Biomagnification refers to the increasing concentration of a substance, such as a toxic chemical, in the tissues of organisms at successively higher levels in a food chain. DDT (Dichlorodiphenyltrichloroethane) is a persistent organic pollutant that is known to biomagnify. Carbon monoxide is a toxic gas but doesn't typically biomagnify in this manner. CFCs (Chlorofluorocarbons) are known for ozone depletion. Manure is organic matter and acts as a fertilizer, it does not biomagnify. \[ \boxed{DDT} \] Quick Tip: Biomagnification occurs with substances that are persistent (do not break down easily), fat-soluble (accumulate in fatty tissues), and biologically active. Common examples of biomagnifying substances include heavy metals (like mercury) and certain pesticides (like DDT).

Energy is lost at each trophic level in a food chain, typically around 90% is lost as heat or used for metabolic processes, and only about 10% is transferred to the next level. Therefore, the shortest food chains are the most efficient in terms of the amount of energy from the producer that reaches the final consumer.
Comparing the given options:
(a) Grass \(\rightarrow\) Grasshopper \(\rightarrow\) Frog \(\rightarrow\) Snake (4 trophic levels)
(b) Plants \(\rightarrow\) Deer \(\rightarrow\) Lion (3 trophic levels)
(c) Plants \(\rightarrow\) Man (2 trophic levels)
(d) Phytoplankton \(\rightarrow\) Zooplankton \(\rightarrow\) Small Fish \(\rightarrow\) Big Fish (4 trophic levels)
The food chain "Plants \(\rightarrow\) Man" has the fewest trophic levels (2 levels), meaning the least amount of energy is lost between the producer (Plants) and the final consumer (Man). \[ \boxed{Plants \rightarrow Man} \] Quick Tip: Energy transfer between trophic levels is inefficient (around 10% rule). Shorter food chains result in a greater proportion of energy from the producers being transferred to the top consumers. The most efficient food chain is generally the one with the fewest trophic levels.

The second shell (L-shell) can hold a maximum of 8 electrons. Assuming the second shell is completely filled for an element progressing to the third shell, it has 8 electrons.
Number of electrons in the third shell = 25% of electrons in the second shell
Number of electrons in the third shell = 0.25 \(\times\) 8 = 2 electrons.
The electronic configuration of element 'M' is:
First shell (K-shell): 2 electrons
Second shell (L-shell): 8 electrons
Third shell (M-shell): 2 electrons
So, the electronic configuration is 2, 8, 2.
The atomic number of element 'M' = Total number of electrons = 2 + 8 + 2 = 12.
The element with atomic number 12 is Magnesium (Mg).
Electronic configurations of the given options:
(a) Sodium (Na, Z=11): 2, 8, 1
(b) Magnesium (Mg, Z=12): 2, 8, 2
(c) Aluminium (Al, Z=13): 2, 8, 3
(d) Calcium (Ca, Z=20): 2, 8, 8, 2
Element 'M' is Magnesium. \[ \boxed{Magnesium} \] Quick Tip: The maximum number of electrons in the first shell (K) is 2, second shell (L) is 8, and third shell (M) is 18 (but it starts filling the fourth shell after 8 electrons in the third shell for elements like K, Ca). Calculate the number of electrons in the third shell based on the second shell's count (usually assumed to be full if the third shell is being filled). Determine the atomic number and identify the element.

When a narrow parallel beam of light passes obliquely through a glass slab, it undergoes refraction at two parallel surfaces. The emergent ray is parallel to the incident ray but is displaced laterally.

A plane mirror reflects light, changing its direction according to the laws of reflection. It does not cause lateral displacement in the context of a transmitted beam.
A convex lens converges a parallel beam of light to a focal point (or diverges if it's a concave lens).
A glass prism disperses white light into its constituent colors and deviates the path of light, but the primary characteristic described (lateral displacement of a parallel emergent beam) is specific to a glass slab.

Therefore, the optical device 'X' is a glass slab. \[ \boxed{glass slab} \] Quick Tip: Lateral displacement is the perpendicular distance between the incident ray and the emergent ray when light passes through a refracting medium with parallel faces (like a glass slab). A prism causes deviation and dispersion, not just lateral displacement of a parallel emergent beam.

Let the original wire have length L and cross-sectional area A. Its resistance is \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity of the material.
When the wire is cut "lengthwise into three identical parts", it means the length of each part becomes L/3, while the cross-sectional area A and resistivity \(\rho\) remain the same for each part.
Resistance of each of the three identical parts, say \(R_1\), will be: \(R_1 = \rho \frac{(L/3)}{A} = \frac{1}{3} \left(\rho \frac{L}{A}\right) = \frac{R}{3}\).
So, each of the three parts has a resistance of \(R/3\).
These three parts are then connected in parallel. The equivalent resistance R' of a parallel combination of three resistors \(R_1, R_1, R_1\) is given by: \(\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_1} + \frac{1}{R_1} = \frac{3}{R_1}\).
Substituting \(R_1 = R/3\): \(\frac{1}{R'} = \frac{3}{(R/3)} = \frac{3 \times 3}{R} = \frac{9}{R}\).
So, \(R' = \frac{R}{9}\).
We need to find the value of R/R'. \(\frac{R}{R'} = \frac{R}{(R/9)} = R \times \frac{9}{R} = 9\). \[ \boxed{9} \] Quick Tip: Resistance is directly proportional to length (\(R \propto L\)) and inversely proportional to cross-sectional area (\(R \propto 1/A\)). If a wire is cut into 'n' identical parts lengthwise, the resistance of each part becomes R/n. For 'n' identical resistors each of resistance \(r_{part}\) connected in parallel, the equivalent resistance is \(R_{eq} = r_{part}/n\).

A chemical reaction involves the formation of new substances with different chemical properties.

(a) Milk left open at room temperature during summer: Bacteria convert lactose (milk sugar) into lactic acid, causing the milk to sour. This is a chemical change (fermentation).
(b) Grapes get fermented: Yeast converts sugars in grapes into ethanol and carbon dioxide. This is a chemical change.
(c) An iron nail is left exposed to humid atmosphere: Iron reacts with oxygen and moisture to form iron oxide (rust). This is a chemical change (corrosion).
(d) Melting of glaciers: Ice (solid water) changes its state to liquid water. The chemical composition (H\(_2\)O) remains the same. This is a physical change.

Therefore, a chemical reaction does not occur during the melting of glaciers. \[ \boxed{Melting of glaciers} \] Quick Tip: Chemical changes involve the formation of new chemical substances and are often irreversible or difficult to reverse. Physical changes involve a change in state or form but not in chemical composition (e.g., melting, boiling, freezing, dissolving).

To prepare dry hydrogen chloride (HCl) gas, a drying agent is needed that absorbs moisture but does not react with HCl. Hydrogen chloride is an acidic gas.

