570 out of 720 — ten below the danger line, eighty below target. A hundred and fifty right, thirty wrong, nothing left blank. This is your first full-length paper and the headline is not the total: it is where the thirty errors sit. Seven of them — almost a quarter — are in a single run of eight questions at the start of Thermodynamics, after which you went twenty-eight in a row without a mistake. That run alone is thirty-five marks. The rest of the paper is the work of someone scoring well above the danger line.
Where the marks went
Difficulty
Right
Out of
Accuracy
Easy
40
45
89%
89%
Moderate
68
80
85%
85%
Hard
42
55
76%
76%
The profile is the right way up for the second paper running: strongest on the easy questions, weakest on the hard ones. The upside-down shape that held for three papers has now gone.
Chapter by chapter
Chapter
Right
Out of
Marks
Against the 128 gate
Morphology of Flowering Plants
31
36
119 / 144
9 short
Anatomy of Flowering Plants
31
36
119 / 144
9 short
Thermodynamics
29
36
109 / 144
19 short
Kinetic Theory
28
36
104 / 144
24 short
Ray Optics
31
36
119 / 144
9 short
Total
150
180
570 / 720
80 below target
Every chapter missed the 128 gate, and the two gas chapters — Thermodynamics and Kinetic Theory — missed it widest. Morphology, Anatomy and Ray Optics are each only nine marks short — that is two questions apiece.
The single biggest thing on this paper
Seven errors · 35 marks · eight consecutive questions
You started Thermodynamics cold
Thermodynamics runs from Q73 to Q108. Every one of your seven errors in that chapter is in Q73–Q80. From Q81 to Q108 — twenty-eight questions — you did not make a single mistake.
Block
Questions
Right
Thermodynamics, first eight
Q73–Q80
1 / 8
Thermodynamics, the rest
Q81–Q108
28 / 28
The clinching detail is inside one section. The first-law section is Q79–Q86. You got Q79 and Q80 wrong and Q81 to Q86 right — same section, same topic, minutes apart. So this is not a gap in thermodynamics. It is the cost of switching from Biology to Physics without a pause. Three of the eight were easy questions.
Before the physics section of a full paper, stop for sixty seconds. Write the first law and the sign convention at the top of the page, then start. You already know this material — the next twenty-eight questions prove it.
Q74 — you chose (2), the answer is (4)What internal energy depends on · easy
The same sample of an ideal gas is taken from one state to another along two completely different paths, one of them much longer than the other. Of the quantities below, the one that comes out the same for both paths is
(1) the heat absorbed less twice the work done
(2) the heat absorbed by the gasYOUR ANSWER
(3) the work done by the gas
(4) the change in internal energyCORRECT
Internal energy is a state function: it depends only on where the gas starts and finishes, never on the route. Heat and work both depend on the path, which is the whole point of the question. You chose the heat absorbed.
Q77 — you chose (2), the answer is (4)A state function · easy
Each of the following is a function of the state of a system, EXCEPT one. Looking at whether the quantity can be fixed once the state is fixed, identify the exception.
(1) The temperature of the system
(2) The internal energy of the systemYOUR ANSWER
(3) The pressure of the system
(4) The work done in reaching that stateCORRECT
The same idea three questions later, asked as a negative stem. Temperature, internal energy and pressure are all fixed once the state is fixed; work is not — it depends on how you got there. You chose internal energy, which is a state function, so it cannot be the exception.
Q74 and Q77 are the same fact asked twice. Getting the first one right would have given you the second.
Q79 — you chose (4), the answer is (1)Applying the first law · hard
A gas absorbs 480 J of heat from its surroundings and in the same process does 190 J of work on them. The change in the internal energy of the gas is
(1) 290 JCORRECT
(2) 480 J
(3) 190 J
(4) 670 JYOUR ANSWER
ΔU = Q − W when W is the work done by the gas. 480 − 190 = 290 J. Your 670 J is 480 plus 190 — the work added instead of subtracted. The gas spent that energy on its surroundings, so it comes off.
Say the sentence out loud before computing: the gas did work, so it lost that energy.
