Read this first. Accuracy on the questions she actually attempted was 80.4%, which is not the problem.
The 17 blanks are: they cost 68 marks of pure opportunity. Attempt all 17 at the same accuracy and this paper is about 547;
get them right and it is 560. So the first fix is the clock, not the chapters.
Section by section
| Section | Correct | Wrong | Blank | Marks | Target | Gap |
|---|---|---|---|---|---|---|
| Botany | 35 | 5 | 5 | 135 | 165 | -30 |
| Zoology | 37 | 6 | 2 | 142 | 160 | -18 |
| Chemistry | 30 | 9 | 6 | 111 | 156 | -45 |
| Physics | 29 | 12 | 4 | 104 | 120 | -16 |
| Total | 131 | 32 | 17 | 492 | 600 | −108 |
Twelve of the 32 wrong answers were tagged easy — direct recall, not reasoning.
Ten of 35 numericals were wrong. Of the 16 Assertion–Reason questions she got 11 right; of the four missed, three were keyed B
(“both true, R does not explain A”). That is the old habit in its narrower form: she now accepts that R is true, but still
attaches it to A when it does not belong.
Error families
Direction and sign 5 questions
The fix. Write the three words before you look at the options: flux → growing or dying → opposing current. Decide the pole first, then the rotation, using clockwise = south seen from outside.
Q141PhysicsFlux and Faraday's lawsSingle correctHard
A circular loop of radius 0.2 m lies in a field of 0.4 T perpendicular to its plane. Its radius shrinks at 5 × 10⁻³ m s⁻¹. The emf induced is about
you marked A · correct D
(A)1.3 mV
(B)0.4 mV
(C)5 mV
(D)2.5 mV
- Why
- The area changes as dA/dt = 2πr dr/dt, and the π is the whole difficulty.
- NCERT / formula
- ε = B dA/dt = B(2πr)(dr/dt)
- Working
- Given B = 0.4 T, r = 0.2 m, dr/dt = 5 × 10⁻³ m s⁻¹
- 2πr = 1.257 m
- ε = 0.4 × 1.257 × 5 × 10⁻³ = 2.5 × 10⁻³ V
- Trap: 1.3 mV forgets the factor 2 in 2πr.
Q171PhysicsInduced current and chargeSingle correctHard
In that graph the induced currents in regions I and III are
you marked C · correct B
(A)in the same sense, the one in region I being the larger
(B)in opposite senses, the one in region I being the larger
(C)equal in magnitude and sense
(D)in opposite senses, the one in region III being the larger
- Why
- A rising and a falling flux drive currents in opposite senses, and the steeper slope gives the larger current.
- NCERT / formula
- ε = −N dΦ/dt
Q172PhysicsInduced current and chargeSingle correctModerate
A coil is connected to a galvanometer and a magnet is moved towards it twice, once slowly and once quickly. The charge that circulates is
you marked B · correct A
(A)the same both times
(B)larger for the quicker motion
(C)zero for the slower motion
(D)larger for the slower motion
- Why
- The current rises and the time falls in the same proportion.
- NCERT / formula
- q = ΔΦ/R
- Working
- q = NΔΦ/R has no time in it
- So the speed of the motion does not change the charge
- Trap: the DEFLECTION is larger for the quicker motion, but that follows the current.
Q175PhysicsEddy currentsMatch the followingEasy
Match Column I with Column II. Column I Column II A Laminated core I Eddy currents melt a metal B Induction furnace II A way of reducing eddy currents C Magnetic braking III Brings a galvanometer coil quickly to rest D Electromagnetic damping IV Eddy currents in the rails oppose the motion
you marked D · correct C
(A)A–I, B–II, C–IV, D–III
(B)A–IV, B–I, C–III, D–II
(C)A–II, B–I, C–IV, D–III
(D)A–II, B–I, C–III, D–IV
- Why
- One remedy for eddy currents and three uses of them.
- NCERT / formula
- Applications of eddy currents
Q178PhysicsAC generatorSingle correctModerate
The coil of a generator turns at 1500 revolutions per minute. Its angular frequency is about
you marked A · correct C
(A)79 rad s⁻¹
(B)1500 rad s⁻¹
(C)157 rad s⁻¹
(D)25 rad s⁻¹
- Why
- Revolutions per minute must be converted to revolutions per second first.
