Step 1: Understanding the Concept:

This problem involves a neutralization reaction between a strong acid (H\(_2\)SO\(_4\)) and a strong base (NaOH). To solve it, we need to use stoichiometry, which relates the quantities of reactants and products in a chemical reaction.


Step 2: Key Formula or Approach:

1. Write the balanced chemical equation for the reaction.

2. Calculate the number of moles of the given substance (H\(_2\)SO\(_4\)).

\[ Moles = \frac{Given Mass}{Molar Mass} \]
3. Use the mole ratio from the balanced equation to find the moles of the substance required (NaOH).

4. Calculate the required mass of NaOH using its molar mass.

\[ Mass = Moles \times Molar Mass \]

Step 3: Detailed Explanation:

Balanced Equation: The reaction between sulfuric acid and sodium hydroxide is:
\[ H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \]
This equation shows that 1 mole of H\(_2\)SO\(_4\) reacts completely with 2 moles of NaOH.


Calculate Moles of H\(_2\)SO\(_4\):

First, we need the molar mass of H\(_2\)SO\(_4\).

Molar Mass of H\(_2\)SO\(_4\) = (2 \(\times\) 1.008) + 32.07 + (4 \(\times\) 16.00) \(\approx\) 98 g/mol.

Now, calculate the moles of H\(_2\)SO\(_4\) in 9.8 g.
\[ Moles of H_2SO_4 = \frac{9.8 g}{98 g/mol} = 0.1 mol \]

Calculate Moles of NaOH Required:

From the mole ratio in the balanced equation (1:2), the moles of NaOH required are twice the moles of H\(_2\)SO\(_4\).
\[ Moles of NaOH = 0.1 mol H_2SO_4 \times \frac{2 mol NaOH}{1 mol H_2SO_4} = 0.2 mol NaOH \]

Calculate Mass of NaOH Required:

The molar mass of NaOH is given as 40 g/mol.
\[ Mass of NaOH = Moles of NaOH \times Molar Mass of NaOH \] \[ Mass of NaOH = 0.2 mol \times 40 g/mol = 8 g \]

Step 4: Final Answer:

Therefore, 8 g of NaOH is required to completely neutralize 9.8 g of H\(_2\)SO\(_4\).
Quick Tip: For any stoichiometry problem, the first and most crucial step is to write a correctly balanced chemical equation. The coefficients in the balanced equation provide the exact mole ratios needed for the calculation.

Step 1: Understanding the Concept:

Canal rays, also known as anode rays, are beams of positive ions that are created in certain types of gas-discharge tubes. They were discovered by Eugen Goldstein in 1886. We need to evaluate the given statements based on the known properties of these rays.


Step 2: Detailed Explanation:

Let's analyze each statement:


(i) They carry positively charged particles.

This statement is correct. Canal rays are formed when high-energy electrons (cathode rays) knock electrons out of the neutral gas atoms in the tube, creating positively charged ions. These ions are then accelerated towards the negative cathode.


(ii) The mass of the particles of these rays does not depend upon the gas present in the cathode ray tube.

This statement is incorrect. The particles of canal rays are the positive ions of the gas used in the discharge tube. Therefore, their mass is dependent on the mass of the gas atoms. For example, if hydrogen gas is used, the canal rays consist of H\(^+\) ions; if helium is used, they consist of He\(^+\) ions.


(iii) The particles behave in a different manner in electric field as those of cathode rays.

This statement is correct. Cathode rays are composed of negatively charged electrons, so they deflect towards the positive plate in an electric field. Canal rays are composed of positively charged ions, so they deflect towards the negative plate. Their behavior is opposite.


(iv) The charge to mass ratio of the particles does not depends on the gas present in the cathode ray tube.

This statement is incorrect. The charge-to-mass (e/m) ratio depends on the gas. Since the mass of the ion changes with the gas (as explained in point ii), the e/m ratio also changes. The e/m ratio is maximum for the lightest gas, hydrogen.


Step 3: Final Answer:

The correct statements are (i) and (iii). Therefore, option (E) is the correct choice.
Quick Tip: A simple way to remember the properties is to contrast them with cathode rays. Cathode rays are fundamental particles (electrons) and their properties are independent of the gas. Canal rays are ionized gas atoms, so their properties (mass, e/m ratio) depend on the gas used.

Step 1: Understanding the Concept:

The de Broglie hypothesis states that all matter exhibits wave-like properties. The wavelength associated with a particle is called the de Broglie wavelength (\(\lambda\)) and is inversely proportional to its momentum (p).

Note: There appears to be a typo in the question's given mass (2.2 x 10\(^{-2}\) kg), which is extremely large for a sub-atomic particle. Based on the correct answer, the mass should be 2.2 x 10\(^{-26}\) kg. We will proceed with the corrected mass to arrive at the given answer.


Step 2: Key Formula or Approach:

The de Broglie wavelength (\(\lambda\)) is calculated using the formula: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
where:

h = Planck's constant (6.6 \(\times\) 10\(^{-34}\) J s)

m = mass of the particle

v = velocity of the particle


Step 3: Detailed Explanation:

Given values:

h = 6.6 \(\times\) 10\(^{-34}\) J s

m = 2.2 \(\times\) 10\(^{-26}\) kg (Corrected mass)

v = 3.0 \(\times\) 10\(^{5}\) m/s


Calculation:

Substitute the values into the de Broglie equation: \[ \lambda = \frac{6.6 \times 10^{-34} J s}{(2.2 \times 10^{-26} kg) \times (3.0 \times 10^{5} m/s)} \]
First, calculate the momentum (mv): \[ p = (2.2 \times 3.0) \times (10^{-26} \times 10^{5}) = 6.6 \times 10^{-21} kg m/s \]
Now, calculate the wavelength: \[ \lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-21}} = 1.0 \times 10^{-13} m \]
The options are given in picometers (pm). We need to convert meters to picometers.

Since 1 pm = 10\(^{-12}\) m, then 1 m = 10\(^{12}\) pm.
\[ \lambda = (1.0 \times 10^{-13} m) \times (10^{12} pm/m) = 0.1 pm \]

Step 4: Final Answer:

The de Broglie wavelength of the particle is 0.1 pm.
Quick Tip: When solving physics or chemistry problems, always perform a sanity check on the given values. A mass of 10\(^{-2}\) kg for a sub-atomic particle is impossible. Recognizing such potential typos can help you work backwards from the options if necessary. Pay close attention to unit conversions, especially between meters, nanometers, and picometers.

Step 1: Understanding the Concept:

Elements in the periodic table can be classified as metals, non-metals, or metalloids. Metalloids (or semi-metals) are elements that have properties intermediate between those of metals and non-metals. They are typically semiconductors.


Step 2: Detailed Explanation:

Let's classify each of the given elements:


(A) Antimony (Sb): It is located in Group 15 of the periodic table. It exhibits properties of both metals (e.g., metallic luster) and non-metals (e.g., it is a poor conductor of heat and electricity). Antimony is a classic example of a metalloid.

(B) Aluminium (Al): It is a post-transition metal in Group 13. It is a good conductor of electricity and heat and is malleable and ductile, which are characteristic properties of metals.

(C) Magnesium (Mg): It is an alkaline earth metal in Group 2. It is a reactive, silvery-white metal with clear metallic properties.

(D) Phosphorus (P): It is a non-metal in Group 15. It exists in several allotropic forms and is a poor conductor of heat and electricity.

(E) Calcium (Ca): It is an alkaline earth metal in Group 2. It is a reactive metal with typical metallic characteristics.



Step 3: Final Answer:

Based on the classification, Antimony (Sb) is the metalloid among the given options.
Quick Tip: Memorize the common metalloids, which form a "staircase" on the periodic table. The most commonly cited metalloids are Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te), and Polonium (Po).