(a) Calcium chloride (anhydrous CaCl\(_2\)): It is a common and effective drying agent that does not react with HCl gas.
(b) Calcium oxide (CaO): It is a basic oxide. It would react with acidic HCl gas: CaO + 2HCl \(\rightarrow\) CaCl\(_2\) + H\(_2\)O. So, it cannot be used.
(c) Calcium hydroxide (Ca(OH)\(_2\)): It is a base. It would react with acidic HCl gas: Ca(OH)\(_2\) + 2HCl \(\rightarrow\) CaCl\(_2\) + 2H\(_2\)O. So, it cannot be used.
(d) Calcium carbonate (CaCO\(_3\)): It is a salt of a weak acid and a strong base, effectively basic. It would react with HCl gas: CaCO\(_3\) + 2HCl \(\rightarrow\) CaCl\(_2\) + H\(_2\)O + CO\(_2\). So, it cannot be used.

The most suitable drying agent from the options is anhydrous calcium chloride. \[ \boxed{Calcium chloride} \] Quick Tip: Drying agents should be chemically inert towards the gas they are drying. Acidic gases (like HCl, SO\(_2\), CO\(_2\)) cannot be dried using basic drying agents (like CaO, NaOH). Basic gases (like NH\(_3\)) cannot be dried using acidic drying agents (like concentrated H\(_2\)SO\(_4\)). Common neutral drying agents include anhydrous calcium chloride, fused calcium chloride, silica gel, and phosphorus pentoxide (P\(_2\)O\(_5\)). Concentrated sulfuric acid is an acidic drying agent.

N/A Quick Tip: Alkanes: Only C--C single bonds Alkenes: At least one C=C bond Alkynes: At least one C\(\equiv\)C bond Propene is the simplest alkene meeting these criteria

Proteins are complex macromolecules composed of amino acids. Amino acids contain carbon, hydrogen, oxygen, and nitrogen. Nitrogen is a key component of the amino group (--NH\(_2\)) in amino acids. Plants absorb nitrogen from the soil, usually in the form of nitrate (NO\(_3^-\)) or ammonium (NH\(_4^+\)) ions, to synthesize amino acids and subsequently proteins.

Phosphorus is essential for nucleic acids (DNA, RNA) and ATP, as well as cell membranes.
Iron is a micronutrient involved in chlorophyll synthesis and enzyme function.
Magnesium is a component of chlorophyll and an activator for many enzymes.

While all these are essential, nitrogen is the defining element for protein synthesis from the options. \[ \boxed{Nitrogen} \] Quick Tip: Proteins are made of amino acids, and all amino acids contain nitrogen. Plants require nitrogen in relatively large amounts (macronutrient) for growth and development, primarily for protein and nucleic acid synthesis.

Each bulb is rated 4V, 6W.
When connected in series, the voltage across the combination is the sum of the voltages across individual bulbs.
To work safely with desired brightness, each bulb should operate at its rated voltage of 4V.
Let 'n' be the minimum number of identical bulbs connected in series.
The total voltage across the 'n' bulbs in series will be \(n \times 4V\).
This total voltage must be equal to the mains supply voltage, which is 240 V.
So, \(n \times 4 = 240\). \(n = \frac{240}{4}\). \(n = 60\).
Therefore, a minimum of 60 bulbs must be connected in series.
When these 60 bulbs are in series, the current through each bulb can be calculated:
Current \(I = \frac{P}{V} = \frac{6W}{4V} = 1.5A\).
The resistance of each bulb \(R_{bulb} = \frac{V}{I} = \frac{4V}{1.5A} = \frac{4}{3/2} = \frac{8}{3} \Omega\).
Or \(R_{bulb} = \frac{V^2}{P} = \frac{(4V)^2}{6W} = \frac{16}{6} = \frac{8}{3} \Omega\).
Total resistance of 60 bulbs in series = \(60 \times \frac{8}{3} = 20 \times 8 = 160 \Omega\).
Current from mains = \(\frac{240V}{160\Omega} = \frac{24}{16} = \frac{3}{2} = 1.5A\). This matches the rated current for each bulb, so they will operate with desired brightness. \[ \boxed{60} \] Quick Tip: In a series circuit, the voltage divides across the components. To operate multiple identical devices at their rated voltage from a higher supply voltage, connect them in series such that the sum of their rated voltages equals the supply voltage. Number of devices = (Total Supply Voltage) / (Rated Voltage of one device).

Given:
Rated Voltage (V) = 220 V
Rated Power (P) = 11 W
The bulb is operating at its rated voltage (power supply of 220 V).
We can use the formula relating power, voltage, and resistance: \(P = \frac{V^2}{R}\).
We need to find the resistance (R). \(R = \frac{V^2}{P}\).
Substitute the given values: \(R = \frac{(220 V)^2}{11 W}\) \(R = \frac{220 \times 220}{11}\) \(R = \frac{48400}{11}\).
To simplify \(\frac{48400}{11}\): \(\frac{44000}{11} = 4000\) and \(\frac{4400}{11} = 400\).
So, \(48400 \div 11 = (44000 + 4400) \div 11 = 4000 + 400 = 4400\).
Alternatively, \(220 / 11 = 20\). So, \(R = 20 \times 220 = 4400\).
The resistance of its filament is 4400 \(\Omega\). \[ \boxed{4400 \Omega} \] Quick Tip: The power rating of an electrical device is usually specified for a particular operating voltage. Use the power formula \(P = V^2/R\) to find resistance when power and voltage are known. Other useful power formulas are \(P = VI\) and \(P = I^2R\).

Let's analyze each statement:

[A.] Lymph vessels carry lymph through the body and finally open into larger arteries. This is FALSE. Lymph is returned to the bloodstream via large veins (e.g., subclavian veins), not arteries.
[B.] Lymph contains some amount of plasma, proteins and blood cells. This is TRUE. Lymph is derived from interstitial fluid, which is similar to plasma but with fewer large proteins. It contains white blood cells (lymphocytes), but generally lacks red blood cells and platelets in significant numbers. So, "blood cells" here refers mainly to white blood cells.
[C.] Lymph contains some amount of plasma, proteins and red blood cells. This is FALSE. Lymph typically does not contain red blood cells. Their presence would indicate damage to blood vessels.
[D.] Lymph vessels carry lymph through the body and finally open into larger veins. This is TRUE. The lymphatic system collects excess interstitial fluid and returns it to the circulatory system by emptying into large veins near the heart.

The true statements are B and D. \[ \boxed{B and D} \] Quick Tip: Lymph is interstitial fluid that has entered lymphatic vessels. It is similar in composition to blood plasma but has fewer proteins and normally lacks red blood cells. It is rich in lymphocytes (a type of white blood cell). The lymphatic system is a one-way system that returns fluid from the tissues to the venous (blood) circulation.

Many cultivated varieties of plants like roses and bananas have been selected or bred for characteristics that favor vegetative propagation. While they still produce flowers and often fruits, their capacity to produce viable seeds is often reduced or lost.

