252 out of 320, twenty-seven clear of your target of 225. Sixty-six right, twelve wrong, two left blank. This is a strong paper and the hard block is again your best: fifteen of sixteen. The twelve errors are not spread evenly across the chapter — seven of them are the same kind of mistake, picking the twin of the right quantity. Fix that one habit and this paper is a 280.
Where the marks went
Difficulty
Right
Out of
Accuracy
Easy
26
32
81%
81%
Moderate
25
32
78%
78%
Hard
15
16
94%
94%
The shape is upside down again. Fifteen of sixteen hard questions, twenty-six of thirty-two easy ones. The questions that force you to slow down are the ones you get.
Easy accuracy, four papers running
Paper
Easy
Moderate
Hard
Morphology of Flowering Plants
26 / 32
—
16 / 16
Anatomy of Flowering Plants
26 / 32
26 / 32
14 / 16
Kinetic Theory (this paper)
26 / 32
25 / 32
15 / 16
Thermodynamics
31 / 32
27 / 32
11 / 16
Twenty-six out of thirty-two on the easy questions, for the third paper in a row. Not twenty-five, not twenty-seven — twenty-six, three times. That is not a knowledge gap, it is a fixed rate of going too fast through the questions that look safe. The Thermodynamics paper you sat alongside this one broke the pattern at 31 of 32, which is the proof it can be broken.
Section by section
Section
Right
Out of
Lost
Molecular Nature of Matter, Avogadro's Number and the Mole
5
6
1 wrong
Behaviour of Gases and the Ideal Gas Equation
9
11
2 wrong
Kinetic Theory and the Pressure of an Ideal Gas
7
10
2 wrong, 1 blank
Kinetic Interpretation of Temperature
11
11
—
Molecular Speeds and the Maxwell Distribution
12
13
1 blank
Degrees of Freedom
7
8
1 wrong
Equipartition of Energy and the Molar Specific Heats of Gases
11
14
3 wrong
Mean Free Path
4
7
3 wrong
Mean free path is the weak block — four of seven, and three of those four losses sit in the last two sections of the paper. Equipartition and the molar specific heats cost another three.
The habit that cost the most
Seven errors · 28 marks of swing
Picking the twin of the right quantity
Every one of these is a pair of quantities that look alike and sit one line apart in the options: Cv and Cp, kB and R, per molecule and per mole, number density and pressure, mean free path and molecular diameter. In every case you knew the physics — you reached for the neighbour. The fix is one line of writing: before you look at the options, write down what the question is asking FOR, with its units. Per what? Per mole, per gram, per molecule? Once that is on the page the twin stops being tempting.
Q70 — you chose (A), the answer is (B)look-alike: Cv vs Cp Ā· easy
A rigid diatomic ideal gas has five degrees of freedom. Its molar specific heat at constant pressure is
(A) 20.75 J molā»Ā¹ Kā»Ā¹YOUR ANSWER
(B) 29.05 J molā»Ā¹ Kā»Ā¹CORRECT
(C) 12.45 J molā»Ā¹ Kā»Ā¹
(D) 33.20 J molā»Ā¹ Kā»Ā¹
This is the clearest example on the paper. 20.75 is Cv; 29.05 is Cp. You dropped the +1 in (f/2 + 1)R. The check takes two seconds: Cp − Cv must come to R = 8.3, and 29.05 − 20.75 = 8.3. Your answer fails that test.
Whenever you compute Cp or Cv, write both and subtract. If the difference is not 8.3, you have the wrong one.
Q68 — you chose (B), the answer is (C)negative stem Ā· look-alike: Cv vs Cp Ā· moderate
Each of the following is correct EXCEPT one. Identify the exception.
(A) Cv = (f/2)R, where f is the number of degrees of freedom
(B) Cp = (f/2 + 1)R for the same gasYOUR ANSWER
(C) Cv = (f/2 + 1)R and Cp = (f/2)RCORRECT
(D) γ falls as the number of degrees of freedom rises
The same pair again, three questions earlier, this time inside a negative stem. The exception is the option with Cv and Cp swapped round. You picked the option that was actually correct.
Q72 — you chose (C), the answer is (B)look-alike: per molecule vs per mole vs total Ā· moderate
Taking kB = 1.38 × 10−23 J K−1, the average energy held in one degree of freedom of one molecule at 300 K is
(A) 4.14 × 10−21 J
(B) 2.07 × 10−21 JCORRECT
(C) 6.21 × 10−21 JYOUR ANSWER
(D) 1245.0 J
The question asks for the energy in one degree of freedom of one molecule — that is ½kBT. You gave 6.21 × 10−21, which is (3/2)kBT: the total over all three translational degrees, not the share of one.