Step 1: Understanding the Concept:

The periodic table is divided into blocks based on the orbital where the highest-energy electron (or differentiating electron) resides. These blocks are s, p, d, and f. The term "main group elements" refers to a specific collection of these blocks.


Step 2: Detailed Explanation:

Let's define the elements in each block:


s-block elements: These are the elements in Groups 1 and 2. Their valence electrons are in the s-orbital.

p-block elements: These are the elements in Groups 13 to 18. Their valence electrons are in the p-orbitals.

d-block elements: These are the elements in Groups 3 to 12, also known as the transition metals. Their differentiating electron enters a d-orbital.

f-block elements: These are the lanthanoids and actinoids, also known as the inner transition metals. Their differentiating electron enters an f-orbital.


The main group elements (also called representative elements) are defined as the elements belonging to the s-block and p-block. These elements show a wide range of properties and follow periodic trends more consistently than the transition metals.


Step 3: Final Answer:

Therefore, the main group elements are composed of both s-block and p-block elements.
Quick Tip: Visualize the periodic table: the two columns on the far left (s-block) and the six columns on the far right (p-block) together constitute the main group elements. The block in the middle is the d-block (transition metals), and the two rows at the bottom are the f-block (inner transition metals).

Step 1: Understanding the Concept:

The octet rule states that atoms tend to bond in such a way that they each have eight electrons in their valence shell. An "expanded octet" occurs when a central atom in a molecule has more than eight electrons in its valence shell. This is possible for elements in the third period and below because they have vacant d-orbitals that can participate in bonding.


Step 2: Detailed Explanation:

Let's draw the Lewis structure for each compound and count the valence electrons around the central atom.


(A) SCl\(_2\): Sulfur (central atom) is in Group 16, so it has 6 valence electrons. Each Chlorine contributes 1 electron for the single bond. The structure is Cl-S-Cl. Sulfur forms 2 single bonds (4 electrons) and has 2 lone pairs (4 electrons). Total around S = 4 + 4 = 8 electrons. It follows the octet rule.


(B) NO\(_2\): Nitrogen (central atom) is in Group 15, with 5 valence electrons. Oxygen is in Group 16, with 6. Total valence electrons = 5 + 2(6) = 17. This is an odd-electron molecule, which is an exception to the octet rule, but it does not have an expanded octet. Nitrogen has fewer than 8 electrons.


(C) NH\(_3\): Nitrogen (central atom) has 5 valence electrons. It forms 3 single bonds with Hydrogen (3 electrons) and has 1 lone pair (2 electrons). Total around N = 3(2) + 2 = 8 electrons. It follows the octet rule.


(D) PCl\(_3\): Phosphorus (central atom) is in Group 15, with 5 valence electrons. It forms 3 single bonds with Chlorine (3 electrons) and has 1 lone pair (2 electrons). Total around P = 3(2) + 2 = 8 electrons. It follows the octet rule.


(E) H\(_2\)SO\(_4\): Sulfur (central atom) is in Period 3 and can have an expanded octet. In the most common Lewis structure for sulfuric acid, sulfur forms two single bonds with the -OH groups and two double bonds with the other two oxygen atoms to minimize formal charges.
In this structure, Sulfur forms 6 bonds (2 single + 2 double). The total number of valence electrons around the central Sulfur atom is 2 \(\times\) (2 for each double bond) + 2 \(\times\) (2 for each single bond) = 4 + 4 = 12 electrons. Since 12 \(>\) 8, sulfur has an expanded octet.


Step 3: Final Answer:

H\(_2\)SO\(_4\) is the compound where the central atom (Sulfur) has an expanded octet.
Quick Tip: To quickly check for an expanded octet, look for central atoms from Period 3 or below (like P, S, Cl, As, Se, Br, etc.) bonded to several highly electronegative atoms (like F, O, Cl). These are the most common candidates for expanded octets.

Step 1: Understanding the Concept:

Molecular geometry describes the three-dimensional arrangement of atoms in a molecule. The VSEPR (Valence Shell Electron Pair Repulsion) theory is used to predict geometry. Tetrahedral geometry occurs when a central atom is bonded to four other atoms and has no lone pairs of electrons. This corresponds to a steric number of 4 and an AX\(_4\) type molecule.


Step 2: Detailed Explanation:

Let's determine the geometry for each species using VSEPR theory.


(A) SF\(_6\):

Central atom: Sulfur (S)
Valence electrons of S: 6
Electrons from 6 F atoms: 6
Total electron pairs: 12 / 2 = 6 pairs
Bonding pairs: 6, Lone pairs: 0
VSEPR type: AX\(_6\)
Geometry: Octahedral


(B) PCl\(_5\):

Central atom: Phosphorus (P)
Valence electrons of P: 5
Electrons from 5 Cl atoms: 5
Total electron pairs: 10 / 2 = 5 pairs
Bonding pairs: 5, Lone pairs: 0
VSEPR type: AX\(_5\)
Geometry: Trigonal bipyramidal


(C) BF\(_3\):

Central atom: Boron (B)
Valence electrons of B: 3
Electrons from 3 F atoms: 3
Total electron pairs: 6 / 2 = 3 pairs
Bonding pairs: 3, Lone pairs: 0
VSEPR type: AX\(_3\)
Geometry: Trigonal planar


(D) BeCl\(_2\):

Central atom: Beryllium (Be)
Valence electrons of Be: 2
Electrons from 2 Cl atoms: 2
Total electron pairs: 4 / 2 = 2 pairs
Bonding pairs: 2, Lone pairs: 0
VSEPR type: AX\(_2\)
Geometry: Linear


(E) NH\(_4^+\):

Central atom: Nitrogen (N)
Valence electrons of N: 5
Electrons from 4 H atoms: 4
Charge: +1 (means we subtract 1 electron)
Total valence electrons = 5 + 4 - 1 = 8
Total electron pairs: 8 / 2 = 4 pairs
Bonding pairs: 4, Lone pairs: 0
VSEPR type: AX\(_4\)
Geometry: Tetrahedral


Step 3: Final Answer:

The ammonium ion, NH\(_4^+\), has a tetrahedral geometry.
Quick Tip: To quickly find the geometry, calculate the steric number = (number of atoms bonded to the central atom) + (number of lone pairs on the central atom). A steric number of 4 with 0 lone pairs always results in a tetrahedral molecular geometry.

Step 1: Understanding the Concept:

Enthalpy change (\(\Delta\)H) represents the heat absorbed or released during a process at constant pressure. A negative enthalpy change (\(\Delta\)H \(<\) 0) indicates an exothermic process, where heat is released. A positive enthalpy change (\(\Delta\)H \(>\) 0) indicates an endothermic process, where heat is absorbed. We are looking for the process that is always exothermic.


Step 2: Detailed Explanation:

Let's analyze the enthalpy change for each process:


(A) Enthalpy of ionisation: This is the energy required to remove an electron from a gaseous atom or ion. Energy must be supplied to overcome the nuclear attraction. This process is always endothermic (\(\Delta\)H \(>\) 0).


(B) Enthalpy of sublimation: This is the energy required to convert a substance from a solid directly to a gas. Energy is needed to break the intermolecular forces holding the solid together. This process is always endothermic (\(\Delta\)H \(>\) 0).


(C) Enthalpy of vapourisation: This is the energy required to convert a substance from a liquid to a gas. Energy is needed to overcome the intermolecular forces in the liquid. This process is always endothermic (\(\Delta\)H \(>\) 0).


(D) Enthalpy of bond dissolution (bond breaking): Breaking a chemical bond always requires an input of energy to overcome the forces holding the atoms together. This process is always endothermic (\(\Delta\)H \(>\) 0).


