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NEET 2027 · Physics · Class 11 Chapter 1 · Question bank, 26 August 2026

Error Notes
Eight from the PYQ Sheets

Questions 6, 7, 9, 15, 21, 24, 26 and 31 from the two uploaded CBSE-PMT sheets, each worked from first principles with every option accounted for.

8
questions
7
dimensional
1
error rule
32
options worked
One thing to note. The marked responses were not included this time, so there is no “you chose X” row. Instead all four options are worked for every question, which is more useful anyway. Send the marked sheet and the per-question diagnosis can be added.

Jump to a question

Q6Q7Q9Q15Q21Q24Q26Q31

What these eight are testing

Skill map of the eight questionsWhat these eight questions are actually testingRead it off the equationthe relation is given — extract dimensionsQ6Q21Q26Straight recalllearn the formula or the unitQ7Q15Compare two thingscompute both, then matchQ9Q24Error arithmeticpower rule on percentagesQ31seven of eight are dimensional — one is error arithmetic
Four skills, unevenly weighted — the bank is heavily dimensional.
The pattern in this selection

Three of the eight — Q6, Q21 and Q26 — hand you the equation and ask you to read dimensions off it. Nothing needs to be recalled; the marks are lost by mis-substituting, not by forgetting. In Q6 it is a term dropped from the denominator, in Q21 it is reading V as a length rather than a volume, in Q26 it is abandoning the algebra when an equation collapses.

Q7 and Q15 are pure recall. Q9 and Q24 want two things computed and compared. Q31 is the power rule on percentage errors.

Two links to the 25 August paper

Q15 is ε0 again. On the last paper the permittivity question was the single dimensional-formulae miss, and the slip was an undercounted length in the denominator. This one asks for the same quantity in unit form, where the same undercount produces a wrong answer again. That makes it the highest-value item in this set.

Q31 is Q39 with different numbers. That one was answered correctly, so this is confirmation rather than repair — useful for confidence, not for diagnosis.

Q6Dimensional formulaeCBSE PMT 93

Everything you need is printed in the question

The questionTurpentine oil flows through a tube of length l and radius r. The pressure difference across the ends is P. The viscosity is given by η = P(r² − x²) ÷ 4vl, where v is the oil's velocity at distance x from the axis. The dimensions of η are:
  1. [M⁰L⁰T⁰]
  2. [MLT⁻¹]
  3. [ML²T⁻²]
  4. [ML⁻¹T⁻¹]
Viscosity dimensions read off the formulaRead the dimensions straight off the formula you were givenη = P (r² − x²) ÷ (4 v l)top[M L⁻¹T⁻²] × [L²]pressure × an areabottom[L T⁻¹] × [L]velocity × a length÷M: 1 − 0 = 1   L: 1 − 2 = −1   T: −2 − (−1) = −1[η] = M L⁻¹ T⁻¹The (r² − x²) bracket is a subtraction of two areas, so it counts as [L²] once.
Top over bottom, one bracket at a time.
Concept to useYou are handed the defining relation, so no recall is required — substitute dimensions for each symbol and simplify.
Formula or ruleη = P(r² − x²) ÷ (4 v l)

Baby steps

  1. [P] = [M L⁻¹ T⁻²]. The bracket subtracts two areas, so it is [L²] — subtraction never changes dimensions.
  2. Top = [M L⁻¹ T⁻²][L²] = [M L T⁻²].
  3. Bottom = [v][l] = [L T⁻¹][L] = [L² T⁻¹]. The 4 is a pure number.
  4. Divide: M gives 1, L gives 1 − 2 = −1, T gives −2 − (−1) = −1.
Answer(D)  [ML⁻¹T⁻¹]
ShortcutViscosity is always [M L⁻¹ T⁻¹], whatever formula it arrives in. SI unit Pa·s = pressure × time gives it in one step.
OptionWhy this option and not the others
A[M⁰L⁰T⁰]Dimensionless. Viscosity has a unit (Pa·s), so it cannot be a pure number.
B[MLT⁻¹][MLT⁻¹] is linear momentum.
C[ML²T⁻²][ML²T⁻²] is energy or torque — this is what you get if the denominator is dropped entirely.
D[ML⁻¹T⁻¹] — correct answer[M L⁻¹ T⁻¹]. Top [MLT⁻²] over bottom [L²T⁻¹].
The fixWrite the top and the bottom as two separate brackets before you divide. The error in questions of this shape is almost always a term left out of the denominator, not a wrong dimension.
Q7Dimensional formulaeCBSE PMT 88