(a) Flowers: Roses and bananas produce flowers.
(b) Buds: Buds are essential for growth and are present.
(c) Seeds: Many cultivated roses are hybrids and produce few or no viable seeds. Commercial bananas are parthenocarpic, meaning they develop fruit without fertilization, and are thus seedless or have only rudimentary, non-viable seeds.
(d) Fruits: Roses produce hips (fruits), and bananas are well-known for their fruits.

The most significant lost capacity, especially in commercially propagated varieties, is the production of viable seeds, leading to reliance on vegetative propagation. \[ \boxed{seeds} \] Quick Tip: Vegetative propagation is common in plants like roses (cuttings) and bananas (suckers). Parthenocarpy is the development of fruit without prior fertilization, often resulting in seedless fruits. Many horticultural varieties are selected for traits that might reduce seed viability to ensure consistency through vegetative propagation.

Let 'T' be the allele for tallness (dominant) and 't' be the allele for dwarfness (recessive).
A pure-tall pea plant has genotype TT.
A pure-dwarf pea plant has genotype tt.

Parental Cross (P generation): TT \(\times\) tt
Gametes from TT parent: T
Gametes from tt parent: t
F\(_1\) Generation: All offspring will have the genotype Tt.
Since T is dominant over t, all F\(_1\) plants will be tall.
Percentage of tall plants in F\(_1\) generation = 100%.

F\(_2\) Generation (Selfing F\(_1\)): Tt \(\times\) Tt
Gametes from Tt parent: T, t
Using a Punnett square:
\begin{tabular{c|cc
& T & t

\hline
T & TT & Tt

t & Tt & tt

\end{tabular
The genotypes in the F\(_2\) generation are TT, Tt, Tt, tt in the ratio 1:2:1.
The phenotypes are:
TT (Tall) - 1 part
Tt (Tall) - 2 parts
tt (Dwarf) - 1 part
So, the ratio of tall to dwarf plants is 3:1.
Percentage of tall plants in F\(_2\) generation = \(\frac{3}{4} \times 100% = 75%\).
Percentage of dwarf plants in F\(_2\) generation = \(\frac{1}{4} \times 100% = 25%\).

Therefore, the percentage of tall pea plants in F\(_1\) is 100% and in F\(_2\) is 75%. \[ \boxed{100% ; 75%} \] Quick Tip: In Mendelian genetics, a monohybrid cross between a homozygous dominant (e.g., TT) and a homozygous recessive (e.g., tt) parent results in an F\(_1\) generation where all offspring are heterozygous (Tt) and express the dominant phenotype. Selfing the F\(_1\) generation (Tt x Tt) results in an F\(_2\) generation with a genotypic ratio of 1 TT : 2 Tt tt and a phenotypic ratio of 3 dominant recessive.

In a bisexual flower, the male reproductive part is the stamen, which consists of the anther and the filament.

(a) Anther: The anther is the part of the stamen that produces pollen grains. Pollen grains contain the male gametes.
(b) Ovary: The ovary is part of the pistil (female reproductive part) and contains ovules, which house the female gamete (egg cell).
(c) Stigma: The stigma is the receptive tip of the pistil where pollen lands.
(d) Filament: The filament is the stalk that supports the anther.

The male gametes are developed within the pollen grains, which are produced in the anther. \[ \boxed{anther} \] Quick Tip: Stamen = Anther + Filament (Male part) Pistil/Carpel = Stigma + Style + Ovary (Female part) Anthers produce pollen grains, which contain male gametes. Ovules (inside the ovary) contain the female gamete (egg cell).

Magnification \(m = -1\).
A negative magnification indicates a real and inverted image. \(m = \frac{v}{u} = -1 \Rightarrow v = -u\).
Since the image is real, it is formed on the opposite side of the lens from the object for a convex lens (which is needed to form a real image on a screen). For a real image formed by a convex lens, \(v\) is positive if \(u\) is negative (standard sign convention).
The magnification \(m=-1\) means the image is real, inverted, and the same size as the object. This occurs when the object is placed at \(2f\) (twice the focal length) from a convex lens. The image is then also formed at \(2f\) on the other side.
Given focal length \(f = 20\) cm.
Object distance \(u\) must be \(2f\).
So, \(u = 2 \times 20 = 40\) cm.
(Using sign convention, \(u = -40\) cm, then \(v = -(-40 cm) = 40\) cm).
The object distance (magnitude) must be 40 cm.
Let's verify with the lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).
If \(u = -40\) cm, then \(v = 40\) cm (since \(v=-u\) and image is real). \(\frac{1}{20} = \frac{1}{40} - \frac{1}{-40} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}\). This is correct. \[ \boxed{40 cm} \] Quick Tip: Magnification \(m = -1\) means the image is real, inverted, and the same size as the object. For a convex lens, this occurs when the object is placed at a distance of \(2f\) from the lens. The image is also formed at \(2f\) on the other side. A real image can be formed on a screen.

Assertion (A): In many reptiles (like crocodiles, some turtles, and lizards), Temperature-dependent Sex Determination (TSD) occurs. The incubation temperature of the eggs during a critical period of development determines whether the offspring will be male or female. This statement is true.
Reason (R): Sex is not genetically determined (e.g., by sex chromosomes like XX/XY or ZZ/ZW) in some animals. TSD is an example of environmental sex determination, where sex is determined by environmental factors rather than solely by genetics. This statement is true.
Explanation: The reason (R) explains why the assertion (A) is true. Because sex is not genetically determined in these reptiles (it's environmentally determined by temperature), the temperature at which eggs are kept decides the sex. \[ \boxed{\small Both, Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).} \] Quick Tip: Temperature-dependent Sex Determination (TSD) is a type of environmental sex determination found in many reptiles and some fish. In TSD, specific temperature ranges during egg incubation lead to the development of males, while other ranges lead to females (or vice-versa, or a mix). This contrasts with Genotypic Sex Determination (GSD) found in mammals and birds, where sex is determined by sex chromosomes.

Assertion (A): In large animals, oxygen cannot reach different parts of the animal's body "easily" just by simple diffusion from the surface. The surface area to volume ratio is too small, and diffusion distances are too large for efficient oxygen supply to all cells. Specialized respiratory and circulatory systems are required. So, the statement "easily" makes the assertion false.
Reason (R): Respiratory pigments (like hemoglobin in vertebrates or hemocyanin in some invertebrates) bind to oxygen in areas of high oxygen concentration (e.g., lungs, gills) and release it in areas of low oxygen concentration (body tissues). They play a crucial role in transporting oxygen. This statement is true.
Explanation: Assertion (A) is false because oxygen transport in large animals is a complex process, not "easy" by simple means. Reason (R) is a true statement about how oxygen is transported, which is precisely why such pigments are needed in large animals where simple diffusion is insufficient. The reason highlights a mechanism that makes oxygen transport possible but doesn't support the idea that it's "easy" without these mechanisms. \[ \boxed{Assertion (A) is false, but Reason (R) is true.} \] Quick Tip: Simple diffusion is only efficient for oxygen transport over very short distances or in small organisms with a high surface area to volume ratio. Large, complex animals require specialized respiratory surfaces (lungs, gills) and circulatory systems with respiratory pigments to transport oxygen effectively to all body tissues.

