Q.Which range of wavelength (in nm) is called photosynthetically active radiation (PAR)?

Step 1 — Test (A): "Each trophic level loses some energy as heat."

Energy flow in an ecosystem obeys the second law of thermodynamics: every transfer and every metabolic transaction dissipates energy as heat. This is precisely why energy flow is unidirectional and why a pyramid of energy is always upright. True. ✓

Step 2 — Test (C): "Energy flow is unidirectional."

Energy enters the ecosystem as sunlight, is fixed by producers, and flows $\text{producers} \rightarrow \text{herbivores} \rightarrow \text{carnivores}$, being progressively lost as heat. It is never recycled back to the sun (unlike nutrients, which cycle). True. ✓

Step 3 — Test (B): the 10 % law.

Lindeman's 10 per cent law states that only about $10\%$ of the energy stored at one trophic level is transferred to the next, the remaining $\approx 90\%$ being lost in respiration, heat and unassimilated matter.

This is the standard textbook statement. True. ✓

Step 4 — Test (D): the fraction of sunlight captured.

The actual numbers are:

• Of the total incident solar radiation, less than 50 % is photosynthetically active radiation (PAR) — the wavelengths plants can use at all.
• Of that PAR, plants capture only about 2–10 %.

So the energy actually trapped is roughly $2\text{–}10\%$ of PAR, i.e. a still smaller share of total incident sunlight — not "about 10 % of the solar energy that falls on leaves." The statement conflates the 10 % transfer law (Step 3, which governs trophic transfers) with the capture efficiency of producers, which is a different quantity. False — hence the answer. ✗

The conceptual point: the 10 % figure belongs to trophic transfer, not to solar capture. Mixing the two is the classic error this question is built on. Despite capturing only a tiny sliver of sunlight, that small trapped fraction sustains the entire living world.

ANSWER: D

Step 1 — Identify the family.

Potato (Solanum tuberosum) belongs to the family Solanaceae — the "potato family", which also includes tomato, brinjal, chilli, tobacco and Datura.

Step 2 — Recall the floral characters of Solanaceae.

• Symmetry: flowers are actinomorphic (radially symmetrical) → symbol $\oplus$.
• Calyx: 5 sepals, united (gamosepalous), persistent → $K_{(5)}$ (brackets denote fusion).
• Corolla: 5 petals, united (gamopetalous) → $C_{(5)}$.
• Androecium: 5 stamens, epipetalous (attached to the petals), alternating with the corolla lobes, free from one another → $A_5$ (no brackets, since they are not fused to each other).
• Gynoecium: bicarpellary, syncarpous (2 fused carpels), ovary superior, bilocular with axile placentation → $G_{(2)}$, underlined for superior ovary.

Step 3 — Match against the options.

• (A) $\oplus\ K_{(5)}, C_5, A_{(9)+1}, G_1$ — the $A_{(9)+1}$ (diadelphous stamens: nine fused + one free) and monocarpellary ovary are the signature of Fabaceae (pea family). Also, Fabaceae flowers are zygomorphic. ✗
• (B) $\oplus\ K_{(5)}, C_{(5)}, A_5, G_{(2)}$ — gamosepalous, gamopetalous, five stamens, bicarpellary syncarpous. Solanaceae. ✓
• (C) $\oplus\ K_{10}, C_{10}, A_{10}, G_2$ — no family in the syllabus has 10 free sepals and 10 free petals; this is a fabricated distractor. ✗
• (D) $\oplus\ P_{3+3}, A_{3+3}, G_{(3)}$ — the perianth ($P$) in two whorls of three, six stamens in two whorls of three, tricarpellary syncarpous ovary — this is Liliaceae (the lily family, a monocot). ✗

Reading tip: the fastest discriminator here is the $P$ vs $K,C$ distinction (Liliaceae has a perianth, not distinct calyx and corolla) and the $A_{(9)+1}$ diadelphous androecium (Fabaceae). Once those two are struck out, only (B) and the nonsense option (C) remain.

ANSWER: B

Step 1 — The standard vocabulary of activity rhythms

Organisms partition the 24-hour cycle, and ecology names each pattern:

Step 2 — Root the word

Diurnal comes from the Latin diurnus → dies, "day" — the same root as "diary" and "journal". So diurnal literally means "belonging to the day", which is exactly what the question asks for.

Step 3 — Screen the options

• (A) Vesporal — a garbled form of vespertine/vesperal, which in any case means evening-active, not day-active. ✗
• (B) Diurnal — day-active. ✓
• (C) Cresporal — a garbled form of crepuscular, which means twilight-active. ✗
• (D) Auroral — from Aurora, goddess of the dawn; refers to dawn activity, not the whole day. ✗

Options (A), (C) and (D) are deliberately built to look like the twilight terms so that a student who has not fixed the vocabulary hesitates. Only diurnal names activity through the daylight hours.

Step 4 — Why it matters ecologically

This is a form of temporal niche partitioning: two species can use the same food resource in the same habitat and still coexist if one is diurnal and the other nocturnal, because they are never competing at the same moment.

ANSWER: B

Step 1 — The concept: what a biome is

A biome is a large regional unit characterised by its major vegetation type and the animals adapted to it. The two abiotic variables that overwhelmingly determine which biome forms at a place are:

Every biome can be placed on a simple plot of these two axes — that is the whole basis of the classic Whittaker biome diagram.

