Full Test — 18 October 2026

Error notes for Aamirah Fathima · EMI + Haloalkanes & Haloarenes + Molecular Basis of Inheritance + Biotechnology · 180 questions, +4 / −1 / 0

Total

565 / 720

Against the 600 target

−35

Correct

149 / 180

Unattempted

Previous (AITS-04)

534

Section	Right	Wrong	Score	Target	Gap
Physics Electromagnetic Induction	32	13	115	120	−5
Chemistry Haloalkanes and Haloarenes	33	12	120	153	−33
Botany Molecular Basis of Inheritance	42	3	165	165	+0
Zoology Biotechnology and its Applications	42	3	165	162	+3
TOTAL	149	31	565	600	−35

The two things that changed

Zero blanks on all 180 questions. This is the first full paper with no unattempted question. The flagged-but-unmarked habit cost 24 marks on Practical Chemistry, 8 on MBI and 4 on Coordination. Today it cost nothing. That rule — clear every circle by two-thirds time — is now closed and does not need to be on the checklist any more.

Biology is finished. Botany 165 and Zoology 165 against targets of 165 and 162, from six losses in ninety questions. The entire 35-mark gap to 600 now sits in Physics (−5) and Chemistry (−33). Nothing in this document asks her to revise biology.

Where the 124 lost marks actually went

Thirty-one losses, but not thirty-one weaknesses. Five clusters account for twenty-three of them:

Cluster	Questions	Marks	What it is
A dropped or spurious factor	Q8, 17, 30, 31, 39	25	The formula was right every time. A square, a ½, a length or a multiplication by L went missing while writing it down.
Classification of halides	Q80, 81, 84	15	Vinylic vs allylic, secondary vs tertiary benzylic, gem vs vic. One NCERT page settles all three.
The halogen falling off the ring	Q72, 73	10	On both electrophilic substitutions she chose the product that had lost its chlorine.
Negative stems	Q11, 67, 174	15	All three times she marked a statement that is true and relevant — the NOT was read but not carried into the options.
Rearranging e = L·dI/dt	Q14, 23	10	Same equation, same inversion, nine questions apart.

Twenty-five marks of Physics — nearly half the section's losses — are one habit, not one topic. She is not short of EMI. She is writing formulae too fast to keep every factor in them.

What reopened

Assertion–Reason (Q42, Q89 — 8 marks). This family ran across four papers in September, closed on the Practical Chemistry re-test (8/8) and again on 8 October (8/8), and has come back on this paper. Both failures were the same shape: she accepted the Reason without testing it as a standalone sentence. On Q89 the Reason says the haloarene carbon is sp³ — a claim she had already contradicted correctly twice on the same page, at Q67 and Q71.

Arrangement and order (Q53). Down to a single loss from a family that once cost 20+ marks a paper. Worth noting as nearly closed rather than as a weakness.

The six biology losses, in one line each

Q104 — Statement II had "exons removed, introns joined". She had answered Q96 on the same definition correctly eight questions earlier.
Q117 & Q118 — counted the stop codon as an amino acid, twice. Then applied the rule correctly on Q131 and Q133 later in the same section.
Q156 — 27 is the basmati figure; 2,00,000 is the rice figure. Two NCERT numbers a paragraph apart.
Q162 — tobacco budworm is the lepidopteran. She got Q163, which names the coleopterans, right on the next page.
Q174 — negative stem; she marked a genuine NCERT advantage.

Five of the six are facts she demonstrably holds, missed under speed. None of them is a reason to re-read a chapter.

What to do before the next test

The factor drill, 20 minutes. Twelve EMI numericals, worked with one rule: write the formula, then circle every exponent and coefficient in it BEFORE substituting a single number. The five questions in cluster A were all lost between reading the formula and writing it.
One NCERT page, 15 minutes. The halide classification page — vinylic, allylic, benzylic, aryl, gem, vic. Fifteen marks sat on it today.
The NOT rule. On any stem containing NOT, INCORRECT or EXCEPT, tick the options that ARE true in the margin first, then mark whatever is left. Three questions, 15 marks, every one of them a true statement chosen.
Assertion–Reason, 10 items. Read the Reason alone, decide true or false, and only then ask whether it explains the Assertion. Both losses died at the first step.
Nothing in biology. 330 of 360 against a 327 target.

On this paper, closing cluster A alone takes her to 590. Adding the halide page and the NOT rule takes her past 600 without touching a new chapter.

Physics · 13 Chemistry · 12 Botany · 3 Zoology · 3

Physics — Electromagnetic Induction

32 correct · 13 wrong · 0 unattempted · 115/180 · −5 against target

Q8 SLIP AC generator — peak emf

An ac generator has a coil of 100 turns, each of area 0.1 m². It rotates at 50 revolutions per second in a uniform magnetic field of 0.5 T. The peak emf generated is nearly

(1) 157 V

(2) 785 V ← she marked

(3) 3140 V

(4) 1570 V ← correct

Q8. The ½ in e = ½Bωl² exists only because a rotating rod's speed varies along its length. A coil has no such gradient. Aamirah's 785 V is exactly half of 1570 V — the ½ was carried across.

She marked(2) 785 V

Correct(4) 1570 V

Given	N = 100, A = 0.1 m², B = 0.5 T, f = 50 rev s⁻¹
Asked	The PEAK emf of the generator
Concept	A rotating coil sweeps the same area on every turn, so there is no velocity gradient to average over. The ½ that appears in the rotating-rod formula has no counterpart here.
Formula / rule	e₀ = N A B ω, with ω = 2πf
Baby steps	ω = 2π × 50 = 100π ≈ 314 rad s⁻¹ N A B = 100 × 0.1 × 0.5 = 5 e₀ = 5 × 314 = 1570 V
Why not the others	157 V uses ω = f instead of 2πf. 3140 V doubles it. 785 V is exactly 1570 ÷ 2 — the ½ from e = ½Bωl² was carried in from Q6 and Q7, which she had just answered correctly.
Shortcut	NABω. If the thing rotating is a COIL, there is no ½. If it is a ROD about one end, there is.