Assertion (A): The standard safety procedure for diluting concentrated acids (like nitric acid, sulfuric acid) is to add the acid slowly to water, with constant stirring, not the other way around. Adding water to concentrated acid can cause vigorous exothermic reaction, leading to splashing and boiling of the acid, which is dangerous. So, the assertion describes an incorrect and unsafe procedure. Thus, Assertion (A) is false.
Reason (R): Concentrated nitric acid is highly soluble in water; it mixes with water in all proportions. This statement is true.
Explanation: Assertion (A) is false as it describes the wrong method for diluting concentrated acid. Reason (R) is true, nitric acid is indeed very soluble in water, and this dissolution is highly exothermic. \[ \boxed{Assertion (A) is false, but Reason (R) is true.} \] Quick Tip: \textbf{Safety Rule for Dilution:} Always add acid to water (A.A.W.) slowly and with stirring. This helps dissipate the heat generated during dilution more effectively. Adding water to concentrated acid can generate a lot of heat rapidly in a small volume, potentially causing the solution to boil and splash, posing a safety hazard.

Assertion (A): When white light passes through a glass prism, it splits into its constituent spectrum of colors (typically observed as violet, indigo, blue, green, yellow, orange, red - VIBGYOR). This phenomenon is called dispersion. The assertion states it splits into seven colors, which is a common way to describe the visible spectrum. This is true.
Reason (R): Different colors of light have different wavelengths. The refractive index of glass varies slightly with the wavelength of light. Violet light (shorter wavelength) has a higher refractive index in glass than red light (longer wavelength). Due to this difference in refractive index, violet light bends (deviates) more than red light when passing through a prism. Red light bends the least. This statement is true.
Explanation: The reason (R) correctly explains why dispersion (Assertion A) occurs. The differential bending of different colors of light due to their different speeds (and hence refractive indices) in the prism material is the cause of white light splitting into its spectrum. \[ \textbf{Both (A) and (R) are true, and (R) correctly explains (A).} \] Quick Tip: Dispersion of light is the splitting of white light into its constituent colors. This occurs because the refractive index of the prism material is different for different wavelengths (colors) of light. Shorter wavelengths (like violet) bend more, and longer wavelengths (like red) bend less.

The compound commonly used in black and white photography is silver chloride (AgCl) or silver bromide (AgBr).
The reaction that occurs is a photochemical decomposition reaction, which is endothermic.
Justification:
The decomposition of silver chloride (or silver bromide) into silver and chlorine (or bromine) occurs when light energy is absorbed. For example: \[ 2AgCl(s) \xrightarrow{Sunlight} 2Ag(s) + Cl_2(g) \]
Since the reaction requires energy (in the form of light) to proceed, it is an endothermic reaction. Energy is absorbed from the surroundings (light) to break the bonds in AgCl. \[ \textbf{Compound: Silver Chloride (AgCl) or Silver Bromide (AgBr).}
\textbf{Reaction: Endothermic.}
\textbf{Justification: Requires light energy for decomposition.} \] Quick Tip: Photochemical reactions are those initiated by the absorption of light. Endothermic reactions absorb energy from their surroundings. Exothermic reactions release energy. Decomposition reactions often require an input of energy (heat, light, or electricity).

N/A

N/A Quick Tip: For ionic bond formation, show electron loss by metal and electron gain by non-metal leading to ion formation and then the ionic compound. Amphoteric oxides react with both acids and bases. Common examples include Al\(_2\)O\(_3\), ZnO, PbO, SnO. When writing reactions of amphoteric oxides, show one reaction with an acid and one with a base.

The tissues that form the vascular bundle in plants are:
1. Xylem
2. Phloem

Functions:
1. Xylem:

Its primary function is the transport of water and dissolved minerals from the roots to all other parts of the plant (stems, leaves).
It also provides mechanical support to the plant due to its lignified cell walls.


2. Phloem:

Its primary function is the transport of food (mainly sugars like sucrose), synthesized during photosynthesis in the leaves, to other parts of the plant where it is needed for growth or storage (e.g., roots, fruits, seeds, growing tips). This process is called translocation.
\[ \textbf{Plant Vascular Tissues:}
\textbf{Xylem (water/minerals/support), Phloem (sugars).} \] Quick Tip: Vascular bundles are the transport system in vascular plants. Xylem transports "sap" upwards (water and minerals). Phloem transports "food" (sugars) both upwards and downwards to where it's needed.

N/A

N/A Quick Tip: For prism refraction, remember: incident ray, refracted ray, emergent ray, angle of incidence (i), angle of refraction (r), angle of emergence (e), and angle of deviation (D). Light bends towards the base of the prism. Presbyopia is the age-related loss of accommodation. Bifocal lenses have two focal lengths. Progressive lenses have a continuous change in focal length.

Flow Chart Explaining Father's Role in Sex Determination:


\fbox{Human Parents
\(\downarrow\)

\begin{tabular{c c c
\fbox{\parbox{3cm{\centering Mother (Female)
Genotype: XX
Produces eggs with X chromosome only & & \fbox{\parbox{3cm{\centering Father (Male)
Genotype: XY
Produces sperms:
50% with X chromosome
50% with Y chromosome

\end{tabular
\(\swarrow \quad \searrow\)

Fertilization
\(\↙ \quad \quad \quad \↘\)

\begin{tabular{c c
\fbox{\parbox{4.5cm{\centering If egg (X) is fertilized by sperm with X chromosome:
Child's Genotype: XX
\(\downarrow\)
Sex of Child: Female & \fbox{\parbox{4.5cm{\centering If egg (X) is fertilized by sperm with Y chromosome:
Child's Genotype: XY
\(\downarrow\)
Sex of Child: Male

\end{tabular

Explanation:
Human females have two X chromosomes (XX) and produce eggs that all carry an X chromosome. Human males have one X and one Y chromosome (XY) and produce two types of sperm: half carry an X chromosome and half carry a Y chromosome.
During fertilization, if an egg (X) is fertilized by a sperm carrying an X chromosome, the resulting zygote will have an XX genotype, developing into a female child.
If an egg (X) is fertilized by a sperm carrying a Y chromosome, the resulting zygote will have an XY genotype, developing into a male child.
Since the mother can only contribute an X chromosome, the sex of the child is determined by the type of chromosome (X or Y) carried by the sperm from the father that fertilizes the egg. Therefore, the father is responsible for the sex of the child. \[ \begin{tabular}{p{\linewidth}} \textbf{Parental Contribution:}
Mother (XX) \(\rightarrow\) Only X eggs
Father (XY) \(\rightarrow\) X or Y sperm (50% each)
\textbf{Offspring Sex Determination:}
\begin{tabular}{cc} X (egg) + X (sperm) & \(\rightarrow\) XX \(\female\)
X (egg) + Y (sperm) & \(\rightarrow\) XY \(\male\)
\end{tabular}
\textbf{Critical Factor:} Paternal sperm chromosome
\end{tabular} \] Quick Tip: Human sex determination is based on XX (female) and XY (male) chromosomes. Females are homogametic (produce only X-bearing gametes). Males are heterogametic (produce X-bearing and Y-bearing gametes). The Y chromosome carries genes responsible for male development.