Build it from mvr rather than trying to remember it

The questionThe dimensional formula of angular momentum is:
  1. [ML²T⁻²]
  2. [ML⁻²T⁻¹]
  3. [MLT⁻¹]
  4. [ML²T⁻¹]
Building the dimensions of angular momentumAngular momentum is built from three pieces: mass, velocity, radiusmM×vL T⁻¹×rLM × L T⁻¹ × L[L] = M L² T⁻¹Planck's constant has the SAME dimensions — memorise the pair
Three factors, three dimensional blocks.
Concept to useAngular momentum is the moment of linear momentum, so it is momentum multiplied by a distance.
Formula or ruleL = m v r  (equivalently L = Iω)

Baby steps

  1. [m] = [M].
  2. [v] = [L T⁻¹].
  3. [r] = [L].
  4. Multiply: [M][L T⁻¹][L] = [M L² T⁻¹].
Answer(D)  [ML²T⁻¹]
ShortcutAngular momentum and Planck's constant are dimensionally identical. Learning that one pair answers this question and several others in the same set.
OptionWhy this option and not the others
A[ML²T⁻²][ML²T⁻²] is energy or torque — one power of T too few.
B[ML⁻²T⁻¹]Negative power of L; no mechanical quantity here has that form.
C[MLT⁻¹][MLT⁻¹] is linear momentum — the r was never multiplied in.
D[ML²T⁻¹] — correct answer[M L² T⁻¹], from m × v × r.
The fixCheck the route through Iω as well: [I] = [ML²] and [ω] = [T⁻¹], which gives the same answer. Two independent routes agreeing is a free check.
Q9Dimensional formulaeCBSE PMT 05

Divide, and only the time survives

The questionThe ratio of the dimensions of Planck's constant to those of moment of inertia has the dimensions of:
  1. time
  2. frequency
  3. angular momentum
  4. velocity
Ratio of Planck constant to moment of inertiaDivide, and everything except the time cancelsM L² T⁻¹Planck constant h÷M L²moment of inertia Ih ÷ I = T⁻¹ = frequencyFlip the ratio by mistake and you get [T] = time — that is the trap option
M and L² cancel; only the time is left.
Concept to useBoth quantities carry M and L², so those cancel completely and the answer is decided entirely by the powers of T.
Formula or ruleh ÷ I, with [h] = [ML²T⁻¹] and [I] = [ML²]

Baby steps

  1. [h] = [M L² T⁻¹].
  2. [I] = [M L²] — a moment of inertia has no time in it at all.
  3. Divide: M cancels, L² cancels.
  4. Left with [T⁻¹], which is a frequency.
Answer(B)  frequency
Shortcut[T⁻¹] is frequency; [T] is time. The two options sit next to each other precisely to catch a flipped ratio.
OptionWhy this option and not the others
Atime[T] is the ratio taken upside down, I ÷ h. This is the intended trap.
Bfrequency — correct answer[T⁻¹] is frequency, since M and L² cancel exactly.
Cangular momentumAngular momentum is [ML²T⁻¹] — that is h itself, not h/I.
DvelocityVelocity is [LT⁻¹]; the L² would have to survive.
The fixBefore dividing, ask which way round the question put it. “Ratio of A to B” always means A ÷ B. Underline the word to in the stem.
Q15UnitsCBSE PMT 04

The same ε0 that cost a mark last time, now in unit form

The questionThe unit of the permittivity of free space ε0 is:
  1. coulomb / newton-metre
  2. newton-metre² / coulomb²
  3. coulomb² / newton-metre²
  4. coulomb² / (newton-metre)²
Units of the permittivity of free spaceRearrange Coulomb's law, then read the units offF = q² ÷ (4π ε0 r²)  →  ε0 = q² ÷ (4π F r²)on topcoulomb²on the bottomnewton × metre²÷coulomb² / (newton × metre²)newton-metre²/coulomb² is 1/ε0, not ε0 — the ratio upside downand coulomb²/(newton-metre)² squares the newton too
Rearrange first, then read the units off the fraction.
Concept to useRearrange Coulomb's law to make ε0 the subject, then read the units straight off the expression.
Formula or ruleF = q² ÷ (4πε0r²) → ε0 = q² ÷ (4π F r²)

Baby steps

  1. Charge squared goes on top: coulomb².
  2. Force and radius squared go underneath: newton × metre².
  3. 4π is a pure number and carries no unit.
  4. So the unit is C² N⁻¹ m⁻².
Answer(C)  coulomb² / newton-metre²
ShortcutCheck against the dimensional form [M⁻¹L⁻³T⁴A²]: one inverse M from the newton, three inverse L (one from the newton, two from m²). That L count is exactly what went wrong on the last paper.
OptionWhy this option and not the others
Acoulomb / newton-metreOnly one power of charge, and no square on the metre. Neither matches.
Bnewton-metre² / coulomb²This is 1/ε0 — the whole expression inverted. The most chosen wrong option.
Ccoulomb² / newton-metre² — correct answercoulomb² / (newton × metre²), directly from ε0 = q²/(4πFr²).
Dcoulomb² / (newton-metre)²(newton-metre)² squares the newton as well as the metre, giving N²m² instead of N m².
The fixThis is the third appearance of ε0 in two weeks and the second slip in the same place — the length count in the denominator. Write ε0 = q²/(4πFr²) from memory once a day until it is automatic, then read either the unit or the dimension off it.
Q21Dimensional analysisCBSE PMT 96

