The pressure (normal thrust per unit area) at any point in a stationary fluid depends only on the depth $d$ of that point below the free surface. Since point $P$ is at depth $d = H-h$, the normal thrust per unit area is $(H-h)\rho g$.

By Pascal's Law, the pressure transmitted is equal: $\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies \frac{m_1 g}{\pi r^2} = \frac{m_2 g}{\pi R^2}$
$$m_1 = m_2 \left(\frac{r}{R}\right)^2 = 1500 \left(\frac{1}{20}\right)^2 = \frac{1500}{400} = 3.75\text{ kg}$$.

The pressure at the base of both vessels is the same ($P = h\rho g$) since the liquid columns have equal height. Because the base areas $A$ are identical, the normal force exerted by the water on the base is equal for both: $F_A = F_B = P \times A = h\rho g A$. However, since vessel A is wider at the top, it holds a greater volume of water, meaning its weight is larger: $W_A > W_B$.

Initial volume of ice is $V_{\text{ice}} = \frac{m}{\text{density of ice}} = \frac{m}{x}$.
Final volume of water is $V_{\text{water}} = \frac{m}{\text{density of water}} = \frac{m}{y}$.
The change in volume is:
$$\Delta V = V_{\text{water}} - V_{\text{ice}} = \frac{m}{y} - \frac{m}{x} = m \left(\frac{1}{y} - \frac{1}{x}\right)$$

Volume of block $V = 5 \times 5 \times 5 = 125\text{ cm}^3$.
Relative density ($RD$) = $7$, so density of steel = $7\text{ g/cm}^3$.
Weight in air = $V \times d_{\text{steel}} = 125 \times 7\text{ gf} = 5 \times 5 \times 5 \times 7\text{ gf}$.
Upthrust (buoyancy force) = $V \times d_{\text{water}} = 125 \times 1\text{ gf} = 5 \times 5 \times 5 \times 1\text{ gf}$.
Apparent weight in water = Weight in Air - Upthrust:
$$W_{\text{apparent}} = 125 \times 7\text{ gf} - 125 \times 1\text{ gf} = 125 \times (7-1)\text{ gf} = 6 \times 5 \times 5 \times 5\text{ gf}$$.

By Torricelli's theorem, the velocity of efflux is $v = \sqrt{2gD}$.
The time taken for the water to fall a distance of $(H-D)$ to the ground is $t = \sqrt{\frac{2(H-D)}{g}}$.
The horizontal range is:
$$x = v \times t = \sqrt{2gD} \times \sqrt{\frac{2(H-D)}{g}} = 2\sqrt{D(H - D)}$$.

The horizontal range of the water is given by $x = 2\sqrt{h(H-h)}$.
To find the maximum range, we differentiate $y = h(H-h) = hH - h^2$ with respect to $h$ and set it to zero:
$$\frac{dy}{dh} = H - 2h = 0 \implies h = \frac{H}{2}$$.

Radius $r = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$.
Area of cross-section $A = \pi r^2 = \pi (2 \times 10^{-3})^2 = 4\pi \times 10^{-6}\text{ m}^2$.
Strain $\frac{\Delta L}{L} = 0.1\% = 10^{-3}$.
From Young's Modulus formula:
$$F = Y \times A \times \frac{\Delta L}{L} = (9 \times 10^{10}) \times (4\pi \times 10^{-6}) \times 10^{-3} = 360\pi\text{ N}$$.

Since Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$, we can write the load $F$ as:
$$F = \left(\frac{Y A}{L}\right) \Delta L$$
The slope of the load ($F$) vs elongation ($\Delta L$) graph is $\frac{Y A}{L}$. For wires of the same material ($Y$) and length ($L$), the slope is directly proportional to the cross-sectional area $A$. Therefore, the thinnest wire (smallest $A$) will have the lowest slope.

