Correct Answer: (A) (a) and (b)

Solution:

In a homologous series:

• Statement (a): Consecutive members differ by a –CH₂ unit (True).
• Statement (b): Melting and boiling points generally increase with molecular mass due to stronger intermolecular forces (True).
• Statement (c): The difference in molecular mass between successive members is 14 u (not 16 u). Hence, this statement is False.
• Statement (d): C₂H₂ (ethyne) and C₃H₄ (propyne) are successive members of the alkyne series. Hence, this statement is False.

Thus, only statements (a) and (b) are correct.

Correct Answer: (D) 4

Solution:

Potassium has an atomic number of 19, meaning it has 19 electrons. The electronic configuration is written as:

Shell 1: 2 electrons (K-shell)
Shell 2: 8 electrons (L-shell)
Shell 3: 8 electrons (M-shell)
Shell 4: 1 electron (N-shell)

Hence, 4 shells are required for its configuration: 2, 8, 8, 1.

Correct Answer: (B) Burning of coal

Solution:

Burning of coal involves carbon (C) combining with oxygen (O₂) to form carbon dioxide (CO₂), which is a single product:

This is a combination reaction.

Correct Answer: (B) Al₂O₃

Solution:

Aluminum oxide (Al₂O₃) is an amphoteric oxide, reacting with acids and bases:

• With HCl: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
• With KOH: Al₂O₃ + 2KOH → 2KAlO₂ + H₂O

Correct Answer: (D) Bronze

Solution:

Bronze is an alloy primarily composed of copper and tin. Other options:

• Nichrome: Nickel, chromium, and iron.
• Brass: Copper and zinc.
• Constantan: Copper and nickel.

Correct Answer: (A) 5.1

Solution:

Tooth decay begins when the pH of the mouth falls below 5.5. Bacteria in the mouth produce acids that dissolve the enamel of the teeth. Among the given options, 5.1 is the pH value below 5.5.

Correct Answer: (D) is exothermic and pH of the solution formed is more than 7.

Solution:

The reaction of calcium oxide (CaO) with water is:

• Exothermic: The liberation of heat indicates that the reaction is exothermic.
• pH > 7: Calcium hydroxide (Ca(OH)₂) is a base. Its solution releases OH^-, making the pH greater than 7 (alkaline).

Correct Answer: (B) lifted ribs and flattened diaphragm.

Solution:

During inhalation:

• Ribs: The rib cage expands, and the ribs are lifted upwards and outwards due to intercostal muscle contraction.
• Diaphragm: The diaphragm contracts and flattens, increasing the thoracic cavity's volume.

The increased chest volume reduces air pressure in the lungs, causing air to flow in.

Correct Answer: (A) Pituitary

Solution:

Pituitary Gland: Located at the brain's base, it is a single structure. In contrast:

• Ovaries: Two ovaries are present in females.
• Testes: Two testes are present in males.
• Adrenal Glands: Two glands, one above each kidney.

Correct Answer: (C) (b) and (c)

Solution:

• (a): Incorrect. The right atrium receives deoxygenated blood from the body through the vena cava, not oxygenated blood.
• (b): Correct. The left atrium pumps oxygenated blood to the left ventricle, which then sends it to the body via the aorta.
• (c): Correct. The right atrium receives deoxygenated blood from the body through the superior and inferior vena cava.
• (d): Incorrect. The left atrium pumps blood to the left ventricle, not directly to the aorta.

Correct Answer: (D) Rhizopus

Solution:

The diagram depicts a fungus with sporangia that contain spores. This is characteristic of Rhizopus, a common bread mold.

• Spirogyra: A filamentous green alga, unrelated to the diagram.
• Planaria: A flatworm, not a fungus.
• Yeast: A single-celled fungus reproducing by budding, not sporangia.
• Rhizopus: Produces spores in sporangia, matching the diagram.

Correct Answer: (C) Tt and tt

Solution:

Using Punnett squares:

• (A) Tt x Tt: Results in 75% tall, 25% short.
• (B) TT x Tt: Results in 100% tall.
• (C) Tt x tt: Results in 50% tall (Tt) and 50% short (tt), matching the observation.
• (D) TT x tt: Results in 100% tall (Tt).

Thus, the parental genotypes must be Tt (heterozygous tall) and tt (homozygous short).