- NCERT / formula
- ω = 2πf
- Working
- Given 1500 rpm = 25 rev s⁻¹
- ω = 2π × 25 = 157 rad s⁻¹
- Trap: 79 rad s⁻¹ uses πf instead of 2πf.
Missing factor — a 2, a π or a squared length 4 questions
The fix. Before arithmetic, say the formula aloud and point at each symbol: is there a ½? is the length squared? does a reversal double ΔΦ? does the rate need 2π?
Q140PhysicsFlux and Faraday's lawsSingle correctEasy
A coil of 60 turns and area 0.05 m² in a field of 0.15 T is turned through 180° in 0.15 s. The average emf is
you marked D · correct C
(A)12 V
(B)1.5 V
(C)6 V
(D)3 V
- Why
- Turning through 180° reverses the flux, so the change is twice the initial value.
- NCERT / formula
- ε = N(2BA)/Δt
- Working
- Given N = 60, B = 0.15 T, A = 0.05 m², Δt = 0.15 s
- BA = 7.5 × 10⁻³ Wb; ΔΦ = 2BA = 0.015 Wb
- ε = 60 × 0.015 / 0.15 = 6 V
- Trap: 3 V takes the change as BA instead of 2BA — the factor of two is the whole question.
Q145PhysicsLenz's law and directionSingle correctEasy
A loop lies in a field directed out of the page whose magnitude is rising. The induced current is
you marked B · correct D
(A)zero
(B)reversing continuously
(C)anticlockwise
(D)clockwise
- Why
- The induced flux must be into the page to oppose the growing outward flux.
- NCERT / formula
- Lenz's law
Q146PhysicsLenz's law and directionDiagram-basedEasy
A magnet is moved away from a coil with its south pole facing the coil. Seen from the magnet, the near face becomes [figure]
you marked A · correct B
(A)a south pole, with the current clockwise
(B)a north pole, with the current anticlockwise
(C)a south pole, with the current anticlockwise
(D)a north pole, with the current clockwise
- Why
- A receding pole is attracted, so the near face must be the unlike pole.
- NCERT / formula
- Lenz's law; clockwise seen from outside = south pole
Q164PhysicsMotional emfSingle correctEasy
A metal disc of radius 0.2 m spins at 5 revolutions per second about its axis in a field of 0.3 T parallel to the axis. The emf between the centre and the rim is about
you marked A · correct D
(A)0.06 V
(B)0.95 V
(C)0.38 V
(D)0.19 V
- Why
- Each radius behaves like a rod rotating about one end.
- NCERT / formula
- ε = ½Bωr², ω = 2πf
- Working
- Given B = 0.3 T, f = 5 rev s⁻¹, r = 0.2 m
- ω = 31.4 rad s⁻¹
- ε = ½ × 0.3 × 31.4 × 0.04 = 0.19 V
- Trap: 0.38 V drops the ½.
Assertion–Reason: a true Reason that does not explain A 4 questions
The fix. Three separate judgements, in order: is A true? is R true on its own? does R explain this A? A true sentence about the same topic is not an explanation.
Q50ZoologyPrinciples of biotechnologyAssertion–ReasonHard
Assertion (A): An alien piece of DNA does not multiply inside a host cell by itself. | Reason (R): The origin of replication also controls the copy number of the linked DNA.
you marked A · correct B
(A)Both A and R are true and R is the correct explanation of A
(B)Both A and R are true but R is NOT the correct explanation of A
(C)A is true but R is false
(D)A is false but R is true
- Why
- Both statements are true, but the copy-number control is not the reason the alien DNA fails to multiply; it fails because it is not linked to an origin at all.
- NCERT / formula
- §11.1 — the need for a replicon and the role of the ori
Q65ZoologyCloning vectorsAssertion–ReasonModerate
Assertion (A): The Ti plasmid used in the laboratory no longer produces tumours in plants. | Reason (R): Agrobacterium tumefaciens is a pathogen of several dicot plants.
you marked D · correct B
(A)Both A and R are true and R is the correct explanation of A
(B)Both A and R are true but R is NOT the correct explanation of A
(C)A is true but R is false
(D)A is false but R is true
- Why
- Both statements are true, but the pathogenicity of the bacterium is not the reason the laboratory plasmid is harmless; it has been modified.