Where it went wrongFormula contamination from the two questions immediately before it. Q6 and Q7 were both ½Bωl² and both right; the ½ travelled into Q8.

Q11 CONCEPT Eddy currents — NEGATIVE stem

Which of the following does NOT make use of eddy currents?

(1) Induction furnace

(2) Magnetic braking in trains ← she marked

(3) Electric power meters

(4) Electric fuse ← correct

She marked(2) Magnetic braking in trains

Correct(4) Electric fuse

Given	Four devices
Asked	The one that does NOT use eddy currents
Concept	Eddy currents are circulating currents induced in a solid conductor by a changing flux. An electric fuse works on the heating effect of a steady conduction current (I²R in a thin wire), with no induction at all.
Formula / rule	— (recall)
Baby steps	Induction furnace — eddy currents heat the metal. USES them. Magnetic braking — eddy currents in the rails oppose motion. USES them. Power meters — the aluminium disc turns because of eddy currents. USES them. Electric fuse — a thin wire melts from ohmic heating. Does NOT.
Why not the others	She marked magnetic braking, which is NCERT's own flagship example of eddy currents — the single most clearly 'uses eddy currents' option on the page.
Shortcut	Three of the four are in the NCERT eddy-current section by name. The fuse is not in that section at all.

Where it went wrongNegative stem. She found a device that uses eddy currents and marked it. The word NOT was read but not held through the option list.

Q14 SLIP Self-inductance from e = L dI/dt

When the current in a coil changes uniformly from 5 A to 1 A in 0.1 s, an emf of 20 V is induced in it. The self-inductance of the coil is

(1) 0.25 H

(2) 0.5 H ← correct

(3) 2 H ← she marked

(4) 4 H

Q14 and Q23 are the same equation. Always build the rate ΔI/Δt FIRST, write it down, then divide the voltage BY it. The inductance is never the bigger number divided by the smaller.

She marked(3) 2 H

Correct(2) 0.5 H

Given	ΔI = 5 − 1 = 4 A, Δt = 0.1 s, e = 20 V
Asked	Self-inductance L
Concept	The induced emf is the inductance multiplied by the RATE of current change. To recover L you divide the voltage by that rate — never the rate by the voltage.
Formula / rule	e = L (ΔI/Δt) → L = e ÷ (ΔI/Δt)
Baby steps	Build the rate first: ΔI/Δt = 4 ÷ 0.1 = 40 A s⁻¹ L = 20 ÷ 40 L = 0.5 H
Why not the others	2 H is 40 ÷ 20 — the division performed upside down. 4 H and 0.25 H are the other two ways of pairing those numbers.
Shortcut	Write the rate on its own line before dividing. 40 A s⁻¹ then 20 ÷ 40.

Where it went wrongRatio inversion. Identical to Q23 twelve questions later, where the same rearrangement failed again.

Q17 SLIP Energy stored — percentage change

If the current through an inductor is increased by 20%, the magnetic energy stored in it increases by

(1) 20% ← she marked

(2) 40%

(3) 44% ← correct

(4) 21%

20 of Physics's 52 lost marks sit in this one table. Not one of these five is a concept she does not know — each is a single factor dropped or added while writing the formula down.

She marked(1) 20%

Correct(3) 44%

Given	I → 1.20 I
Asked	Percentage increase in stored energy
Concept	Magnetic energy goes as the SQUARE of the current. A percentage change in I is therefore not the percentage change in U.
Formula / rule	U = ½ L I²
Baby steps	New current = 1.2 I New energy ∝ (1.2)² = 1.44 Increase = 1.44 − 1 = 0.44 = 44%
Why not the others	20% is the answer if the square is ignored. 40% doubles the percentage instead of squaring the factor. 21% squares the percentage (0.2² = 4%) rather than the multiplier.
Shortcut	Convert % to a multiplier, square the multiplier, subtract 1. 1.2 → 1.44 → 44%.

Where it went wrongThe square was not applied. Same root as Q30 and Q31 on this paper.

Q19 CONCEPT Series and parallel inductors

Two inductors connected in series give an equivalent inductance of 9 H and, when connected in parallel, 2 H (mutual inductance neglected). Their individual inductances are

(1) 4 H and 5 H ← she marked

(2) 3 H and 6 H ← correct

(3) 2 H and 7 H

(4) 1 H and 8 H

She marked(1) 4 H and 5 H

Correct(2) 3 H and 6 H

Given	L₁ + L₂ = 9 H; L₁L₂/(L₁+L₂) = 2 H
Asked	The individual inductances
Concept	Two conditions were given. A candidate pair has to satisfy BOTH, and the series one is satisfied by every option on the page — so it carries no information.
Formula / rule	series: L₁ + L₂ · parallel: L₁L₂/(L₁+L₂)
Baby steps	From the parallel value: L₁L₂ = 2 × 9 = 18 With L₁ + L₂ = 9 and L₁L₂ = 18, solve x² − 9x + 18 = 0 x = 3 and x = 6
Why not the others	Every one of the four options sums to 9. Only 3 and 6 also give a product of 18: 4×5 = 20 → parallel 2.22 H; 2×7 = 14 → 1.56 H; 1×8 = 8 → 0.89 H.
Shortcut	When all four options pass one test, that test is decoration. Go straight to the second condition.

Where it went wrongShe checked the sum, found 4 + 5 = 9, and stopped. The option list was built so that the first check could not discriminate.

Q23 SLIP Mutual inductance from e = M dI/dt

The current in one of two adjacent coils changes from 12 A to 4 A in 0.4 s, and an emf of 80 V is induced in the other coil. The mutual inductance of the pair is

(1) 16 H

(2) 8 H

(3) 2 H ← she marked

(4) 4 H ← correct

Q14 and Q23 are the same equation. Always build the rate ΔI/Δt FIRST, write it down, then divide the voltage BY it. The inductance is never the bigger number divided by the smaller.