Q80 — you chose (2), the answer is (3)The sign convention · easy
In the convention used in the text, heat given to the system and work done by the system are counted as positive. On that convention, a gas that is compressed while giving out heat has
(1) Q negative and W positive, so its internal energy certainly falls
(2) Q positive and W negative, so its internal energy certainly risesYOUR ANSWER
(3) both Q and W negative, so its internal energy may rise or fallCORRECT
(4) both Q and W positive, so its internal energy certainly rises
On this convention heat given out is Q negative, and compression means work is done on the gas, so W is negative too. ΔU = Q − W then has a negative and a positive contribution, so it may rise or fall. You read the compression as W positive.
Q73 — you chose (2), the answer is (3)The zeroth law · moderate
Two bodies A and B are each found to be in thermal equilibrium with a third body C, although A and B have never been placed in contact with each other. What follows from this, and the law that allows it to be stated, are
(1) A and B are at the same temperature; the first law
(2) A and B must have equal internal energies; the zeroth lawYOUR ANSWER
(3) A and B are at the same temperature; the zeroth lawCORRECT
(4) nothing can be said about A and B until they are brought together
Thermal equilibrium means equal temperature, not equal internal energy — a large cold body and a small hot one can hold the same internal energy at different temperatures. The law that lets you say it is the zeroth. You had the right law and the wrong quantity.
Q75 — you chose (3), the answer is (2)Reading the free expansion · hard
A gas is confined to one half of a rigid insulated vessel, the other half being evacuated, and the partition is then pulled out. Reading what the walls are made of and then what lies beyond the partition, the correct statement about the expansion is that
(1) the gas does work against the vacuum, so its temperature falls
(2) no heat enters or leaves and no work is done, so the internal energy does not changeCORRECT
(3) no heat enters or leaves but the gas does work, so the internal energy fallsYOUR ANSWER
(4) heat enters from the walls and the gas does work, so the internal energy rises
Rigid walls mean no work can be done on the surroundings; insulated walls mean no heat crosses them; and a vacuum offers nothing to push against. So Q = 0 and W = 0, and the internal energy is unchanged. You chose the option that has the gas doing work — but there is nothing there for it to work against.
Section by section
Section
Questions
Right
The root and its modifications
Q1–8
8 / 8
The stem and its modifications
Q9–15
6 / 7
The leaf
Q16–22
6 / 7
The inflorescence and the flower
Q23–30
6 / 8
The fruit, the seed and the families
Q31–36
5 / 6
Meristems and the simple tissues
Q37–43
7 / 7
The complex tissues
Q44–50
5 / 7
The epidermal tissue system
Q51–56
5 / 6
The root and the stem in section
Q57–65
7 / 9
The leaf in section
Q66–72
7 / 7
Thermal equilibrium and internal energy
Q73–78
1 / 6
The first law of thermodynamics
Q79–86
6 / 8
Specific heats of a gas
Q87–92
6 / 6
Isothermal and adiabatic processes
Q93–100
8 / 8
Cyclic processes and the second law
Q101–108
8 / 8
The mole and the ideal gas equation
Q109–115
7 / 7
The kinetic interpretation of pressure
Q116–122
6 / 7
Temperature and molecular energy
Q123–129
7 / 7
The distribution of molecular speeds
Q130–137
4 / 8
Degrees of freedom and the mean free path
Q138–144
4 / 7
Reflection at plane surfaces
Q145–152
6 / 8
Spherical mirrors
Q153–157
5 / 5
Refraction at plane surfaces and apparent depth
Q158–165
7 / 8
Total internal reflection
Q166–172
5 / 7
Lenses, the prism and dispersion
Q173–180
8 / 8
The opening thermodynamics section stands out at one of six. The two weakest after it — the distribution of molecular speeds, and degrees of freedom with the mean free path — cover the same ground as the weakest section of the standalone Kinetic Theory paper, so that is a genuine topic gap rather than a slow start.
Reading a figure, then reading it the wrong way round
Nine errors · 45 marks
Two-clue figures — 38 right out of 47
This is the largest family on the paper and the one that has carried over from every previous script. Two of these are worth separating out, because they are not reading failures at all: you identified both items correctly and then wrote them in the wrong order. When a stem ends in respectively or in that order, the order is part of the answer.