- NCERT / formula
- §11.1.2 — the modified Ti plasmid
Q103ChemistryFunctional group testsAssertion–ReasonModerate
Assertion (A): Acetone gives an orange-yellow precipitate with 2,4-dinitrophenylhydrazine. | Reason (R): Tollens' reagent distinguishes an aldehyde from a ketone.
you marked C · correct B
(A)Both A and R are true and R is the correct explanation of A
(B)Both A and R are true but R is NOT the correct explanation of A
(C)A is true but R is false
(D)A is false but R is true
- Why
- Both statements are true, but the ability of Tollens' reagent to make the distinction is not the reason acetone reacts with 2,4-DNP.
- NCERT / formula
- Lab manual: 2,4-DNP and Tollens' tests
Q173PhysicsInduced current and chargeAssertion–ReasonModerate
Assertion (A): The charge that flows through a circuit during a flux change does not depend on how fast the change is made. | Reason (R): The induced current is inversely proportional to the resistance of the circuit.
you marked D · correct B
(A)Both A and R are true and R is the correct explanation of A
(B)Both A and R are true but R is NOT the correct explanation of A
(C)A is true but R is false
(D)A is false but R is true
- Why
- Both statements are true, but the dependence on resistance is not the reason the charge is independent of the time.
- NCERT / formula
- q = NΔΦ/R ; I = ε/R
Titration arithmetic 3 questions
The fix. Fix the n-factor first, then the mole ratio: oxalate needs 2 MnO₄⁻ : 5 C₂O₄²⁻, iron(II) needs 1 : 5. Equivalent mass = molar mass ÷ n-factor.
Q130ChemistryTitration chemistrySingle correctEasy
A 0.2 N solution of sodium carbonate has a molarity of
you marked A · correct D
(A)0.2 M
(B)0.4 M
(C)0.05 M
(D)0.1 M
- Why
- Normality is molarity times the n-factor, which is 2 for sodium carbonate.
- NCERT / formula
- N = M × n-factor
- Working
- Given N = 0.2 N, n-factor = 2
- M = N/2 = 0.1 M
- Trap: 0.4 M multiplies instead of dividing.
Q131ChemistryTitration chemistrySingle correctHard
20 mL of 0.05 M oxalic acid is titrated against 0.02 M potassium permanganate in acid medium. The volume of permanganate needed is
you marked B · correct C
(A)10 mL
(B)50 mL
(C)20 mL
(D)8 mL
- Why
- Two moles of permanganate oxidise five moles of oxalate.
- NCERT / formula
- 2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O
- Working
- Given 20 mL of 0.05 M oxalic acid = 1 mmol of oxalate
- KMnO₄ needed = (2/5) × 1 = 0.4 mmol
- V = 0.4 / 0.02 = 20 mL
- Trap: 8 mL uses the 1 : 5 ratio of the iron titration.
Q132ChemistryTitration chemistrySingle correctHard
20 mL of 0.1 M Mohr's salt is titrated against 0.05 M potassium permanganate. The volume needed is
you marked C · correct D
(A)40 mL
(B)4 mL
(C)20 mL
(D)8 mL
- Why
- One mole of permanganate oxidises five moles of iron(II).
- NCERT / formula
- MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O
- Working
- Given 20 mL of 0.1 M Mohr's salt = 2 mmol of Fe²⁺
- KMnO₄ needed = 2/5 = 0.4 mmol
- V = 0.4 / 0.05 = 8 mL
- Trap: 20 mL uses the 2 : 5 ratio of the oxalate titration.
Reagent, colour or test swapped 5 questions
The fix. Learn these as pairs, not singly: ceric reagent ↔ alcohol, neutral FeCl₃ ↔ phenol; Fehling ↔ aliphatic aldehyde only; xanthoproteic yellow→orange ↔ protein.
Q91ChemistryDetection of extra elementsSingle correctEasy
The sodium fusion extract is prepared by dissolving the fused mass in
you marked A · correct D
(A)dilute nitric acid
(B)ethanol
(C)concentrated sulphuric acid
(D)distilled water
- Why
- The ionic salts formed during fusion are extracted with water and kept alkaline.
- NCERT / formula
- Lab manual: preparation of the sodium fusion extract
Q94ChemistryDetection of extra elementsSingle correctEasy
A black precipitate with lead acetate after acidifying the sodium extract indicates
you marked A · correct C
(A)a halogen
(B)nitrogen
(C)sulphur
(D)phosphorus
- Why
- Lead sulphide is black.
