WEEKLY REVIEW · SEPTEMBER 7–13, 2026
Three questions we revisited this week—and what needed clarifying
Reviewing this week's recorded answers led us back to three questions. It also exposed problems in our wording and explanations, and a botanical source distinction worth making clear. Below, we keep the original questions visible and separate them from what we can now explain more clearly.
Partial launch coverage: September 7–13, 2026 (IST). Our retained records begin September 8, and exact question forms could be recovered only for September 12–13. This is a review of that recoverable sample, not a complete weekly difficulty ranking.
The order below preserves the three highest rates of answers that differed from the stored keys among qualifying exact versions. Each had at least 10 answered accounts. Skips are separate. The defects mean these rates must not be read as proof of student misunderstanding.
1. Resistance: an unstated reference
24/29 answered attempts differed from the stored key (82.76%); 5 matched, 5 explicit skips. Most-selected key-mismatching option B: 11/29 answered (37.93%), or 11/24 mismatches (45.83%).
Read original question and options
A metallic wire has a resistance of 10 Ω at 20°C. If the temperature coefficient of resistance of the metal is 0.004 °C⁻¹, its resistance at 100°C will be:
- A: 13.2 Ω
- B: 14.0 Ω
- C: 10.32 Ω
- D: 6.8 Ω
Stored key: A
| Option | Count |
|---|---|
| A | 5 |
| B | 11 |
| C | 7 |
| D | 6 |
What needed clarifying: the original did not specify the coefficient's reference temperature. We should have stated it. The stored A answer assumes a coefficient referenced to 20°C.
Clarified calculation, with an added assumption: use α₂₀ = 0.004 °C⁻¹ and the linear model. Then R₁₀₀ ≈ 10[1 + 0.004(100 − 20)] = 13.2 Ω. If that numerical coefficient instead referred to 0°C, the same model gives 10 × 1.4/1.08 ≈ 12.96 Ω, absent from the options. This is our substitution into the reference-temperature model.
A choice of 14.0 Ω can result from using 100 rather than the change of 80 under the clarified assumption. It does not prove that each person who chose B did that. A defensible objection to the missing reference deserves checking.
Next time: write the reference temperature next to the coefficient before calculating.
2. Doppler effect: units and explanations
14/17 answered attempts differed from the stored key (82.35%); 3 matched, 9 explicit skips. Most-selected key-mismatching option B: 8/17 answered (47.06%), or 8/14 mismatches (57.14%).
Read original question and options
Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]:
- A: 350 Hz
- B: 361 Hz
- C: 411·Hz
- D: 448 Hz
Stored key: D
| Option | Count |
|---|---|
| A | 4 |
| B | 8 |
| C | 2 |
| D | 3 |
What needed correcting: our original English question says “16.5 m”, a length, while Hindi says “16.5 m/s”, a speed. The records do not retain the display language, so the counts describe a bilingual artifact with unequal wording, not a verified population who all saw equivalent questions. The English option C also contains an extra dot in its unit (411·Hz).
Clarified example, not the original English wording: take the second car's speed as 16.5 m/s. Assume still air and mutual approach along the line of sound travel. The first car is the source at 22 m/s; the second is the observer.
f′ = f(v + vₒ)/(v − vₛ)
f′ = 400(340 + 16.5)/(340 − 22)
f′ ≈ 448.43 Hz → 448 Hz, stored option D
Here v is sound speed, vₒ observer speed and vₛ source speed. These signs apply to approach in the stated setup. OpenStax Doppler relation.
Our stored explanation also assigned incorrect calculations to the other options. Swapping the two speeds gives about 447.60 Hz; using only source motion gives 427.67 Hz; using only observer motion gives 419.41 Hz. Using both receding signs gives 357.46 Hz. These do not support the stored claims about 350, 361 or 411 Hz. We should not invent a reason someone selected B.
Next time: label source and observer, check the speed units, then substitute. We have not authenticated the stored claim that this exact wording and options came from an official 2017 paper.
3. Orchid roots: check the named plant
21/26 answered attempts differed from the stored key (80.77%); 5 matched, 5 explicit skips. Most-selected key-mismatching option B: 17/26 answered (65.38%), or 17/21 mismatches (80.95%).
Read original question and options
Which of the following is an example of a root modification adapted for absorption of moisture from the atmosphere?
- A: Velamen roots of epiphytic orchids
- B: Pneumatophores of Rhizophora
- C: Prop roots of Ficus benghalensis
- D: Fasciculated roots of Asparagus
Stored key: A
| Option | Count |
|---|---|
| A | 5 |
| B | 17 |
| C | 2 |
| D | 2 |
What is supported: the intended key is A. Missouri Botanical Garden describes velamen as a moisture-absorbing covering of tropical orchid aerial roots. That supports the intended orchid association; it does not establish a detailed mechanism of direct vapour uptake or a claim about every orchid. Botanical definition of velamen.
The useful distinction is the function: the official NCERT Class XI Biology lab manual, printed page 43 (PDF page 2, 2025–26 footer), uses Rhizophora mangle as an example of roots for gaseous exchange. B follows that educational convention, but gaseous exchange does not answer this question about moisture absorption; A remains the intended answer. NCERT lab manual, page 43.
The 17 B selections do not tell us why those respondents chose it. For your next attempt, underline the function being asked for before comparing the options.
Why another botanical source may look different
Kew's Flora Zambesiaca description records prop roots and no pneumatophores for Rhizophora. This differs from the NCERT lab manual's convention; it is not enough to label B a confirmed content error. Kew's description. The current main textbook chapter, Reprint 2026–27, omits the root-modifications subsection, so we attribute this example to the lab manual. Current textbook, printed page 59. If you bring a conflicting passage, include its source and edition.
How we counted
We used retained daily-set answers within September 7–13 IST, with a data cutoff of September 14, 09:22 IST. We retained the first recorded answer per account and exact question/key version within the period. Known staff, bot and fixture records were excluded. Account counts are not verified unique-person counts.
There were 698 records with recoverable forms; 524 unknown-form records were excluded from the statistical selection. Late or missing ingestion remains possible. Unattempted questions are not counted as explicit skips. Rates use answered attempts only; ties favour a larger answered denominator, then stable item ID. These are key-mismatch statistics, not a language-specific or complete weekly assessment.
The order of these three candidates is retained for transparency. We have not replaced defective items with filler or silently rewritten what students saw. Question and explanation corrections are being handled separately; this article does not claim that historical scoring or app content has already changed.
Disagree with our answer? Explain your reasoning—we'll check it. Include the question, assumption, calculation or source passage. We will verify objections and correct confirmed errors.
Question versions
2026.09.09-1411
- itm_3efd6a596a8ac5a7 · daily-2026-09-13 · form 93bb11cf4ea79aae
- itm_39df3008b41f170e · daily-2026-09-12 · form 8b2dec305fa008d6
- itm_ad91fc016a44fe90 · daily-2026-09-12 · form 8b2dec305fa008d6