Potential energy of a spring is $U = \frac{1}{2} k x^2$, where $x$ is the elongation.
Therefore, potential energy is proportional to the square of elongation ($U \propto x^2$).
Ratio of P.E. is:
$$\frac{U_2}{U_1} = \left(\frac{x_2}{x_1}\right)^2 = \left(\frac{10}{2}\right)^2 = 5^2 = 25 \implies U_2 = 25 V$$.

The terminal velocity ($v_t$) of a spherical body of radius $r$ falling through a viscous medium is given by Stokes' Law:
$$v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta}$$
where $\rho$ and $\sigma$ are the densities of the sphere and the fluid respectively, and $\eta$ is the coefficient of viscosity. This shows that terminal velocity is directly proportional to the square of the radius of the sphere ($v_t \propto r^2$).

When the ball falls at terminal velocity $v = 10\text{ cm/s}$, the downward effective weight $W_{\text{eff}}$ is balanced by the upward viscous drag force $F_v$: $F_v = W_{\text{eff}} = 6\pi \eta r (10)$.
When pulled upwards with an external force $F = 2W_{\text{eff}}$, the net upward force on the ball is:
$$F_{\text{net}} = F - W_{\text{eff}} = 2W_{\text{eff}} - W_{\text{eff}} = W_{\text{eff}}$$
Since the net upward driving force is exactly equal to the downward effective weight, the upward terminal velocity must be identical to the downward terminal velocity, which is $10\text{ cm/s}$.

Because the battery is disconnected, the charge $Q$ on the plates must remain constant.
Capacitance is $C = \frac{\epsilon_0 A}{d}$. As the distance $d$ increases, the capacitance $C$ decreases.
Since $V = Q/C$, the voltage $V$ increases.
The electrostatic energy stored is $U = \frac{Q^2}{2C}$. Since $Q$ is constant and $C$ decreases, the energy $U$ increases (the external work done in pulling the plates apart is stored as electrostatic potential energy).

Total charge $Q_{\text{total}} = Q_1 + Q_2 = 150\mu\text{C} + 150\mu\text{C} = 300\mu\text{C} = 3 \times 10^{-4}\text{ C}$.
Total capacitance $C_{\text{total}} = C_1 + C_2 = 4\pi \epsilon_0 R_1 + 4\pi \epsilon_0 R_2 = 4\pi \epsilon_0 (R_1 + R_2)$.
Since $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2$:
$$C_{\text{total}} = \frac{R_1 + R_2}{9 \times 10^9} = \frac{0.2 + 0.1}{9 \times 10^9} = \frac{0.3}{9 \times 10^9} = \frac{1}{3 \times 10^{10}}\text{ F}$$
Common potential $V = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{3 \times 10^{-4}}{1 / (3 \times 10^{10})} = 9 \times 10^6\text{ V}$.

Area $A = 400\text{ cm}^2 = 400 \times 10^{-4}\text{ m}^2 = 4 \times 10^{-2}\text{ m}^2$.
Distance $d = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$.
Capacitance $C = \frac{\epsilon_0 A}{d} = \frac{(8.854 \times 10^{-12}) \times (4 \times 10^{-2})}{2 \times 10^{-3}} = 1.77 \times 10^{-10}\text{ F}$.
Charge $Q = C V = (1.77 \times 10^{-10}\text{ F}) \times 200\text{ V} = 3.54 \times 10^{-8}\text{ C}$.

When one plate is given a charge $+q$ and the other is earthed, a charge of $-q$ is induced on the inner surface of the earthed plate. Outside the plates (at $P_1$ and $P_2$), the electric fields produced by the $+q$ and $-q$ charges are equal in magnitude and opposite in direction, so they cancel out completely ($E = 0$). Inside the plates (at $P$), the fields due to both plates point in the same direction and add up to $E = \frac{\sigma}{\epsilon_0} \neq 0$.

