🏠 NEET Home

NEET 2027 · Physics · Class 11 Chapter 1 · Test of 25 August 2026

Error Notes
Units and Measurement

Six questions cost marks out of forty-five. This file works all six, names the error in each, and ends with an eight-question drill aimed only at the cluster that actually appeared.

153
of 180
39
correct
3
wrong
3
blank
92.9%
accuracy

Where the marks went

Accuracy here means correct as a share of attempted, so blanks do not flatter the number.

Topic accuracy bar chartAccuracy on attempted questions, by topicErrors in measurement100%n=9Dimensional analysis100%n=10Units & conversion100%n=5Dimensional formulae93%n=15Significant figures67%n=6one topic sits well below the rest
Four topics are at or near ceiling. One is not, and it is not the one we expected to be weak.
What went right

Errors in measurement scored 9 out of 9. That is the block flagged as a syllabus gap — the material NTA examines but the rationalised NCERT chapter no longer contains. It included the r² trap in Q42, where the margin working 1 + 1 + 4 + 3 → 9 shows the squared radius was handled deliberately rather than guessed.

Dimensional analysis and unit conversion were also clean on everything attempted. The predicted weak spot was not the weak spot.

The one cluster

Significant figures: 4 of 6, and both misses are the same error. Q33 kept three figures when two were justified; Q35 counted a leading zero and kept one when two were justified. In both the physics and the arithmetic were correct and the loss came entirely at the last step, deciding how many digits to write down.

This is a single rule applied inconsistently, not a gap in understanding. It maps to toolkit modules M7 (counting) and M9 (arithmetic).

A shape worth naming

Q6 and Q35 are the same kind of failure in different clothing: the difficult step was completed correctly and a simple count was then botched. In Q6 the time exponent 2 − (−2) = 4 was right and an L was dropped. In Q35 the division was right and a digit was miscounted.

That is the right-content-wrong-arrangement pattern already logged across the other papers. At 92.9% accuracy the remaining marks are not in new content — they are in the last read-back before committing an answer.

The three wrong

Q6Dimensional formulaeWrong

One L lost in the denominator

The questionThe dimensional formula of the permittivity of free space ε0 is
Exponent ledger for permittivityε0 = q² ÷ (F r²) — subtract the bottom exponents from the toptop: q²bottom: F r²resultM01−1L03−3T2-24A202[ε0] = M⁻¹ L⁻³ T⁴ A²F already carries one L. r² adds two more — that is 3, not 2.
How the answer is built, one step at a time.

The hard part of this question is the time exponent, 2 − (−2) = 4, and she got that right. What slipped was the length count: the force in the denominator already carries one L before r² adds two more.

OptionWhy this option and not the others
A[M−1 L−2 T4 A2] — markedOnly r² was counted in the denominator. F = [M L T⁻²] contributes an L too, so the total is L³, giving L⁻³ on inversion.
B[M−1 L−3 T2 A2]This is the classic trap — adding the denominator's T⁻² instead of subtracting it. She avoided this one.
C[M L3 T−4 A−2]Every sign reversed. This is 1/ε0, not ε0.
D[M−1 L−3 T4 A2] — correct answerq² = [A²T²] over F r² = [M L³ T⁻²], so M⁻¹, L⁻³, T²⁻⁻⁻²⁾ = T⁴, A².
The fixBefore dividing, write the denominator out as a single bracket and count each base letter once. Here: F r² = [M L T⁻²][L²] = [M L³ T⁻²]. Count the L's on the page, not in your head.
Q33Significant figuresWrong

Kept three figures when only two were measured

The questionThe product 1.2 × 2.31 × 4.567, to the correct number of significant figures, is
Weakest-link rule for multiplicationMultiplying: the answer copies the input with the FEWEST significant figures1.22 sig figs▼ weakest2.313 sig figs4.5674 sig figs1.2 × 2.31 × 4.567 = 12.6597…12.7 — 3 sig figs→13 — 2 sig figsRounding to 12.7 keeps three figures. Only two were ever measured.
How the answer is built, one step at a time.