She marked(3) 2 H

Correct(4) 4 H

Given	ΔI = 12 − 4 = 8 A, Δt = 0.4 s, e = 80 V
Asked	Mutual inductance M
Concept	Mutual inductance obeys exactly the same equation as self-inductance, with the emf in one coil and the current change in the other.
Formula / rule	e = M (ΔI/Δt) → M = e ÷ (ΔI/Δt)
Baby steps	ΔI/Δt = 8 ÷ 0.4 = 20 A s⁻¹ M = 80 ÷ 20 M = 4 H
Why not the others	2 H and 8 H are what you get from mis-pairing the numbers; 16 H uses Δt as a multiplier.
Shortcut	Rate first, then divide the voltage by it. The same two keystrokes as Q14.

Where it went wrongSecond failure of the same rearrangement on one paper. Together Q14 and Q23 cost 10 marks.

Q28 CONCEPT Induced electric field inside the region

A uniform magnetic field confined to a circular region of radius R changes at a constant rate. Inside the region (r < R), the magnitude of the induced electric field at a distance r from the centre

(1) is proportional to r ← correct

(2) is proportional to 1/r ← she marked

(3) is proportional to r²

(4) is independent of r

Q28 and Q29 are the two halves of one graph. Inside the field region E grows with r; outside it falls as 1/r. Reading 'r' before deciding which side of R you are on fixes both.

She marked(2) is proportional to 1/r

Correct(1) proportional to r

Given	Field confined to radius R, dB/dt constant, point at r < R
Asked	How E varies with r INSIDE the region
Concept	Faraday's law round a circle of radius r: the emf equals the rate of change of the flux enclosed. Inside, the enclosed area grows as r², while the path length grows only as r — so E rises linearly. Outside, the enclosed flux stops growing and E falls as 1/r.
Formula / rule	E (2πr) = (dB/dt)(πr²) → E = (r/2)(dB/dt) for r < R
Baby steps	Enclosed flux = B × πr² (the whole circle is inside the field) E × 2πr = πr² (dB/dt) E = (r/2)(dB/dt) → E ∝ r
Why not the others	1/r is the law OUTSIDE the region, where the enclosed flux is frozen at B×πR² and E = (R²/2r)(dB/dt).
Shortcut	Inside, E climbs with r. Outside, it falls. Decide which side of R the point is on before choosing.

Where it went wrongShe used the outside law for an inside point — and then used the outside law correctly on Q29, which genuinely was an outside point. The physics is there; the region test is not.

Q30 SLIP Terminal speed of a falling wire

A horizontal wire of mass 0.1 kg and length 0.5 m slides without friction on two vertical conducting rails. The circuit resistance is 2 Ω, and a uniform horizontal magnetic field of 1 T acts perpendicular to the plane of the rails. The terminal speed of the falling wire is

(1) 2 m s⁻¹

(2) 4 m s⁻¹ ← she marked

(3) 8 m s⁻¹ ← correct

(4) 16 m s⁻¹

20 of Physics's 52 lost marks sit in this one table. Not one of these five is a concept she does not know — each is a single factor dropped or added while writing the formula down.

She marked(2) 4 m s⁻¹

Correct(3) 8 m s⁻¹

Given	m = 0.1 kg, l = 0.5 m, R = 2 Ω, B = 1 T, g = 10 m s⁻²
Asked	Terminal (constant) speed
Concept	At terminal speed the magnetic retarding force exactly balances the weight. The retarding force carries l twice — once in the emf and once in the force on the current.
Formula / rule	mg = B²l²v / R → v = mgR / (B²l²)
Baby steps	emf = Blv, so current I = Blv/R Force on the wire = BIl = B²l²v/R Set equal to mg: v = (0.1 × 10 × 2) ÷ (1² × 0.5²) = 2 ÷ 0.25 = 8 m s⁻¹
Why not the others	4 m s⁻¹ is 2 ÷ 0.5 — the length used once instead of twice. That single missing factor of l halves the answer.
Shortcut	Build the force in two steps (emf, then BIl) and l appears twice on its own.

Where it went wrongl instead of l². Third member of the same factor-dropping family as Q17 and Q31.

Q31 SLIP Power to keep a rod moving

A rod of length 0.4 m is pulled at a constant speed of 5 m s⁻¹ along two parallel rails connected by a 2 Ω resistor, in a perpendicular magnetic field of 0.5 T. The power that must be supplied to keep the rod moving is

(1) 0.1 W

(2) 0.5 W ← correct

(3) 1 W ← she marked

(4) 0.25 W

She marked(3) 1 W

Correct(2) 0.5 W

Given	l = 0.4 m, v = 5 m s⁻¹, B = 0.5 T, R = 2 Ω
Asked	The power that must be supplied
Concept	At constant speed all the supplied power ends up dissipated in the resistor, so the question is really 'what is the power in R?'
Formula / rule	e = Blv, P = e²/R
Baby steps	e = 0.5 × 0.4 × 5 = 1 V I = e/R = 1/2 = 0.5 A P = eI = 1 × 0.5 = 0.5 W
Why not the others	1 W is the value of the EMF in volts, not the power. The last division by R was never made.
Shortcut	emf → current → power. Three lines, and the answer cannot come out as the emf.

Where it went wrongShe computed e = 1 V and read that number straight off as the answer. The emf and the power happened to look alike, which is exactly what the option list was built for.

