, question 13 — option C
  4. IAT 2026, Biology, question 13 — option D
Answer & worked solution
Answer A
The structure contains uracil rather than thymine and ribose sugars with the RNA hydroxyl pattern. It is shown as a paired, double-stranded molecule. Those two observations together identify double-stranded RNA, option A. Double-stranded RNA can be processed into small interfering RNAs that guide silencing of a complementary messenger RNA.

Question 14

IAT 2026 · Biology
The pBR322 cloning vector has genes coding for tetracycline and ampicillin resistance. A foreign DNA to be cloned is inserted into the tetracycline resistance gene and the recombinant plasmid is then transformed into E. coli cells. Which one of the following choices is the most likely outcome of this cloning reaction?
  1. The cells with the recombinant plasmid can grow in the presence of ampicillin but not tetracycline
  2. The cells with the recombinant plasmid can grow in the presence of both ampicillin and tetracycline
  3. The cells with the non-recombinant plasmid can grow in the presence of ampicillin but not tetracycline
  4. The cells with the non-recombinant plasmid can grow in the presence of tetracycline but not ampicillin
Answer & worked solution
Answer A
Insertion into the tetracycline-resistance region disrupts that marker while leaving the ampicillin-resistance region intact. Recombinant cells therefore grow in ampicillin but fail to grow in tetracycline. This is the insertional-inactivation pattern in A.

Question 15

IAT 2026 · Biology
The graphs below depict the number of individuals of two organisms, named P and Q, when grown independently, and together. Based on these growth patterns, which one of the following statements is correct?
IAT 2026, Biology, question 15 — question diagram
  1. Q is a predator of P
  2. P and Q exhibit mutualism
  3. P is a parasite of Q
  4. Q is a commensal of P
Answer & worked solution
Answer A
Compare the separate-population curves with the mixed-culture outcome: P declines when Q is present, while Q can benefit from P as a resource. Among the interactions listed, Q acting as predator on P best matches this direction of effect. The graph supports the choice among the given alternatives; it is not, by itself, a universal proof of the feeding mechanism.

Chemistry

Question 1

IAT 2026 · Chemistry
What is the ratio of the velocity of an electron in the fourth orbit of Be3+ to the velocity of the electron in the second orbit of He+?
  1. 1:1
  2. 1:2
  3. 6:1
  4. 3:2
Answer & worked solution
Answer A
In the Bohr model for a one-electron ion, vnZ/nv_n\propto Z/n. For Be3+\mathrm{Be^{3+}} in n = 4, Z/n=4/4=1Z/n=4/4=1; for He+\mathrm{He^+} in n = 2, Z/n=2/2=1Z/n=2/2=1. Their velocities are equal, so the ratio is 1:1.

Question 2

IAT 2026 · Chemistry
What is the order of bond energy between C=S & C=Te, and between Cl–Cl & F–F?
  1. C=Te > C=S and F–F > Cl–Cl
  2. C=S > C=Te and F–F > Cl–Cl
  3. C=S > C=Te and Cl–Cl > F–F
  4. C=Te > C=S and Cl–Cl > F–F
Answer & worked solution
Answer C
C=S has more effective orbital overlap than C=Te because tellurium is much larger, so C=S has the greater bond energy. F–F is unusually weak because the small fluorine atoms bring lone pairs into strong repulsion; Cl–Cl is stronger. Both comparisons are combined in C.

Question 3

IAT 2026 · Chemistry
Which one of the following molecules shows an increase in bond order after loss of an electron from the highest occupied molecular orbital?
  1. N2
  2. C2
  3. F2
  4. B2
Answer & worked solution
Answer C
Removing an electron from an antibonding orbital raises bond order by one-half; removing one from a bonding orbital lowers it. The highest occupied orbital of F2 is antibonding, whereas the relevant HOMOs of B2, C2 and N2 are bonding. Thus F2 shows the required increase.

Question 4

IAT 2026 · Chemistry
An ideal gas goes through a reversible isothermal expansion (solid line) followed by a reversible adiabatic expansion (dashed line). Which of the following diagram(s) closely depict(s) the entire process?
IAT 2026, Chemistry, question 4 — question diagram
  1. (ii) and (iv) only
  2. (i) only
  3. (i), (ii), and (iii) only
  4. (i) and (iii) only
Answer & worked solution
Answer D
During isothermal expansion, T stays constant and P decreases as 1/V1/V. The following adiabatic expansion cools the gas and obeys PVγ=constantPV^\gamma=\text{constant}, with a steeper fall than the isotherm in a P–V plot. These conditions are satisfied by i and iii. Diagram ii incorrectly makes temperature rise in the adiabatic expansion; iv does not trace the required expansion branch in P versus 1/V.

Question 5

IAT 2026 · Chemistry
Consider the following silica-gel based thin-layer chromatogram of compounds A and B. Which one of the following statements is correct?
IAT 2026, Chemistry, question 5 — question diagram
  1. A is more polar than B; B has 𝑅𝑓=0.33.
  2. B is more polar than A; A has 𝑅𝑓=0.75.
  3. B is less polar than A; B has 𝑅𝑓=0.75.
  4. B is more polar than A; A has 𝑅𝑓=0.25.
Answer & worked solution
Answer B
The solvent front moves 5 cm. Hence Rf(A)=3.75/5=0.75R_f(A)=3.75/5=0.75 and Rf(B)=1.25/5=0.25R_f(B)=1.25/5=0.25. Silica gel is polar, so the compound more strongly retained by it moves less far. B is therefore more polar, and A has Rf=0.75R_f=0.75.