She multiplied correctly and landed on 12.6597. The loss came at the rounding step: she rounded to 12.7, which still shows three significant figures. The 1.2 in the question only ever justified two.

OptionWhy this option and not the others
A12.7 — markedCorrect arithmetic, wrong precision. 12.7 has three significant figures; the weakest input, 1.2, allows only two.
B13 — correct answer1.2 has 2 sig figs, 2.31 has 3, 4.567 has 4. The fewest is 2, so 12.6597 rounds to 13.
C12.66Four significant figures — the untouched calculator value, trimmed.
D12.6597The raw product with no rounding at all.
The fixAfter every × or ÷, circle the input with the fewest significant figures before you touch the calculator, and write that count in the margin. Round to that count, not to something that merely looks tidier.
Q35Significant figuresWrong

Counted the leading zero, so stopped one digit early

The questionThe mass of a body is 2.42 g and its volume is 4.6 cm3. Its density, to the correct number of significant figures, is
Which digits are significant in 0.5261The leading zero is a placeholder — counting starts at the 50.5261placeholder1st significant figure2nd significant figure0.5 — only 1 sig fig→0.53 — 2 sig figsThe next digit is 6, so the 2 rounds up to 3.
How the answer is built, one step at a time.

2.42 ÷ 4.6 = 0.5260869… and two significant figures were needed. She reported 0.5. That is one significant figure, because the zero before the point is a placeholder and does not count.

OptionWhy this option and not the others
A0.5261 g cm−3Four significant figures — the raw quotient.
B0.526 g cm−3Three significant figures; 4.6 allows only two.
C0.53 g cm−3 — correct answerThe two significant digits are 5 and 2. The next digit is 6, so the 2 rounds up to 3.
D0.5 g cm−3 — markedOnly one significant figure. The leading zero was counted as though it were one of the two.
The fixSay the digits out loud starting at the first non-zero one. In 0.5260869 that is “five, two” — so the answer must show two digits after the point, not one.

The three left blank

Two were left empty and one was written then struck out with nothing put back. None of the three is a hard question; all three are worth four marks each.

Q9Dimensional formulaeNot attempted

Left blank — but it is a two-line question

The questionStefan's constant σ has the dimensional formula
Cancelling area in Stefan's constantStefan's E is power PER UNIT AREA, so the L² must cancelM L² T⁻³power÷L²area=M T⁻³then divide by K⁴ for the fourth-power temperature[σ] = M T⁻³ K⁻⁴keep the L² and you get option (b) — the standard trap
How the answer is built, one step at a time.

Stefan's constant is one of the standard six dimensional formulae. The only subtlety is that E in E = σT⁴ is power per unit area, not power. Once the L² cancels, the rest is immediate.

OptionWhy this option and not the others
A[M T−2 K−4]T⁻² instead of T⁻³; power carries three inverse times.
B[M L2 T−3 K−4]The standard trap: E treated as total power, so the L² never cancels.
C[M L T−3 K−4]One stray L; this is thermal conductivity's L, not Stefan's.
D[M T−3 K−4] — correct answer[M L² T⁻³] ÷ [L²] ÷ [K⁴] = [M T⁻³ K⁻⁴].
The fixAdd σ to the memorised list alongside h, G, η and ε0. The recall cue is the SI unit W m−2 K−4, which reads off as the answer directly.
Q24Dimensional analysisNot attempted

Started, struck it out, moved on

The questionThe frequency n of a stretched string depends on its length L, tension F and mass per unit length μ. Dimensional analysis gives n ∝
Order of solving the exponent equationsAlways solve the equation that has only ONE unknown first1T:−2y = −1y = +½only y appears — start here2M:y + z = 0z = −½y is known now3L:x + y − z = 0x = −1x + ½ + ½ = 0n ∝ (1/L) √(F/μ)The minus in front of z is where this one usually breaks: −(−½) = +½.
How the answer is built, one step at a time.

A letter was written into the grid and then crossed out with nothing put back. The method here is identical to the pendulum derivation she has already done correctly — the only difference is which equation you solve first.