Q36 NCERT LINE Dimensional formula of inductance

The dimensional formula of self-inductance is

(1) [ML²T⁻²A⁻²] ← correct

(2) [ML²T⁻³A⁻¹] ← she marked

(3) [MLT⁻²A⁻²]

(4) [ML²T⁻¹A⁻²]

She marked(2) [ML²T⁻³A⁻¹]

Correct(1) [ML²T⁻²A⁻²]

Given	—
Asked	Dimensions of L
Concept	Derive it from the energy expression rather than memorising it: inductance is energy per current squared.
Formula / rule	U = ½LI² → [L] = [energy] / [current]²
Baby steps	[energy] = ML²T⁻² [current]² = A² [L] = ML²T⁻² A⁻²
Why not the others	[ML²T⁻³A⁻¹] is the dimensional formula of the VOLT (potential difference). That is the one she marked.
Shortcut	U = ½LI². Energy over amp squared. Two seconds, no recall needed.

Where it went wrongRecall rather than derivation — and the recalled formula belonged to a different quantity she had been using all through the paper.

Q39 SLIP Self-induced emf from a current function

The current in a coil of self-inductance 0.5 H varies as I = (2t² + 3t) A. The magnitude of the self-induced emf at t = 2 s is

(1) 5.5 V ← correct

(2) 11 V ← she marked

(3) 7 V

(4) 2.75 V

20 of Physics's 52 lost marks sit in this one table. Not one of these five is a concept she does not know — each is a single factor dropped or added while writing the formula down.

She marked(2) 11 V

Correct(1) 5.5 V

Given	L = 0.5 H, I = 2t² + 3t, t = 2 s
Asked	Magnitude of the self-induced emf
Concept	Differentiate to get the rate, then multiply by the inductance. The differentiation is only half the job.
Formula / rule	e = L (dI/dt)
Baby steps	dI/dt = 4t + 3 At t = 2: dI/dt = 8 + 3 = 11 A s⁻¹ e = 0.5 × 11 = 5.5 V
Why not the others	11 V is dI/dt itself — correct differentiation, then the multiplication by L was skipped. 2.75 V halves it again.
Shortcut	The answer to 'find the emf' can never be the derivative on its own. L is still waiting.

Where it went wrongShe did the harder step (the calculus) and dropped the trivial one. Same family as Q17, Q30, Q31.

Q40 CONCEPT What the induced emf depends on

The magnitude of the emf induced in a coil depends on (a) the number of turns in the coil, (b) the rate of change of magnetic flux through it, (c) the resistance of the coil

(1) (a) only ← she marked

(2) (a) and (b) only ← correct

(3) (b) and (c) only

(4) (a), (b) and (c)

She marked(1) (a) only

Correct(2) (a) and (b) only

Given	Three candidate factors
Asked	Which the induced EMF depends on
Concept	Faraday's law contains N and dΦ/dt and nothing else. Resistance decides the induced CURRENT, not the induced emf — this is the standard emf-versus-current separation.
Formula / rule	e = −N (dΦ/dt)
Baby steps	(a) N appears explicitly. YES. (b) dΦ/dt is the whole law. YES. (c) R does not appear. It sets I = e/R afterwards. NO.
Why not the others	Marking (a) only drops the rate of change of flux — the one term that Faraday's law is actually about.
Shortcut	Write the law out. Whatever is not in it, is not a dependence.

Where it went wrongOver-restriction. On a multi-statement question she rejected a statement that is literally in the formula, which is the mirror image of her usual habit of accepting too much.

Q42 CONCEPT Assertion–Reason: magnet in a copper tube

Assertion (A): A magnet dropped through a long vertical copper tube falls more slowly than it would in free fall.
Reason (R): Eddy currents induced in the walls of the tube oppose the motion of the magnet.

(1) Both (A) and (R) are true and (R) is the correct explanation of (A) ← correct

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A) ← she marked

(3) (A) is true but (R) is false

(4) (A) is false but (R) is true

Q42. The eddy-current force IS the whole mechanism of the slow fall — there is no second cause. When the Reason states the mechanism named in the Assertion, the answer is (1).

She marked(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)

Correct(1)

Given	A standard NCERT demonstration and its mechanism
Asked	Whether R explains A
Concept	Test the Reason by asking: if the Reason were removed, would the Assertion still happen? Remove the eddy currents and the magnet falls at g. So the Reason is not merely true alongside A — it is the entire cause of A.
Formula / rule	—
Baby steps	A is true: the slow fall is the classic copper-tube demonstration. R is true: eddy currents in the walls oppose the changing flux (Lenz). Is R the cause of A? Remove eddy currents → no retardation → A fails. So yes. Both true, R explains A → option (1)
Why not the others	Option (2) would mean there is some OTHER reason for the slow fall and the eddy currents are a coincidence. There is no other reason.
Shortcut	Delete the Reason in your head. If the Assertion collapses, the answer is (1).

Where it went wrongAssertion–Reason instability. This family was closed on the 8 October paper (8/8) and has reopened here, together with Q89 in Chemistry.

Chemistry — Haloalkanes and Haloarenes

33 correct · 12 wrong · 0 unattempted · 120/180 · −33 against target

Q49 NCERT LINE PBr₃ with alcohols

In the reaction 3ROH + PBr₃ → 3RBr + A, the product A is

(1) H₃PO₄ ← she marked

(2) H₃PO₃ ← correct

(3) POBr₃

(4) HPO₃

She marked(1) H₃PO₄

Correct(2) H₃PO₃

Given	3ROH + PBr₃ → 3RBr + A
Asked	The phosphorus-containing by-product
Concept	Phosphorus is +3 in PBr₃ and stays +3 in the by-product. Phosphorous acid, H₃PO₃, is the +3 oxoacid; phosphoric acid, H₃PO₄, is +5 and would require an oxidation that nothing here supplies.
Formula / rule	3ROH + PBr₃ → 3R–Br + H₃PO₃
Baby steps	Count the oxidation state of P in PBr₃: +3 No oxidising agent is present, so P stays +3 The +3 oxoacid is H₃PO₃ — phosphorous acid
Why not the others	H₃PO₄ requires P in +5. POBr₃ is the product of PBr₅ with a trace of water, not of this reaction.
Shortcut	PBr₃ → H₃PO₃. PCl₅ → POCl₃ + HCl. Match the oxidation state and the answer picks itself.