Question 6

IAT 2026 · Chemistry
Which one of the following molecules is chiral?
  1. IAT 2026, Chemistry, question 6 — option A
  2. IAT 2026, Chemistry, question 6 — option B
  3. IAT 2026, Chemistry, question 6 — option C
  4. IAT 2026, Chemistry, question 6 — option D
Answer & worked solution
Answer A
Compare the two sides of each ring while retaining the wedge-and-dash orientations. In B, C and D the matching left and right substituent arrangements permit an internal reflection symmetry. A has opposite orientations on the two side carbons and lacks that symmetry; its mirror image cannot be superposed by rotating the whole molecule. A is the chiral member.

Question 7

IAT 2026 · Chemistry
For two pure volatile liquids X and Y, attractive intermolecular interactions of both X-X and Y-Y are weaker than those of X-Y. The total vapour pressure of an equimolar solution of X and Y is 𝑝𝑡𝑜𝑡𝑎𝑙. The vapour pressure of pure X and pure Y are 𝑝𝑋0 and 𝑝𝑌0, respectively. Which one of the following relations is correct?
  1. 𝑝𝑡𝑜𝑡𝑎𝑙=(𝑝𝑋0+𝑝𝑌0)/2
  2. 𝑝𝑡𝑜𝑡𝑎𝑙>(𝑝𝑋0+𝑝𝑌0)/2
  3. 𝑝𝑡𝑜𝑡𝑎𝑙=𝑝𝑋0+𝑝𝑌0
  4. 𝑝𝑡𝑜𝑡𝑎𝑙<(𝑝𝑋0+𝑝𝑌0)/2
Answer & worked solution
Answer D
Stronger X–Y attractions hold molecules in the liquid more effectively than the interactions in either pure liquid. The solution therefore shows negative deviation from Raoult's law. For an equimolar ideal solution the pressure would be (pX0+pY0)/2(p_X^0+p_Y^0)/2; the actual total pressure is lower, giving D.

Question 8

IAT 2026 · Chemistry
The rate constant of a reaction at 600 K with an activation energy of 191.47 kJ mol1 is 5.0×105 s1. What is the temperature at which the half-life of the reaction becomes 152 s? [Consider pre-exponential factor and activation energy to be independent of temperature. 𝑅=8.314 J K1mol1]
  1. 680 K
  2. 760 K
  3. 720 K
  4. 640 K
Answer & worked solution
Answer A
The units s1\mathrm{s^{-1}} indicate first-order kinetics, so the target rate constant is k2=ln2/1524.560×103 s1k_2=\ln2/152\approx4.560\times10^{-3}\ \mathrm{s^{-1}}. Apply ln(k2/k1)=(Ea/R)(1/6001/T2)\ln(k_2/k_1)=(E_a/R)(1/600-1/T_2). With Ea=191470 Jmol1E_a=191470\ \mathrm{J\,mol^{-1}}, this gives T2680T_2\approx680 K. This higher temperature is consistent with the much shorter half-life.

Question 9

IAT 2026 · Chemistry
Metal-ligand 𝜋-bond formation in Mn2(CO)10 and [MnO4] requires electron-pair donation between metal and ligand orbitals. Which one of the following represents the direction of electron-pair donation?
  1. Mn2(CO)10: metal orbital ligand orbital
    [MnO4]: ligand orbital metal orbital
  2. Mn2(CO)10: metal orbital ligand orbital
    [MnO4]: metal orbital ligand orbital
  3. Mn2(CO)10: ligand orbital metal orbital
    [MnO4]: ligand orbital metal orbital
  4. Mn2(CO)10: ligand orbital metal orbital
    [MnO4]: metal orbital ligand orbital
Answer & worked solution
Answer A
CO is a π-acceptor: filled metal d orbitals donate into vacant CO π* orbitals in the carbonyl. In permanganate, filled oxygen orbitals donate π electron density towards the electron-deficient Mn(VII) centre. The π-donation directions are therefore metal → ligand for Mn2(CO)10\mathrm{Mn_2(CO)_{10}} and ligand → metal for MnO4\mathrm{MnO_4^-}.

Question 10

IAT 2026 · Chemistry
Which one of the following octahedral complexes has the highest spin-only magnetic moment?
  1. [Co(NH3)4Cl2]
  2. [V(H2O)4I2]+
  3. [Fe(NH3)4(CN)2]+
  4. [Cr(H2O)4(OH)2]
Answer & worked solution
Answer D
Determine oxidation states first. The complexes contain Co(II), V(III), Fe(III) and Cr(II), giving d7, d2, d5 and d4 respectively. The aqua/hydroxo Cr(II) complex is weak-field high-spin d4 with four unpaired electrons, μ=4(4+2)=24\mu=\sqrt{4(4+2)}=\sqrt{24} BM. This exceeds high-spin Co(II)'s three, V(III)'s two, and the strong-field Fe(III) complex's low-spin count of one.

Question 11

IAT 2026 · Chemistry
Which one is an INCORRECT statement with regard to the following reaction?
IAT 2026, Chemistry, question 11 — question diagram
  1. The reaction rate decreases upon changing the solvent from ethyl alcohol to 1:1 mixture of ethyl alcohol and water.
  2. The reaction rate does not change upon increasing the concentration of hydroxide ion.
  3. The rate determining step is the dissociation of tert-butylbromide.
  4. The reaction rate is proportional to the concentration of tert-butylbromide.
Answer & worked solution
Answer A
The tertiary halide undergoes the SN1 process described by the other statements. Its slow step is C–Br ionisation, so the rate is proportional to tert-butyl bromide concentration and independent of hydroxide concentration. Adding water makes the solvent more favourable for stabilising the ions and increases, rather than decreases, the ionisation rate. A is the incorrect statement.