- NCERT / formula
- Lab manual: lead acetate test for sulphur
Q108ChemistryCarbohydrate, protein and fat testsSingle correctEasy
A protein solution gives a violet colour with dilute NaOH and a few drops of CuSO₄. The test is
you marked A · correct B
(A)the xanthoproteic test
(B)the biuret test
(C)Millon's test
(D)the ninhydrin test
- Why
- Copper(II) forms a coloured complex with the peptide linkages.
- NCERT / formula
- Lab manual: biuret test
Q110ChemistryCarbohydrate, protein and fat testsStatement countHard
How many of the following statements are correct? | A. Molisch's test is a general test for carbohydrates. | B. Iodine gives a blue-black colour with starch. | C. The acrolein test detects the glycerol part of a fat. | D. The biuret test gives a yellow colour that turns orange with ammonia.
you marked D · correct C
(A)Only one
(B)Only two
(C)Only three
(D)All four
- Why
- The yellow-to-orange change belongs to the xanthoproteic test.
- NCERT / formula
- Lab manual: tests for carbohydrates, proteins and fats
Q111ChemistryCation analysisSingle correctEasy
The group reagent for Group VI cations is
you marked C · correct D
(A)dilute hydrochloric acid
(B)ammonium carbonate
(C)H₂S in ammoniacal medium
(D)disodium hydrogen phosphate
- Why
- Magnesium is precipitated as magnesium ammonium phosphate.
- NCERT / formula
- Lab manual: Group VI — Na₂HPO₄
Botany recall and replication mechanics 5 questions
The fix. The single rule under most of these: a nucleotide joins only a free 3′-OH, so every new strand grows 5′→3′ and the leading-strand template runs 3′→5′ towards the fork.
Q8BotanySearch for the genetic materialSingle correctEasy
In the mixed injection of Griffith's experiment, living S bacteria were recovered from the dead mice. This showed that
you marked C · correct B
(A)the mice made S bacteria of their own
(B)the change in the R bacteria was heritable
(C)heat-killed S bacteria had revived
(D)R bacteria had killed the S bacteria
- Why
- The transformed cells behaved as S type and bred true.
- NCERT / formula
- §6.1 — 'He recovered living S bacteria from the dead mice.'
Q25BotanyDNA replication and its enzymesSingle correctEasy
Replication is described as semiconservative because each daughter molecule has
you marked D · correct C
(A)two parental strands
(B)two new strands
(C)one parental and one newly made strand
(D)strands that are part old and part new along their length
- Why
- The parental strands are shared out, one to each daughter.
- NCERT / formula
- §6.4 — semiconservative replication
Q29BotanyDNA replication and its enzymesSingle correctModerate
If the two templates of a fork are compared, the one carrying the Okazaki fragments runs
you marked C · correct A
(A)5′→3′ towards the fork
(B)3′→5′ towards the fork
(C)3′→5′ away from the fork
(D)5′→3′ away from the fork
- Why
- That template points the wrong way for continuous 5′→3′ synthesis.
- NCERT / formula
- §6.4 — continuous and discontinuous synthesis
Q32BotanyDNA replication and its enzymesSingle correctHard
A bacterial genome of 4.6 × 10⁶ base pairs is copied in about 18 minutes. The average rate of polymerisation is about
you marked A · correct B
(A)2000 base pairs per minute
(B)2000 base pairs per second
(C)4600 base pairs per second
(D)46 000 base pairs per minute
- Why
- Dividing the genome size by the time in seconds gives the rate.
- NCERT / formula
- §6.4 — 'approximately 2000 bp/second'
Q36BotanyDNA replication and its enzymesSingle correctHard
A cell with 4 pg of DNA completes replication but does not divide. Its DNA content is then
you marked D · correct A
(A)8 pg
(B)2 pg
(C)4 pg
(D)16 pg
- Why
- Every molecule has become two and none has been given away.
- NCERT / formula
- §6.4 — replication and the cell cycle
Biotechnology recall 4 questions
The fix. Read the chapter's defining sentences again — they are short and they are what is asked: the two core techniques, the vector's four features, the three steps of PCR.