(A diagram should be drawn here. Description below.)
Diagram Description:

Draw a straight vertical line segment representing the straight conductor.
Mark an arrow along this line segment to indicate the direction of current (e.g., upwards). Label it 'Current (I)'.
Around the conductor, draw several concentric circles in a plane perpendicular to the conductor. These circles represent the magnetic field lines.
On these circular field lines, mark arrows to indicate the direction of the magnetic field. If the current is upwards, the magnetic field lines will be anti-clockwise when viewed from above (or following the right-hand thumb rule).


Pattern of Magnetic Field Lines:
The magnetic field lines due to a current-carrying straight conductor are concentric circles around the conductor. The plane of these circles is perpendicular to the conductor, and the conductor is at the center of these circles.

Direction of Current and Magnetic Field:

Direction of Current (I): Should be clearly marked on the straight conductor (e.g., an arrow pointing upwards or downwards).
Direction of Magnetic Field (B): Should be marked with arrows on the concentric circular field lines. The direction is tangential to the field line at any point.


Rule to Determine Direction:
The rule that helps determine the direction of magnetic field lines in this case is the Right-Hand Thumb Rule (also known as Maxwell's Corkscrew Rule).
Statement of the Right-Hand Thumb Rule: If you imagine holding the current-carrying straight conductor in your right hand such that your thumb points in the direction of the current, then the direction in which your fingers curl around the conductor gives the direction of the magnetic field lines. \[ \begin{tabular{p{\linewidth}} \textbf{Current-Carrying Conductor:}
• Shows \(\vec{B}\) field lines as concentric circles
• Direction follows right-hand rule
\textbf{Right-Hand Rule:}
Thumb \(\parallel\) current (\(I\)), fingers curl \(\parallel\) \(\vec{B}\)
\end{tabular} \]
% Image placeholder:
% \includegraphics[path/to/magnetic_field_straight_wire.png Quick Tip: Magnetic field lines around a straight current-carrying wire are concentric circles. The Right-Hand Thumb Rule is used to find the direction of these circular magnetic field lines. The strength of the magnetic field decreases as the distance from the conductor increases (field lines are further apart).

To obtain a virtual and magnified image using a concave mirror, the object must be placed between the pole (P) and the principal focus (F) of the mirror.
Given focal length \(f = -18\) cm (concave mirror).
Given magnification \(m = +2\). A positive magnification indicates a virtual and erect image.
Magnification \(m = -\frac{v}{u}\).
So, \(+2 = -\frac{v}{u} \Rightarrow v = -2u\).

Using the mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\).
Substitute \(f = -18\) cm and \(v = -2u\): \(\frac{1}{-18} = \frac{1}{-2u} + \frac{1}{u}\) \(\frac{1}{-18} = \frac{-1 + 2}{2u}\) \(\frac{1}{-18} = \frac{1}{2u}\) \(2u = -18\) \(u = -9\) cm.

The object should be placed 9 cm in front of the concave mirror.
This object distance \(u = -9\) cm is between the pole (0 cm) and the focus (-18 cm), i.e., \(|u| < |f|\), which is consistent with the condition for forming a virtual, erect, and magnified image by a concave mirror.

Justification: The calculated object distance is \(u = -9\) cm.
The focal length is \(f = -18\) cm.
Since \(|u| = 9\) cm is less than \(|f| = 18\) cm, the object is placed between the pole and the focus. This position results in a virtual, erect, and magnified image, which matches the required magnification of +2. \[ \boxed{Object distance u = -9 cm (9 cm in front of the mirror). Object placed between P and F.} \] Quick Tip: For a concave mirror to form a virtual, erect, and magnified image, the object must be between P and F (\(u < f\)). Mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). Magnification: \(m = -\frac{v}{u} = \frac{h_i}{h_o}\). Sign conventions: For a concave mirror, f is negative. u is negative if object is in front. v is positive for virtual image, negative for real image. m is positive for virtual/erect, negative for real/inverted.

The most widely used method for refining impure metals is electrolytic refining.

Description of Electrolytic Refining:
Electrolytic refining is a process used to obtain high-purity metals from impure metals. It is based on the principle of electrolysis.

Setup:

Anode: A thick block of the impure metal is made the anode.
Cathode: A thin strip of pure metal is made the cathode.
Electrolyte: A solution of a soluble salt of the metal being refined is used as the electrolyte.

Process:
When an electric current is passed through the electrolytic solution:

At the anode (impure metal), metal atoms from the impure block lose electrons (oxidize) and go into the electrolyte solution as positive ions.
Example: \(M(impure) \rightarrow M^{n+}(aq) + ne^-\)
These metal ions (\(M^{n+}\)) from the electrolyte move towards the cathode (pure metal strip).
At the cathode, the metal ions gain electrons (reduce) and get deposited as pure metal on the cathode.
Example: \(M^{n+}(aq) + ne^- \rightarrow M(pure)\)

Impurities:

Soluble impurities from the anode dissolve in the electrolyte.
Insoluble impurities from the anode settle down below the anode as anode mud or anode slime. These often contain valuable metals like gold, silver, platinum.


This method is very effective for refining metals like copper, zinc, tin, nickel, silver, gold, etc., to a high degree of purity. \[ \begin{tabular}{p{\linewidth}} \textbf{Electrolytic Refining Process:}
Anode (Impure): M → M\(^{n+}\) + ne\(^-\) (Oxidation)
Cathode (Pure): M\(^{n+}\) + ne\(^-\) → M (Reduction)
\textbf{Impurity Fate:}
• Soluble: Remain in electrolyte
• Insoluble: Anode mud (Au, Ag, Pt in Cu refining)
\end{tabular} \] Quick Tip: Remember \textbf{OIL RIG}: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Or \textbf{LEO says GER}: Loss of Electrons is Oxidation, Gain of Electrons is Reduction. In electrolytic refining: Anode = Impure metal (gets thinner), Cathode = Pure metal (gets thicker). Anode mud can be a valuable byproduct.

The first step of cellular respiration is Glycolysis.
It occurs in the cytoplasm of the cell.