Correct Answer: (B) direction of the current flowing through it

Solution:

The strength of a solenoid's magnetic field depends on:

• Number of turns: More turns create a stronger field.
• Radius: A smaller radius increases field strength.
• Core material: A ferromagnetic core enhances the field.

The direction of current determines the direction of the field (using the right-hand rule) but not its strength.

Correct Answer: (C) ohm • metre

Solution:

Resistivity formula:

Units:

• R = ohms (Ω)
• A = m²
• l = m

Resulting unit: ρ = Ω • m (ohm-meter).

Correct Answer: (C) 2 Ω

Solution:

For minimum resistance, connect all resistors in parallel:

For 5 resistors of 10 Ω each:

R_eq = 2 Ω

Correct Answer: (A) (a) and (b)

Solution:

(a): True. The human eyeball's diameter is approximately 2.3 cm.

(b): True. The iris controls the amount of light entering the eye by adjusting the size of the pupil.

(c): False. Most refraction occurs at the cornea, not the crystalline lens.

(d): False. The distance between the lens and the retina does not change. The ciliary muscles alter the lens's shape for focusing.

Therefore, only statements (a) and (b) are correct.

Correct Answer: (D) (A) is false, but (R) is true.

Solution:

Assertion (A): False. Deflection of a compass needle increases as the current increases because a stronger magnetic field is produced.

Reason (R): True. The strength of the magnetic field around a current-carrying conductor is directly proportional to the current's magnitude.

Correct Answer: (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Solution:

(A): True. Human females have two identical sex chromosomes (XX).

(R): True. The sex chromosome contributed by the male (X or Y) determines the child's sex. However, this is unrelated to females having a "perfect pair" of sex chromosomes.

Correct Answer: (C) (A) is true, but (R) is false.

Solution:

(A): True. Myopia causes light to focus in front of the retina, blurring distant objects.

(R): False. Myopia is corrected using diverging (concave) lenses, not converging lenses.

Correct Answer: (C) (A) is true, but (R) is false.

Solution:

(A): True. Moderately reactive metals are often found as sulfides or carbonates in ores.

(R): False. The process described (calcination for carbonate ores and roasting for sulphide ores) does not explain why these metals are found as sulphides or carbonates.

Solution:

A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more simpler substances. This breakdown can occur due to heat (thermal decomposition), light (photodecomposition), or electricity (electrolysis).

The thermal decomposition of ferrous sulphate crystals (FeSO₄·7H₂O) is shown by the equation:

2FeSO₄·7H₂O(s) → Fe₂O₃(s) + SO₂(g) + SO₃(g) + 14H₂O(g)

Ferrous sulphate heptahydrate (green crystals) decomposes upon heating to produce ferric oxide (reddish-brown solid), sulfur dioxide, sulfur trioxide (gases), and water vapor.

Solution:

A balanced chemical equation is a representation of a chemical reaction in which the number of atoms of each element is equal on both sides of the equation. This is achieved by adjusting the coefficients (numbers in front of chemical formulas) without changing the actual chemical formulas of the reactants or products.

Why it is necessary:

• A balanced equation reflects the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.
• It ensures that the total mass of reactants equals the total mass of products, maintaining the equality of atoms on both sides.
• Balancing chemical equations is essential for accurately predicting the amounts of reactants and products involved in a reaction.

(a) Test tube B will show a color change. Iodine solution turns blue-black in the presence of starch.

(b) Saliva contains the enzyme amylase, which breaks down starch into simpler sugars. In test tube A, the amylase in the saliva digests the starch. Therefore, when iodine is added, there is no starch left to react with, and no color change occurs. Test tube B, which did not receive saliva, still contains starch, and thus turns blue-black when iodine is added. This demonstrates the digestive action of saliva on starch.

Solution:

The two types of germ cells (also known as gametes or sex cells) in human beings are:

• Sperm cells (spermatozoa): Male germ cells.
• Egg cells (ova): Female germ cells.

Two structural differences between sperm and egg cells:

• Size: Egg cells are significantly larger than sperm cells. The egg cell is one of the largest cells in the human body, while sperm cells are tiny and motile.
• Motility: Sperm cells are motile; they have a flagellum (tail) that allows them to swim. Egg cells are non-motile and are transported by cilia and muscle contractions in the female reproductive tract.