Q16 — you chose (A), the answer is (C)Ā°C given, kelvin needed Ā· look-alike: k_B vs R Ā· hard
A sealed flask holds 1.0 litre of an ideal gas at a pressure of 1.0 × 105 Pa and a temperature of 27 °C. Taking kB = 1.38 × 10−23 J K−1, the number of molecules in the flask is closest to
(A) 2.42 × 1025YOUR ANSWER
(B) 4.02 × 10−2
(C) 2.42 × 1022CORRECT
(D) 2.68 × 1023
Two conversions were needed before any physics: 27 °C → 300 K, and 1.0 litre → 1.0 × 10−3 m3. You did the temperature and left the volume in litres, which is exactly a factor of 1000 — your 2.42 × 1025 against the correct 2.42 × 1022. The digits match; only the exponent is wrong.
Convert every quantity to SI before the first line of working, not during it.
Q24 — you chose (B), the answer is (A)look-alike: a ratio written the right way round vs upside down Ā· moderate
A gas of density 1.2 kg m−3 exerts a pressure of 1.0 × 105 Pa. Its root mean square speed is closest to
(A) 500.0 m sā»Ā¹CORRECT
(B) 408.2 m sā»Ā¹YOUR ANSWER
(C) 250000.0 m sā»Ā¹
(D) 166.7 m sā»Ā¹
vrms = √(3P/ρ) = √(3 × 105 ÷ 1.2) = √(2.5 × 105) = 500 m s−1. Your 408.2 is √(2P/ρ) — the 3 became a 2.
Q26 — you chose (B), the answer is (D)moderate
A gas whose root mean square speed is 500 m s−1 exerts a pressure of 2.0 × 105 Pa. Its density is
(A) 1200.00 kg mā»Ā³
(B) 1.60 kg mā»Ā³YOUR ANSWER
(C) 0.27 kg mā»Ā³
(D) 2.40 kg mā»Ā³CORRECT
The same formula read backwards: ρ = 3P / v̄2. You halved instead of using the square, giving 1.60 against the correct 2.40.
Q75 — you chose (D), the answer is (C)reference shifted Ā· look-alike: number density vs number of molecules Ā· moderate
NCERT gives the mean free path as λ = 1/(√2 nπd2), in which n is the number of molecules per unit volume. If the pressure of a gas is doubled while its temperature is held fixed, the mean free path becomes
(A) four times what it was
(B) twice what it was
(C) half of what it wasCORRECT
(D) unchangedYOUR ANSWER
The formula carries n, the number of molecules per unit volume — not P. Doubling the pressure at fixed temperature doubles n, and λ goes as 1/n, so λ halves. You answered unchanged, which is what you get if you look at the formula, see no P in it, and conclude pressure cannot matter.
When a quantity is not in the formula, ask what in the formula it controls.
Q2 — you chose (C), the answer is (A)two statements Ā· look-alike: number density vs number of molecules Ā· easy
Statement I: Two sealed flasks of equal volume, one holding hydrogen and the other holding oxygen at the same temperature and pressure, contain the same number of molecules. / Statement II: The two flasks nevertheless hold different masses of gas. / In the light of the above statements, choose the correct answer.
(A) Both Statement I and Statement II are correctCORRECT
(B) Both Statement I and Statement II are incorrect
(C) Statement I is correct but Statement II is incorrectYOUR ANSWER
(D) Statement I is incorrect but Statement II is correct
Both statements are true. Equal volumes at the same temperature and pressure hold equal NUMBERS of molecules (Avogadro), and the two flasks still hold different MASSES, because an oxygen molecule is sixteen times heavier. Number and mass are the look-alike pair here, and both statements can be true at once.
Assertion–Reason: much better, with one gap left
Six of eight, and on this paper you started marking the Assertion and the Reason separately in the margin — that is new, and it is working. But look at where the two losses are. You answered (A) on seven of the eight. When R really does explain A you are perfect. When R is true but is not the explanation, you still say (A).
What the question was
How many
You got
R genuinely explains A — answer (A)
5
5 / 5
R true but NOT the explanation — answer (B)
3
1 / 3
Q78 — you chose (A), the answer is (B)assertionāreason Ā· look-alike: mean free path vs molecular diameter Ā· easy
Assertion (A): A gas at low pressure has a longer mean free path than the same gas at high pressure and the same temperature. / Reason (R): The mean free path varies inversely with the square of the diameter of a molecule.A: T / F — ____ R true on its own? Y / N — ____ A because R? ____
(A) Both A and R are true and R is the correct explanation of AYOUR ANSWER
(B) Both A and R are true but R is NOT the correct explanation of ACORRECT
(C)A is true but R is false
(D)A is false but R is true
A is true: lower pressure, fewer molecules per unit volume, longer mean free path. R is also true: λ does vary as 1/d2. But R is about the diameter of the molecule, and A is about the pressure. R is a true sentence that answers a different question, so it cannot be the explanation of A. The answer is (B).
Third question, every time: does R mention the same quantity A is about? Here one says diameter and the other says pressure — so (B).