Q9 — you chose (1), the answer is (2)Reading the underground parts · hard
Four underground structures are drawn above. The one in which the stem itself is reduced to a short disc bearing fleshy scale leaves, and the one in which the stem is condensed but stands vertically and is itself fleshy, are
(1) X and YYOUR ANSWER
(2) Y and XCORRECT
(3) Y and Z
(4) W and X
You picked out the right two structures — X and Y are exactly the bulb and the corm. But the stem asks for the disc-like one first and the vertical fleshy one second, which is Y and X. The answer was in your hands and went down reversed.
When a stem asks for two things in a stated order — respectively, in that order, or simply the one that… and the one that… are — number them 1 and 2 in the margin before you look at the options.
Q132 — you chose (4), the answer is (1)Reading the three marked speeds · hard
Three speeds are marked P, Q and R on one distribution above, in order of increasing speed. Taking them in that order, they are
(1) the most probable, the average and the root mean square speedCORRECT
(2) the root mean square, the average and the most probable speed
(3) the most probable, the root mean square and the average speed
(4) the average, the most probable and the root mean square speedYOUR ANSWER
Again the right three speeds, in the wrong order. The curve peaks at the most probable speed, the average sits a little to its right, and the root mean square is the largest of the three. You began with the average.
Q145 — you chose (4), the answer is (2)Reading the two angles · moderate
A ray meets a plane mirror as drawn above. Reading where the normal has been put and then which two angles are marked from it, the law of reflection states that
(1) i and r add up to ninety degrees
(2) i and r are equal, both being measured from the normalCORRECT
(3) i is always the larger of the two
(4) i and r are equal, both being measured from the mirror surfaceYOUR ANSWER
Both angles in the law of reflection are measured from the normal, not from the mirror surface. The figure draws the normal in and marks the two angles from it — that is the second clue the question is pointing at.
Q59 — you chose (3), the answer is (2)Naming the first bundle · hard
Of the same three bundles, the one marked P has its xylem and phloem standing on different radii and alternating with each other. Such a bundle is described as
(1) conjoint and open, and is the kind found in a dicot stem
(2) radial, and is the kind found in a rootCORRECT
(3) conjoint and closed, and is the kind found in a monocot stemYOUR ANSWER
(4) radial, and is the kind found in a monocot stem
(This is the second of a pair — it reuses the figure of three bundles printed with the previous question.) Xylem and phloem on different radii, alternating, is the definition of a radial bundle, and radial bundles are what roots have. You answered conjoint and closed, which is the monocot stem — the arrangement you were being asked to rule out.
Q135 — you chose (1), the answer is (3)Reading the straight line · hard
The root mean square speed of a gas is plotted against the square root of its absolute temperature above. Reading that the graph is a straight line through the origin, what follows is that the speed varies as
(1) the reciprocal of the absolute temperatureYOUR ANSWER
(2) the square of the absolute temperature
(3) the square root of the absolute temperatureCORRECT
(4) the absolute temperature itself
A straight line through the origin on a plot of v against √T means v is proportional to √T. You read it as a reciprocal relation, which would have been a curve, not a line through the origin.
The match questions: the same swap three times
Three errors · 15 marks · three out of three
Rows A and B right, rows C and D swapped
There were three match questions on this paper. You got all three wrong, and on every one of them you matched A and B correctly and then exchanged C and D. In each case you chose the option ending C-III, D-I where the answer ended C-I, D-III. That is not three separate slips — it is one habit: once the first two rows look settled, the last two get assigned by elimination rather than checked.
Do the row you are least sure of first, not the one you are most sure of. The planted swap is always in the rows you would otherwise fill in last.
Q150 — you chose (4), the answer is (2)Matching the mirror terms · easy
Match Column I with Column II.
Column I
Column II
A
Pole
I
Half the radius of curvature of a spherical mirror
B
Centre of curvature
II
The mid-point of the reflecting surface
C
Focal length
III
The line joining the pole to the centre of curvature
D
Principal axis
IV
The centre of the sphere of which the mirror is a part
(1) A-IV, B-II, C-I, D-III
(2) A-II, B-IV, C-I, D-IIICORRECT
(3) A-I, B-III, C-II, D-IV
(4) A-II, B-IV, C-III, D-IYOUR ANSWER
Pole is the mid-point of the reflecting surface (A–II) and the centre of curvature matches IV — both of which you had. Focal length is half the radius of curvature, which is I, and the principal axis is the line in III. You put focal length against III and the axis against I.