(E) Enthalpy of combustion: This is the heat released when a substance undergoes complete combustion with oxygen. Combustion reactions, like burning fuel, are by definition processes that release a significant amount of energy in the form of heat and light. This process is always exothermic (\(\Delta\)H \(<\) 0).


Step 3: Final Answer:

The enthalpy of combustion is the process for which the enthalpy change is always negative.
Quick Tip: Remember the general rule: processes that involve breaking bonds or overcoming forces of attraction (ionization, phase changes to less ordered states, bond breaking) are endothermic (\(\Delta\)H \(>\) 0). Processes that involve forming stable bonds or products (combustion, most synthesis reactions) are exothermic (\(\Delta\)H \(<\) 0).

Step 1: Understanding the Concept:

The spontaneity of a reaction is determined by the change in Gibbs free energy (\(\Delta\)G). A reaction is spontaneous if \(\Delta\)G \(<\) 0, non-spontaneous if \(\Delta\)G \(>\) 0, and at equilibrium if \(\Delta\)G = 0. The Gibbs free energy is related to enthalpy (\(\Delta\)H) and entropy (\(\Delta\)S) by the equation:
\[ \Delta G = \Delta H - T\Delta S \]
where T is the temperature in Kelvin.


Step 2: Key Formula or Approach:

We need to find the signs of \(\Delta\)H and \(\Delta\)S such that \(\Delta\)G is negative at low T and positive at high T. We can analyze the equation \(\Delta G = \Delta H - T\Delta S\) for each option.


Step 3: Detailed Explanation:

The condition is:

At low temperature (T \(\to\) 0), the reaction is spontaneous (\(\Delta\)G \(<\) 0).
At high temperature, the reaction is non-spontaneous (\(\Delta\)G \(>\) 0).


Let's examine the equation \(\Delta G = \Delta H - T\Delta S\).

At very low T, the term \(T\Delta S\) becomes negligible. So, \(\Delta G \approx \Delta H\). For the reaction to be spontaneous at low T, \(\Delta\)H must be negative (\(\Delta\)H \(<\) 0). This eliminates options (A), (C), and (E).


Now we are left with options (B) and (D). Let's test them.


Case (B): \(\Delta\)H \(<\) 0 and \(\Delta\)S \(>\) 0
\[ \Delta G = (negative value) - T(positive value) \]
In this case, both terms are negative. \(\Delta\)G will be negative at all temperatures. The reaction is always spontaneous. This does not match the condition.


Case (D): \(\Delta\)H \(<\) 0 and \(\Delta\)S \(<\) 0
\[ \Delta G = (negative value) - T(negative value) \] \[ \Delta G = (negative value) + T(positive value) \]
Here we have a competition between the negative enthalpy term and the positive entropy term (\(T|\Delta S|\)).

At low T: The \(T|\Delta S|\) term is small. The negative \(\Delta\)H term dominates, so \(\Delta\)G \(<\) 0. The reaction is spontaneous.
At high T: The \(T|\Delta S|\) term becomes large. It will eventually overcome the negative \(\Delta\)H term, making \(\Delta\)G \(>\) 0. The reaction becomes non-spontaneous.

This scenario perfectly matches the conditions given in the question.


Step 4: Final Answer:

The correct thermodynamic conditions are \(\Delta\)H \(<\) 0 and \(\Delta\)S \(<\) 0.
Quick Tip: You can remember the conditions with this logic: For spontaneity to depend on temperature, \(\Delta\)H and \(\Delta\)S must have the same sign. If both are positive, it's spontaneous at high T ("entropy-driven"). If both are negative, it's spontaneous at low T ("enthalpy-driven").

Step 1: Understanding the Concept:

This problem involves the equilibrium of a weak monobasic acid (HA). We are given the pH and initial concentration of the acid and asked to find its pKa. The pKa is a measure of the acid's strength.


Step 2: Key Formula or Approach:

1. Calculate the hydrogen ion concentration [H\(^+\)] from the given pH.

\[ [H^+] = 10^{-pH} \]
2. Write the expression for the acid dissociation constant, Ka.

\[ HA \rightleftharpoons H^+ + A^- \]
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
3. Calculate Ka using the equilibrium concentrations.

4. Calculate pKa from Ka.

\[ pK_a = -\log(K_a) \]

Step 3: Detailed Explanation:

1. Calculate [H\(^+\)]:

Given pH = 3. \[ [H^+] = 10^{-3} M \]

2. Determine Equilibrium Concentrations:

The dissociation equilibrium is HA \(\rightleftharpoons\) H\(^+\) + A\(^-\).
From the stoichiometry, at equilibrium, [H\(^+\)] = [A\(^-\)].
So, [A\(^-\)] = 10\(^{-3}\) M.

The equilibrium concentration of the undissociated acid [HA] is the initial concentration minus the amount that dissociated. \[ [HA]_{eq} = [HA]_{initial} - [H^+] \] \[ [HA]_{eq} = 0.1 - 10^{-3} = 0.1 - 0.001 = 0.099 M \]
Since the amount dissociated (0.001 M) is very small compared to the initial concentration (0.1 M), we can approximate [HA]\(_{eq}\) \(\approx\) 0.1 M.


3. Calculate Ka:

Now, substitute the equilibrium concentrations into the Ka expression: \[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(10^{-3})(10^{-3})}{0.1} \] \[ K_a = \frac{10^{-6}}{10^{-1}} = 10^{-5} \]

4. Calculate pKa:

Finally, calculate the pKa. \[ pK_a = -\log(K_a) = -\log(10^{-5}) \] \[ pK_a = 5 \]

Step 4: Final Answer:

The pKa of the acid is 5.
Quick Tip: For weak acid problems, if the percent dissociation is less than 5%, you can safely approximate the equilibrium concentration of the acid, [HA]\(_{eq}\), as being equal to its initial concentration. In this case, dissociation is \((0.001/0.1) \times 100% = 1%\), so the approximation is valid and simplifies the calculation.

Step 1: Understanding the Concept:

Limiting molar conductance (\(\Lambda_m^\circ\)) of an ion is its molar conductivity at infinite dilution. It is a measure of how efficiently an ion conducts electricity through a solution. It depends on factors like the charge of the ion and its ionic mobility (how fast it can move through the solvent).


Step 2: Detailed Explanation:

In aqueous solutions, most ions move through the water via diffusion, where their speed is limited by their size (including the shell of hydrating water molecules) and the viscosity of the water. However, the H\(^+\) ion (which exists as the hydronium ion, H\(_3\)O\(^+\)) and the OH\(^-\) ion have exceptionally high limiting molar conductances.


This is due to a unique conduction mechanism called the Grotthuss mechanism or proton jumping. Instead of a single H\(_3\)O\(^+\) ion physically moving through the solution, a proton can be transferred from one water molecule to an adjacent one, creating a new H\(_3\)O\(^+\) ion. This creates the effect of a proton moving very rapidly through the solution without the need for large-scale physical diffusion.
\[ H_3O^+ + H_2O \rightarrow H_2O + H_3O^+ \]
This "hopping" mechanism is much faster than the movement of other ions like Na\(^+\), K\(^+\), Ca\(^{2+}\), and Mg\(^{2+}\), which must physically push their way through the water molecules. As a result, the H\(^+\) ion has the highest limiting molar conductance of any cation in water.


Step 3: Final Answer:

The H\(^+\) ion has the highest limiting molar conductance due to the Grotthuss (proton jumping) mechanism.
Quick Tip: In any question asking to compare the limiting molar conductance of common ions in water, always remember that H\(^+\) and OH\(^-\) are exceptionally high. H\(^+\) has the highest among cations, and OH\(^-\) has the highest among anions. Their values are significantly larger than all other ions.