Question 12

IAT 2026 · Chemistry
What are the numbers of protons (H+) and electrons (e), respectively, required for the reduction of [Cr2O7]2 to Cr3+ under an aqueous acidic condition?
  1. 7,3
  2. 14,6
  3. 7,6
  4. 6,14
Answer & worked solution
Answer B
Balance oxygen with seven water molecules and hydrogen with fourteen protons. Then balance charge with six electrons: Cr2O72+14H++6e2Cr3++7H2O\mathrm{Cr_2O_7^{2-}+14H^++6e^-\to2Cr^{3+}+7H_2O}. There are therefore 14 protons and 6 electrons per dichromate ion.

Question 13

IAT 2026 · Chemistry
What are X and P in the following reaction sequence?
IAT 2026, Chemistry, question 13 — question diagram
  1. IAT 2026, Chemistry, question 13 — option A
  2. IAT 2026, Chemistry, question 13 — option B
  3. IAT 2026, Chemistry, question 13 — option C
  4. IAT 2026, Chemistry, question 13 — option D
Answer & worked solution
Answer D
The neighbouring aldehyde groups undergo an intramolecular Cannizzaro reaction. Transfer of H from the CHO group is favoured over transfer of D, so that group is oxidised to carboxylate while the labelled aldehyde is reduced to CH(D)OH. Acidification converts the salt to the hydroxy acid, which closes to the five-membered lactone. D shows both the carboxylate/alcohol intermediate and the lactone retaining D on the reduced carbon.

Question 14

IAT 2026 · Chemistry
100 mL of 1.0 M aqueous NaOH solution was diluted to 1.0 L by adding water. Half of this solution was discarded. A new 100 mL of 0.5 M aqueous NaOH solution was added to the remaining solution. What is the concentration of the final aqueous NaOH solution?
  1. 0.50 M
  2. 0.33 M
  3. 0.17 M
  4. 0.10 M
Answer & worked solution
Answer C
Initially there are 0.100×1.0=0.1000.100\times1.0=0.100 mol NaOH. Dilution changes the volume, not the amount; discarding half then leaves 0.050 mol in 0.500 L. The added solution contributes 0.100×0.5=0.0500.100\times0.5=0.050 mol. Final concentration is 0.100/(0.500+0.100)=0.16670.100/(0.500+0.100)=0.1667 M, approximately 0.17 M.

Question 15

IAT 2026 · Chemistry
What is the order of p𝐾𝑏 for the following molecules in an aqueous medium?
IAT 2026, Chemistry, question 15 — question diagram
  1. S<P<R<Q
  2. S<P<Q<R
  3. P<R<Q<S
  4. P<Q<R<S
Answer & worked solution
Answer C
A smaller pKb means a stronger base. In benzylamine P, nitrogen's lone pair is separated from the phenyl ring by CH2 and is not delocalised into the ring, making P much more basic. In the aniline series, methyl donation gives the aqueous order R > Q > S in basicity for these listed compounds. Reversing that order for pKb gives P<R<Q<SP<R<Q<S.

Mathematics

Question 1

IAT 2026 · Mathematics
Let 𝑝(𝑥) be a quadratic polynomial such that 𝑝(1)=𝑝(1)=0. What is the coefficient of 𝑥 in 𝑝(𝑥)?
  1. 1
  2. 0
  3. 1
  4. 2
Answer & worked solution
Answer B
With roots +1 and −1, the quadratic has the form p(x)=a(x1)(x+1)=a(x21)p(x)=a(x-1)(x+1)=a(x^2-1), where a is nonzero. There is no linear term, so the coefficient of x is zero.

Question 2

IAT 2026 · Mathematics
Consider the following sets of points in the complex plane
𝐴={cos(2𝑛𝜋5)+𝑖sin(2𝑛𝜋5):𝑛} and
𝐵={cos(2𝑛5)+𝑖sin(2𝑛5):𝑛}.
Which of the following statements is TRUE?
  1. 𝐴 is infinite but 𝐵 is finite.
  2. 𝐴 is infinite and 𝐵 is also infinite.
  3. 𝐴 is finite and 𝐵 is also finite.
  4. 𝐴 is finite but 𝐵 is infinite.
Answer & worked solution
Answer D
The points of A are fifth roots of unity and repeat when n increases by 5, so A contains exactly five points. Equality of two points of B would require 2(nm)/5=2πk2(n-m)/5=2\pi k for an integer k. Irrationality of π forces k = 0 and n = m. Thus B has infinitely many distinct points.

Question 3

IAT 2026 · Mathematics
Let 𝑟,𝑙 be two integers such that 𝑟𝑙3. What is the total number of functions
𝑓:{1,2,,𝑟}{1,2,,𝑟}
such that 𝑓(1),𝑓(2),,𝑓(𝑙) are all distinct?
  1. 𝑟(𝑟1)(𝑟2)(𝑟𝑙+1)
  2. 𝑟𝑟
  3. 𝑟𝑟𝑙(𝑟1)(𝑟2)(𝑟𝑙+1)
  4. 𝑟𝑟𝑙+1(𝑟1)(𝑟2)(𝑟𝑙+1)
Answer & worked solution
Answer D
Assign distinct images to the first l inputs in r(r1)(rl+1)r(r-1)\cdots(r-l+1) ways. Each of the remaining r−l inputs can map independently to any of r values, contributing rrlr^{r-l}. Multiplication gives rrl+1(r1)(rl+1)r^{r-l+1}(r-1)\cdots(r-l+1), which is D.