The two laws of refraction of light are:

1. The incident ray, the refracted ray, and the normal to the interface at the point of incidence, all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media. This constant is known as the refractive index of the second medium with respect to the first. (Snell's Law: sin i / sin r = μ)

The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in that medium.

μ = c / v

Where:

• μ is the absolute refractive index
• c is the speed of light in vacuum (3 × 10⁸ m/s)
• v is the speed of light in the medium

Given that the absolute refractive index of glass is 1.5, we can find the speed of light in glass:

1.5 = (3 × 10⁸) / v

v = (3 × 10⁸) / 1.5 = 2 × 10⁸ m/s

Therefore, the speed of light in glass is 2 × 10⁸ m/s.

Solution:

Given:

Resistances in series:
R_s = R₁ + R₂ + R₃ = 1 Ω + 2 Ω + 3 Ω = 6 Ω

Current through the circuit:
I = V / R = 2 V / 6 Ω = 1/3 A

Potential difference across the 3 Ω resistor:
V = I × R = (1/3 A) × 3 Ω = 1 V

The term used for materials that cannot be broken down by biological processes is non-biodegradable or persistent materials.

Two ways non-biodegradable materials harm various components of an ecosystem are:

1. Pollution: Non-biodegradable substances accumulate in the environment, causing pollution of soil, water, and air. This can harm organisms directly (e.g., plastic ingestion by marine animals) or indirectly by affecting the quality of their habitat.
2. Disruption of Food Chains: Non-biodegradable materials can enter the food chain, accumulating in organisms at higher trophic levels through biomagnification. This can lead to toxicity and disruption of food webs.

Solution:

1. Alveoli are richly supplied with blood capillaries to facilitate efficient gas exchange. The extensive capillary network maximizes the surface area for the diffusion of oxygen from the alveoli into the blood and carbon dioxide from the blood into the alveoli. A large surface area and short diffusion distance enable rapid gas exchange.
2. Respiratory pigment (hemoglobin) in human blood has a much higher affinity for oxygen than for carbon dioxide. Hemoglobin readily binds oxygen in the lungs (where oxygen partial pressure is high) and releases it in the tissues (where oxygen partial pressure is low). While hemoglobin does carry some carbon dioxide, its primary role is oxygen transport.
3. During anaerobic respiration (in the absence of oxygen), pyruvate (a 3-carbon molecule) is converted to lactic acid in human muscle cells. This is an alternative metabolic pathway to aerobic respiration, where pyruvate is completely oxidized to carbon dioxide in the presence of oxygen. Lactic acid buildup causes muscle fatigue.

Solution:

1. Movements in the gut: The movements that occur all along the gut in the human digestive system are called peristalsis. Peristalsis involves rhythmic contractions and relaxations of the smooth muscles in the walls of the digestive tract, propelling food through the gut.
   • Mixing: The churning action mixes food with digestive enzymes, ensuring efficient digestion.
   • Movement: Peristalsis moves food through the digestive tract, allowing it to be processed in each section.
2. Storage of bile juice: Bile juice is stored in the gallbladder.
   • Emulsification of fats: Bile salts break down large fat globules into smaller droplets, increasing the surface area available for the action of lipases (fat-digesting enzymes).
   • Absorption of fat-soluble vitamins: Bile helps in the absorption of fat-soluble vitamins (A, D, E, and K) in the small intestine.

Solution:

Fertilisation in angiosperms is highly dependent on pollination. Pollination involves the transfer of pollen grains from the anther to the stigma of a flower. If pollination does not occur, the male gametes (sperm cells) cannot reach the ovules to fuse with the female gametes (egg cells), making fertilisation impossible.

After fertilisation, the zygote is located inside the ovule within the ovary. The zygote undergoes multiple cell divisions to form an embryo. Over time, the ovule develops into a seed, and the ovary matures into a fruit.

Additional Information: Angiosperms exhibit two types of pollination: self-pollination and cross-pollination. Cross-pollination often involves external agents such as wind, water, or pollinators like bees. Successful fertilisation triggers post-fertilisation changes, including seed and fruit formation.

Solution:

The anther produces pollen grains, which contain the male gametes (sperm cells), similar to how testes produce sperm in animals.

The ovary in a flower contains ovules, which are fertilised by male gametes to form seeds. This is similar to the function of the ovary in animals, which produces eggs.