Step 1: Understanding the Concept:

This question relates to Faraday's laws of electrolysis. To deposit an element from its molten salt, its ions must be reduced at the cathode. The amount of substance deposited is directly proportional to the quantity of electric charge passed through the electrolyte.


Step 2: Key Formula or Approach:

1. Write the balanced half-reaction for the reduction of the metal ion.

2. Determine the number of moles of electrons required to deposit one mole of the metal.

3. Use the Faraday constant (F), which is the charge of one mole of electrons (96500 C/mol e\(^-\)), to calculate the total charge required.

\[ Total Charge = (moles of electrons) \times F \]

Step 3: Detailed Explanation:

1. Half-Reaction:

In fused (molten) magnesium chloride (MgCl\(_2\)), magnesium exists as Mg\(^{2+}\) ions. At the cathode, these ions gain electrons and are reduced to metallic magnesium. \[ Mg^{2+} + 2e^- \rightarrow Mg(s) \]

2. Moles of Electrons:

The balanced half-reaction shows that for every 1 mole of solid magnesium (Mg) deposited, 2 moles of electrons (e\(^-\)) are required.


3. Calculate Total Charge:

We are given that 1 Faraday (F) is the charge of 1 mole of electrons, and 1 F = 96500 C.
Since we need 2 moles of electrons, the total charge (Q) required is: \[ Q = (moles of e^-) \times F \] \[ Q = 2 mol e^- \times \frac{96500 C}{1 mol e^-} \] \[ Q = 193000 C \]
To express this in scientific notation: \[ Q = 1.93 \times 10^5 C \]

Step 4: Final Answer:

The quantity of charge required is 1.93 x 10\(^5\) C.
Quick Tip: A quick shortcut for these problems is the formula: Charge required (in Coulombs) = n \(\times\) F, where 'n' is the magnitude of the charge on the ion being deposited. For Mg\(^{2+}\), n=2. For Al\(^{3+}\), n=3. For Na\(^+\), n=1.

Step 1: Understanding the Concept:

The solubility of a gas in a liquid depends on the intermolecular forces between the gas molecules and the solvent molecules. The principle of "like dissolves like" is a good guide. Water is a highly polar solvent that can form hydrogen bonds. Therefore, gases that are also polar or can form hydrogen bonds with water will have the highest solubility.


Step 2: Detailed Explanation:

Let's analyze the properties of each gas and its potential interaction with water:


(A) Formaldehyde (HCHO):

The molecule has a polar carbonyl group (C=O).
The oxygen atom has lone pairs and can act as a hydrogen bond acceptor with water molecules.
Due to its polarity and ability to form hydrogen bonds, formaldehyde is very soluble in water. In fact, an aqueous solution of formaldehyde is sold as formalin.


(B) Methane (CH\(_4\)):

This is a nonpolar molecule (tetrahedral geometry with symmetrical C-H bonds).
It can only interact with water through weak London dispersion forces.
Its solubility in water is very low.


(C) CO\(_2\):

Although the C=O bonds are polar, the molecule is linear and symmetrical, making it nonpolar overall.
It has higher solubility than other nonpolar gases like methane because it can react with water to a small extent to form carbonic acid (H\(_2\)CO\(_3\)), which aids its dissolution. However, its solubility is much lower than that of formaldehyde.


(D) Vinyl chloride (CH\(_2\)=CHCl):

The presence of the electronegative chlorine atom makes the molecule slightly polar.
However, it cannot form hydrogen bonds with water. Its solubility is low.


(E) Argon (Ar):

This is a noble gas, existing as individual atoms. It is nonpolar.
It interacts with water only through very weak London dispersion forces and has extremely low solubility.


Step 3: Final Answer:

Comparing the options, formaldehyde is the most polar molecule and the one most capable of forming strong intermolecular interactions (hydrogen bonds) with water. Therefore, it has the highest solubility.
Quick Tip: When asked about solubility in water, look for the substance with the greatest polarity and, most importantly, the ability to form hydrogen bonds. Gases like NH\(_3\), SO\(_2\), HCl, and formaldehyde (HCHO) are highly soluble in water for these reasons.

Step 1: Understanding the Concept:

The rate law for a chemical reaction expresses the rate of reaction in terms of the concentrations of the reactants and a proportionality constant, k, known as the rate constant. The units of the rate constant depend on the overall order of the reaction.


Step 2: Key Formula or Approach:

The general formula for the units of the rate constant, k, for a reaction of order 'n' is: \[ Units of k = (Concentration)^{1-n} (Time)^{-1} \]
In chemistry, concentration is typically expressed in mol L\(^{-1}\) (or M), and time is usually in seconds (s).
So, the formula becomes: \[ Units of k = (mol L^{-1})^{1-n} s^{-1} \]

Step 3: Detailed Explanation:

For a second-order reaction, the overall order n = 2.

Substitute n = 2 into the general formula: \[ Units of k = (mol L^{-1})^{1-2} s^{-1} \] \[ Units of k = (mol L^{-1})^{-1} s^{-1} \]
Now, apply the exponent -1 to both parts inside the parenthesis: \[ Units of k = (mol)^{-1} (L^{-1})^{-1} s^{-1} \] \[ Units of k = mol^{-1} L^{1} s^{-1} \]
This can be written as mol\(^{-1}\) L s\(^{-1}\).


Alternatively, consider a simple second-order rate law: \[ Rate = k[A]^2 \]
Rearrange to solve for the units of k: \[ Units of k = \frac{Units of Rate}{Units of [A]^2} \]
The units are:

Rate: mol L\(^{-1}\)s\(^{-1}\)
[A]\(^2\): (mol L\(^{-1}\))\(^2\) = mol\(^2\) L\(^{-2}\)
\[ Units of k = \frac{mol L^{-1}s^{-1}}{mol^2 L^{-2}} = mol^{1-2} L^{-1-(-2)} s^{-1} = mol^{-1} L^{1} s^{-1} \]

Step 4: Final Answer:

The unit of the rate constant for a second-order reaction is mol\(^{-1}\) L s\(^{-1}\).
Quick Tip: You can quickly determine the units for any order. For a zero-order (n=0), it's mol L\(^{-1}\)s\(^{-1}\). For a first-order (n=1), it's s\(^{-1}\). For a second-order (n=2), it's L mol\(^{-1}\)s\(^{-1}\). Notice that the unit of time (s\(^{-1}\)) is always present.

Step 1: Understanding the Concept:

This question requires an understanding of the key characteristics of zero-order and first-order reactions, including their rate laws, integrated rate laws, half-lives, and the effect of temperature on reaction rates. We need to identify the statement that is factually incorrect.


Step 2: Detailed Explanation:

Let's evaluate each statement:


(A) For a zero-order reaction, the rate law is Rate = k[A]\(^0\) = k. The rate is constant and does not depend on the concentration of the reactant [A]. This statement is correct.


(B) For a first-order reaction, the half-life (t\(_{1/2}\)) is given by the formula t\(_{1/2}\) = 0.693/k. This equation shows that the half-life depends only on the rate constant (k) and is independent of the initial concentration of the reactant. This statement is correct.


(C) The relationship between the rate constant (k) and temperature (T) is described by the Arrhenius equation, k = A\(e^{-E_a/RT}\). According to this equation, as temperature increases, the value of the rate constant k also increases, leading to a faster reaction rate. This statement is correct.


(D) For a zero-order reaction, the integrated rate law is [A]\(_t\) = -kt + [A]\(_0\). This equation is in the form of a straight line, y = mx + c, where y = [A]\(_t\), x = t, the slope m = -k (a negative slope), and the y-intercept c = [A]\(_0\). Thus, a plot of reactant concentration versus time is a straight line with a negative slope. This statement is correct.


(E) For a first-order reaction, 75% completion means that the reactant concentration has dropped to 25% of its initial value.