Since the capacitor is disconnected from the battery, the charge must remain constant ($q' = q$).
Inserting a dielectric slab increases the capacitance to $C' = kC$.
The new potential difference is $V' = q/C' = q/(kC) = V/k$, which represents a decrease.
The new stored energy is $E' = \frac{q^2}{2C'} = \frac{q^2}{2kC} = E/k$, which also decreases.

When plates with charges $Q_1$ and $Q_2$ are placed parallel to each other, the charges distribute on the outer and inner surfaces. The charge on the outer surfaces of both plates is $\frac{Q_1 + Q_2}{2}$.
The charge on the inner surface of the first plate is $Q_1 - \frac{Q_1 + Q_2}{2} = \frac{Q_1 - Q_2}{2}$.
The charge on the inner surface of the second plate is $Q_2 - \frac{Q_1 + Q_2}{2} = -\frac{Q_1 - Q_2}{2}$.
Since the electric field inside is due solely to the inner charges, the potential difference is:
$$V = \frac{q_{\text{inner}}}{C} = \frac{Q_1 - Q_2}{2C}$$.

In a series connection, the charge $Q$ on both capacitors is the same. Since $V = Q/C$, the potential difference across a capacitor is inversely proportional to its capacitance ($V \propto 1/C$).
Thus, the ratio of potential differences is:
$$\frac{V_4}{V_6} = \frac{C_6}{C_4} = \frac{6}{4} = \frac{3}{2}$$.

To exhibit keto-enol tautomerism, a carbonyl compound must have at least one $\alpha$-hydrogen atom attached to an $sp^3$ hybridised carbon. In Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$), the carbonyl carbon is directly bonded to an $sp^2$ carbon of the benzene ring, which has no $\alpha$-hydrogen. Thus, it cannot undergo tautomerism.

During the electrolysis of sodium succinate, the succinate anions discharge at the anode to form succinate radicals. Decarboxylation of this radical produces a biradical ($\cdot\text{CH}_2-\text{CH}_2\cdot$), which rapidly undergoes internal radical coupling to form a double bond, yielding ethylene ($\text{CH}_2=\text{CH}_2$) gas.

Protonation of the alkene yields the secondary carbocation $(\text{CH}_3)_3\text{C}-\text{CH}^+-\text{CH}_3$. To increase stability, a $1,2$-methyl shift occurs from the adjacent quaternary carbon, forming a highly stable tertiary carbocation: $(\text{CH}_3)_2\text{C}^+-\text{CH(CH}_3)_2$. Attack of the chloride nucleophile then gives the rearranged major product, $2$-chloro-$2,3$-dimethylbutane.

Electrophilic addition of halogens to alkenes proceeds via a cyclic halonium (specifically, a cyclic bromonium) ion intermediate. This bulky three-membered ring blocks one face of the molecule, preventing syn-addition. The halide ion (acting as a nucleophile) must attack from the opposite face in a backside attack, resulting exclusively in anti-addition.

The rate-determining step in electrophilic addition is the formation of the carbocation. Electron-donating groups stabilize the carbocation intermediate via resonance ($+M$ effect) or inductive effects. The amino group ($-\text{NH}_2$ in R) is a stronger electron-donating resonance group than the methoxy group ($-\text{OCH}_3$ in Q), both of which are far superior to the hyperconjugative electron donation of the alkyl group in P. Thus, the stability of the intermediate carbocation (and the reaction rate) follows the order: R > Q > P.

Hydroboration-Oxidation is a clean method for the **anti-Markovnikov** addition of water across a double bond without carbocation rearrangements. Oxymercuration-Demercuration represents a highly efficient **Markovnikov** addition of water, also proceeding without skeletal rearrangements.

Addition of halogen acids to conjugated dienes is temperature-dependent. At low temperatures (kinetic control), the reaction rate is determined by the activation energy. The $1,2$-addition product ($3$-chlorobut-1-ene) forms much faster due to the proximity of the chloride ion to the newly formed carbocation, making it the major product at low temperatures.