Equation for the process of breakdown of glucose in a human cell:

(i) In the presence of oxygen (Aerobic Respiration):
Glucose (C\(_6\)H\(_{12}\)O\(_6\)) is first broken down into pyruvate through glycolysis in the cytoplasm. Then, in the presence of oxygen, pyruvate enters the mitochondria and is completely oxidized.
The overall equation for aerobic respiration is: \[ C_6H_{12}O_6 (Glucose) + 6O_2 (Oxygen) \xrightarrow{In mitochondria} 6CO_2 (Carbon dioxide) + 6H_2O (Water) + Energy (ATP) \]
More detailed steps:

Glycolysis (in cytoplasm): Glucose \(\rightarrow\) 2 Pyruvate + Energy (ATP, NADH)
Link reaction & Krebs Cycle (in mitochondria): Pyruvate \(\rightarrow\) Acetyl-CoA \(\rightarrow\) CO\(_2\) + Energy (ATP, NADH, FADH\(_2\))
Oxidative Phosphorylation (in mitochondria): Uses O\(_2\) to produce a large amount of ATP.


(ii) Due to lack of oxygen (Anaerobic Respiration in human muscle cells - Lactic Acid Fermentation):
When there is a lack of oxygen, such as during strenuous exercise, human muscle cells undergo anaerobic respiration. Pyruvate formed during glycolysis is converted into lactic acid.
Equation: \[ C_6H_{12}O_6 (Glucose) \xrightarrow{Glycolysis in cytoplasm} 2 Pyruvate \xrightarrow{Lack of O_2 in cytoplasm} 2 Lactic Acid + Small amount of Energy (ATP) \]
Specifically for the conversion of pyruvate: \[ Pyruvic acid \xrightarrow{Lack of O_2} Lactic acid + Energy \]
The overall from glucose would be: \[ C_6H_{12}O_6 (Glucose) \xrightarrow{In cytoplasm, lack of O_2} 2C_3H_6O_3 (Lactic Acid) + Energy (2 ATP) \] \[ \begin{tabular}{p{\linewidth}} \textbf{1. Glycolysis:} Cytoplasmic glucose breakdown
\textbf{2. Aerobic:} C\(_6\)H\(_{12}\)O\(_6\) + 6O\(_2\) \(\rightarrow\) 6CO\(_2\) + 6H\(_2\)O + ATP
\textbf{3. Anaerobic:} C\(_6\)H\(_{12}\)O\(_6\) \(\rightarrow\) 2C\(_3\)H\(_6\)O\(_3\) (lactate) + 2ATP
\end{tabular} \] Quick Tip: Glycolysis is common to both aerobic and anaerobic respiration. Aerobic respiration yields significantly more ATP than anaerobic respiration. Anaerobic respiration in human muscles produces lactic acid, which can cause muscle fatigue.

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N/A Quick Tip: Balancing equations ensures the number of atoms of each element is conserved. Precipitation reactions are often double displacement reactions where one product is insoluble. Use solubility rules to predict precipitates (e.g., most chlorides are soluble, except AgCl, PbCl\(_2\), Hg\(_2\)Cl\(_2\)).

Decomposers:
Decomposers are organisms, primarily bacteria and fungi, that break down dead organic matter (like dead plants and animals, and waste products) into simpler inorganic substances. They obtain nutrients for their own growth and release these simpler substances back into the ecosystem.

Two Examples of Decomposers:
1. Bacteria (e.g., Bacillus subtilis, \textit{Pseudomonas species)
2. Fungi (e.g., Mushrooms, Molds like \textit{Aspergillus, Yeast)

How Decomposers Maintain Balance in an Ecosystem:
Decomposers play a crucial role in maintaining the balance in an ecosystem in several ways:
1. Nutrient Cycling: They break down complex organic compounds in dead organisms into simpler inorganic nutrients (like nitrates, phosphates, sulfates, carbon dioxide). These nutrients are released back into the soil, water, and air, making them available for producers (plants) to use again. This recycling of nutrients is essential for the continuous functioning of the ecosystem, ensuring that essential elements are not locked up in dead matter.
2. Cleaning the Environment: By decomposing dead organic matter and waste, decomposers prevent the accumulation of dead bodies and waste products, thus keeping the environment clean and preventing the spread of diseases.
3. Soil Fertility: The decomposition process enriches the soil with humus and essential nutrients, improving soil structure and fertility, which supports plant growth (the base of most food webs).
Without decomposers, nutrients would remain locked in dead organisms, and ecosystems would eventually collapse due to a lack of available nutrients for producers. \[ \begin{tabular{p{\linewidth}} \textbf{Decomposers:} Microorganisms (bacteria/fungi) that mineralize organic matter
\textbf{Functions:}
• Convert detritus → inorganic nutrients
• Clean ecosystems by waste breakdown
• Enhance soil structure \& fertility
\end{tabular} \] Quick Tip: Decomposers are the "recyclers" of the ecosystem. They are saprophytes, meaning they feed on dead and decaying organic matter. Their activity is vital for biogeochemical cycles (e.g., carbon cycle, nitrogen cycle).

Definition of Potential Difference:
Potential difference (V) between two points in an electric circuit carrying current is defined as the work done (W) to move a unit positive charge (q) from one point to the other.
Mathematically, \(V = \frac{W}{q}\).

S.I. Unit of Potential Difference:
The S.I. unit of potential difference is the Volt (symbol: V).

Definition of Volt:
One volt is defined as the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
So, \(1 Volt = \frac{1 Joule}{1 Coulomb}\) or \(1 V = 1 J/C\).

Expression in terms of S.I. unit of work and charge:
As per the definition, potential difference (V) is work done (W) per unit charge (q).
The S.I. unit of work is the Joule (J).
The S.I. unit of charge is the Coulomb (C).
Therefore, the S.I. unit of potential difference (Volt) can be expressed as: \[ Volt (V) = \frac{Joule (J)}{Coulomb (C)} \] \[ [\(\bullet\)] \textbf{Definition:} Work done per unit charge [\(\bullet\)] \textbf{Mathematical Form:} \(\Delta V = \frac{W}{q}\) [\(\bullet\)] \textbf{S.I. Unit:} Volt (V) [\(\bullet\)] \textbf{Dimensional Formula:} [ML\(^2\)T\(^{-3}\)I\(^{-1}\)] [\(\bullet\)] \textbf{Base Unit Equivalence:} 1 V = 1 J/C \] Quick Tip: Potential difference is also known as voltage. It is the "electrical pressure" or "force" that causes current to flow. Remember the relationship: \(V = W/q\).

Two Limitations of Electrical Impulses in Multicellular Organisms:
1. Limited Reach/Specificity: Electrical impulses can only reach cells that are directly connected by neurons (nerve cells). They cannot reach every single cell in the body. Also, while fast, the transmission is limited to the neuronal network.
2. Refractory Period/Cannot be Continuously Generated: Once an electrical impulse is generated in a neuron and transmitted, the neuron takes some time (refractory period) to reset before it can generate and transmit another impulse. This limits the frequency of impulse transmission and prevents continuous stimulation.