After one half-life (t\(_{1/2}\)), the concentration becomes 50% of the initial value.
After a second half-life, the concentration becomes 50% of the previous value, which is 0.5 \(\times\) 50% = 25% of the initial value.

Therefore, the time required for 75% completion (t\(_{75%}\)) is equal to two half-lives (t\(_{75%}\) = 2 \(\times\) t\(_{1/2}\)). The statement says it is thrice the half-life period. This statement is incorrect.


Step 3: Final Answer:

The incorrect statement is (E).
Quick Tip: For first-order reactions, remember the pattern of concentration reduction: 100% \(\xrightarrow{t_{1/2}\) 50% \(\xrightarrow{t_{1/2}}\) 25% \(\xrightarrow{t_{1/2}}\) 12.5%, and so on. The time to reach 75% completion (i.e., 25% remaining) is exactly two half-lives.

Step 1: Understanding the Concept:

The color of transition metal ions in solution is typically due to the absorption of light, which promotes an electron from a lower energy d-orbital to a higher energy d-orbital. This phenomenon is known as a d-d transition. For a d-d transition to occur, the ion must have a partially filled d-subshell (i.e., between d\(^1\) and d\(^9\) electrons). Ions with an empty d-subshell (d\(^0\)) or a completely filled d-subshell (d\(^{10}\)) cannot undergo d-d transitions and are therefore generally colorless.


Step 2: Detailed Explanation:

Let's determine the electronic configuration of the d-orbitals for each ion.


(A) Ca\(^{2+}\): Calcium (Z=20) has the configuration [Ar] 4s\(^2\). Ca\(^{2+}\) has the configuration [Ar]. It has no d-electrons. While it is colorless, Calcium is an alkaline earth metal (s-block), not a transition metal.


(B) Cr\(^{3+}\): Chromium (Z=24) is [Ar] 3d\(^5\) 4s\(^1\). Cr\(^{3+}\) is formed by removing three electrons, resulting in the configuration [Ar] 3d\(^3\). Since it has 3 d-electrons (partially filled), it is colored (typically green or violet).


(C) Ti\(^{4+}\): Titanium (Z=22) is [Ar] 3d\(^2\) 4s\(^2\). Ti\(^{4+}\) is formed by removing all four valence electrons, resulting in the configuration [Ar] 3d\(^0\). Since its d-subshell is empty, no d-d transition is possible. Therefore, Ti\(^{4+}\) is a colorless ion.


(D) Fe\(^{2+}\): Iron (Z=26) is [Ar] 3d\(^6\) 4s\(^2\). Fe\(^{2+}\) has the configuration [Ar] 3d\(^6\). With a partially filled d-subshell, it is colored (typically pale green).


(E) Fe\(^{3+}\): Fe\(^{3+}\) has the configuration [Ar] 3d\(^5\). With a partially filled d-subshell, it is colored (typically yellow or brown).


Step 3: Final Answer:

Among the given transition metal ions, Ti\(^{4+}\) has a d\(^0\) configuration and is therefore colorless.
Quick Tip: To quickly determine if a transition metal ion is colored, find its d-electron count. If the count is 1 to 9, it is almost always colored. If the count is 0 (like Sc\(^{3+}\), Ti\(^{4+}\)) or 10 (like Cu\(^+\), Zn\(^{2+}\)), it is colorless.

Step 1: Understanding the Concept:

The maximum oxidation state of 3d transition metals is often achieved when they combine with the most electronegative elements, namely oxygen and fluorine. We need to identify which of the given metals can be oxidized to a +6 state by fluorine.


Step 2: Detailed Explanation:

Let's examine the highest known fluorides for each metal:


(A) Titanium (Ti): Electronic configuration [Ar] 3d\(^2\) 4s\(^2\). Its maximum oxidation state is +4, as seen in TiF\(_4\).

(B) Chromium (Cr): Electronic configuration [Ar] 3d\(^5\) 4s\(^1\). It has 6 valence electrons and can achieve a +6 oxidation state. It forms chromium hexafluoride, CrF\(_6\), a volatile solid. In this compound, chromium is in the +6 oxidation state.

(C) Vanadium (V): Electronic configuration [Ar] 3d\(^3\) 4s\(^2\). Its maximum oxidation state is +5, forming vanadium pentafluoride, VF\(_5\).

(D) Manganese (Mn): Electronic configuration [Ar] 3d\(^5\) 4s\(^2\). Although it shows a +7 oxidation state with oxygen (in MnO\(_4^-\)), its highest fluoride is MnF\(_4\). The formation of higher fluorides is limited because the large size of multiple fluorine atoms around a small Mn ion leads to steric hindrance.

(E) Cobalt (Co): Electronic configuration [Ar] 3d\(^7\) 4s\(^2\). Its common oxidation states are +2 and +3. The highest known fluoride is CoF\(_3\), though some sources suggest CoF\(_4\) may exist. A +6 state is not achieved.



Step 3: Final Answer:

Chromium is the 3d metal from the list that forms a stable fluoride (CrF\(_6\)) in the +6 oxidation state.
Quick Tip: The highest oxidation state for the 3d elements from Scandium to Manganese generally corresponds to the total number of 4s and 3d electrons. This trend holds well for oxides (e.g., Mn\(_2\)O\(_7\)). With fluorides, steric hindrance can sometimes prevent the highest possible state from being reached (e.g., MnF\(_4\) vs Mn\(_2\)O\(_7\)).

Step 1: Understanding the Concept:

The industrial production of sulfuric acid is primarily done through the Contact Process. This process involves several steps, one of which is the catalytic oxidation of sulfur dioxide (SO\(_2\)) to sulfur trioxide (SO\(_3\)). This is the key and rate-determining step, requiring a specific catalyst to be efficient.


Step 2: Detailed Explanation:

The key reaction in the Contact Process is: \[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
This reaction is reversible and exothermic. To achieve a high yield at a reasonable rate, a catalyst is essential. The industry standard catalyst for this reaction is Vanadium(V) oxide, with the chemical formula V\(_2\)O\(_5\).


The other options are catalysts for different processes:

Nickel and Nickel(II) oxide are commonly used as catalysts in hydrogenation reactions (e.g., converting vegetable oils into margarine).
Chromium oxides are used in various catalytic processes, but not as the primary catalyst for the Contact Process.


Step 3: Final Answer:

Vanadium(V) oxide (V\(_2\)O\(_5\)) is the transition metal oxide used as a catalyst in the manufacture of sulfuric acid.
Quick Tip: It is highly beneficial to memorize the catalysts for major industrial processes: \textbf{Haber-Bosch Process} (Ammonia synthesis): Iron (Fe) catalyst with promoters like K\(_2\)O and Al\(_2\)O\(_3\). \textbf{Contact Process} (Sulfuric acid): Vanadium(V) oxide (V\(_2\)O\(_5\)). \textbf{Ostwald Process} (Nitric acid): Platinum-Rhodium (Pt-Rh) gauze.

Step 1: Understanding the Concept:

Ionic radius is the measure of an ion's size in a crystal lattice. For a series of isovalent ions (ions with the same charge) across a period in the periodic table, the ionic radius generally decreases. This is due to the increase in effective nuclear charge.


Step 2: Detailed Explanation:

All the ions given are from the 3d transition series and have a +3 charge (Sc\(^{3+}\), Ti\(^{3+}\), V\(^{3+}\), Cr\(^{3+}\), Mn\(^{3+}\)).

The elements are arranged in the periodic table in the order: Sc, Ti, V, Cr, Mn.
As we move from left to right (from Sc to Mn), the atomic number increases, meaning the number of protons in the nucleus increases.
The electrons are being added to the same (n-1)d subshell, which does not provide very effective shielding.
Consequently, the effective nuclear charge (the net positive charge experienced by the outer electrons) increases across the series.
This stronger pull from the nucleus draws the electron cloud closer, causing a steady decrease in the ionic radius.