Question 4

IAT 2026 · Mathematics
Let 𝑎1,𝑎2,𝑎3, be a geometric progression of positive integers such that 𝑎1=3 and 𝑎𝑛+22𝑎𝑛=𝑎𝑛+1 for all positive integers 𝑛. What is the value of 𝑎1+𝑎2+𝑎3+𝑎4+𝑎5?
  1. 255
  2. 99
  3. 93
  4. 120
Answer & worked solution
Answer C
Let the common ratio be q. The recurrence gives anq22an=anqa_nq^2-2a_n=a_nq, so q2q2=(q2)(q+1)=0q^2-q-2=(q-2)(q+1)=0. Positivity excludes q = −1, leaving q = 2. The first five terms are 3, 6, 12, 24 and 48, whose sum is 93.

Question 5

IAT 2026 · Mathematics
Let 𝑛=2026. What is the remainder when 49𝑛+41𝑛+10𝑛 is divided by 100?
  1. 1
  2. 49
  3. 90
  4. 2
Answer & worked solution
Answer D
Since n is even, 49n=(492)n/21(mod100)49^n=(49^2)^{n/2}\equiv1\pmod{100}. Also 4151(mod100)41^5\equiv1\pmod{100} and n is a multiple of 5, so 41n141^n\equiv1. Finally n is a multiple of 10, making 10n010n\equiv0. The required remainder is 1+1+0=21+1+0=2.

Question 6

IAT 2026 · Mathematics
Consider the data of scores obtained by students in an examination. If the score of every student is increased by 2 marks, then which of the following statements is TRUE?
  1. The mean deviation about the median is increased by 2.
  2. The variance is increased by 2.
  3. The mean deviation about the mean is increased by 2.
  4. The mean deviation about the mean does not change.
Answer & worked solution
Answer D
Adding 2 to every score also adds 2 to the mean. Each deviation remains (xi+2)(xˉ+2)=xixˉ(x_i+2)-(\bar x+2)=x_i-\bar x, so the average absolute deviation about the mean is unchanged. Variance and deviation about the median are translation-invariant for the same reason, rather than increasing by 2.

Question 7

IAT 2026 · Mathematics
For a 2×2 matrix 𝐴, whose elements are real numbers, denote by 𝐴𝑚 the product 𝐴𝐴𝐴𝑚 times, where 𝑚 is a positive integer. Define 𝑥0=0, 𝑥1=1, 𝑥𝑛=𝑥𝑛1+𝑥𝑛2, for all 𝑛2 and
𝐴𝑛=[𝑥𝑛+1𝑥𝑛𝑥𝑛𝑥𝑛1], for all 𝑛1.
Which of the following statements is TRUE for all 𝑚3?
  1. 𝐴𝑚𝐴𝑚1[1001]=[0000]
  2. 𝐴1𝑚=𝐴1𝑚1+𝐴1𝑚2
  3. det(𝐴𝑚)=1
  4. 𝐴1𝑚𝐴1𝑚1+[1001]=[0000]
Answer & worked solution
Answer B
From the initial Fibonacci values, A1=(1110)A_1=\begin{pmatrix}1&1\\1&0\end{pmatrix}. Direct multiplication gives A12=A1+IA_1^2=A_1+I. Multiply by A1m2A_1^{m-2} to obtain A1m=A1m1+A1m2A_1^m=A_1^{m-1}+A_1^{m-2} for every m ≥ 3. This is B; in particular the determinant alternates with m and is not always −1.

Question 8

IAT 2026 · Mathematics
Let 𝑓: be the function given by
𝑓(𝑥)=|𝑥2|+3|𝑥1|+||𝑥2|1|.
What is the number of points where 𝑓 is NOT differentiable?
  1. 2
  2. 0
  3. 1
  4. 3
Answer & worked solution
Answer A
Split at the candidate breakpoints 1, 2 and 3. The function simplifies to 65x6-5x for x < 1, 3x23x-2 for 1 < x < 2 and also for 2 < x < 3, and 5x85x-8 for x > 3. Slopes change at 1 and 3 but agree across 2. Thus there are exactly two non-differentiable points.

Question 9

IAT 2026 · Mathematics
For real numbers 𝑎 and 𝑏, consider the function 𝑓: given by
𝑓(𝑥)={𝑎𝑥𝑏if 𝑥<1,5𝑥+1if 1𝑥1,𝑎2𝑥+3𝑏if 𝑥>1.
How many pairs (𝑎,𝑏) are there for which 𝑓 is continuous at every point of ?
  1. 2
  2. 1
  3. infinitely many
  4. 0
Answer & worked solution
Answer D
Continuity at −1 requires ab=4a-b=-4, giving b=a+4b=a+4. Continuity at +1 requires a2+3b=6a^2+3b=6. Substitution yields a2+3a+6=0a^2+3a+6=0, whose discriminant is 924=159-24=-15. There is no real a, hence no real pair (a,b).

Question 10

IAT 2026 · Mathematics
Let C be the set of all the circles in a plane. If
R={(𝐶1,𝐶2)C×C𝐶1 and 𝐶2 intersect},
then which of the following statements is TRUE?
  1. R is reflexive and transitive but not symmetric.
  2. R is not a relation.
  3. R is symmetric and transitive but not reflexive.
  4. R is reflexive and symmetric but not transitive.
Answer & worked solution
Answer D
Every circle intersects itself, and intersection is symmetric. Transitivity fails: two separated circles can each intersect a larger third circle without intersecting one another. For example, two unit circles with centres 4 units apart are disjoint, while a circle centred midway with radius 2 intersects both. Hence R is reflexive and symmetric, but not transitive.