Q64 — you chose (4), the answer is (3)Matching root and stem · moderate
Match Column I with Column II.
Column I
Column II
A
Dicot root
I
Conjoint open bundles standing in a ring
B
Monocot root
II
Two to four xylem bundles and little or no pith
C
Dicot stem
III
Scattered conjoint closed bundles in an undifferentiated ground tissue
D
Monocot stem
IV
More than six xylem bundles and a large pith
(1) A-I, B-III, C-II, D-IV
(2) A-IV, B-II, C-I, D-III
(3) A-II, B-IV, C-I, D-IIICORRECT
(4) A-II, B-IV, C-III, D-IYOUR ANSWER
A and B again correct. Dicot stem takes conjoint open bundles in a ring (I) and monocot stem the scattered closed bundles (III). You had them the other way round.
Q35 — you chose (4), the answer is (3)Matching the families · moderate
Match Column I with Column II.
Column I
Column II
A
Fabaceae
I
A perianth of six members standing in two whorls of three
B
Solanaceae
II
A legume, from a monocarpellary superior ovary
C
Liliaceae
III
Stamens attached to the corolla rather than to the thalamus
D
The epipetalous condition
IV
A berry or a capsule, from a bicarpellary superior ovary
(1) A-IV, B-II, C-I, D-III
(2) A-I, B-III, C-II, D-IV
(3) A-II, B-IV, C-I, D-IIICORRECT
(4) A-II, B-IV, C-III, D-IYOUR ANSWER
Fabaceae and Solanaceae placed correctly. Liliaceae has the perianth of six in two whorls of three (I); the stamen character (III) belongs to the fourth row. Reversed again.
Assertion–Reason
Eleven of sixteen. The old habit has not gone: you chose option (1) nine times, when only six of the sixteen were (1), and four of your five errors are (1) answers. One is a shape worth naming — in Q53 the reason was simply false and you accepted it.
What the question was
How many
You got
R explains A — answer (1)
6
5 / 6
R true but not the explanation — answer (2)
6
3 / 6
A true but R false — answer (3)
4
3 / 4
Q53 — you chose (1), the answer is (3)Why a root has no cuticle · moderate
Assertion (A): The epidermis of a root is not covered by a cuticle, although the epidermis of a stem is. Reason (R): The cuticle is a layer of cellulose laid down on the inner face of each epidermal cell.
(1) Both A and R are true and R is the correct explanation of AYOUR ANSWER
(2) Both A and R are true but R is NOT the correct explanation of A
(3)A is true but R is falseCORRECT
(4)A is false but R is true
The reason is false as a sentence. A cuticle is made of cutin, a waxy material, and it lies on the outer face of the epidermis, not a layer of cellulose on the inner face. Once R is false the answer can only be (3), whatever you think of A.
Read R on its own, as if A were not there, and ask only: is this sentence true? Do that before you think about whether it explains anything.
Q143 — you chose (1), the answer is (2) — not the (3) the paper printedHalving the number density · moderate
Assertion (A): Pumping half the gas out of a closed vessel at constant temperature makes the mean free path of the remaining molecules longer. Reason (R): The mean free path is inversely proportional to the square of the diameter of a molecule.
(1) Both A and R are true and R is the correct explanation of AYOUR ANSWER
(2) Both A and R are true but R is NOT the correct explanation of ACORRECT
(3)A is true but R is false
(4)A is false but R is true
My mistake, not yours — this question was printed with the wrong key. R is in fact true: the mean free path really does go as 1/d². So the honest answer is (2) — both true, but R is about the diameter of a molecule while A is about how many molecules are left in the vessel, so R cannot explain A. Your (1) says R does explain A, which is still not right, so the mark is unchanged. But the paper told you R was false, and it is not.
The reasoning to take away is the one that was being tested: R was true and irrelevant. That is the (2) shape.
Q163 — you chose (1), the answer is (2)Why a stick looks bent · moderate
Assertion (A): A straight stick held obliquely in water looks bent at the surface. Reason (R): The speed of light in water is smaller than its speed in vacuum.