Where it went wrongStraight recall miss, and the near-miss option differs by a single oxygen.

Q53 CONCEPT Boiling points of isomeric bromobutanes

The correct decreasing order of boiling points of the isomeric bromobutanes is

(1) 2-bromo-2-methylpropane > 2-bromobutane > 1-bromobutane

(2) 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane ← correct

(3) 2-bromobutane > 1-bromobutane > 2-bromo-2-methylpropane ← she marked

(4) All three have the same boiling point

Q53. Branching pulls the molecule towards a sphere, shrinking the contact area for van der Waals forces. Straight chain boils highest, the fully branched one lowest — every time.

She marked(3) 2-bromobutane > 1-bromobutane > 2-bromo-2-methylpropane

Correct(2) 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane

Given	Three isomers of C₄H₉Br — identical molar mass
Asked	Decreasing order of boiling point
Concept	With molar mass held constant, only the SHAPE can differ. Branching pushes a molecule towards a sphere, shrinking the surface available for van der Waals contact, which lowers the boiling point.
Formula / rule	— (NCERT: 'boiling points decrease with increase in branching')
Baby steps	1-bromobutane — unbranched, largest contact surface → 102 °C 2-bromobutane — one branch point → 91 °C 2-bromo-2-methylpropane — fully branched, nearly spherical → 73 °C
Why not the others	She put 2-bromobutane at the top, which would need the middle isomer to beat the straight chain.
Shortcut	Straight chain always wins among isomers. Rank by how spherical the molecule is.

Where it went wrongArrangement/order. Her longest-running error family: she got the content right and the sequence wrong.

Q60 CONCEPT What the SN1 rate depends on

The rate of an S_N1 reaction depends on

(1) the concentration of the nucleophile only ← she marked

(2) the concentration of the alkyl halide only ← correct

(3) the concentrations of both the alkyl halide and the nucleophile

(4) neither concentration

Q60. The rate law reads off the SLOW step. In SN1 the slow step is the halide breaking apart on its own, so the nucleophile's concentration cannot appear in it.

She marked(1) the concentration of the nucleophile only

Correct(2) the concentration of the alkyl halide only

Given	An SN1 reaction
Asked	Which concentrations control the rate
Concept	A rate law reads off the SLOW step only. In SN1 the slow step is the halide ionising on its own — the nucleophile is nowhere near it, and only joins in the fast second step.
Formula / rule	rate = k [R–X] (first order)
Baby steps	SN1 step 1 (slow): R–X → R⁺ + X⁻ — only the halide takes part SN1 step 2 (fast): R⁺ + Nu⁻ → R–Nu Rate is set by step 1 → depends on [R–X] alone
Why not the others	Option (3) is the SN2 rate law. Option (1) — the one she marked — names the one species that the slow step never meets.
Shortcut	SN1: the '1' IS the number of species in the rate law, and that species is the substrate.

Where it went wrongSN1 and SN2 swapped. She chose not merely the wrong one but the species that is absent from the rate-determining step entirely.

Q66 NCERT LINE Iodobenzene from a diazonium salt

Iodobenzene is conveniently prepared from benzenediazonium chloride by treating it with

(1) I₂ in the presence of FeCl₃ ← she marked

(2) NaI in dry acetone

(3) HI

(4) KI ← correct

She marked(1) I₂ in the presence of FeCl₃

Correct(4) KI

Given	Benzenediazonium chloride → iodobenzene
Asked	The reagent
Concept	Iodine is the one halogen that needs no copper catalyst: potassium iodide alone displaces nitrogen from the diazonium salt. This is the exception NCERT prints immediately after Sandmeyer and Gattermann.
Formula / rule	C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂ + KCl
Baby steps	Cl and Br need CuCl/CuBr (Sandmeyer) or Cu powder (Gattermann) F needs the Balz–Schiemann route via the fluoroborate I needs only KI — simply warm the diazonium salt with it
Why not the others	I₂/FeCl₃ is direct iodination of benzene, which is reversible and needs an oxidising agent. NaI in dry acetone is Finkelstein — that works on alkyl halides, not on a diazonium salt.
Shortcut	Cl, Br → copper. F → fluoroborate. I → just KI.

Where it went wrongShe marked the Friedel-Crafts-looking reagent. The Finkelstein option sitting at (2) also pulled, because the chapter had just drilled it.

Q67 CONCEPT Why aryl halides resist substitution — NEGATIVE stem

Which of the following is NOT a reason for the low reactivity of aryl halides towards nucleophilic substitution?

(1) Partial double bond character of the C–X bond due to resonance

(2) The carbon attached to the halogen is sp² hybridised ← she marked

(3) The phenyl cation formed by self-ionisation is not stabilised by resonance

(4) The C–X bond in aryl halides is longer than in alkyl halides ← correct

She marked(2) The carbon attached to the halogen is sp² hybridised

Correct(4)

Given	Four statements about aryl halide reactivity
Asked	The one that is NOT a reason
Concept	Three of these are NCERT's actual reasons. The fourth is a reversed fact: the aryl C–X bond is SHORTER and stronger than the alkyl one, because of both resonance and the sp² carbon's greater s-character.
Formula / rule	—
Baby steps	Resonance gives partial double bond character — a real reason. sp² carbon is more electronegative, holds the pair tighter, shorter bond — a real reason. Phenyl cation is not resonance-stabilised, so SN1 is shut off — a real reason. 'C–X is longer' is simply false. It is shorter. → this is the answer.
Why not the others	She marked (2), which is one of the three genuine reasons NCERT lists by name.
Shortcut	On a NOT stem, mark each option TRUE or FALSE in the margin first. The odd one out here is false, not merely irrelevant.

Where it went wrongNegative stem again — the third on this paper, with Q11 and Q174. She selected a true, relevant statement.