Question 11

IAT 2026 · Mathematics
Consider the function 𝑓: defined by 𝑓(𝑥)=sin2(7𝑥)sin2(5𝑥). Which of the following statements is NOT TRUE?
  1. 𝑓 is increasing on (3𝜋2,2𝜋).
  2. 𝑓(𝑥)>0, for all 𝑥(0,𝜋48).
  3. 𝑓(𝑥+𝜋2)+𝑓(𝑥)=0, for all 𝑥.
  4. 𝑓(𝜋12)=0
Answer & worked solution
Answer A
Use sin2usin2v=sin(u+v)sin(uv)\sin^2u-\sin^2v=\sin(u+v)\sin(u-v) to obtain f(x)=sin(12x)sin(2x)f(x)=\sin(12x)\sin(2x). This is positive on (0,π/48)(0,\pi/48), vanishes at π/12\pi/12, and changes sign on shifting x by π/2. It is not increasing throughout (3π/2,2π)(3\pi/2,2\pi): its derivative at x=5π/3x=5\pi/3 is 63-6\sqrt3. Thus A is false.

Question 12

IAT 2026 · Mathematics
What is the value of 12min{1𝑥,1𝑥3}𝑑𝑥?
  1. 1
  2. 0
  3. 1
  4. 2
Answer & worked solution
Answer A
Compare the two expressions through x3xx^3-x. The smaller is 1x31-x^3 on [−1,0], 1x1-x on [0,1], and 1x31-x^3 on [1,2]. Their integrals are respectively 5/45/4, 1/21/2 and 11/4-11/4. Adding them gives −1.

Question 13

IAT 2026 · Mathematics
Consider the points 𝐴(4𝑖ˆ+𝑗ˆ+3𝑘ˆ), 𝐵(2𝑗ˆ) and 𝐶(4𝑖ˆ+3𝑗ˆ3𝑘ˆ). Which of the following statements is TRUE?
  1. 𝐴𝐵×𝐵𝐶=𝑖ˆ+𝑗ˆ+𝑘ˆ.
  2. 𝐴𝐵+3𝐵𝐶 is perpendicular to 𝐴𝐶.
  3. 𝐴𝐵, 𝐵𝐶 and 𝐶𝐴 are mutually perpendicular.
  4. 𝐴, 𝐵 and 𝐶 are collinear.
Answer & worked solution
Answer D
In coordinates, A = (4,1,3), B = (0,2,0) and C = (−4,3,−3). Their midpoint check gives (A+C)/2=(0,2,0)=B(A+C)/2=(0,2,0)=B. Therefore B lies on segment AC and all three points are collinear.

Question 14

IAT 2026 · Mathematics
Let 𝑙1 be the line joining (1,1,1) and (3,1,3) and let 𝑙2 be the line joining (0,2,1) and (2,0,3). What is the angle between 𝑙1 and 𝑙2?
  1. 60𝑜
  2. 45𝑜
  3. 30𝑜
  4. 90𝑜
Answer & worked solution
Answer C
Direction vectors are u=(2,0,2)\vec u=(2,0,2) and v=(2,2,4)\vec v=(2,-2,4). Their dot product is 12 and their lengths are 8\sqrt8 and 24\sqrt{24}. Thus cosθ=12/192=3/2\cos\theta=12/\sqrt{192}=\sqrt3/2, giving the acute angle 30°.

Question 15

IAT 2026 · Mathematics
Suppose there are two boxes 𝐵1 and 𝐵2, each having 3 red and 4 black balls. One ball is drawn at random from 𝐵1. If it is red, 4 red balls are put into 𝐵2, otherwise 3 black balls are put into 𝐵2. Then one ball is randomly drawn from 𝐵2. If this ball is red, what is the conditional probability that the ball drawn from 𝐵1 was also red?
  1. 3557
  2. 37
  3. 3353
  4. 99257
Answer & worked solution
Answer A
Let R1 denote a red draw from the first box and R2 a red draw from the second. We have P(R1)=3/7P(R1)=3/7, P(R2R1)=7/11P(R2|R1)=7/11 and P(R2R1c)=3/10P(R2|R1^c)=3/10. Bayes' rule gives P(R1R2)=(3/7)(7/11)(3/7)(7/11)+(4/7)(3/10)=35/57P(R1|R2)=\frac{(3/7)(7/11)}{(3/7)(7/11)+(4/7)(3/10)}=35/57.

Physics

Question 1

IAT 2026 · Physics
The acceleration of a point particle is given by the equation
𝑑2𝐱𝑑𝑡2=𝛼𝐱|𝐱|7+𝛽𝑑𝐱𝑑𝑡
where 𝐱 denotes position and 𝑡 denotes time. Which of the following relations show the correct dimensions for 𝛼 and 𝛽?
  1. [𝛼]=[𝑀1𝐿6𝑇2], [𝛽]=[𝑀0𝐿0𝑇3]
  2. [𝛼]=[𝑀0𝐿6𝑇1], [𝛽]=[𝑀0𝐿1𝑇2]
  3. [𝛼]=[𝑀0𝐿7𝑇2], [𝛽]=[𝑀0𝐿0𝑇1]
  4. [𝛼]=[𝑀0𝐿7𝑇2], [𝛽]=[𝑀0𝐿0𝑇0]
Answer & worked solution
Answer C
Each term on the right must have acceleration dimensions LT2LT^{-2}. Since x/x7x/|x|^7 has dimensions L6L^{-6}, α must have dimensions L7T2L^7T^{-2}. Since velocity has dimensions LT1LT^{-1}, β must have dimensions T1T^{-1}. Both are independent of mass, as in C.