The general trend for the radius of these +3 ions is: \[ Sc^{3+} > Ti^{3+} > V^{3+} > Cr^{3+} > Mn^{3+} \]
However, there can be slight irregularities due to electron-electron repulsion and crystal field stabilization effects. Let's consider the configurations:

Sc\(^{3+}\): 3d\(^0\)
Ti\(^{3+}\): 3d\(^1\)
V\(^{3+}\): 3d\(^2\)
Cr\(^{3+}\): 3d\(^3\) (a stable, half-filled t\(_{2g}\) configuration in an octahedral field)
Mn\(^{3+}\): 3d\(^4\)

The trend of decreasing radius holds firmly from Sc\(^{3+}\) to Cr\(^{3+}\). The Cr\(^{3+}\) ion is particularly small due to the combination of high effective nuclear charge and a stable d\(^3\) configuration. While the nuclear charge is even higher for Mn\(^{3+}\), the addition of the fourth d-electron can introduce repulsions (or occupation of a higher energy e\(_g\) orbital in a high-spin complex) that slightly counteract the contraction. Comparing actual data, Cr\(^{3+}\) (approx. 61.5 pm) is indeed smaller than high-spin Mn\(^{3+}\) (approx. 64.5 pm). Therefore, among the given options, Cr\(^{3+}\) has the least ionic radius.


Step 3: Final Answer:

Based on the periodic trend of increasing effective nuclear charge, Cr\(^{3+}\) is the ion with the least ionic radius among the choices.
Quick Tip: For isovalent ions across a period (e.g., M\(^{3+}\) ions in the 3d series), the primary rule is that radius decreases as atomic number increases. Remember this trend as it holds true for the majority of cases and is a fundamental concept in periodic properties.

Step 1: Understanding the Concept:

The question asks for a reagent that can oxidize Manganese from a +2 oxidation state (in Mn(II) salts like MnSO\(_4\)) to a +7 oxidation state (in permanganate, MnO\(_4^-\)). This is a significant increase in oxidation state and requires a very powerful oxidizing agent.


Step 2: Detailed Explanation:

Let's evaluate the oxidizing power of the given reagents in this context.

The target reaction involves the transformation: Mn\(^{2+}\) \(\rightarrow\) MnO\(_4^-\).



(A) KClO\(_3\) (Potassium chlorate): A strong oxidizing agent, but typically not strong enough to convert Mn\(^{2+}\) to MnO\(_4^-\) quantitatively in solution.
(B) K\(_2\)S\(_2\)O\(_8\) (Potassium peroxodisulfate or persulfate): The peroxodisulfate ion (S\(_2\)O\(_8^{2-}\)) is an extremely powerful oxidizing agent, especially in the presence of a catalyst like Ag\(^+\) ions. It is well-known for its ability to oxidize Mn\(^{2+}\) to the deep purple permanganate ion. The balanced reaction in acidic solution is:

\[ 2Mn^{2+} + 5S_2O_8^{2-} + 8H_2O \rightarrow 2MnO_4^{-} + 10SO_4^{2-} + 16H^+ \]
(C) KNO\(_3\) (Potassium nitrate): A moderate oxidizing agent, insufficient for this transformation.
(D) KClO\(_4\) (Potassium perchlorate): Although the chlorine is in its highest oxidation state (+7), the perchlorate ion is a surprisingly sluggish oxidizing agent at room temperature in solution.
(E) KOH (Potassium hydroxide): This is a base, not an oxidizing agent.


Step 3: Final Answer:

Potassium peroxodisulfate (K\(_2\)S\(_2\)O\(_8\)) is the powerful oxidizing agent required to convert Mn(II) ions to permanganate.
Quick Tip: In qualitative analysis, the test for manganese ions often involves oxidation with strong agents like peroxodisulfate (S\(_2\)O\(_8^{2-}\)), lead dioxide (PbO\(_2\)), or sodium bismuthate (NaBiO\(_3\)) to form the characteristic purple MnO\(_4^-\) ion. Remembering these specific reagents is useful.

Step 1: Understanding the Concept:

The electronic configurations of lanthanoids involve the filling of the 4f subshell. Generally, electrons are added to the 4f orbitals as the atomic number increases. However, there are anomalies in this filling pattern, particularly at configurations that result in a half-filled (4f\(^7\)) or completely filled (4f\(^{14}\)) subshell, which are exceptionally stable.


Step 2: Detailed Explanation:

Let's look at the electronic configurations of the lanthanoids listed:


The general configuration is [Xe] 4f\(^n\) 5d\(^0\) 6s\(^2\) or [Xe] 4f\(^n\) 5d\(^1\) 6s\(^2\).
Europium (Eu, Z=63): To achieve the stable half-filled f-orbital configuration, its configuration is [Xe] 4f\(^7\)6s\(^2\). It does not have a 5d electron.
Gadolinium (Gd, Z=64): This element comes immediately after Europium. The next electron to be added could either go into the 4f orbital (to make 4f\(^8\)) or into the 5d orbital. To preserve the high stability of the half-filled 4f\(^7\) subshell, the differentiating electron enters the 5d orbital. Therefore, the configuration of Gadolinium is [Xe] 4f\(^7\)5d\(^1\)6s\(^2\). This matches the configuration given in the question.

For completeness, the configurations of the other options are:

(A) Neodymium (Nd, Z=60): [Xe] 4f\(^4\)6s\(^2\)
(B) Samarium (Sm, Z=62): [Xe] 4f\(^6\)6s\(^2\)
(E) Holmium (Ho, Z=67): [Xe] 4f\(^{11}\)6s\(^2\)


Step 3: Final Answer:

The lanthanoid with the outer electronic configuration 4f\(^7\)5d\(^1\)6s\(^2\) is Gadolinium.
Quick Tip: The stability of half-filled and fully-filled orbitals is a key principle in determining electron configurations. For the lanthanoids, this is most prominent at Europium (4f\(^7\)), Gadolinium (which keeps the 4f\(^7\) core), Ytterbium (4f\(^{14}\)), and Lutetium (which keeps the 4f\(^{14}\) core). Memorizing these anomalies is very helpful for exams.

Step 1: Understanding the Concept:

The question requires us to apply the rules of IUPAC nomenclature for coordination compounds to name the given complex. This involves correctly identifying and naming the ligands, the central metal ion, its oxidation state, and the counter-ion.


Step 2: Key Formula or Approach (IUPAC Rules):

1. Cation first, then Anion: Name the positive ion (cation) before the negative ion (anion).
2. Naming the Complex Ion:
a. Ligands: Name the ligands first, in alphabetical order. Use prefixes (di-, tri-, etc.) to indicate the number of each ligand.
b. Metal: Name the central metal atom.
c. Oxidation State: Indicate the oxidation state of the central metal with a Roman numeral in parentheses.
3. Naming the Counter-ion: Name the counter-ion. Prefixes are not used for simple counter-ions.


Step 3: Detailed Explanation:

Let's apply these rules to [Co(NH\(_3\))\(_{3}\)(H\(_2\)O)\(_{3}\)]Cl\(_{3}\).


