We know that $\text{Volume} = \frac{\text{Mass}}{\text{Density}}$.

1. Initial volume of $m\text{ gm}$ of ice is:
$$V_{\text{ice}} = \frac{m}{x}\text{ cc}$$
2. Final volume of the resulting water after melting is:
$$V_{\text{water}} = \frac{m}{y}\text{ cc}$$
3. The change in volume is:
$$\Delta V = V_{\text{water}} – V_{\text{ice}} = \frac{m}{y} – \frac{m}{x} = m\left(\frac{1}{y} – \frac{1}{x}\right)\text{ cc}$$

According to Archimedes’ principle, the apparent loss in weight of a submerged body is equal to the weight of the water displaced by it.

1. Apparent loss of weight in water is:
$$\text{Loss of Weight} = W_{\text{air}} – W_{\text{water}} = 60\text{ N} – 40\text{ N} = 20\text{ N}$$
2. The specific gravity (relative density) of the material is given by:
$$\text{Specific Gravity} = \frac{\text{Weight in Air}}{\text{Apparent Loss of Weight in Water}} = \frac{60\text{ N}}{20\text{ N}} = 3$$

Using Hooke’s Law and the definition of Young’s Modulus ($Y$):
$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} \implies F = Y \cdot A \cdot \frac{\Delta L}{L}$$
Given:
* Diameter $d = 4\text{ mm} \implies \text{Radius } r = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$
* Area $A = \pi r^2 = \pi (2 \times 10^{-3})^2 = 4\pi \times 10^{-6}\text{ m}^2$
* Strain $\frac{\Delta L}{L} = 0.1\% = \frac{0.1}{100} = 10^{-3}$
* $Y = 9 \times 10^{10}\text{ N/m}^2$

Substituting these values:
$$F = (9 \times 10^{10}) \times (4\pi \times 10^{-6}) \times 10^{-3} = 360\pi\text{ N}$$

The elastic potential energy stored in a spring is directly proportional to the square of its elongation:
$$U = \frac{1}{2} k x^2 \implies U \propto x^2$$
1. For the first stretch ($x_1 = 2\text{ cm}$):
$$V = \frac{1}{2} k (2)^2 = 2k$$
2. For the second stretch ($x_2 = 10\text{ cm}$):
$$U_2 = \frac{1}{2} k (10)^2 = 50k$$
3. Comparing the two equations:
$$\frac{U_2}{V} = \frac{50k}{2k} = 25 \implies U_2 = 25V$$

When two charged conductors are connected by a conducting wire, charges redistribute until they reach a common potential ($V_{\text{common}}$):
$$V_{\text{common}} = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{Q_1 + Q_2}{C_1 + C_2}$$
1. Total charge:
$$Q_1 + Q_2 = 150\ \mu\text{C} + 150\ \mu\text{C} = 300\ \mu\text{C} = 3 \times 10^{-4}\text{ C}$$
2. Total capacitance of two isolated spherical conductors is:
$$C_1 + C_2 = 4\pi\varepsilon_0 R_1 + 4\pi\varepsilon_0 R_2 = 4\pi\varepsilon_0 (R_1 + R_2)$$
Given $R_1 = 0.2\text{ m}$ and $R_2 = 0.1\text{ m}$, and using $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2$:
$$C_1 + C_2 = \frac{0.2 + 0.1}{9 \times 10^9} = \frac{0.3}{9 \times 10^9} = \frac{1}{3 \times 10^{10}}\text{ F}$$
3. Calculating the common potential:
$$V_{\text{common}} = \frac{3 \times 10^{-4}}{\frac{1}{3 \times 10^{10}}} = 3 \times 10^{-4} \times 3 \times 10^{10} = 9 \times 10^6\text{ V}$$

1. First, calculate the capacitance ($C$) of the parallel plate air capacitor:
$$C = \frac{\varepsilon_0 A}{d}$$
Given:
* Area $A = 400\text{ cm}^2 = 400 \times 10^{-4}\text{ m}^2 = 0.04\text{ m}^2$
* Separation $d = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$
* Permittivity of free space $\varepsilon_0 \approx 8.854 \times 10^{-12}\text{ F/m}$

$$C = \frac{8.854 \times 10^{-12} \times 0.04}{2 \times 10^{-3}} = 1.7708 \times 10^{-10}\text{ F}$$
2. Now, find the charge ($Q$) stored on the capacitor plates under $V = 200\text{ V}$:
$$Q = C \times V = (1.7708 \times 10^{-10}\text{ F}) \times 200\text{ V} = 3.5416 \times 10^{-8}\text{ C}$$