Halohydrin formation involves anti-addition. When unsymmetrical groups (like $-\text{Br}$ and $-\text{OH}$) add in an anti fashion to a symmetrical *trans*-alkene, it results in the formation of a pair of non-superimposable enantiomers in equal amounts, creating a **racemic mixture**.

Ketones react with excess monohydric alcohols (like ethanol) under acidic conditions (dry $\text{HCl}$ gas) to undergo nucleophilic addition followed by dehydration. This replaces the carbonyl oxygen with two alkoxy groups, forming a stable **ketal** (which is a subclass of acetals) named $2,2$-diethoxypropane.

Reactivity of carbonyl compounds towards nucleophilic addition is governed by steric and electronic factors. Sterically, bulkier groups around the carbonyl carbon hinder the approach of the nucleophile. Electronically, electron-donating groups (like methyl in II and III, and phenyl in IV) reduce the partial positive charge on the carbonyl carbon, making it less electrophilic. Therefore, formaldehyde is the most reactive, and acetophenone is the least reactive: I > II > III > IV.

Primary amines react with carbonyls to form imines (containing a $\text{C}=\text{N}$ bond). Secondary amines ($\text{R}_2\text{NH}$) lack a second hydrogen on the nitrogen atom to form a carbon-nitrogen double bond. Instead, after nucleophilic attack, elimination of water occurs by abstracting an $\alpha$-hydrogen from the adjacent carbon, creating a carbon-carbon double bond, yielding an **enamine**.

The central carbonyl group at C-2 is flanked on both sides by strongly electron-withdrawing terminal carbonyl groups (at C-1 and C-3). The strong $-I$ effect of these adjacent groups makes the C-2 carbon highly electrophilic (electron-deficient). In addition, hydration of the C-2 carbon relieves steric and electrostatic repulsion between the three adjacent polar carbonyl groups, making the central C-2 hydrate extremely stable.

A hemiacetal is a carbon compound characterized by a single carbon atom that is simultaneously bonded to one hydroxyl group ($-\text{OH}$) and one ether group ($-\text{OR}$). An acetal has two ether groups on the same carbon ($\text{R-CH(OR')}_2$).

The Wurtz reaction of mixed alkyl halides yields a complex mixture of alkanes (ethane, propane, butane) because of random coupling. Corey-House synthesis is specifically designed for the coupling of two different alkyl groups to synthesize unsymmetrical alkanes in high purity: $\text{R-X} + \text{R'}_2\text{CuLi} \rightarrow \text{R-R'} + \text{R'-Cu} + \text{LiX}$.

The thermal stability of alkaline earth metal sulfates increases down the group as the size of the metal cation increases. A larger cation has lower charge density and lower polarizing power, causing less distortion (polarization) of the large sulfate anion, which makes the ionic lattice thermally more stable.

For Group 2 metal hydroxides, the lattice energy decreases much faster than the hydration energy down the group as cation size increases. This causes the solubility of hydroxides to increase down the group ($\text{Be(OH)}_2 < \text{Mg(OH)}_2 < \text{Ca(OH)}_2 < \text{Sr(OH)}_2 < \text{Ba(OH)}_2$). Their thermal stability also increases down the group as their basic character increases.

Thermal stability of carbonates increases down the group as polarizing power of the cation decreases. Therefore, $\text{BeCO}_3$ decomposes at a much lower temperature than $\text{MgCO}_3$ or $\text{CaCO}_3$. This means the **ease of oxide formation on heating** is highest for $\text{BeCO}_3$ and decreases down: $\text{BeCO}_3 > \text{MgCO}_3 > \text{CaCO}_3$. Thus, the order in (3) is incorrect.

Potassium phosphite ($\text{K}_2\text{HPO}_3$) contains a direct phosphorous-hydrogen ($\text{P-H}$) bond, which is non-polar and cannot participate in hydrogen bonding. Potassium hydrogen phosphate ($\text{K}_2\text{HPO}_4$) contains an active polar $-\text{O-H}$ group that participates in H-bonding.