Q72 CONCEPT Nitration of chlorobenzene

The major product of the nitration of chlorobenzene with concentrated HNO₃ and H₂SO₄ is

(1) 1-chloro-2-nitrobenzene

(2) 1-chloro-4-nitrobenzene ← correct

(3) 1-chloro-3-nitrobenzene

(4) nitrobenzene ← she marked

Q72 and Q73. Halogen is deactivating but ortho/para directing: it slows the ring down and steers the incoming group to the para position. It does not leave. Only a nucleophile under forcing conditions displaces it.

She marked(4) nitrobenzene

Correct(2) 1-chloro-4-nitrobenzene

Given	Chlorobenzene + conc. HNO₃/H₂SO₄
Asked	The MAJOR product
Concept	This is electrophilic substitution: NO₂⁺ replaces a ring HYDROGEN. The chlorine is deactivating but ortho/para directing, and the para product dominates because the ortho positions are crowded by the halogen.
Formula / rule	C₆H₅Cl + HNO₃/H₂SO₄ → p-O₂N–C₆H₄–Cl (major)
Baby steps	Identify the reaction type: electrophilic substitution The halogen directs ortho/para and stays on the ring Para is the major product on steric grounds → 1-chloro-4-nitrobenzene
Why not the others	Option (4), nitrobenzene, is the product you would write if the nitro group DISPLACED the chlorine. Nothing in electrophilic substitution removes a halogen from a ring.
Shortcut	Electrophilic substitution swaps an H. The substituent already there only chooses WHERE.

Where it went wrongSee Q73 — the same misconception, back to back, for 8 marks.

Q73 CONCEPT Friedel–Crafts acylation of chlorobenzene

The major product of the reaction of chlorobenzene with CH₃COCl in the presence of anhydrous AlCl₃ is

(1) 4-chloroacetophenone ← correct

(2) 3-chloroacetophenone

(3) 2-chloroacetophenone

(4) acetophenone ← she marked

She marked(4) acetophenone

Correct(1) 4-chloroacetophenone

Given	Chlorobenzene + CH₃COCl / anhydrous AlCl₃
Asked	The MAJOR product
Concept	Friedel–Crafts acylation — another electrophilic substitution. CH₃CO⁺ attacks the ring and replaces an H at the para position, para again being favoured over the crowded ortho.
Formula / rule	C₆H₅Cl + CH₃COCl —AlCl₃→ p-CH₃CO–C₆H₄–Cl
Baby steps	Acylium ion CH₃CO⁺ forms with AlCl₃ It attacks the ring — at the positions the Cl directs it to Para dominates → 4-chloroacetophenone
Why not the others	Acetophenone, which she marked, has no chlorine at all. That product requires the chlorine to leave, which an electrophile cannot make it do.
Shortcut	If your product has lost the halogen after an electrophilic reaction, the answer is wrong.

Where it went wrongOne misconception, two questions, 8 marks. On both Q72 and Q73 she chose the option where the halogen falls off the ring. Fixing this single idea is worth more than any other hour of chemistry revision this week.

Q77 NCERT LINE Freon-12

Freon-12, used as a refrigerant, is

(1) CCl₃F ← she marked

(2) CClF₃

(3) CHClF₂

(4) CCl₂F₂ ← correct

She marked(1) CCl₃F

Correct(4) CCl₂F₂

Given	Freon-12
Asked	Its formula
Concept	Freon-12 is dichlorodifluoromethane, CCl₂F₂ — the one NCERT names as the common refrigerant and the chief chlorofluorocarbon released into the atmosphere.
Formula / rule	CCl₂F₂
Baby steps	Freon-12 = dichlorodifluoromethane Two chlorines, two fluorines on one carbon CCl₂F₂
Why not the others	CCl₃F, the option she marked, is Freon-11 — the neighbouring compound in the same family.
Shortcut	The '12' sits beside the two Cl and two F. CCl₂F₂.

Where it went wrongNear-neighbour recall. One chlorine-for-fluorine away from correct.

Q80 CONCEPT Vinylic halide

Which of the following is a vinylic halide?

(1) CH₂=CH–CH₂Br ← she marked

(2) C₆H₅CH₂Br

(3) CH₂=CH–Br ← correct

(4) CH₃CH₂Br

Q80, Q81 and Q84 are one question asked three ways: which carbon carries the halogen? Three losses, 12 marks, and a single line of NCERT settles all of them.

She marked(1) CH₂=CH–CH₂Br

Correct(3) CH₂=CH–Br

Given	Four halides
Asked	The vinylic one
Concept	A vinylic halide has the halogen bonded DIRECTLY to a doubly-bonded (sp²) carbon. If the halogen sits on the saturated carbon NEXT to the double bond, it is allylic instead.
Formula / rule	vinylic: C=C–X · allylic: C=C–C–X
Baby steps	CH₂=CH–CH₂Br — Br on the carbon next to C=C → ALLYLIC C₆H₅CH₂Br — Br on the carbon next to the ring → BENZYLIC CH₂=CH–Br — Br on the sp² carbon itself → VINYLIC ✓ CH₃CH₂Br — plain primary alkyl halide
Why not the others	She marked the allylic halide, which is the option the question was built to catch — it looks vinylic because the double bond is visible in the formula.
Shortcut	Put your finger on the halogen. Is the carbon it touches part of the double bond, or next to it?

Where it went wrongClassification by eye rather than by the carbon the halogen is actually bonded to. See Q81 and Q84.

Q81 CONCEPT Secondary benzylic halide

Which of the following is a secondary benzylic halide?