When alkyne vapors are passed through a **red hot iron tube** at around $873\text{ K}$, they undergo cyclic trimerisation:

* Acetylene ($\text{HC}\equiv\text{CH}$) trimerises to form **Benzene** ($\text{C}_6\text{H}_6$).
* Propyne ($\text{CH}_3-\text{C}\equiv\text{CH}$) trimerises such that three methyl groups arrange symmetrically around the benzene ring, producing **Mesitylene** ($1,3,5\text{-trimethylbenzene}$):
$$3\text{ CH}_3\text{-C}\equiv\text{CH} \xrightarrow{\text{Red Hot Fe Tube}} 1,3,5\text{-C}_6\text{H}_3(\text{CH}_3)_3$$

Grignard reagents ($\text{R-MgX}$) are highly reactive organometallic bases and nucleophiles. They react instantly with any active proton or deuteron donor species like $\text{H}_2\text{O}$ or $\text{D}_2\text{O}$ via an acid-base pathway:

$$\text{R}^{\delta-}\text{-MgX}^{\delta+} + \text{D-OD} \longrightarrow \text{R-D} + \text{Mg(OD)X}$$
Here, $\text{R} = \text{Me}_3\text{C}$ (tert-butyl group). The carbanion $\text{Me}_3\text{C}^-$ abstract a deuteron ($\text{D}^+$) from heavy water to produce:
$$\text{Me}_3\text{C-D} \quad (\text{Deuterated isobutane}) + \text{Mg(OD)Cl}$$

Carboxylic acids with a carbonyl group at the $\beta$-position (known as **$\beta$-keto acids** like $\text{CH}_3\text{COCH}_2\text{COOH}$) undergo decarboxylation exceptionally easily upon mild heating.

* **Mechanism:** Decarboxylation proceeds via a concerted, cyclic six-membered transition state that does not require strong alkaline conditions (like Soda lime). The product initially formed is an enol, which immediately tautomerises to the much more stable ketone (Acetone):
$$\text{CH}_3\text{-CO-CH}_2\text{-COOH} \xrightarrow{\Delta} \text{CH}_3\text{-CO-CH}_3 + \text{CO}_2$$

Viscosity depends on the strength of intermolecular forces of attraction (like hydrogen bonding).

* **$p$-Nitrophenol** exhibits strong **intermolecular** hydrogen bonding, allowing molecules to associate in extensive chains, which increases friction and viscosity.
* **$o$-Nitrophenol** exhibits **intramolecular** hydrogen bonding (chelation within a single molecule). Consequently, its molecules stay discrete, lowering its cohesive viscosity.

Therefore, the correct viscosity order is: **$p$-nitrophenol $> o$-nitrophenol**. Statement (3) is incorrect.

According to **Fajan’s Rules**, the covalent character of an ionic bond increases with:
1. High charge of the cation ($q^+$).
2. Small size of the cation ($r^+$).
3. High charge and large size of the anion.

* Comparing the cations here: $\text{Na}^+$ ($+1$), $\text{Hg}^{2+}$ ($+2$), $\text{Al}^{3+}$ ($+3$), and $\text{Si}^{4+}$ ($+4$).
* $\text{Si}^{4+}$ carries the highest positive charge ($+4$) and has the smallest ionic radius, which maximizes its polarising power to distort the electron cloud of the chloride ($\text{Cl}^-$) anions. Thus, **$\text{SiCl}_4$** exhibits the maximum covalent character (and is a liquid at room temperature).

For a salt to dissolve readily in water, the release of energy through hydration must exceed the energy required to break the crystal lattice:
$$\text{Hydration Energy (H.E.)} > \text{Lattice Energy (L.E.)}$$
* In the alkaline earth metal sulfates ($\text{MSO}_4$), solubility **decreases** down the group ($\text{MgSO}_4 > \text{CaSO}_4 > \text{SrSO}_4 > \text{BaSO}_4$).
* This occurs because the large size of the sulfate anion ($\text{SO}_4^{2-}$) makes the lattice energy relatively constant down the group. However, the hydration energy drops sharply as the cation size increases from $\text{Mg}^{2+}$ to $\text{Ba}^{2+}$.
* Due to the small size and high charge density of the $\text{Mg}^{2+}$ ion, its hydration energy is exceptionally high and easily overcomes the lattice energy of **$\text{MgSO}_4$**, making it highly soluble.

Lithium (Group 1) and Magnesium (Group 2) exhibit a **diagonal relationship** due to their similar ionic radii ($r_{\text{Li}^+} = 76\text{ pm}$, $r_{\text{Mg}^{2+}} = 72\text{ pm}$) and similar electronegativities.