Step 2 — Where each biome sits

Step 3 — Test each statement

• (A) High temperature and minimum rainfall help to form grasslands. — True. Grasslands/savannas occupy the warm, comparatively dry band: enough rain for grasses, too little to sustain a closed forest canopy.
• (B) Variation in temperature and mean precipitation accounts for the major biomes. — True. This is precisely the defining principle in Step 1.
• (C) Low temperature and less rainfall is a characteristic of Tundra biomes. — True. The Arctic tundra is intensely cold and receives very little precipitation; it is sometimes called a "cold desert" for that reason.
• (D) More rainfall and low temperature is the characteristic of deserts. — FALSE. This is the exact inverse of reality. Deserts (e.g. the Thar, the Sahara) are defined by scanty rainfall (typically $< 25$ cm/yr) together with high daytime temperatures and extreme diurnal temperature range. "More rainfall and low temperature" would describe something closer to a temperate/coniferous forest — never a desert.

Step 4 — Answer the question as asked

The question asks for the incorrect statement, so we select (D).

ANSWER: D

Step 1 — Trace the energy from the Sun

NCERT (Ecosystem) lays out the chain explicitly:

1. Incident solar radiation. Of the total solar radiation falling on the Earth, less than 50% is Photosynthetically Active Radiation (PAR) — the wavelengths (roughly $400$–$700$ nm) that chlorophyll can actually absorb. The rest is largely infrared and ultraviolet, useless for photosynthesis.

2. Capture by producers. Of that PAR, plants (the primary producers) capture only about

This captured fraction becomes the Gross Primary Productivity (GPP) of the ecosystem.

3. What's left over. Producers respire away part of the GPP; the remainder is Net Primary Productivity (NPP) — the energy actually available to herbivores:

Step 2 — Why the number is so small

Most PAR is reflected, transmitted through the leaf, or lost as heat; the light-reaction machinery itself is not perfectly efficient. The striking point NCERT makes is that this tiny 2–10% capture supports the entire living world — every consumer, decomposer and human being ultimately draws on it.

Step 3 — Screen the options

• (A) 60–70% ✗ — vastly too high; no natural plant community comes close.
• (B) 20–30% ✗ — this is nearer the assimilation figures used elsewhere, not PAR capture.
• (C) 12–20% ✗ — still far above the observed range.
• (D) 2–10% ✓ — the value stated in NCERT.

Step 4 — The consequence

This inefficiency at the very base, compounded by the 10% law of energy transfer at each subsequent trophic level, is why food chains are short (rarely more than 4–5 links) and why an energy pyramid is always upright.

ANSWER: D

The key concept here is annual ring formation in temperate trees. In temperate regions, the cambium (the layer that produces new wood) is highly sensitive to seasonal changes. Spring brings abundant water and warmth, so the cambium produces large, thin-walled xylem vessels — this is spring wood (early wood), which is lighter and less dense. As summer transitions to autumn, water becomes scarce and temperatures drop, so the cambium produces smaller, thick-walled vessels — autumn wood (late wood), which is darker and denser. The abrupt shift from one type to the next creates a visible boundary.

Now, why do the other options fail?

1. Option (A): "The water stress is more."

   Water stress does contribute to the formation of autumn wood, but it is not the sole cause of the clear demarcation. The demarcation is a result of the combination of temperature, light, and water changes — not just water stress alone. In tropical regions, water stress can also occur (e.g., during dry seasons), but without the accompanying temperature drop, the demarcation is often less distinct.

2. Option (B): "The climatic conditions are not uniform throughout the year."

   This is correct. In temperate regions, the four seasons bring dramatic shifts in temperature, rainfall, and day length. These non-uniform conditions cause the cambium to alternate between producing spring wood and autumn wood, creating a sharp boundary. This is why temperate trees show clear annual rings.

3. Option (C): "The climatic conditions are uniform throughout the year."

   This is false. Uniform conditions (as in tropical rainforests) lead to continuous, uniform wood growth with little to no visible annual rings. So this would prevent clear demarcation, not cause it.

4. Option (D): "The temperature is high."

   High temperature alone does not cause demarcation. In fact, consistently high temperatures (as in tropics) reduce seasonal variation. The demarcation requires a contrast between warm, wet springs and cool, dry autumns — not just high temperature.

Step 1 — The definition.

The Respiratory Quotient (RQ), or respiratory ratio, is

both measured over the same period of respiration.

Step 2 — Substitute the given numbers.

The balanced equation for tripalmitin quoted in NCERT is:

So $\text{O}_2$ consumed $= 145$ and $\text{CO}_2$ released $= 102$:

Step 3 — Evaluate.

Step 4 — Check it makes biological sense (the concept).

Why is RQ less than 1 for a fat? Fats are highly reduced and contain very little oxygen relative to their carbon and hydrogen. Oxidising them therefore requires much more atmospheric $\text{O}_2$ per $\text{CO}_2$ produced. Hence the denominator (145) is much larger than the numerator (102), pushing RQ down to $\approx 0.7$.

The standard reference values are worth memorising:

Our computed $0.70$ lands exactly on the fat value — a perfect consistency check, since tripalmitin is a fat.

Step 5 — Eliminate the others.

• (B) 1.0 is the carbohydrate value.
• (C) 0.5 and (D) 1.4 do not match $102/145$ and correspond to no standard substrate here.

ANSWER: A