Q52ZoologyRestriction enzymesDiagram-basedModerate
In the sequence shown, the heavy lines mark the points at which the two backbones are cut. The ends produced are [figure]
you marked C · correct D
(A)blunt ends, because the enzyme cuts outside the recognition site
(B)blunt ends, because both strands are cut at the same point
(C)sticky ends, because the cuts are made at the centre of the palindrome
(D)sticky ends, because single-stranded stretches are left over
- Why
- Cutting off-centre but symmetrically leaves overhanging single strands.
- NCERT / formula
- §11.1.1 — sticky ends produced by restriction enzymes
Q55ZoologyRestriction enzymesSingle correctHard
A linear DNA molecule with three recognition sites for one restriction enzyme, when digested to completion, gives
you marked D · correct B
(A)six fragments
(B)four fragments
(C)three fragments
(D)two fragments
- Why
- Each cut on a linear molecule adds one fragment, so three cuts give four pieces.
- NCERT / formula
- §11.1.1 — restriction digestion and the fragments produced
Q61ZoologyCloning vectorsSingle correctHard
Recombinants selected by insertional inactivation of the tetracycline resistance gene are first plated on
you marked D · correct A
(A)ampicillin and then transferred to tetracycline
(B)a medium with neither antibiotic
(C)a medium with both antibiotics together
(D)tetracycline and then transferred to ampicillin
- Why
- The intact marker is used first to select transformants; the inactivated one then separates the recombinants.
- NCERT / formula
- §11.1.2 — selection on two media
Q79ZoologyPCRStatement countModerate
How many of the following statements are correct? | A. PCR uses a thermostable polymerase. | B. One cycle has three steps. | C. The amplified fragment can be cloned into a vector. | D. PCR requires a restriction enzyme in every cycle.
you marked D · correct C
(A)Only one
(B)Only two
(C)Only three
(D)All four
- Why
- No restriction enzyme takes part in the reaction; cutting is a separate step.
- NCERT / formula
- §11.2.2 — the polymerase chain reaction
Graph and charge reading 2 questions
The fix. Emf is N × the slope of the flux–time graph; charge is NΔΦ/R and contains no time at all.
Q162PhysicsMotional emfSingle correctHard
A rod of length 1.2 m rotates about one end at 10 rad s⁻¹ in a field of 0.5 T. The potential difference between the mid-point and the far end is
you marked C · correct B
(A)3.6 V
(B)2.7 V
(C)0.9 V
(D)1.8 V
- Why
- The emf of a segment goes as the difference of the squares of the radii.
- NCERT / formula
- ε = ½Bω(r₂² − r₁²)
- Working
- Given B = 0.5 T, ω = 10 rad s⁻¹, r₁ = 0.6 m, r₂ = 1.2 m
- r₂² − r₁² = 1.44 − 0.36 = 1.08 m²
- ε = ½ × 0.5 × 10 × 1.08 = 2.7 V
- Trap: 1.8 V takes half the whole-rod emf of 3.6 V — the squared length is the whole difficulty.
Q179PhysicsAC generatorSingle correctHard
A generator of peak emf 100 V runs at 25 Hz. The instantaneous emf 60° after the zero of emf, and the time from a zero to the next peak, are
you marked B · correct A
(A)86.6 V and 10 ms
(B)50 V and 10 ms
(C)50 V and 20 ms
(D)86.6 V and 20 ms
- Why
- The instantaneous value follows the sine, and a zero to the next peak is a quarter period.
- NCERT / formula
- ε = ε₀ sin θ ; t = T/4
- Working
- Given ε₀ = 100 V, θ = 60°, ν = 25 Hz
- ε = 100 sin 60° = 86.6 V
- T = 40 ms, so T/4 = 10 ms
- Trap: 50 V uses sin 30°, and 20 ms is the half period.
Left blank — 17 questions, 68 marks of opportunity
Q148 and Q149 carry her bracket in the margin: flagged, never returned to. Six of the 17 are in Chemistry.
On a +4/−1 paper a blank and a wrong answer differ by only 5 marks — at 80% accuracy, guessing after eliminating one option is clearly worth it.
Q5BotanyStructure of DNASingle correctModerate
According to Chargaff, in double-stranded DNA
left blank — correct answer A
(A)the amount of adenine equals that of thymine and guanine that of cytosine
(B)the sum of purines is always greater than that of pyrimidines
(C)the ratio A+T to G+C is the same in every species
(D)adenine equals guanine and thymine equals cytosine
- Why
- Base pairing makes each purine equal to its partner pyrimidine.