1. Identify Cation and Anion:

The complex is an ionic compound. The part in square brackets is the cation, and the Cl atoms are the anions.
Cation: [Co(NH\(_3\))\(_{3}\)(H\(_2\)O)\(_{3}\)]\(^{3+}\)
Anion: Cl\(^-\) (three of them)


2. Name the Complex Cation:

a. Name and Order the Ligands:
- NH\(_3\) is named ammine.
- H\(_2\)O is named aqua.
- Alphabetically, "ammine" comes before "aqua".
- There are three of each, so we use the prefix "tri-".
- The ligand part of the name is: triamminetriaqua.

b. Name the Metal:
- The central metal is Cobalt (Co). Since the complex ion is a cation, the name remains cobalt.

c. Determine the Oxidation State:
- Let the oxidation state of Co be 'x'.
- NH\(_3\) and H\(_2\)O are neutral ligands (charge = 0).
- Each chloride counter-ion has a charge of -1. There are three chlorides, for a total of -3.
- The overall compound is neutral.
- So, x + 3(0) + 3(0) + 3(-1) = 0
- x - 3 = 0 \(\Rightarrow\) x = +3
- The oxidation state is (III).

d. Combine the parts of the cation name: triamminetriaquacobalt(III)


3. Name the Counter-ion:

The anion is Cl\(^-\), which is named chloride. We do not use the prefix "tri-" for the counter-ion, even though there are three of them.


4. Full Name:

Combining the cation and anion names, we get: triamminetriaquacobalt(III) chloride.


Step 4: Final Answer:

This matches option (B).
Quick Tip: A common mistake in naming coordination compounds is the alphabetical ordering of ligands. Remember to alphabetize the ligand names themselves (e.g., ammine vs. aqua), not the numerical prefixes (e.g., triammine vs. triaqua).

Step 1: Understanding the Concept:

Isomers are compounds that have the same chemical formula but different arrangements of atoms. In coordination chemistry, there are several types of structural isomerism. Ionization isomerism is a specific type where the difference lies in the exchange of a ligand within the coordination sphere with a counter-ion outside the coordination sphere.


Step 2: Detailed Explanation:

Let's analyze the two given complexes:

Complex 1: [Co(NH\(_3\))\(_5\)Br]SO\(_4\)

Coordination Sphere: [Co(NH\(_3\))\(_5\)Br]\(^{2+}\)
Ligands: Five ammine (NH\(_3\)) ligands and one bromo (Br\(^-\)) ligand.
Counter-ion: Sulfate ion (SO\(_4^{2-}\)).

Complex 2: [Co(NH\(_3\))\(_5\)SO\(_4\)]Br

Coordination Sphere: [Co(NH\(_3\))\(_5\)SO\(_4\)]\(^+\)
Ligands: Five ammine (NH\(_3\)) ligands and one sulfato (SO\(_4^{2-}\)) ligand.
Counter-ion: Bromide ion (Br\(^-\)).


Both complexes have the same overall chemical formula: Co(NH\(_3\))\(_5\)BrSO\(_4\).

Comparing the two structures, we see that the bromide ion (Br\(^-\)), which is a ligand in the first complex, has become the counter-ion in the second complex. Conversely, the sulfate ion (SO\(_4^{2-}\)), which is the counter-ion in the first complex, has become a ligand in the second complex.

This exchange of an anionic ligand with a counter-ion is the definition of ionization isomerism. These isomers give different ions when dissolved in water. For example, the first complex will give a positive test for SO\(_4^{2-}\) ions (e.g., with BaCl\(_2\)), while the second will give a positive test for Br\(^-\) ions (e.g., with AgNO\(_3\)).


Step 3: Final Answer:

The two complexes are ionization isomers because they differ by the exchange of a ligand (Br\(^-\)) with a counter-ion (SO\(_4^{2-}\)). Therefore, option (D) is correct.
Quick Tip: To quickly identify ionization isomerism, check if a counter-ion in one complex appears as a ligand in the other complex. If you can swap an anionic ligand with a counter-ion to get the other isomer, it's ionization isomerism.

Step 1: Understanding the Concept:

Wilkinson's catalyst is a well-known coordination complex used for the homogeneous hydrogenation of alkenes. This question is a direct test of knowledge about the composition of this specific catalyst.


Step 2: Detailed Explanation:

Wilkinson's catalyst is the common name for the chemical compound chloridotris(triphenylphosphine)rhodium(I).

Its chemical formula is [RhCl(PPh\(_3\))\(_3\)].


The central metal ion is Rhodium (Rh).
The oxidation state of Rhodium in this complex is +1.
The ligands are one chloride ion (Cl\(^-\)) and three triphenylphosphine (PPh\(_3\)) molecules.

It is a square planar, 16-electron complex that appears as a reddish-brown solid. It gained prominence for its ability to catalyze the addition of H\(_2\) to alkenes and alkynes at room temperature and atmospheric pressure.


Step 3: Final Answer:

The metal present in Wilkinson's catalyst, [RhCl(PPh\(_3\))\(_3\)], is Rhodium. Therefore, option (D) is correct.
Quick Tip: Certain catalysts are frequently asked about in exams. It's wise to memorize the metal ion and formula for key catalysts: \textbf{Wilkinson's Catalyst:} Rhodium (Rh) \textbf{Ziegler-Natta Catalyst:} Titanium and Aluminium (e.g., TiCl\(_4\)/Al(C\(_2\)H\(_5\))\(_3\)) \textbf{Haber's Process Catalyst:} Iron (Fe) \textbf{Contact Process Catalyst:} Vanadium (V)

Step 1: Understanding the Concept:

A "spin free" complex is another term for a high-spin complex. In Crystal Field Theory, this occurs when the crystal field splitting energy (\(\Delta_o\)) is smaller than the electron pairing energy (P). As a result, electrons will occupy the higher energy e\(_g\) orbitals before pairing up in the lower energy t\(_{2g}\) orbitals. This typically happens with weak-field ligands. A "spin paired" or low-spin complex occurs with strong-field ligands, where \(\Delta_o >\) P, and electrons pair up in the t\(_{2g}\) orbitals first.


Step 2: Detailed Explanation:

Let's analyze each complex:


(A) [Ni(CO)\(_4\)]: Nickel is in the 0 oxidation state (Ni\(^0\)), with a d\(^{10}\) configuration ([Ar] 3d\(^8\)4s\(^2\) \(\rightarrow\) 3d\(^{10}\)). CO is a very strong-field ligand. The complex is tetrahedral. Since the d-shell is full, there are no unpaired electrons. It is diamagnetic, not high-spin.


(B) [Co(NH\(_3\))\(_6\)]\(^{3+}\): Cobalt is in the +3 oxidation state (Co\(^{3+}\)), with a d\(^6\) configuration. NH\(_3\) is generally a strong-field ligand, especially with Co\(^{3+}\). This leads to a low-spin complex (\(\Delta_o >\) P). The electrons will pair up in the t\(_{2g}\) orbitals. Configuration: t\(_{2g}^6\)e\(_g^0\). There are 0 unpaired electrons. This is a low-spin complex.


(C) [Ni(CN)\(_4\)]\(^{2-}\): Nickel is in the +2 oxidation state (Ni\(^{2+}\)), with a d\(^8\) configuration. CN\(^-\) is a very strong-field ligand. The complex is square planar. The strong field causes pairing of electrons. The configuration results in 0 unpaired electrons. This is a low-spin complex.


(D) [CoF\(_6\)]\(^{3-}\): Cobalt is in the +3 oxidation state (Co\(^{3+}\)), with a d\(^6\) configuration. F\(^-\) is a classic weak-field ligand. Therefore, the crystal field splitting will be small (\(\Delta_o <\) P). The d\(^6\) electrons will fill the orbitals to maximize spin (Hund's rule), resulting in a high-spin configuration: t\(_{2g}^4\)e\(_g^2\). This configuration has 4 unpaired electrons. This is a spin-free (high-spin) complex.


(E) [Mn(CN)\(_6\)]\(^{3-}\): Manganese is in the +3 oxidation state (Mn\(^{3+}\)), with a d\(^4\) configuration. CN\(^-\) is a very strong-field ligand. This will be a low-spin complex (\(\Delta_o >\) P). The d\(^4\) electrons will all occupy the t\(_{2g}\) orbitals. Configuration: t\(_{2g}^4\)e\(_g^0\). This has 2 unpaired electrons, but it is a spin-paired configuration.