(1) Both A and R are true and R is the correct explanation of AYOUR ANSWER
(2) Both A and R are true but R is NOT the correct explanation of ACORRECT
(3)A is true but R is false
(4)A is false but R is true
Both true, but they do not connect. The stick looks bent because rays from it change direction at the surface, and the apparent depth is raised. That the speed of light is lower in water is a true fact and the reason refraction happens at all, but it is not the explanation of the bending you see. (2).
Q22 — you chose (1), the answer is (2)Why a phyllode is not a leaf · moderate
Assertion (A): The flattened green organ of the Australian Acacia carries on the photosynthesis of the plant. Reason (R): The Australian Acacia bears pinnately compound leaves while it is young.
(1) Both A and R are true and R is the correct explanation of AYOUR ANSWER
(2) Both A and R are true but R is NOT the correct explanation of ACORRECT
(3)A is true but R is false
(4)A is false but R is true
Both true. But R talks about what young Acacia plants bear, and A is about the flattened green organ doing the photosynthesis. R is a true statement about the same plant that does not explain A. (2).
Numericals: the dropped factor
Twenty-two of twenty-seven. Two of the five losses are the same slip — a factor left out of a square root, landing you on a number that was printed as an option precisely because that is the usual mistake. Of the other three, one is a mis-set angle, one is a term left out of a sum, and the third is the first-law sign slip in Q79 above.
Q130 — you chose (1), the answer is (2)Root mean square speed · hard
The molar mass of nitrogen is 28 g mol−1. Taking R = 8.3 J mol−1 K−1, the root mean square speed of its molecules at 350 K is closest to
(1) 322 m s−1YOUR ANSWER
(2) 558 m s−1CORRECT
(3) 1116 m s−1
(4) 17.6 m s−1
vrms = √(3RT/M) = √(3 × 8.3 × 350 ÷ 0.028) = 558 m s−1. Your 322 is √(RT/M) — the 3 left out. The ratio between them is √3, which is the signature of that slip.
Q141 — you chose (3), the answer is (1)The mean free path · hard
A gas has 3 × 1025 molecules per cubic metre and its molecules have a diameter of 2 × 10−10 m. Its mean free path is closest to
(1) 1.88 × 10−7 mCORRECT
(2) 1.88 × 10−5 m
(3) 2.65 × 10−7 mYOUR ANSWER
(4) 1.33 × 10−7 m
λ = 1/(√2 nπd²) = 1.88 × 10−7 m. Your 2.65 × 10−7 is the same sum without the √2 — and 2.65 ÷ 1.88 is exactly √2.
After any root-mean-square or mean-free-path calculation, check which constants are inside the square root. The options are built so that dropping one lands on a printed answer.
Q168 — you chose (1), the answer is (4)From the critical angle backwards · hard
The critical angle is usually worked out from the refractive index. Asked the other way round — a liquid is found to have a critical angle of 48° with air — its refractive index is closest to
(1) 1.49YOUR ANSWER
(2) 0.74
(3) 1.6
(4) 1.35CORRECT
μ = 1 ÷ sin C = 1 ÷ sin 48° = 1 ÷ 0.743 = 1.35. Your 1.49 is 1 ÷ sin 42° — the complement of the angle. The critical angle is measured from the normal, so it goes straight into the sine.
Q28 — you chose (2), the answer is (1)Counting the floral parts · hard
A flower of the family Solanaceae has 5 sepals, 5 petals, 5 stamens and a gynoecium of 2 carpels. Counting every floral leaf in the flower, that is every sepal, petal, stamen and carpel together, the total is
(1) 17CORRECT
(2) 15YOUR ANSWER
(3) 34
(4) 10
5 sepals + 5 petals + 5 stamens + 2 carpels = 17. Your 15 is the first three added and the carpels left out — but the gynoecium is made of floral leaves too, which is what the stem spells out.
Five-statement questions: the same two statements, twice
Seventeen of twenty, which is solid. But two of the three losses are identical in shape: you rejected statement A and accepted statement E, when the answer was A, B, C and D. In both cases A was a definition you know and E was the planted false one.