V is a volume, so V² is L⁶ and not L²

The questionFor the equation (P + a/V²) = bθ/V, where P is pressure, V is volume and θ is absolute temperature, the dimensions of the constant a are:
  1. [ML⁻⁵T⁻¹]
  2. [ML⁵T¹]
  3. [ML⁵T⁻²]
  4. [M⁻¹L⁵T²]
Reading a constant out of a bracketAnything ADDED to pressure must itself be a pressure( P + a / V² )so  a = P × V²V = [L] → V² = [L²]wrong — V is a volumeV = [L³] → V² = [L⁶]right[a] = [M L⁻¹ T⁻²] × [L⁶][a] = M L⁵ T⁻²
The whole question turns on V being a volume.
Concept to useOnly quantities with identical dimensions can be added, so a/V² must itself be a pressure.
Formula or rulea = P × V²

Baby steps

  1. a/V² is added to P, so [a/V²] = [P] = [M L⁻¹ T⁻²].
  2. Therefore [a] = [P][V²].
  3. V is a volume: [V] = [L³], so [V²] = [L⁶].
  4. [a] = [M L⁻¹ T⁻²][L⁶] = [M L⁻¹⁺⁶ T⁻²].
Answer(C)  [ML⁵T⁻²]
ShortcutThe T exponent is inherited unchanged from pressure, so any option without T⁻² can be struck out before you do any length arithmetic.
OptionWhy this option and not the others
A[ML⁻⁵T⁻¹]Negative power of L. Multiplying pressure by a volume squared can only make the L exponent larger, never smaller.
B[ML⁵T¹]L⁵ is right but T¹ is not; pressure contributes T⁻².
C[ML⁵T⁻²] — correct answer[M L⁵ T⁻²], from [ML⁻¹T⁻²] × [L⁶].
D[M⁻¹L⁵T²]M and T both inverted — this is 1/a rather than a.
The fixCircle the letter V in the stem and write [L³] above it before touching the algebra. The same habit fixes the van der Waals version of this question, which uses the identical bracket.
Q24Dimensional formulaeCBSE PMT 08

Sort all five first, then look for a pair that is actually offered

The questionWhich of these five parameters have the same dimensions? 1 energy density · 2 refractive index · 3 dielectric constant · 4 Young's modulus · 5 magnetic field
  1. 2 and 4
  2. 3 and 5
  3. 1 and 4
  4. 1 and 5
Sorting five quantities into dimension bucketsSort all five into dimension buckets, then look for an offered pairM L⁻¹ T⁻²1 energy density4 Young's modulusdimensionless2 refractive index3 dielectric constantM T⁻² A⁻¹5 magnetic fieldanswer: 1 and 42 and 3 are also dimensionally identical — both are pure numbersbut no option offers that pair, so it is not the answer here
Every quantity sorted before any option is considered.
Concept to useWork out each of the five independently. Do not try to spot the answer from the option pairs, because more than one true pair may exist.
Formula or ruleenergy density = E/V · Young's modulus = stress/strain

Baby steps

  1. Energy density = [ML²T⁻²] ÷ [L³] = [M L⁻¹ T⁻²].
  2. Young's modulus = stress ÷ strain, and strain is dimensionless, so it keeps stress's [M L⁻¹ T⁻²].
  3. Refractive index and dielectric constant are both pure ratios — dimensionless.
  4. Magnetic field is [M T⁻² A⁻¹], which matches nothing else here.
Answer(C)  1 and 4
ShortcutEnergy density and any elastic modulus are always the same as pressure: [ML⁻¹T⁻²]. That single fact settles this question.
OptionWhy this option and not the others
A2 and 4Refractive index is dimensionless, Young's modulus is not.
B3 and 5Dielectric constant is dimensionless, magnetic field is [MT⁻²A⁻¹].
C1 and 4 — correct answerBoth are [M L⁻¹ T⁻²], the dimensions of pressure.
D1 and 5Energy density is [ML⁻¹T⁻²]; magnetic field is not.
The fixNote the genuine oddity here: items 2 and 3 are also dimensionally identical, since both are dimensionless. That pair simply is not offered. If two options had looked correct, the answer would be the one that is listed — but check all five anyway, because a differently-worded version of this question could offer 2 and 3.
Q26Dimensional analysisCBSE PMT 92