* A major shared characteristic is their ability to combine directly with atmospheric nitrogen ($\text{N}_2$) on heating to form stable ionic nitrides:
$$6\text{ Li} + \text{N}_2 \xrightarrow{\Delta} 2\text{ Li}_3\text{N} \quad (\text{Lithium nitride})$$
$$3\text{ Mg} + \text{N}_2 \xrightarrow{\Delta} \text{Mg}_3\text{N}_2 \quad (\text{Magnesium nitride})$$
Other alkali metals ($\text{Na, K, Rb, Cs}$) do not form nitrides directly under these conditions.

* **Statement (1) is correct:** Because of the exceptionally small size and high polarising power of $\text{Be}^{2+}$, $\text{BeCl}_2$ has significant covalent character.
* **Statement (2) and (3) are correct:** Monomeric $\text{BeCl}_2$ has only 4 valence electrons around Be, making it electron-deficient. To complete its octet, it polymerises. In the vapour phase at intermediate temperatures, it exists as a planar **dimer** ($(\text{BeCl}_2)_2$) with bridge coordinate bonds where Be is $sp^2$ hybridised.
* **Statement (4) is incorrect:** In the **solid state**, Beryllium Chloride exists as an infinite linear polymer chain containing chlorine bridge coordinate bonds. Here, each $\text{Be}$ atom forms 4 coordinate covalent bonds and is tetrahedrally coordinated, resulting in **$sp^3$ hybridisation** (not $sp^2$).

The biological unit controlling heredity is the **Gene**.

* A gene is a specific sequence of nucleotides in DNA (or RNA, in some viruses) that encodes the synthesis of a gene product, either RNA or protein, which ultimately dictates phenotypic expression. Chromosomes serve as the structural carriers of these genes.

According to Mendel’s First Law (the **Law of Segregation**):

* Allele pairs separate (segregate) during gamete formation (meiosis). Each gamete receives only one allele for any given gene.
* Because gametes are haploid ($n$), they contain exactly **one allele of a gene**. When fertilization occurs, the diploid state ($2n$) containing two alleles is restored in the zygote.

ABO blood grouping in humans is controlled by a single gene ($I$) with three alleles ($I^A$, $I^B$, and $i$).

1. A child with blood group ‘O’ must be homozygous recessive with the genotype **$ii$**. This child must have inherited one recessive allele **$i$** from the mother and one recessive allele **$i$** from the father.
2. Since the father has blood group ‘B’, his phenotype must be determined by the dominant allele $I^B$.
3. Because the father must have passed on a recessive allele ‘$i$’ to his child, his genotype is heterozygous **$I^Bi$**. (If he were homozygous $I^BI^B$, he could only produce gametes containing the $I^B$ allele, making it impossible to have an O-group child).

The number of genetically distinct gametes produced by an individual is given by the formula:
$$\text{Number of Gametes} = 2^n$$
where **$n$** is the number of **heterozygous gene loci** in the genotype.

* Let’s analyze the genotype $\text{AABbcc}$:
* Locus 1: $\text{AA}$ is homozygous ($0$).
* Locus 2: $\text{Bb}$ is heterozygous ($1$).
* Locus 3: $\text{cc}$ is homozygous ($0$).
* Thus, $n = 1$. The number of different gametes is:
$$2^1 = 2\text{ types}$$
* The two types of gametes are: **$\text{ABc}$** and **$\text{Abc}$**.

Red-green colour blindness is an **X-linked recessive** genetic disorder.

1. Genotypes of the parents:
* Colour-blind father: $X^c Y$
* Homozygous normal mother: $X^C X^C$
2. Cross Analysis:
* A father always passes his **$Y$ chromosome** to his sons, which contains no gene for colour vision.
* A mother always passes her **$X^C$ chromosome** (normal dominant allele) to her sons.
3. The resulting genotype of all sons will be **$X^C Y$** (completely normal). Therefore, the probability of their son being colour-blind is **$0$**.
* Note: All daughters from this cross will be carriers ($X^C X^c$), but phenotypically normal.

The number of **linkage groups** in an organism is equal to its **haploid chromosome number** ($n$).

* The garden pea (*Pisum sativum*) has a diploid chromosome number of $2n = 14$.
* Therefore, its haploid chromosome number is:
$$n = \frac{14}{2} = 7$$
* This means *Pisum sativum* has **7 linkage groups** (corresponding to its 7 pairs of homologous chromosomes).