(1) C₆H₅CH₂Cl

(2) C₆H₅C(CH₃)₂Cl ← she marked

(3) p-ClC₆H₄CH₃

(4) C₆H₅CH(Cl)CH₃ ← correct

She marked(2) C₆H₅C(CH₃)₂Cl

Correct(4) C₆H₅CH(Cl)CH₃

Given	Four chlorides
Asked	The SECONDARY benzylic one
Concept	Two tests must pass together. Benzylic means the halogen sits on the carbon attached to the ring. Secondary means that same carbon touches exactly two other carbons — the ring carbon plus one alkyl group.
Formula / rule	benzylic carbon + two carbon substituents = secondary benzylic
Baby steps	C₆H₅CH₂Cl — benzylic, but the carbon has only the ring → PRIMARY C₆H₅C(CH₃)₂Cl — benzylic with ring + two methyls → TERTIARY p-ClC₆H₄CH₃ — Cl is ON the ring → ARYL, not benzylic at all C₆H₅CH(Cl)CH₃ — ring + one methyl → SECONDARY benzylic ✓
Why not the others	She marked the tertiary benzylic halide. The benzylic half of the test passed; the degree was not checked.
Shortcut	Count the carbons touching the C–X carbon: 1 = primary, 2 = secondary, 3 = tertiary.

Where it went wrongSecond of three classification losses. She stopped at 'benzylic' and did not run the second test.

Q84 NCERT LINE Ethylidene chloride

Ethylidene chloride is

(1) CH₃–CHCl₂ ← correct

(2) CH₂Cl–CH₂Cl

(3) CH₂=CHCl

(4) CHCl=CHCl ← she marked

She marked(4) CHCl=CHCl

Correct(1) CH₃–CHCl₂

Given	The common name 'ethylidene chloride'
Asked	The structure
Concept	Ethylidene means both halogens sit on the SAME carbon — a geminal (gem) dihalide. Ethylene chloride puts one on each carbon — a vicinal (vic) dihalide.
Formula / rule	ethylidene = CH₃–CHCl₂ (gem) · ethylene = CH₂Cl–CH₂Cl (vic)
Baby steps	'-ylidene' = two bonds from one carbon So both Cl go on C-1: CH₃–CHCl₂ That is 1,1-dichloroethane
Why not the others	She marked CHCl=CHCl, which is dichloroethene — not even a saturated dihalide. CH₂Cl–CH₂Cl at (2) is ethylene chloride, the intended near-miss.
Shortcut	-ylidene → same carbon (gem). -ene → next carbon (vic).

Where it went wrongThird classification loss. Q80, Q81 and Q84 together are 12 marks from one NCERT page on how halides are named by position.

Q89 CONCEPT Assertion–Reason: haloarene hybridisation

Assertion (A): Haloarenes are less reactive than haloalkanes towards nucleophilic substitution.
Reason (R): In haloarenes, the carbon atom bonded to the halogen is sp³ hybridised.

(1) Both (A) and (R) are true and (R) is the correct explanation of (A) ← she marked

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)

(3) (A) is true but (R) is false ← correct

(4) (A) is false but (R) is true

She marked(1) Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct(3) (A) is true but (R) is false

Given	A true assertion paired with a factually wrong reason
Asked	The correct option
Concept	Check the Assertion and the Reason as SEPARATE factual claims before worrying about whether one explains the other. Here the Reason fails on its own terms: the carbon in a haloarene is part of the aromatic ring, so it is sp², not sp³.
Formula / rule	—
Baby steps	A: haloarenes are indeed far less reactive. TRUE. R: the ring carbon bonded to X is sp² hybridised, not sp³. FALSE. A true, R false → option (3)
Why not the others	She marked (1), accepting a statement that contradicts the hybridisation she had already used correctly on Q67 and Q71 of the same paper.
Shortcut	Read the Reason as a standalone sentence first. A false Reason ends the question immediately.

Where it went wrongAssertion–Reason, with Q42 in Physics. Both times the failure was accepting the Reason without testing it alone.

Botany — Molecular Basis of Inheritance

42 correct · 3 wrong · 0 unattempted · 165/180 · +0 against target

Q104 NCERT LINE Splicing — statement reversed

Statement I: hnRNA contains both exons and introns.
Statement II: During splicing, the exons are removed and the introns are joined together.

(1) Both Statement I and Statement II are correct

(2) Both Statement I and Statement II are incorrect

(3) Statement I is correct but Statement II is incorrect ← correct

(4) Statement I is incorrect but Statement II is correct ← she marked

Q104. The two statements were a matched pair — one true, one deliberately reversed. Reading Statement II's verbs in order (removed / joined) against the figure settles it in four seconds.

She marked(4) Statement I is incorrect but Statement II is correct

Correct(3)

Given	Two statements on hnRNA processing
Asked	Which are correct
Concept	Introns are the INTERVENING sequences — they are the ones removed. Exons are EXPRESSED — they are the ones retained and joined. Statement II has the two swapped.
Formula / rule	—
Baby steps	Statement I: hnRNA does carry exons and introns. TRUE. Statement II: splicing removes INTRONS and joins EXONS — the statement says the opposite. FALSE. I correct, II incorrect → option (3)
Why not the others	She marked (4), which calls Statement I wrong and Statement II right — the exact inverse of the truth on both counts.
Shortcut	Intron = INTervening = taken out. Exon = EXpressed = kept. She answered Q96 ('exons are...') correctly eight questions earlier.

Where it went wrongReversed-pair statement. She knew the content — Q96 proves it — but did not check the verbs against the definition.

Q117 SLIP Counting amino acids from an mRNA

How many amino acids will be present in the polypeptide translated from the mRNA 5'-AUG UUU GGC CAA UGC UAA-3'?

(1) 6 ← she marked

(2) 18

(3) 5 ← correct

(4) 4

Q117 and Q118 are the same subtraction. She applied it correctly on Q131 and Q133 in the same section, which makes this a speed slip, not a gap.