Why Chemical Communication is Better than Electrical Impulses (in some aspects):
Chemical communication (e.g., via hormones) is often considered "better" or more advantageous than electrical impulses for certain types of communication in multicellular organisms due to the following reasons:
1. Wider Reach and Broader Effects: Chemical signals (hormones) are transported through the bloodstream and can reach all cells in the body. This allows for widespread and coordinated responses across different organs and tissues, even those not directly innervated.
2. Persistent and Long-Lasting Effects: Chemical signals can have more sustained and longer-lasting effects compared to the transient nature of electrical impulses. Hormones can regulate slow, ongoing processes like growth, development, and metabolism over extended periods.
3. Specificity through Receptors: While chemical signals can reach many cells, they typically only affect target cells that possess specific receptors for them. This allows for targeted responses despite widespread distribution.
4. No Refractory Period for the Signal Itself: Unlike neurons, glands can often release chemical signals continuously or in pulses as needed, without the same kind of immediate refractory limitation on signal generation, allowing for more varied patterns of stimulation.
5. Graded Responses: The intensity of response to chemical signals can often be graded depending on the concentration of the hormone, allowing for finer control over physiological processes.

While electrical impulses are extremely fast and ideal for rapid, precise responses (like reflexes or muscle control), chemical communication complements this by providing a slower, more widespread, and often longer-lasting mode of regulation essential for many physiological processes. \[ [\(\circ\)] \textbf{Electrical Signaling:} Restricted to directly connected neurons Recovery period required (refractory phase) [\(\circ\)] \textbf{Chemical Signaling:} Paracrine/endocrine distribution Longer-lasting postsynaptic effects Ligand-receptor specificity Diverse modulation possibilities Response proportional to signal \] Quick Tip: Electrical impulses (nervous system) are fast, targeted, and short-lived. Chemical signals (endocrine system - hormones) are generally slower, widespread (via bloodstream), and can have longer-lasting effects. Both systems are crucial and work together to coordinate bodily functions.

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N/A Quick Tip: For 34(A)(i): Use Right-Hand Thumb Rule for each wire. At midpoint between wires with opposite currents, fields add up. For 34(A)(ii): Live wire (red/brown) at high potential, Neutral (black/blue) at zero potential, Earth (green/yellow-green) for safety. For 34(B): Force on a current-carrying conductor in a magnetic field is the basis of electric motors. Fleming's Left-Hand Rule helps determine the direction of this force. Force is maximum when current and magnetic field are perpendicular.

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N/A Quick Tip: For 35(A): Pollination \(\rightarrow\) Fertilization \(\rightarrow\) Ovule to Seed \(\rightarrow\) Ovary to Fruit. For 35(B): Placenta is the lifeline between mother and fetus. If no fertilization, the uterine lining sheds (menstruation).

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N/A Quick Tip: For 36(A): Work backwards from combustion products. Dehydration of alcohols gives alkenes. Hydrogenation of alkenes gives alkanes. For 36(B): Ethanoic acid is C\(_2\)H\(_4\)O\(_2\). Esterification: Acid + Alcohol \(\rightleftharpoons\) Ester + Water. Hydrolysis is the reverse. Acids react with carbonates to give CO\(_2\).

(i) Give any one more instance in which this type of spectrum is observed.
One more instance where a spectrum similar to that produced by a prism is observed is a rainbow formed in the sky after rainfall when sunlight passes through raindrops. (Other examples: spectrum formed by a CD, soap bubbles).

(ii) What happens to white light in the above case (passing through a prism)?
When white light passes through a glass prism, it undergoes dispersion. This means the white light splits into its constituent colors (the spectrum, typically VIBGYOR - Violet, Indigo, Blue, Green, Yellow, Orange, Red). This happens because different colors of light bend by different amounts as they pass through the prism, with violet light bending the most and red light bending the least.

(iii) (A) List two conditions necessary to observe a rainbow.
Two conditions necessary to observe a rainbow are:
1. Presence of Raindrops: There must be water droplets (rain, mist, or spray) suspended in the air.
2. Position of the Sun and Observer: The sun must be shining, and the observer must be positioned with their back towards the sun, looking towards the region with raindrops. The sun should typically be relatively low in the sky.

OR

(iii) (B) Draw a ray diagram to show the formation of a rainbow. Mark on it, points (a), (b) and (c) as given below :
(a) Where dispersion of light occurs.
(b) Where light gets reflected internally.
(c) Where final refraction occurs.
(A diagram should be drawn here. Description below.)
Diagram Description (Single raindrop acting as a prism):

Draw a large circle representing a spherical raindrop.
Draw a ray of sunlight entering the raindrop from one side (e.g., left).
(a) Where dispersion of light occurs: As the sunlight enters the raindrop (at the air-water interface), it refracts and disperses into its constituent colors. Show the white light splitting into at least two colors (e.g., red on top, violet at bottom, relative to the path inside). Mark this point of entry and splitting as (a).
The dispersed rays travel through the raindrop to the opposite inner surface.
(b) Where light gets reflected internally: At this inner surface, the light undergoes total internal reflection (or partial reflection that contributes significantly to the primary rainbow). Show the rays reflecting off this back surface. Mark this point of reflection as (b).
The reflected rays travel to another point on the front surface of the raindrop.
(c) Where final refraction occurs: As the light rays emerge from the raindrop back into the air (at the water-air interface), they undergo another refraction, bending away from the normal and further separating the colors. Mark this point of exit and final refraction as (c).
Show the emergent rays (e.g., red and violet) directed towards an observer's eye (observer not usually shown, but implied direction).

% Image placeholder for rainbow formation:
\[ [\(\bullet\)] \textbf{(i) Observation: Circular colored arc (42° radius) Primary rainbow (double reflection creates secondary) [\(\bullet\)] \textbf{(ii) Key Mechanism:} Wavelength-dependent refraction (Snell's Law) Angular separation: \(\delta\lambda \approx 1.5°\) (red-violet) [\(\bullet\)] \textbf{(iii) Necessary Conditions:} A: Spherical water droplets Anti-solar point alignment B: (a) Refraction at air-water interface (b) 138° total internal reflection (c) Final refraction (spectrum visible) \] Quick Tip: Dispersion is the splitting of white light due to different refractive indices for different colors. Rainbows are formed by dispersion, refraction, and total internal reflection of sunlight by raindrops. For a primary rainbow, there is one internal reflection. Red light is on the outer arc, violet on the inner.

(i) Name the acid and the base from which common salt can be obtained.
Common salt (Sodium chloride, NaCl) can be obtained from the reaction of:

Acid: Hydrochloric acid (HCl)
Base: Sodium hydroxide (NaOH)

Reaction: HCl(aq) + NaOH(aq) \(\rightarrow\) NaCl(aq) + H\(_2\)O(l)

(ii) State the nature (acidic/basic/neutral) of sodium chloride. Give reason for the justification for your answer.
Sodium chloride (NaCl) is a neutral salt.
Reason: Sodium chloride is formed from the neutralization reaction between a strong acid (HCl) and a strong base (NaOH). Salts formed from a strong acid and a strong base are neutral because the ions formed (Na\(^+\) from NaOH and Cl\(^-\) from HCl) do not hydrolyze significantly in water to produce an excess of H\(^+\) or OH\(^-\) ions. Therefore, an aqueous solution of NaCl has a pH of approximately 7.