- NCERT / formula
- §6.3 — Chargaff's rules
Q10BotanySearch for the genetic materialStatement countEasy
How many of the following statements are correct? | A. Radioactive phosphorus labels the DNA of the phage. | B. Radioactive sulphur labels the protein coat. | C. Nitrogen is unsuitable as a label because it occurs in both DNA and protein. | D. Radioactive carbon was used to label the bacterial wall.
left blank — correct answer C
(A)Only one
(B)Only two
(C)Only three
(D)All four
- Why
- No carbon label was used; the experiment turned on the two exclusive isotopes.
- NCERT / formula
- §6.1 — choice of ³²P and ³⁵S
Q11BotanySearch for the genetic materialSingle correctModerate
After blending and centrifuging, radioactivity in the pellet of the ³²P experiment shows that
left blank — correct answer B
(A)the phage coats entered the cells
(B)the phage DNA entered the bacterial cells
(C)the phage did not attach to the bacteria
(D)the bacteria were killed by the isotope
- Why
- Only the material inside the bacteria is carried down with them.
- NCERT / formula
- §6.1 — 'Bacteria which were infected with viruses that had radioactive DNA were radioactive'
Q17BotanyProperties of genetic materialSingle correctEasy
The genetic material of tobacco mosaic virus is
left blank — correct answer A
(A)RNA
(B)protein
(C)a polysaccharide
(D)DNA
- Why
- A few viruses use RNA as their hereditary molecule.
- NCERT / formula
- §6.2 — 'In some viruses RNA is the genetic material'
Q31BotanyDNA replication and its enzymesSingle correctModerate
In prokaryotes, the whole circular chromosome is copied from
left blank — correct answer D
(A)many origins spaced along the molecule
(B)one origin for each gene
(C)the replication fork itself
(D)a single origin of replication
- Why
- The bacterial chromosome has one defined starting point.
- NCERT / formula
- §6.4 — origin of replication
Q70ZoologyCompetent host and gene transferSingle correctEasy
Transformed cells carrying an ampicillin resistance gene are selected by plating them on
left blank — correct answer B
(A)a medium containing a chromogenic substrate only
(B)a medium containing ampicillin
(C)a medium containing tetracycline only
(D)a medium free of any antibiotic
- Why
- Only the transformants survive the antibiotic.
- NCERT / formula
- §11.2.3 — 'only transformants will grow while non-transformants will die'
Q86ZoologyBioreactors and downstream processingSingle correctEasy
A curved base is given to a stirred-tank reactor to
left blank — correct answer B
(A)let the product settle
(B)help the mixing of the contents
(C)increase the volume
(D)reduce the cost
- Why
- The curve avoids corners where the contents would not mix.
- NCERT / formula
- §11.2.4 — 'with a curved base to facilitate the mixing of the reactor contents'
Q106ChemistryCarbohydrate, protein and fat testsSingle correctEasy
A red-brown precipitate with Fehling's solution is given by
left blank — correct answer C
(A)starch
(B)sucrose
(C)a reducing sugar
(D)a protein
- Why
- Only sugars with a free carbonyl group reduce Fehling's solution.
- NCERT / formula
- Lab manual: Fehling's test for reducing sugars
Q107ChemistryCarbohydrate, protein and fat testsSingle correctHard
Sucrose gives a negative Fehling's test but a positive one after
left blank — correct answer B
(A)boiling with sodium hydroxide
(B)hydrolysis with dilute acid
(C)treatment with iodine
(D)addition of α-naphthol
- Why
- Hydrolysis liberates the reducing monosaccharides.
- NCERT / formula
- Lab manual: reducing and non-reducing sugars
Q112ChemistryCation analysisDiagram-basedModerate
In the scheme shown, the cations precipitated at the stage marked Q are those whose sulphides are insoluble in [figure]
left blank — correct answer D
(A)concentrated nitric acid
(B)hot water
(C)ammoniacal solution
(D)dilute acid
- Why
- Group II sulphides are precipitated in acid medium and are the least soluble.
- NCERT / formula
- Lab manual: Group II — H₂S in dilute HCl
Q120ChemistryCation analysisSingle correctEasy
Nessler's reagent gives a brown precipitate with
left blank — correct answer C
(A)the zinc ion
(B)the lead ion
(C)the ammonium ion
(D)the barium ion
- Why
- The brown precipitate confirms the Group 0 cation.