OptionWhy this option and not the others
AL√(F/μ)x = +1 instead of −1. A longer string would then vibrate faster, which is wrong.
B(1/L)√(F/μ) — correct answerT gives y = +½, then M gives z = −½, then L gives x = −1.
C(1/L)√(μ/F)y and z swapped — tension in the wrong place.
D(1/L2)√(F/μ)x = −2, from mis-solving the L equation.
The fixScan the three exponent equations and start with whichever has a single unknown. Here that is T: −2y = −1. Never start with the L equation, which usually has two unknowns in it.
Q26Units & conversionNot attempted

Left blank — the sign flip is the whole question

The questionA calorie equals 4.2 J, where 1 J = 1 kg m2 s−2. In a new system the unit of mass is α kg, of length β m and of time γ s. The magnitude of a calorie in the new units is
The sign flip on the time factorA negative exponent on a reciprocal flips it back upn₂ = n₁ [M₁/M₂]¹ [L₁/L₂]² [T₁/T₂]⁻²energy is [M L² T⁻²], so a = 1, b = 2, c = −2(1/α)¹↓α⁻¹(1/β)²↓β⁻²(1/γ)⁻²↓γ+24.2 α⁻¹ β⁻² γ²
How the answer is built, one step at a time.

This is NCERT Exercise 1.3, so it is fair game and worth having automatic. Energy is [M L² T⁻²], so the exponents are 1, 2 and −2. The negative time exponent acting on a reciprocal is what makes γ come out positive.

OptionWhy this option and not the others
A4.2 α−1 β2 γ2β exponent should be −2, not +2.
B4.2 α β2 γ−2Every sign inverted — the ratio was taken upside down.
C4.2 α−1 β−2 γ−2The trap answer: γ left negative, because the double negative was not resolved.
D4.2 α−1 β−2 γ2 — correct answer(1/γ)−2 = γ+2. Only the time factor flips up.
The fixWrite the three exponents a, b, c above the formula before substituting. Then only the one with a negative exponent can end up with a positive power — which identifies the answer without finishing the algebra.

Drill — eight questions, no calculator

Every item targets the significant-figure cluster or the exponent count from Q6. Write each answer down before revealing.

D012.5 × 3.42 = ?  (correct significant figures)

8.6. The product is exactly 8.55. 2.5 has 2 sig figs, so keep 2. The dropped digit is exactly 5 and the preceding 5 is odd, so it rounds up to 8.6.

D02How many significant figures in 0.0450?

3. The leading zeros are placeholders; 4, 5 and the trailing 0 after the decimal point all count.

D034.6 ÷ 1.23 = ?  (correct significant figures)

3.7. The raw value is 3.7398. 4.6 has 2 sig figs, so keep 2.

D0412.5 + 0.0087 = ?  (correct significant figures)

12.5. This is addition, so count DECIMAL PLACES, not significant figures. 12.5 has 1 decimal place, so the answer has 1. The raw sum is 12.5087.

D05A body of mass 7.11 g occupies 2.0 cm³. Density?

3.6 g cm⁻³. 7.11 ÷ 2.0 = 3.555. The 2.0 allows only 2 sig figs, and the dropped 5 with an odd 5 before it rounds up.

D06How many significant figures in 0.004500?

4. Leading zeros never count; 4, 5 and both trailing zeros do.

D07Round 0.5260869 to two significant figures.

0.53. The two significant digits are 5 and 2; the next digit is 6, so the 2 becomes 3. The leading zero is not one of the two.

D08In ε0 = q²/(F r²), what is the exponent of L in the denominator?

3. F = [M L T⁻²] carries one L and r² carries two, so the denominator is [M L³ T⁻²] and ε0 has L⁻³.

What to do next

Priority order

1. The eight drill items above. Ten minutes. If all eight are right, the cluster is closed and nothing further is needed on significant figures.

2. Memorise σ. Add Stefan's constant to the recall list with h, G, η and ε0. Cue: the SI unit W m−2 K−4 reads off as the answer.

3. Re-do Q24 and Q26 on paper. Both were skipped rather than failed. Q26 is NCERT Exercise 1.3 verbatim, so it is fair game in any paper.

4. Do not re-read the chapter. Nothing here indicates a content gap. Four of five topics are at ceiling.