1. Testis: Anther
2. Ovary: Ovary

Solution:

When an electric current is passed through acidulated water:

1. Gas bubbles are observed at both electrodes.
2. The volume of gas collected at the cathode (hydrogen gas) is approximately twice that at the anode (oxygen gas).

Additional Information: This process is called electrolysis. Acidulated water (water with dilute sulfuric acid) is used to enhance conductivity. At the cathode, hydrogen ions gain electrons to form hydrogen gas. At the anode, hydroxide or oxygen ions lose electrons to form oxygen gas.

The ratio of the mass of gas collected at the cathode to the mass of the gas collected at the anode is approximately 1:8. This is based on the molar masses of hydrogen (2 g/mol) and oxygen (32 g/mol), with hydrogen gas being lighter.

Solution:

Diagram:

Key Components:

• Anode: Impure copper rod.
• Cathode: Thin sheet of pure copper.
• Electrolyte: Copper(II) sulfate solution (CuSO₄(aq)).
• Power Source: A direct current (DC) power source.

Process:

When a DC current passes through the electrolyte:

• At the anode (impure copper): Copper atoms lose electrons and enter the solution as Cu²⁺ ions: Cu(s) → Cu²⁺(aq) + 2e^-.
• At the cathode (pure copper): Cu²⁺ ions gain electrons and deposit as pure copper: Cu²⁺(aq) + 2e^- → Cu(s).
• Impurities settle as anode mud or dissolve into the solution depending on their reactivity.

Solution:

The mirror formula is:

1/f = 1/v + 1/u

• f: -12 cm (concave mirror).
• u: -8 cm (object distance).

Substituting values:

1/(-12) = 1/v + 1/(-8)

1/v = -1/12 + 1/8

1/v = (−2 + 3)/24 = 1/24

v = 24 cm.

The image is real, inverted, magnified, and formed 24 cm on the same side as the object.

Labelled Ray Diagram:

Solution:

Magnification (m) = -1:

m = -v/u

Substitute values:

-1 = -30/u

u = -30 cm.

The object is placed 30 cm from the mirror. Magnification -1 indicates equal distances for object and image, confirming the object is at 2F.

Solution:

New object distance: u' = -30 + 15 = -15 cm.

Using the mirror formula:

1/f = 1/v' + 1/u'

1/(-30) = 1/v' + 1/(-15)

1/v' = -1/30 + 2/30 = 1/30

v' = 30 cm.

The new image is real, inverted, and of the same size as the object. It forms 30 cm from the mirror.

Labelled Ray Diagram:

Solution:

Fleming's Left-Hand Rule:

If you stretch the thumb, forefinger, and middle finger of your left hand such that they are mutually perpendicular, then:

• The forefinger points in the direction of the magnetic field.
• The middle finger points in the direction of the current.
• The thumb points in the direction of the force experienced by the conductor.

Application to the Given Scenario:

• The magnetic field is directed to the right (horizontal direction).
• The current flows vertically upwards through the conductor AB.

Using Fleming's left-hand rule, the thumb will point out of the plane of the paper. Hence, the force experienced by the conductor AB is directed outwards, perpendicular to both the magnetic field and the current.

Solution:

When an electron enters the magnetic field in the same direction as the current flowing through conductor AB, it experiences a force due to the interaction between its motion and the magnetic field. According to the concept of Lorentz force:

F = q(v × B)

Where:

• q is the charge of the particle (negative for an electron).
• v is the velocity of the electron.
• B is the magnetic field.

Explanation:

For an electron, the force direction will be opposite to the force experienced by a positive charge (like the current in the conductor). Since the current in AB flows upwards and the magnetic field is directed horizontally, the electron will experience a force directed into the plane of the paper (opposite to the force experienced by the conductor).

Reason: The force direction is determined by Fleming's left-hand rule. As the electron moves opposite to the conventional current, it experiences a force that is opposite to the force on the conductor. This causes the electron to deviate into the plane of the paper.

Solution:

The phenomenon involved is biomagnification (or biological magnification).