Question 2

IAT 2026 · Physics
The position of a particle of mass 1 kg at time 𝑡 is given by 𝐫=𝑡𝑖ˆ+𝑗ˆ+2𝑡2𝑘ˆ, where 𝑡 is in seconds and the coefficients have the proper units for 𝐫 to be in metres. What is the component of the angular momentum (with respect to the origin) in kg m2s1 along the vector (𝑖ˆ+𝑗ˆ)?
  1. 4𝑡2𝑡2
  2. 12(4𝑡+6𝑡2)
  3. 12(4𝑡2𝑡2)
  4. 4𝑡+6𝑡2
Answer & worked solution
Answer C
Differentiate position to obtain v=(1,0,4t)\vec v=(1,0,4t). With unit mass, L=r×v=(4t,2t2,1)\vec L=\vec r\times\vec v=(4t,-2t^2,-1). The unit vector along i^+j^\hat i+\hat j is (1,1,0)/2(1,1,0)/\sqrt2, so the scalar component is (4t2t2)/2(4t-2t^2)/\sqrt2.

Question 3

IAT 2026 · Physics
A planet is revolving in a circular orbit with a time period 𝑇 around the center of a star solely under the gravity of the star. Suppose the distance between the star and the planet is halved. The individual radii of the star and the planet are also halved, keeping their uniform mass densities unchanged. What will be the time period of the new orbit of the planet?
  1. 𝑇4
  2. 𝑇2
  3. 2𝑇
  4. 𝑇
Answer & worked solution
Answer D
For an orbit controlled by the star, Tr3/MT\propto\sqrt{r^3/M}. Halving the star's radius at unchanged density reduces M to M/8, while orbital radius becomes r/2. Therefore T/T=(1/8)/(1/8)=1T'/T=\sqrt{(1/8)/(1/8)}=1. The planet's mass does not enter this expression.

Question 4

IAT 2026 · Physics
An exotic spherical jellyfish has a bulk modulus 𝐵. Close to the surface of the sea (depth 𝑑=0), its radius is 𝑅. When it dives to a depth 𝑑 (𝑑𝑅), its radius is reduced by Δ𝑅>0. Given the density of the incompressible sea water 𝜌, and the uniform acceleration due to gravity 𝑔 such that 𝜌𝑔𝑑𝐵, what is Δ𝑅𝑅?
  1. [1(1𝜌𝑔𝑑𝐵)2/3]
  2. [1(1𝜌𝑔𝑑𝐵)1/3]
  3. [(1+𝜌𝑔𝑑𝐵)1/31]
  4. [(1+𝜌𝑔𝑑𝐵)2/31]
Answer & worked solution
Answer B
The pressure increase is ΔP=ρgd\Delta P=\rho gd. In the small-strain bulk-modulus relation, ΔV/V=ρgd/B\Delta V/V=-\rho gd/B. Therefore (RΔR)3/R3=1ρgd/B(R-\Delta R)^3/R^3=1-\rho gd/B, giving ΔR/R=1(1ρgd/B)1/3\Delta R/R=1-(1-\rho gd/B)^{1/3}. Its leading term is ρgd/(3B)\rho gd/(3B), consistent with a small positive shrinkage.

Question 5

IAT 2026 · Physics
Consider two Carnot engines of efficiencies 𝜂1 and 𝜂2. The first engine absorbs heat 𝑄1 from a heat reservoir 𝐴 and releases heat 𝑄2 to a heat reservoir 𝐵. The second engine takes heat 𝑄2 from 𝐵 and releases heat 𝑄3 to a heat reservoir 𝐶. If 𝑄1>𝑄2>𝑄3, what is the net efficiency of this combination of the two Carnot engines?
  1. 𝜂1+𝜂2𝜂1𝜂2
  2. 𝜂1+𝜂2+𝜂1𝜂2
  3. 𝜂1𝜂2
  4. 𝜂1+𝜂2
Answer & worked solution
Answer A
The first engine passes on Q2=(1η1)Q1Q_2=(1-\eta_1)Q_1. The second rejects Q3=(1η2)Q2Q_3=(1-\eta_2)Q_2. Overall efficiency is 1Q3/Q1=1(1η1)(1η2)=η1+η2η1η21-Q_3/Q_1=1-(1-\eta_1)(1-\eta_2)=\eta_1+\eta_2-\eta_1\eta_2.

Question 6

IAT 2026 · Physics
A jar is filled with two monoatomic non-interacting gases 𝐴 and 𝐵 with total masses 𝑀𝐴 and 𝑀𝐵, respectively. The molar mass of 𝐴 is double the molar mass of 𝐵. If the jar is kept at temperature 𝑇, what is the ratio of the total pressure of the combined gas to the partial pressure due to the gas 𝐴?
  1. 1+12𝑀𝐵𝑀𝐴
  2. 1+2𝑀𝐵𝑀𝐴
  3. 1+12𝑀𝐴𝑀𝐵
  4. 1+2𝑀𝐴𝑀𝐵
Answer & worked solution
Answer B
Partial pressure is proportional to mole number at fixed volume and temperature. If A has molar mass 2μ and B has μ, then nB/nA=(MB/μ)/(MA/(2μ))=2MB/MAn_B/n_A=(M_B/\mu)/(M_A/(2\mu))=2M_B/M_A. Thus Ptotal/PA=1+nB/nA=1+2MB/MAP_{\rm total}/P_A=1+n_B/n_A=1+2M_B/M_A.