Two equations do all the work; the third collapses to 0 = 0

The questionP is radiation pressure, c is the speed of light and S is the radiation energy striking unit area per second. The non-zero integers x, y, z for which PˣSʸcᶻ is dimensionless are:
  1. x=1, y=1, z=1
  2. x=−1, y=1, z=1
  3. x=1, y=−1, z=1
  4. x=1, y=1, z=−1
Solving for dimensionless exponentsThree equations — but only two of them carry informationPˣ Sʸ cᶻ must be M⁰ L⁰ T⁰M:x + y = 0y = −xusableL:−x + z = 0z = +xusableT:−2x − 3y − z = 00 = 0collapses — no new informationso every solution has the shape (x, −x, x)smallest non-zero: 1, −1, 1P and S both carry M, so their exponents must be equal and opposite.
The third equation carries no information.
Concept to useSet the product's M, L and T exponents each to zero. Here the T equation turns out to be a consequence of the other two, which is why the answer is a family rather than a single point.
Formula or rule[P] = [ML⁻¹T⁻²], [S] = [MT⁻³], [c] = [LT⁻¹]

Baby steps

  1. S is energy per unit area per unit time, so [S] = [ML²T⁻²] ÷ [L²][T] = [M T⁻³].
  2. M: x + y = 0, so y = −x.
  3. L: −x + z = 0, so z = +x.
  4. T: −2x − 3y − z = −2x + 3x − x = 0 for every x — it gives nothing new. So the family is (x, −x, x), and the smallest non-zero integers are 1, −1, 1.
Answer(C)  x=1, y=−1, z=1
ShortcutP and S are the only two carrying M, so their exponents must be equal and opposite immediately. That single observation kills two options on sight.
OptionWhy this option and not the others
Ax=1, y=1, z=1x + y = 2, so M does not cancel.
Bx=−1, y=1, z=1M cancels, but L gives −(−1) + 1 = 2, so L does not.
Cx=1, y=−1, z=1 — correct answery = −x and z = x are both satisfied, and the T equation holds automatically.
Dx=1, y=1, z=−1x + y = 2 again, so M survives.
The fixWhen one exponent equation reduces to 0 = 0, that is information, not a mistake — it means the answer is a whole family and you should report the smallest integers. Do not assume you have made an algebra error and start over.
Q31Errors in measurementCBSE PMT 89

You already have this one — same rule, different numbers

The questionThe density of a cube is found by measuring its mass and the length of its side. If the maximum errors are 3% in the mass and 2% in the length, the maximum error in the density is:
  1. 12%
  2. 14%
  3. 7%
  4. 9%
Error in the density of a cubeρ = m / L³ — the cube turns one length error into three3 %mass+2 %2 %2 %side, counted three times=9 %7 % comes from squaring instead of cubing — that would be an area12 % comes from using the 3 % mass error for the side as well
One length error, counted three times.
Concept to useDensity is mass over volume, and the volume of a cube is the cube of the side, so the length error counts three times.
Formula or ruleρ = m/L³ → Δρ/ρ = Δm/m + 3ΔL/L

Baby steps

  1. Mass appears to the first power: contributes 3%.
  2. The side appears cubed: contributes 3 × 2% = 6%.
  3. Errors always add, never cancel.
  4. Total = 3 + 6 = 9%.
Answer(D)  9%
ShortcutThe exponent in the formula becomes the multiplier on the percentage error. Identical to Q39 on the 45-question paper, which was answered correctly.
OptionWhy this option and not the others
A12%12% is 3 + 3(3) — the 3% mass error used for the side as well.
B14%14% does not correspond to a single clean slip; it is there to catch arithmetic drift.
C7%7% is 3 + 2(2) — squaring instead of cubing, which would be the error in an area, not a volume.
D9% — correct answer3% + 3 × 2% = 9%.
The fixNothing to fix here if it was answered correctly — this is the version of the rule already mastered. If it was missed, the cause is the exponent, and the cure is to write ρ = m/L³ out in full before differencing.

What to carry forward

Three habits, in priority order

1. Split top from bottom. For Q6, Q15 and Q21, write the numerator and the denominator as two separate bracketed expressions before dividing. Every miss in this group is a term that never made it into the denominator.

2. Write [L³] above every V. Volume symbols are the most reliable source of a lost factor in this chapter, and the same bracket appears in the van der Waals question.

3. Memorise ε0 = q²/(4πFr²). One line, written from memory once a day. Both the unit and the dimensional formula come off it, and it has now cost marks twice in the same place.