- NCERT / formula
- Lab manual: detection of NH₄⁺
Q126ChemistryAnion analysisAssertion–ReasonHard
Assertion (A): A salt giving a white precipitate with barium chloride may not contain a sulphate. | Reason (R): Barium sulphite is also a white precipitate.
left blank — correct answer A
(A)Both A and R are true and R is the correct explanation of A
(B)Both A and R are true but R is NOT the correct explanation of A
(C)A is true but R is false
(D)A is false but R is true
- Why
- The sulphite precipitate is the very reason the barium chloride test alone is not conclusive.
- NCERT / formula
- Lab manual: sulphate and sulphite with barium chloride
Q134ChemistryTitration chemistrySingle correctModerate
The strength in grams per litre of a 0.05 N oxalic acid solution is
left blank — correct answer B
(A)1.575 g L⁻¹
(B)3.15 g L⁻¹
(C)0.05 g L⁻¹
(D)6.3 g L⁻¹
- Why
- Strength equals normality times the equivalent mass.
- NCERT / formula
- Strength = N × equivalent mass
- Working
- Given N = 0.05, equivalent mass = 63
- Strength = 0.05 × 63 = 3.15 g L⁻¹
- Trap: 6.3 g L⁻¹ uses the molar mass instead of the equivalent mass.
Q148PhysicsLenz's law and directionSingle correctEasy
A ring is held coaxially above a solenoid and the steady current in the solenoid is switched off. The ring is
left blank — correct answer A
(A)attracted towards the solenoid
(B)set spinning
(C)unaffected
(D)repelled from the solenoid
- Why
- The dying flux is supported, so the ring's current is parallel to that of the solenoid.
- NCERT / formula
- Lenz's law; parallel currents attract
Q149PhysicsLenz's law and directionSingle correctHard
A closed loop and an identical open loop are both moved into the same field region. Compared with the closed loop, the open loop has
left blank — correct answer C
(A)a smaller emf and the same current
(B)no emf and no current
(C)the same emf but no induced current
(D)a larger emf and a smaller current
- Why
- An emf appears across the ends of any conductor in which the flux changes, but current needs a closed path.
- NCERT / formula
- ε = −dΦ/dt ; I = ε/R
Q163PhysicsMotional emfSingle correctEasy
A rod of length 0.5 m rotates about one end at 240 revolutions per minute in a field of 0.1 T. The emf is about
left blank — correct answer D
(A)0.63 V
(B)0.05 V
(C)1.26 V
(D)0.31 V
- Why
- The rate must be converted to radian per second first.
- NCERT / formula
- ω = 2πf ; ε = ½Bωl²
- Working
- Given f = 240 rpm = 4 rev s⁻¹
- ω = 2π × 4 = 25.1 rad s⁻¹
- ε = ½ × 0.1 × 25.1 × 0.25 = 0.31 V
- Trap: 0.05 V leaves out the 2π.
Q176PhysicsEddy currentsSingle correctHard
The heat generated per second by eddy currents in a plate moving through a field varies with the speed as
left blank — correct answer A
(A)v²
(B)v³
(C)v
(D)1/v
- Why
- The emf goes as v and the power as its square.
- NCERT / formula
- P = ε²/R with ε ∝ v
- Working
- ε ∝ v
- P = ε²/R ∝ v²
- Trap: answering 'v' applies the emf dependence directly to the power.
Plan for 3 October
| Order | What | Why |
|---|---|---|
| 1 | Timed section drill — 45 questions in 40 minutes, every row marked before any revisit | 17 blanks = 68 marks; this is the single biggest gain available |
| 2 | Chemistry: the seven titration workings in the Mock #2 key, then the six blank questions above | Chemistry is −45, the worst section |
| 3 | The 30-question EMI re-test (direction + factor traps) | Physics is −16 and 9 of its 12 errors are direction or factor |
| 4 | Q50, Q65, Q103, Q173 — read all four Assertion–Reason items again side by side | Three were the same keyed-B case |
| 5 | Nothing on Zoology | 142/160 with only two blanks — the healthiest section on the paper |
One sheet-filling note for Sunday. Q81 was marked twice — B struck out, C ticked. It was right either way here,
but a real OMR reader voids that row. One mark per row, no corrections.