She marked(1) 6

Correct(3) 5

Given	Six codons: AUG UUU GGC CAA UGC UAA
Asked	Number of amino acids in the polypeptide
Concept	A stop codon is read by a release factor, not by a tRNA. It terminates the chain and contributes no residue.
Formula / rule	amino acids = codons − (1 stop codon)
Baby steps	Count the codons: 6 UAA is a stop codon — it codes for no amino acid 6 − 1 = 5 amino acids
Why not the others	6 counts the stop codon as a residue. 18 is the number of nucleotides. 4 subtracts the start codon as well, which is wrong — AUG does code for methionine.
Shortcut	Subtract the stop, keep the start.

Where it went wrongSee Q118 — the same subtraction, missed twice in a row.

Q118 SLIP Amino acids from a nucleotide count

The coding region of an mRNA, from the start codon to the stop codon (both included), has 1500 nucleotides. The number of amino acids in the polypeptide it codes for is

(1) 500 ← she marked

(2) 498

(3) 1500

(4) 499 ← correct

She marked(1) 500

Correct(4) 499

Given	1500 nucleotides, start codon to stop codon inclusive
Asked	Number of amino acids
Concept	The phrase 'both included' is the whole question. It guarantees the stop codon is inside the 1500, so one of the 500 codons produces no amino acid.
Formula / rule	amino acids = (nucleotides ÷ 3) − 1
Baby steps	1500 ÷ 3 = 500 codons One of them is the stop codon 500 − 1 = 499 amino acids
Why not the others	500 is the codon count, not the residue count. 498 subtracts the start codon too.
Shortcut	Divide by three, then subtract one — whenever the stop codon is said to be included.

Where it went wrongSecond miss of the same rule in two consecutive questions. She then applied it CORRECTLY on Q131 and Q133 later in the same section, which makes this speed, not a gap.

Zoology — Biotechnology and its Applications

42 correct · 3 wrong · 0 unattempted · 165/180 · +3 against target

Q156 NCERT LINE Rice varieties in India

The estimated number of varieties of rice in India alone is

(1) 27 ← she marked

(2) 1,000

(3) 50,000

(4) 2,00,000 ← correct

She marked(1) 27

Correct(4) 2,00,000

Given	India's rice varieties
Asked	The estimated number
Concept	NCERT gives two different rice figures a few lines apart, and this question separates them. 2,00,000 is the estimated number of rice VARIETIES in India. 27 is the number of documented BASMATI varieties grown in India, quoted in the basmati patent case.
Formula / rule	—
Baby steps	Biopiracy section: 'There are 2,00,000 varieties of rice in India' Basmati patent paragraph: 'There are 27 documented varieties of basmati grown in India' The question asks about rice overall → 2,00,000
Why not the others	27 is the basmati figure — the trap this question exists for.
Shortcut	27 belongs to basmati. 2,00,000 belongs to rice.

Where it went wrongTwo adjacent NCERT numbers swapped. Worth writing both on one line in her notes so they are learned as a pair.

Q162 NCERT LINE Tobacco budworm

The tobacco budworm, which is killed by Bt toxin, belongs to the insect order

(1) Coleoptera ← she marked

(2) Diptera

(3) Lepidoptera ← correct

(4) Hymenoptera

She marked(1) Coleoptera

Correct(3) Lepidoptera

Given	Tobacco budworm
Asked	Its insect order
Concept	NCERT lists the Bt-susceptible orders with one example each: lepidopterans (tobacco budworm, armyworm), coleopterans (beetles), dipterans (flies, mosquitoes). The budworm is the headline lepidopteran.
Formula / rule	—
Baby steps	Lepidoptera — tobacco budworm, armyworm Coleoptera — beetles Diptera — flies, mosquitoes Budworm → Lepidoptera
Why not the others	Coleoptera is beetles. She answered Q163 — which names beetles as coleopterans — correctly on the very next page.
Shortcut	Learn the NCERT line as three pairs: budworm/armyworm → Lepidoptera, beetles → Coleoptera, flies/mosquitoes → Diptera.

Where it went wrongThe pairing is in her head (Q163 proves it) but the budworm half was attached to the wrong order.

Q174 CONCEPT Advantages of GM crops — NEGATIVE stem

Which of the following is NOT an advantage of GM crops?

(1) Tolerance to abiotic stresses such as cold, drought and heat

(2) Reduced reliance on chemical pesticides ← she marked

(3) Increased post-harvest losses ← correct

(4) Increased efficiency of mineral usage by plants

She marked(2) Reduced reliance on chemical pesticides

Correct(3) Increased post-harvest losses

Given	Four statements about GM crops
Asked	The one that is NOT an advantage
Concept	NCERT's list of GM benefits includes REDUCED post-harvest losses. Option (3) reverses that word and so becomes the only disadvantage on the page.
Formula / rule	—
Baby steps	Abiotic stress tolerance — NCERT advantage ✓ Reduced pesticide reliance — NCERT advantage ✓ Increased post-harvest losses — the reverse of NCERT's 'reduced post-harvest losses' ✗ Better mineral usage — NCERT advantage ✓
Why not the others	She marked (2), which is one of the five benefits NCERT lists verbatim.
Shortcut	On a NOT stem, tick the three that ARE true and the answer is what is left.

Where it went wrongThird negative-stem loss of the paper, with Q11 and Q67. All three times she marked a true, correct statement — the stem's NOT was not carried through to the options.

Trajectory

Paper	Date	Physics	Chemistry	Botany	Zoology	Total	Blanks
ILTS	20 Sep	—	—	—	—	526	—
Full Mock #1	30 Sep	—	—	—	—	596	—
AITS-04	11 Oct	125	148	155	106	534	—
Full Test	18 Oct	115	120	165	165	565	0

Zoology has moved 106 → 165 in seven days. Physics and Chemistry are now the whole of the remaining gap, and on this paper both were lost to habits rather than to syllabus.

Answer key verified in Python for every numerical and every counting item (Q1–Q45 computed from first principles; Q86, Q117, Q118, Q131, Q132, Q133 recomputed). Marking: +4 correct, −1 wrong, 0 unattempted.