Q144 — you chose (1), the answer is (4)Review of equipartition · moderate
Given below are five statements. A. Each translational degree of freedom contributes one half of kBT to the energy of a molecule. B. A monatomic gas has three degrees of freedom and no more. C. A rigid diatomic molecule has two rotational degrees of freedom. D. The mean free path grows shorter as the number of molecules per unit volume rises. E. Each vibrational mode contributes one half of kBT to the energy.
(1) B, C, D and E onlyYOUR ANSWER
(2) A, C, D and E only
(3) A, B, C and E only
(4) A, B, C and D onlyCORRECT
A is true: each translational degree of freedom carries ½kBT — that is the equipartition theorem itself. You dropped it and took E instead.
Q172 — you chose (3), the answer is (4)Review of total internal reflection · moderate
Given below are five statements. A. Total internal reflection occurs only when light goes from a denser to a rarer medium. B. At the critical angle the refracted ray grazes along the boundary. C. A diamond sparkles because its critical angle is small. D. In an optical fibre the light is guided by repeated total internal reflection. E. A mirage is formed by ordinary reflection at the surface of the hot road.
(1) A, B, C and E only
(2) A, C, D and E only
(3) B, C, D and E onlyYOUR ANSWER
(4) A, B, C and D onlyCORRECT
A is true: total internal reflection happens only going from denser to rarer. Dropped again, with E taken in its place.
When a five-statement set offers you a combination that excludes A, be suspicious — A is usually the plain definition.
Q137 — you chose (2), the answer is (4)Review of the speeds · moderate
Given below are five statements. A. The most probable speed is the speed at which the distribution curve peaks. B. The root mean square speed is the largest of the three speeds quoted. C. At a given temperature a lighter gas has the higher root mean square speed. D. The root mean square speed of a gas is proportional to its absolute temperature. E. The distribution of speeds in a gas was worked out by Maxwell.
(1) B, C, D and E only
(2) A, B, C and D onlyYOUR ANSWER
(3) A, C, D and E only
(4) A, B, C and E onlyCORRECT
Here it was D you let through and E you dropped. D is false: the root mean square speed goes as √T, not as T, so doubling the absolute temperature multiplies the speed by √2 and not by two. E is true — the speed distribution is Maxwell's. You had A, B and C right and then swapped the one false statement for the one true one.
vrms ∝ √T. Any statement that makes a speed proportional to T itself is false.
What is now solid
Question type
Right
Out of
Negative stems (EXCEPT, NOT)
17
18
was a weak point two papers ago
Reference shift / asked the other way round
10
11
was four of eight errors on Morphology
Multi-statement sets
17
20
was the dominant family on Anatomy
Questions left blank
0
180
nothing abandoned over 200 minutes
Three of the four families that used to cost you the most are now strengths. Negative stems in particular have gone from a reliable trap to seventeen out of eighteen.
The response grid
Page 30 of this script is the response grid, with all 180 rows printed. There is not one mark on it. That is now six papers. On NEET the OMR sheet is the only thing that is marked — a perfect question paper with an empty grid scores zero. At 180 questions this also stops being a formality: transferring 180 answers at the end is a ten-minute job you will not have time for.
What to do before the next full paper
Sixty seconds at every subject change.
Before the physics block, write down the first law and the sign convention. Seven errors in eight questions, then twenty-eight clean — this is the cheapest thirty-five marks on the paper.
Match questions: hardest row first.
Three out of three lost to the same C/D swap. Start with the row you are least sure of, and check the last two rows instead of filling them by elimination.
Mark the order the stem asks for.
Twice you picked the right items and wrote them in the wrong order. Whenever a stem asks for two or three things in sequence, number them in the margin before reading the options.
Read R on its own first.
In Assertion–Reason, ask whether R is a true sentence before asking whether it explains A. One of your five losses had a reason that was simply false — and watch the pull towards (1), you chose it nine times out of sixteen.
Check what is inside the square root.
√3 in vrms, √2 in the mean free path. Both missing factors landed you on a printed option.
Fill the grid as you go.
Sixth paper. Start on the next one.
Marked from the red-pen script. Every tick was read by its vertex, and all 180 answers were transcribed twice — once by a geometric ink-and-vertex pipeline, once by independent visual reading — with no disagreement between the two.