Biomagnification refers to the increasing concentration of a substance (like a pesticide) in organisms at successively higher trophic levels of a food chain. It happens because:

1. Pesticides in the Environment: Pesticides applied to crops can enter the environment through runoff, spray drift, etc.
2. Organisms at Lower Trophic Levels: Organisms at lower trophic levels (e.g., plants, primary consumers) absorb the pesticides from the environment. The concentration of the pesticide in these organisms may be relatively low.
3. Bioaccumulation: Organisms tend to accumulate pesticides in their tissues because they don't easily metabolize or excrete them.
4. Higher Trophic Levels: When an organism at a higher trophic level consumes multiple organisms from a lower trophic level, the accumulated pesticide concentration in the predator is amplified, leading to a higher concentration than in the prey. This process repeats at each trophic level.
5. Impact on Humans: Humans, being at a high trophic level in many food chains, are particularly vulnerable to the effects of biomagnification. High concentrations of pesticides in the human body can lead to various health problems.

Solution:

Image Characteristics (Half-Covered Lens):

• Position: At 2F on the right side of the lens (same as the uncovered lens).
• Nature: Real, inverted, and of the same size as the object. The image will be slightly dimmer due to half the lens being covered. The brightness will be half of the image formed by the full lens.

Difference with Uncovered Lens: If the lens is uncovered, the image will be brighter and sharper because all parts of the lens contribute to image formation.

Reason: When only the lower half of the lens is used, the rays from the top half of the object are blocked. However, sufficient rays from the bottom half of the object still refract through the lens to form a complete image. The intensity will be slightly less because fewer rays contribute, but the image's location and characteristics remain the same. This demonstrates that even when part of the lens is obscured, image formation is still possible, provided a significant portion of the lens remains functional. With a full lens, more light contributes to the image resulting in a brighter, sharper image.

Solution:

The lens formula is:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

where:

• $u\; =\; -30\; \backslash ,\; \backslash text\{cm\}$ (negative for object on the left side)
• $f\; =\; -15\; \backslash ,\; \backslash text\{cm\}$ (negative for a concave lens)
• $v\; =\; ?$ (image distance to be calculated)

Substituting the values:

$\frac{1}{v}-\frac{1}{\left(}=\frac{1}{\left(}$

Simplifying:

$\frac{1}{v}=-\frac{1}{15}-\frac{1}{30}$

$\frac{1}{v}=-\frac{3}{30}=-\frac{1}{10}$

$v\; =\; -10\; \backslash ,\; \backslash text\{cm\}$

Conclusion: The image is formed at a distance of 10 cm on the same side as the object (negative sign indicates virtual image).

Solution:

(i)

Carbon has four valence electrons. To achieve a stable octet, it would need to either gain four electrons (forming C⁴⁻) or lose four electrons (forming C⁴⁺). However:

• The energy required to remove four electrons is very high, making the formation of C⁴⁺ highly unfavorable.
• Similarly, gaining four electrons would lead to a highly unstable, negatively charged ion due to significant electron-electron repulsion.

Therefore, carbon avoids forming ionic compounds and instead shares its four electrons through covalent bonding, forming stable covalent compounds.

(ii)

A homologous series is a group of organic compounds with the same functional group and similar chemical properties. Members of a homologous series differ by a –CH₂ unit in their molecular formula.

For aldehydes (RCHO), two consecutive members are:

• Methanal (HCHO)
• Ethanal (CH₃CHO)

(iii)

The structure of cyclohexane (C₆H₁₂) is a six-membered carbon ring with each carbon atom bonded to two other carbon atoms and two hydrogen atoms.

Solution:

(i)

A commercially important carbon compound with a –OH functional group is ethanol. Its molecular formula is C₂H₅OH.

(ii)

Reactions of ethanol:

2C₂H₅OH(l) + 2Na(s) → 2C₂H₅ONa(aq) + H₂(g)

Product: Sodium ethoxide

C₂H₅OH(l) → C₂H₄(g) + H₂O(l)

Product: Ethene (dehydration reaction)

C₂H₅OH(l) + CH₃COOH(l) → CH₃COOC₂H₅(l) + H₂O(l)

Product: Ethyl ethanoate (an ester)

3C₂H₅OH(l) + 2K₂Cr₂O₇(aq) + 8H₂SO₄(aq) → 3CH₃COOH(aq) + 2Cr₂(SO₄)₃(aq) + 2K₂SO₄(aq) + 11H₂O(l)

Product: Ethanoic acid

1. Reaction with sodium metal:
2. Reaction with excess concentrated sulfuric acid:
3. Reaction with ethanoic acid in the presence of an acid catalyst (esterification):
4. Reaction with acidified potassium dichromate (oxidation):

Solution:

(i) Hormonal Co-ordination: Plants vs. Animals

• Transport: Plant hormones are transported through vascular tissues (xylem and phloem), while animal hormones travel through the bloodstream.
• Response Time: Plant responses are slower compared to animals due to different transport mechanisms.
• Effect: Plant hormones often have broader effects, while animal hormones target specific cells or tissues.