Question 7

IAT 2026 · Physics
A simple pendulum of length 𝐿, mass 𝑚 and electric charge 𝑞 on its bob is oscillating with a time period 𝑇 under uniform gravity which is in the 𝑧ˆ direction. Upon applying a uniform electric field |𝐸|𝑛ˆ (where 𝑛ˆ is a unit vector in the plane of oscillation), the time period of the pendulum decreases. Which of the following statements is NOT correct?
  1. 𝑞 is positive and 𝑛ˆ=𝑧ˆ
  2. 𝑞 is positive and 𝑛ˆ𝑧ˆ=12
  3. 𝑞 is negative and 𝑛ˆ=𝑧ˆ
  4. 𝑞 is positive and 𝑛ˆ=𝑧ˆ
Answer & worked solution
Answer A · qualification below

The expected branch assumes a weak upward electric force.

For small oscillations about equilibrium, T=2πL/geffT'=2\pi\sqrt{L/g_{\rm eff}}, where geff=gz^+(q/m)Eg_{\rm eff}=|-g\hat z+(q/m)\vec E|. The intended hanging, weak-field case in A has an upward electrical force, reducing the restoring acceleration and increasing the period; the other choices reinforce gravity's magnitude. A is therefore the keyed answer. The wording omits a strength restriction: if the upward electrical acceleration exceeds 2g, a new upward equilibrium can also have a shorter period. This limitation must accompany the conventional answer.

Question 8

IAT 2026 · Physics
A sphere and a cube of equal masses on a horizontal frictionless floor, are confined between two vertical walls, as shown in the figure. The cube is attached to the wall by a massless spring. At the equilibrium position of the spring, the sphere just touches the cube. The cube is moved towards the left by a small amount from its equilibrium position, compressing the spring and is released at 𝑡=0. The system keeps returning to its initial configuration as that of 𝑡=0 with a time period 𝑇. If all the collisions are elastic, which of the following statements is correct?
IAT 2026, Physics, question 8 — question diagram
  1. If increases, 𝑇 increases.
  2. If increases, 𝑇 does not change.
  3. The sphere never moves.
  4. If increases, 𝑇 decreases.
Answer & worked solution
Answer D
Let ω=k/m\omega=\sqrt{k/m} and let D be the sphere's fixed one-way free-travel distance to the right wall. At first contact, the cube has speed ωl\omega l; equal-mass elastic collision transfers that speed to the sphere and leaves the cube at rest. The sphere's round trip takes 2D/(ωl)2D/(\omega l). The two spring-motion quarter cycles together take π/ω\pi/\omega, so T=π/ω+2D/(ωl)T=\pi/\omega+2D/(\omega l) decreases as l increases.

Question 9

IAT 2026 · Physics
Three infinite plane sheets which have uniform positive surface charge densities 𝜎, 𝜎 and 2𝜎, are arranged parallel to each other with a separation of 𝑑 as shown in the figure. A spherical Gaussian surface 𝑆 of radius 𝑑/2 has its center on the middle sheet. Which of the following statements regarding the electric flux Φ𝐿 through the left hemisphere and the electric flux Φ𝑅 through the right hemisphere of the Gaussian surface is correct?
IAT 2026, Physics, question 9 — question diagram
  1. Φ𝐿=Φ𝑅.
  2. Φ𝐿=2Φ𝑅.
  3. Φ𝐿<Φ𝑅.
  4. Φ𝐿>Φ𝑅.
Answer & worked solution
Answer D
Take rightward field as positive. In the left half of the sphere, the three sheet contributions sum to σ/(2ε0)σ/(2ε0)2σ/(2ε0)=σ/ε0\sigma/(2\varepsilon_0)-\sigma/(2\varepsilon_0)-2\sigma/(2\varepsilon_0)=-\sigma/\varepsilon_0. In the right half they sum to zero. The leftward field has positive outward flux through the left hemisphere, while the right flux is zero. Thus ΦL>ΦR\Phi_L>\Phi_R.

Question 10

IAT 2026 · Physics
A particle of mass 𝑚1 and electric charge 𝑞 starts from rest under the influence of a uniform external electric field 𝐄 to travel a distance 𝑑 in time 𝑡1. If the particle had mass 𝑚2, it would take time 𝑡2 to travel the same distance. What is the ratio 𝑡1𝑡2?
  1. 𝑚1𝑚2
  2. 𝑚2𝑚1
  3. 𝑚2𝑚1
  4. 𝑚1𝑚2
Answer & worked solution
Answer A
The acceleration magnitude is qE/m|qE|/m. Starting from rest over distance d gives d=12(qE/m)t2d=\frac12(|qE|/m)t^2, hence t=2dm/qEt=\sqrt{2dm/|qE|}. Holding the other quantities fixed yields t1/t2=m1/m2t_1/t_2=\sqrt{m_1/m_2}.

Question 11

IAT 2026 · Physics
A long solenoid of initial radius 𝑅0 is put in a region of uniform magnetic field 𝐁 with the axis of the solenoid aligned along the magnetic field. The solenoid is a part of a closed circuit that has no initial current running through it. If the radius of the solenoid starts increasing at a uniform rate, how do the magnetic field strength 𝐵𝑖𝑛 and the associated magnetic energy 𝑈𝑖𝑛 inside the solenoid change?
  1. 𝐵𝑖𝑛 increases, 𝑈𝑖𝑛 decreases.
  2. 𝐵𝑖𝑛 decreases, 𝑈𝑖𝑛 decreases.
  3. 𝐵𝑖𝑛 decreases, 𝑈𝑖𝑛 increases.
  4. 𝐵𝑖𝑛 increases, 𝑈𝑖𝑛 increases.
Answer & worked solution
Answer B · qualification below

The answer uses the ideal flux-conserving solenoid model; see the qualification below.