Q54 — you chose (A), the answer is (B)assertionāreason Ā· easy
Assertion (A): A monatomic molecule is credited with three degrees of freedom and no more. / Reason (R): The law of equipartition gives every degree of freedom an average energy of (1/2)kBT.A: T / F — ____ R true on its own? Y / N — ____ A because R? ____
(A) Both A and R are true and R is the correct explanation of AYOUR ANSWER
(B) Both A and R are true but R is NOT the correct explanation of ACORRECT
(C)A is true but R is false
(D)A is false but R is true
A is true and R is true, but R is the equipartition theorem, which tells you how much energy each degree of freedom gets. It does not tell you why a monatomic molecule has three and no more — that comes from its having no rotation or vibration to speak of. True, relevant, and still not the explanation: (B).
Reading the figure only once
Q8 — you chose (B), the answer is (D)figure Ā· needs a second clue Ā· moderate
Taking a single vertical line drawn across all three curves and comparing the volumes at which it cuts them, the three temperatures are ordered as
(A) TQ > TR > TP
(B) all three are at the same temperatureYOUR ANSWER
(C) TP > TQ > TR
(D) TR > TQ > TPCORRECT
On a P–V diagram the isotherm further from the origin is the hotter one, so TR > TQ > TP. You answered that all three are at the same temperature — which would make them the same curve. You drew the vertical line the question asked for, in red, and then did not read off what it cut.
You drew the line. The next step was to compare the three volumes where it crossed, and the bigger volume at the same pressure is the hotter gas.
The two you left blank
Both are easy, and both were worth four marks at no real risk. Eight marks sitting on the table.
Q23 — left blank, the answer is (C)look-alike: average velocity (zero) vs average speed (not zero) Ā· easy
In a gas at rest in a closed vessel the average velocity of the molecules is zero, yet the average speed is not. The statement that best accounts for this is that
(A) the average speed is defined over a longer time than the average velocity
(B) the molecules are at rest on average and only appear to move
(C) velocity carries a direction and the directions cancel, while speed has no direction to cancelCORRECT
(D) speed is a vector and velocity is a scalar
This is a definition question, not a calculation. Velocity has a direction, so in a gas at rest the velocities cancel to zero; speed has no direction, so it cannot cancel. Nothing to compute — worth a guess even on a bad day.
Q45 — left blank, the answer is (C)match Ā· planted swap in rows C & D Ā· look-alike: v_rms vs v_mean vs v_mp Ā· easy
Match Column I with Column II.Column IColumn IIAvrmsI√(2RT/M)BvmeanII√3 : √(8/π) : √2CvmpIII√(3RT/M)Dvrms : vmean : vmpIV√(8RT/πM)
(A) A-III, B-IV, C-II, D-I
(B) A-IV, B-III, C-I, D-II
(C) A-III, B-IV, C-I, D-IICORRECT
(D) A-I, B-IV, C-III, D-II
You drew a line in the table from row D to item II, which is correct, and then did not pick an option. Having done the hard part, choose: the row you had already matched rules out every option except (C).
If you have matched even one row with certainty, it usually eliminates two or three of the four options. Always convert the work into a choice.
The paired question
Q77 — you chose (C), the answer is (B)pair 4, second question Ā· moderate
[Pair 4, second question] The pressure of that same gas is now doubled while the temperature is held fixed. Using your answer to the previous question, the mean free path becomes
(A) 1.47 × 10−7 m
(B) 1.04 × 10−7 mCORRECT
(C) 2.08 × 10−7 mYOUR ANSWER
(D) 4.17 × 10−7 m
You got Q76 right, so the mean free path you carried in was correct. Doubling the pressure at fixed temperature doubles n, so λ halves: 2.08 → 1.04 × 10−7 m. You left it unchanged at 2.08. This is the same slip as Q75, two questions apart — the same idea tested twice, missed twice.
The response grid, again
Page 13 of this script is the printed response grid. There is not one mark on it, and the same is true of the Thermodynamics paper. That is now five papers with every answer ticked against the options and nothing transferred. In the hall there is an OMR sheet and nothing else — the marks come from the bubbles, not from the question paper. Start filling it as you go, one bubble per question as you answer it, not in a block at the end.
What to do before the next paper
The units line.
Before looking at the options on any numerical question, write what is being asked for and its unit: per mole? per gram? per molecule? That single line would have saved Q16, Q70, Q72 and four more — twenty-eight marks.
Mayer's check.
Any time you write Cp or Cv, write the other one too and subtract. It must be 8.3.
The third A–R question.
You are now judging A and R separately — good. Add the third: does R talk about the same quantity A talks about? If not, the answer is (B), however true R is.
Slow the first pass.
Twenty-six of thirty-two easy, three papers running. The hard questions are fine. Give the easy ones the same reading speed you give the hard ones.
Fill the grid as you go.
Every paper from here, including practice ones.
Marked from the red-pen script. Every tick was read by its vertex and the whole paper was transcribed twice, by two independent methods, with no disagreement on any of the 80 answers.