(a) Justification with examples:
Metals in the middle of reactivity series (e.g., Fe, Zn, Pb) are extracted by displacement reactions because they are less reactive than carbon but more reactive than their oxides.
Example 1: Extraction of iron: Fe₂O₃ + 3CO → 2Fe + 3CO₂ (displacement by CO)
Example 2: Extraction of zinc: ZnO + C → Zn + CO (displacement by carbon)
Carbon displaces the metal from its oxide because carbon is more reactive.

(b) Reason for highly reactive metals:
Metals high in reactivity series (Na, Ca, Al) have great affinity for oxygen. Carbon cannot displace them because carbon is less reactive. These metals are extracted by electrolytic reduction of their molten oxides/chlorides.

(a) Activity to show rusting conditions:
Take three test tubes with iron nails:
– Test tube A: Nail + water + air (open)
– Test tube B: Nail + boiled water (no air) + oil layer
– Test tube C: Nail + anhydrous CaCl₂ (no moisture) + cork
After few days, only nail in A rusts. This shows both air (oxygen) and moisture are essential for rusting.

OR (b)(i) Metals reacting violently with cold water: Sodium (Na) and Potassium (K).
Observations: (1) Metal moves rapidly on water surface, (2) Fizzing/bubbles (H₂ gas), (3) Heat generated (may catch fire).
(ii) Test for gas: Bring a burning matchstick near the gas – it burns with a pop sound, confirming hydrogen.

(a) Communication: Electrical impulses (action potentials) are transmitted from the touched leaf to other leaves.

(b) Observable response: Changes in water content due to osmosis – cells lose turgor, causing leaves to fold.

(c) Difference: Touch-me-not movement is nastic movement (response to touch, direction independent). Tendril movement is tropic movement (thigmotropism) – growth towards support, direction dependent.

(a) Chromosomes: Thread-like structures in nucleus, made of DNA and proteins, carrying genes.

(b) Stability of DNA content: Meiosis halves chromosome number (gametes are haploid). Fertilization restores diploid number. Thus, DNA content remains constant across generations.

(i) Object at infinity: Image at focus (F), behind mirror, virtual, erect, highly diminished.

(ii) Object between infinity and P: Image between P and F, behind mirror, virtual, erect, diminished.

[Ray diagram description: Two rays from object – one parallel to axis reflects as if from F; another directed towards centre of curvature reflects back.]

Given circuit: 10Ω and 15Ω in parallel, then series with 5Ω and 2Ω? (Assuming typical circuit).

Step 1: Parallel combination 10Ω || 15Ω: \( \frac{1}{R_p} = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} \) ⇒ R_p = 6Ω.

Step 2: Total resistance = 6Ω + 5Ω + 2Ω = 13Ω (assuming series).

Step 3: Total current I = V/R = 26V/13Ω = 2A (assuming source 26V).

Step 4: Voltage across parallel combination = I × R_p = 2A × 6Ω = 12V.

Note: Actual values depend on circuit diagram.

(a) Relationship: \( R = \rho \frac{l}{A} \), where ρ = resistivity, l = length, A = cross-section area.

SI unit of resistivity: \( \rho = R \times \frac{A}{l} \). Unit of R = ohm (Ω), A = m², l = m ⇒ ρ unit = Ω m²/m = Ωm (ohm-meter).

(b) Why alloys in heating devices? Alloys have high resistivity, high melting point, and do not oxidize easily at high temperatures (e.g., nichrome).