As the solenoid expands, the externally applied flux tends to increase, so the induced current opposes it. In the ideal flux-conserving circuit model underlying the key, BinπR2=BπR02B_{\rm in}\pi R^2=B\pi R_0^2, giving BinR2B_{\rm in}\propto R^{-2}. For fixed solenoid length, Uin=Bin2(πR2)/(2μ0)R2U_{\rm in}=B_{\rm in}^2(\pi R^2\ell)/(2\mu_0)\propto R^{-2}. Both decrease, so B is correct. This concerns the total internal field, not the magnitude of the induced field alone. A general resistive-circuit treatment needs circuit parameters not specified in the question.

Question 12

IAT 2026 · Physics
A spherical concave mirror of focal length 10 cm and a double convex lens of focal length 5 cm are arranged on the common principal axis as shown in the figure. A small object is placed on the principal axis between the focal points 𝐹1 and 𝐹2 of the mirror and the lens, respectively. If two real and mutually inverted images are formed by the lens at the same location on the principal axis, what is the distance of the object from the mirror on the principal axis?
IAT 2026, Physics, question 12 — question diagram
  1. 25 cm
  2. 12 cm
  3. 20 cm
  4. 30 cm
Answer & worked solution
Answer C
The lens maps a given object position to a unique image position. Thus the direct rays and the mirror-reflected rays can form images at the same location only if the mirror's intermediate image coincides with the object. A concave mirror does that at its centre of curvature, 2fm=202f_m=20 cm. The mirror reverses that intermediate image, so the two lens images have opposite orientations as required.

Question 13

IAT 2026 · Physics
An experimental study of the photoelectric effect involves a metal of work function 𝜙0. What is the smallest wavelength of the incident photon to photoemit an electron of mass 𝑚 which has the same de Broglie wavelength as that of the incident photon? [Given is the Planck's constant, 𝑐 is the speed of light, and 𝜙0𝑚𝑐2]
  1. 𝑚𝑐(112𝜙0𝑚𝑐2)1
  2. 𝑚𝑐(1+12𝜙0𝑚𝑐2)1
  3. 𝑚𝑐(11𝜙0𝑚𝑐2)1
  4. 𝑚𝑐(1+1𝜙0𝑚𝑐2)1
Answer & worked solution
Answer B · qualification below

The formal nonrelativistic result has a self-consistency limitation at the stated short wavelength.

Using the nonrelativistic photoelectric equation and equal wavelengths gives hc/λ=ϕ0+h2/(2mλ2)hc/\lambda=\phi_0+h^2/(2m\lambda^2). Solving the quadratic in 1/λ1/\lambda gives λ=hmc[1±12ϕ0/(mc2)]1\lambda=\frac h{mc}[1\pm\sqrt{1-2\phi_0/(mc^2)}]^{-1}. The plus sign in the denominator gives the smaller wavelength, option B, matching the exam key. There is a model limitation: for ϕ0mc2\phi_0\ll mc^2 this branch implies momentum near 2mc, so a nonrelativistic kinetic-energy formula is not physically self-consistent. The stated answer is the formal exam-model result.

Question 14

IAT 2026 · Physics
Consider normal incidence of a monochromatic beam of photons of power 𝑃 on a flat surface. Of the incident beam, 10% gets absorbed, 10% gets transmitted, and the rest is reflected by the flat surface. If 𝑐 is the speed of light, what is the force exerted on the flat surface by the beam?
  1. 1.8𝑃𝑐
  2. 0.9𝑃𝑐
  3. 1.6𝑃𝑐
  4. 1.7𝑃𝑐
Answer & worked solution
Answer D
Absorption transfers the incident momentum, reflection reverses it and transfers twice as much, and unchanged transmission transfers none. With reflected fraction 0.8, F=(0.1+2×0.8)P/c=1.7P/cF=(0.1+2\times0.8)P/c=1.7P/c. Absorbed photons transfer their energy and momentum to the surface; they should not be described as stationary photons.

Question 15

IAT 2026 · Physics
The three numbers: (number of protons, number of neutrons, the radius) characterize a nucleus. What is the value of 𝑟1𝑟2 for two nuclei characterized by (1,0,𝑟1) and (4,4,𝑟2)?
  1. 8
  2. 18
  3. 12
  4. 2
Answer & worked solution
Answer C
The nuclear-radius model gives r=r0A1/3r=r_0A^{1/3}, with A equal to protons plus neutrons. The mass numbers are 1 and 8, so r1/r2=(1/8)1/3=1/2r_1/r_2=(1/8)^{1/3}=1/2.

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Analytics uses cookies and information such as the page visited, device, approximate location and referral source. We record the paper year, subject and question number when you use a solution. We do not send the words you type in the paper search, names, email addresses or phone numbers in our study events.

We keep event data for up to 14 months. Advertising personalisation and Google signals are disabled for this setup. Our hosting provider may keep routine security and access logs independently of optional analytics.

Your choice is remembered on this browser for six months. You can change it here at any time. Declining does not affect access to questions or solutions. Links to other sites follow their own privacy practices.

How Google uses data from sites using its services. For a privacy query, contact Mathiit using the details on our About & contact page.