Lattice energy ($U$) is proportional to the product of ionic charges ($q_1 q_2$) and inversely proportional to the interionic distance. For $\text{Al}_2\text{O}_3$, the charges are $+3$ and $-2$, giving a charge product of $3 \times 2 = 6$, which is much higher than $\text{AlF}_3$ ($3 \times 1 = 3$), $\text{Na}_2\text{O}$ ($2$), or $\text{CaF}_2$ ($2$).

Due to the small size and high polarizing power of Beryllium, $\text{BeH}_2$ is covalent. In the solid state, it polymerizes into an infinite chain where beryllium atoms are bridged by hydrogen atoms (three-center two-electron bonds).

Limestone ($\text{CaCO}_3$) is thermally decomposed in a kiln to produce $\text{CO}_2$ gas (used in the carbonating tower) and calcium oxide (CaO, quicklime). Quicklime is then slaked with water to form slaked lime, which recovers ammonia from ammonium chloride.

Due to their diagonal relationship, Lithium (Group 1) and Magnesium (Group 2) have small ionic radii and high charge densities. This allows them to overcome the high bond energy of the nitrogen triple bond ($N\equiv N$) and directly form stable solid nitrides ($\text{Li}_3\text{N}$ and $\text{Mg}_3\text{N}_2$).

Beryllium chloride ($\text{BeCl}_2$) is indeed covalent and electron-deficient. In the vapour phase at high temperatures, it exists as a planar chloro-bridged dimer ($\text{Be}_2\text{Cl}_4$). In the solid state, it polymerizes to form a chain structure where beryllium is tetrahedrally coordinated and $sp^3$ hybridized. Thus, statement (3) is incorrect.

Solubility of Group 2 hydroxides increases down the group. Conversely, the solubility of Group 2 sulfates decreases down the group. Electronegativity and ionization potentials also decrease down the group due to increasing atomic radius and shielding effect.

Ammonium dichromate decomposes on heating to yield nitrogen gas, green chromium(III) oxide, and water vapor:
$$\text{(NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta} \text{N}_2 + \text{Cr}_2\text{O}_3 + 4\text{H}_2\text{O}$$
No oxygen is released. The other salts decompose to release oxygen.

A test cross involves crossing an individual of dominant phenotype with a homozygous recessive individual. If the dominant parent is homozygous dominant (WW), $100\%$ of the offspring will be heterozygous violet (Ww). If the dominant parent is heterozygous (Ww), the offspring will segregate in a $1:1$ ratio ($\text{Ww}$ violet and $\text{ww}$ white).

Segregation of alleles occurs during gamete formation in meiosis. Specifically, homologous chromosomes (each carrying one allele of a gene) separate and move to opposite poles during **Anaphase I of meiosis**, ensuring that each gamete receives only one of the two alleles.

A child of blood group 'O' has the homozygous recessive genotype $ii$. The child must inherit one recessive allele $i$ from the mother and one from the father. Since the father has blood group 'B', he must possess at least one dominant $I^B$ allele. Therefore, the father's genotype must be heterozygous $I^B i$.

Recombination frequencies are directly proportional to map distances on a chromosome. Since B and C have the highest recombination frequency ($26\%$), they are the furthest apart. Gene A must lie between them since the distance B-A ($9\%$) plus A-C ($17\%$) equals the total distance B-C ($26\%$, map units/centiMorgans). Thus, the linear order is B – A – C.

Myotonic dystrophy is an autosomal dominant muscular disorder. Phenylketonuria is autosomal recessive. Haemophilia is a sex-linked (X-linked) recessive disorder, and Turner's syndrome is an aneuploidy (monosomy, 45,XO).

In 1891, German biologist Hermann Henking observed a specific nuclear structure during insect spermatogenesis that was received by only half of the sperm cells. He called this structure the **'X-body'**. Later scientists identified this as the X-chromosome, establishing its role in genetic sex determination.