(ii) Brain Regions and Functions:

• Intelligence: Cerebrum (cerebral cortex) controls higher cognitive functions like memory, learning, and reasoning.
• Riding a Bicycle: Cerebellum coordinates muscle movements and balance.
• Vomiting: Medulla oblongata regulates this reflex action.
• Controlling Hunger: Hypothalamus monitors energy balance and regulates appetite.

(iii) Protection of Brain and Spinal Cord:

• Bony Encasement: The brain is protected by the cranium, and the spinal cord by the vertebral column.
• Meninges: Three protective membranes (dura mater, arachnoid mater, pia mater) cushion the brain and spinal cord.
• Cerebrospinal Fluid (CSF): This fluid acts as a shock absorber and minimizes impact from external forces.

Solution:

(i) Tropic Movements:

Tropic movements are directional growth responses of plants to external stimuli such as light (phototropism), gravity (geotropism), and touch (thigmotropism).

• Inhibits Growth: Abscisic acid (ABA)
• Promotes Cell Division: Cytokinins

(ii) Directional Movement of Tendrils:

In response to touch (thigmotropism), cells on the side of the tendril in contact with the support stop growing, while cells on the opposite side elongate. This differential growth causes the tendril to bend and coil around the support. The hormone responsible for this movement is auxin.

Solution:

(a) Two Properties of Heating Elements:

1. High Resistivity: Heating elements are made of materials with high electrical resistivity to ensure efficient conversion of electrical energy into heat.
2. High Melting Point: Heating elements must withstand high temperatures without melting or degrading, ensuring durability during operation.

(b) Two Properties of Electric Fuse:

1. Low Melting Point: The fuse wire has a low melting point, enabling it to melt and break the circuit under high current conditions.
2. High Conductivity: The fuse wire is made from a material with high conductivity to minimize resistance during normal operation.

(c) Principle and Functioning of an Electric Fuse:

An electric fuse works on the heating effect of electric current. When a high current flows through the fuse wire due to a fault (like a short circuit), the wire heats up. If the current exceeds the fuse's rated capacity, the heat generated is sufficient to melt the wire, breaking the circuit. This action prevents high current from reaching appliances, thus protecting them from damage.

Solution:

We use the formula for power:

P = V × I

Given:

• P = 1100 W
• V = 220 V

Calculating the current (I):

I = P / V = 1100 W / 220 V = 5 A

The current flowing in the circuit is 5 A. If a 5 A fuse is connected, it will be at its maximum rating. Any further increase in current will cause the fuse to melt and break the circuit, protecting the heater and other components from damage.

Solution:

(a) Acid and Base for Sodium Chloride Formation:

Sodium chloride (NaCl) is formed from the reaction of:

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

Reaction: HCl + NaOH → NaCl + H₂O

(b) Cation and Anion in Calcium Sulphate:

The cation in calcium sulphate (CaSO₄) is calcium (Ca²⁺), and the anion is sulphate (SO₄²⁻).

(c) Justification for Sodium Chloride and Washing Soda:

Sodium chloride (NaCl) and washing soda (Na₂CO₃) belong to the same family of salts because both contain the same cation, sodium (Na⁺). This classification is based on the common cation present in their chemical structure.

Solution:

(a) Regeneration in Cut Pieces of Planaria:

In Planaria A, all three pieces (L, M, and N) can regenerate into complete organisms. In Planaria B, both halves (O and P) can regenerate into complete organisms. Planaria have a remarkable capacity for regeneration due to their pluripotent stem cells.

(b) Example of Another Organism:

Another organism that exhibits regeneration is the starfish. A severed arm with part of the central disc can regenerate into a complete starfish. Hydra also shows similar regeneration capabilities.

(c) Development in Regeneration:

'Development' in regeneration refers to the process where the lost body part is regrown through cell proliferation, differentiation, and tissue reorganization. This involves the formation of a fully functional organ or body structure.