(iii) (A) What happens when electric current is passed through an aqueous solution of sodium chloride (called brine)? Name the products obtained along with the corresponding places in the electrolytic cell where each of these products is obtained.
When electric current is passed through an aqueous solution of sodium chloride (brine), it undergoes electrolysis. This process is known as the Chlor-alkali process.
The products obtained are:
1. Sodium hydroxide (NaOH): Obtained near the cathode.
2. Chlorine gas (Cl\(_2\)): Obtained at the anode.
3. Hydrogen gas (H\(_2\)): Obtained at the cathode.
Equation for electrolysis of brine: \[ 2NaCl(aq) + 2H_2O(l) \xrightarrow{Electrolysis} 2NaOH(aq) + Cl_2(g) + H_2(g) \]
At cathode: \(2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)\) (and Na\(^+\) ions combine with OH\(^-\) to form NaOH)
At anode: \(2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-\)

OR

(iii) (B) How is washing soda obtained from sodium chloride? Give chemical equation of the reactions involved in the process.
Washing soda (Sodium carbonate decahydrate, Na\(_2\)CO\(_3\)\(\cdot\)10H\(_2\)O) is obtained from sodium chloride using the Solvay process (or Ammonia-Soda process).
The main steps involved are:
1. Formation of Ammoniacal Brine: Ammonia gas is passed through a saturated solution of sodium chloride (brine) to get ammoniacal brine.
2. Carbonation of Ammoniacal Brine: Carbon dioxide gas is passed through the ammoniacal brine. Sodium bicarbonate (NaHCO\(_3\)), being sparingly soluble, precipitates out.
\[ NaCl(aq) + NH_3(g) + H_2O(l) + CO_2(g) \rightarrow NaHCO_3(s) + NH_4Cl(aq) \]
(Sodium bicarbonate)
3. Filtration and Calcination of Sodium Bicarbonate: The precipitated sodium bicarbonate is filtered and then heated (calcined) to obtain anhydrous sodium carbonate (soda ash).
\[ 2NaHCO_3(s) \xrightarrow{Heat} Na_2CO_3(s) + H_2O(l) + CO_2(g) \]
(Soda ash)
4. Recrystallization to form Washing Soda: Anhydrous sodium carbonate (soda ash) is dissolved in water and recrystallized to get crystals of washing soda (sodium carbonate decahydrate).
\[ Na_2CO_3(s) + 10H_2O(l) \rightarrow Na_2CO_3 \cdot 10H_2O(s) \]
(Washing soda)
(Note: The Solvay process also involves recovery of ammonia from ammonium chloride). \[ [\(\bullet\)] \textbf{(i) Reactants:} HCl (pH <7, proton donor) NaOH (pH >7, proton acceptor) [\(\bullet\)] \textbf{(ii) Neutralization Product:} NaCl (pH-neutral salt) No hydrolysis occurs [\(\bullet\)] \textbf{(iii) Industrial Methods:} [A] \textbf{Chlor-alkali:} Cathode: 2H\(_2\)O + 2e\(^-\) \(\rightarrow\) H\(_2\) + 2OH\(^-\) Anode: 2Cl\(^-\) \(\rightarrow\) Cl\(_2\) + 2e\(^-\) [B] \textbf{Solvay Process:} NH\(_3\) + H\(_2\)O + CO\(_2\) \(\rightarrow\) NH\(_4^+\) + HCO\(_3^-\) Na\(^+\) + HCO\(_3^-\) \(\rightarrow\) NaHCO\(_3\)↓ Thermal decomposition to soda ash \] Quick Tip: Common salt (NaCl) is neutral. Electrolysis of brine (Chlor-alkali process) produces NaOH, Cl\(_2\), and H\(_2\). Washing soda (Na\(_2\)CO\(_3\)\(\cdot\)10H\(_2\)O) is manufactured by the Solvay process, starting with NaCl.

(i) Name the action by which we protect ourself in the situation mentioned above and define it.
The action is called a Reflex Action.
Definition: A reflex action is a rapid, automatic, and involuntary response to a stimulus, which is not under the conscious control of the brain. It is mediated by a neural pathway called a reflex arc.

(ii) Write the role of (a) motor and (b) relay neuron.

(a) Motor Neuron (Efferent Neuron): Its role is to transmit nerve impulses from the central nervous system (brain or spinal cord) to an effector organ (like a muscle or a gland) to bring about a response. For example, causing a muscle to contract.
(b) Relay Neuron (Interneuron or Association Neuron): Its role is to connect sensory neurons to motor neurons within the central nervous system (spinal cord or brain). It processes information from the sensory neuron and transmits the impulse to the appropriate motor neuron, often involved in reflex arcs and complex neural pathways.


(iii) (A) What are the two types of nervous system in human body? Name the components of each of them.
The two main types (divisions) of the nervous system in the human body are:
1. Central Nervous System (CNS):
Components:

Brain
Spinal Cord

2. Peripheral Nervous System (PNS):
Components:

Nerves arising from the brain (cranial nerves) and spinal cord (spinal nerves).
The PNS can be further divided into the Somatic Nervous System (controls voluntary movements) and the Autonomic Nervous System (controls involuntary functions), which in turn has Sympathetic and Parasympathetic divisions.


OR

(iii) (B) Which part of the human brain is responsible for :
(a) thinking
The Cerebrum (specifically the cerebral cortex) is primarily responsible for thinking, reasoning, memory, intelligence, and other higher cognitive functions.
(b) picking up a pencil
Picking up a pencil involves voluntary muscle movement coordinated by the Cerebellum (for precision, coordination, and balance) and initiated by the motor areas of the Cerebrum.
(c) controlling blood pressure
Controlling blood pressure (an involuntary action) is primarily regulated by the Medulla Oblongata (part of the hindbrain/brainstem).
(d) controlling hunger
The Hypothalamus (part of the forebrain, located below the thalamus) is responsible for controlling hunger, thirst, and body temperature. \[ [\(\bullet\)] \textbf{(i) Reflex Characteristics:} No conscious thought Spinal cord-mediated Protective function [\(\bullet\)] \textbf{(ii) Neural Circuits:} Motor: Multipolar, long axons Relay: Short, CNS-localized [\(\bullet\)] \textbf{(iii) System Architecture:} [A] \textbf{Structural Division:} CNS: Grey/white matter PNS: Somatic/Autonomic [B] \textbf{Functional Centers:} a) Cognition/Memory (cerebrum) b) Movement precision (cerebellum) c) Vital reflexes (medulla) d) Neuroendocrine (hypothalamus) \] Quick Tip: Reflex arcs involve a sensory receptor, sensory neuron, relay neuron (in CNS), motor neuron, and effector. Cerebrum: Largest part, higher functions. Cerebellum: Coordination, balance. Medulla: Involuntary vital functions. Hypothalamus: Homeostasis, drives.