Step 3: Final Answer:

The complex [CoF\(_6\)]\(^{3-}\) is formed with a weak-field ligand (F\(^-\)), which leads to a high-spin (spin-free) electron configuration. Therefore, option (D) is correct.
Quick Tip: To identify spin-free (high-spin) complexes, look for weak-field ligands. Remember the general spectrochemical series: (Weak) I\(^-\) < Br\(^-\) < Cl\(^-\) < F\(^-\) < H\(_2\)O < NH\(_3\) < en < CN\(^-\) < CO (Strong). Halide ions (F\(^-\), Cl\(^-\), etc.) almost always form high-spin complexes. CN\(^-\) and CO almost always form low-spin complexes.

Step 1: Understanding the Concept:

The hybridization of a carbon atom can be determined by counting the number of sigma (\(\sigma\)) bonds and lone pairs around it (its steric number).

Steric Number 4 (e.g., 4 single bonds) \(\rightarrow\) sp\(^3\) hybridization (tetrahedral geometry).
Steric Number 3 (e.g., 1 double bond, 2 single bonds) \(\rightarrow\) sp\(^2\) hybridization (trigonal planar geometry).
Steric Number 2 (e.g., 1 triple bond, 1 single bond OR 2 double bonds) \(\rightarrow\) sp hybridization (linear geometry).


Step 2: Detailed Explanation:

Let's analyze the molecule CH\(_3\)--CH=CH--C\(\equiv\)N by numbering the carbons from left to right.
\[ \underset{(1)}{CH_3} - \underset{(2)}{CH} = \underset{(3)}{CH} - \underset{(4)}{C}\equivN \]

Carbon 1 (in CH\(_3\)):

This carbon is bonded to three hydrogen atoms via single bonds and one carbon atom via a single bond. It has a total of 4 sigma bonds and 0 lone pairs.
Steric Number = 4. Hybridization is sp\(^3\).


Carbon 2 (in -CH=):

This carbon is bonded to one hydrogen (single bond), carbon 1 (single bond), and carbon 3 (double bond). It has 3 sigma bonds and 1 pi bond.
Steric Number = 3. Hybridization is sp\(^2\).


Carbon 3 (in =CH-):

This carbon is bonded to one hydrogen (single bond), carbon 2 (double bond), and carbon 4 (single bond). It has 3 sigma bonds and 1 pi bond.
Steric Number = 3. Hybridization is sp\(^2\).


Carbon 4 (in -CN):

This carbon is bonded to carbon 3 (single bond) and to a nitrogen atom (triple bond). It has 2 sigma bonds and 2 pi bonds.
Steric Number = 2. Hybridization is sp.


Step 3: Final Answer:

The sequence of hybridization from left to right (C1 to C4) is sp\(^3\), sp\(^2\), sp\(^2\), sp. This corresponds to option (C).
Quick Tip: A quick way to determine hybridization for hydrocarbons: Carbon in an alkane (all single bonds) is \textbf{sp\(^3\)}. Carbon in an alkene (one C=C double bond) is \textbf{sp\(^2\)}. Carbon in an alkyne (one C\(\equiv\)C triple bond) is \textbf{sp}. Apply this to each carbon atom individually based on the bonds it forms.

Step 1: Understanding the Concept:

The Resonance effect (or Mesomeric effect) describes the delocalization of \(\pi\) electrons in a conjugated system. It can be electron-donating (+R or +M effect) or electron-withdrawing (-R or -M effect).

+R effect: A group donates electrons to the conjugated system. This is typically seen in groups with lone pairs of electrons on the atom directly attached to the system (e.g., -OH, -NH\(_2\), -OR, halogens).
-R effect: A group withdraws electrons from the conjugated system. This is typically seen in groups that have a multiple bond (double or triple) between atoms of different electronegativity, where the atom attached to the system is less electronegative (e.g., -NO\(_2\), -CN, -CHO, -COOH).


Step 2: Detailed Explanation:

Let's analyze the effect of each group when attached to a conjugated system like a benzene ring:


(A) --OH (Hydroxyl): The oxygen atom has lone pairs of electrons which it can donate to the ring through resonance. This is a +R effect.


(B) --OCOR (Ester): The oxygen atom directly attached to the system has lone pairs that it can donate. This group exhibits a +R effect.


(C) --NH\(_2\) (Amino): The nitrogen atom has a lone pair of electrons that it can donate to the ring via resonance. This is a strong +R effect.


(D) --CN (Cyano): This group has a carbon triple-bonded to a more electronegative nitrogen atom (C\(\equiv\)N). When attached to a conjugated system, the \(\pi\) electrons can be pulled from the system towards the nitrogen atom. This withdrawal of electron density via resonance is a -R effect.

For example, in cyanobenzene, the electrons from the ring can be delocalized as follows:

\chemfig{C_6H_5-C_(-[::-90])~N_(-[::-90]) \(\leftrightarrow\) \chemfig{(=[::+60]-[::-60]=[:-120]-[::+120]=)_((+))-[::-60]C_(-[::-90])=N_(-[::-90])^{(-)


Step 3: Final Answer:

The --CN group is an electron-withdrawing group that operates through the -R effect. The other groups show the +R effect. Therefore, option (D) is correct.
Quick Tip: A simple rule to identify +R and -R groups: \textbf{+R Groups}: The atom directly bonded to the conjugated system has a lone pair (e.g., \textbf{O}H, \textbf{N}H\(_2\), \textbf{Cl}). \textbf{-R Groups}: The group contains a double or triple bond, and the atom directly bonded to the system is less electronegative than the atom it's multiple-bonded to (e.g., -\textbf{C}HO, -\textbf{C}N, -\textbf{N}O\(_2\)).

Step 1: Understanding the Concept:

This question tests the IUPAC nomenclature for alkanes, specifically the root word for a hydrocarbon with 20 carbon atoms. First, we need to determine the type of hydrocarbon from its molecular formula.


Step 2: Detailed Explanation:

1. Determine the Hydrocarbon Type:

The given molecular formula is C\(_20\)H\(_42\). Let's check if it fits the general formula for saturated alkanes, which is C\(_n\)H\(_{2n+2}\).

For n = 20, the formula should be C\(_20\)H\(_{2 \times 20 + 2}\) = C\(_20\)H\(_{42}\).

The formula matches, so the compound is an alkane. The name should therefore end with the suffix "-ane".


2. Determine the IUPAC Root Word:

We need the IUPAC prefix for a chain of 20 carbon atoms.

1: meth-
2: eth-
10: dec-
11: undec-
12: dodec-
20: eicos-


3. Construct the Name:

Combining the root word for 20 carbons ("eicos-") with the suffix for an alkane ("-ane") gives the name Eicosane.


4. Analyze the Other Options:

(A, B, C, D) These names are incorrect. "Decane" refers to 10 carbons (C\(_10\)), "dodecane" to 12 carbons (C\(_12\)). The prefixes "di-" or "do-" are not used in this way to signify 20. "Didocene" suggests an alkene ("-ene") which is incorrect.


Step 3: Final Answer:

The correct IUPAC name for the alkane C\(_20\)H\(_42\) is Eicosane. Therefore, option (E) is correct.
Quick Tip: Memorizing the IUPAC prefixes for carbon chains beyond 10 is crucial for organic chemistry questions. 11: Undec- 12: Dodec- 13: Tridec- 20: Eicos- 30: Triacont- First, always check the C-to-H ratio to determine if it's an alkane, alkene, or alkyne.

Step 1: Understanding the Concept:

The reaction described is a key industrial process known as catalytic reforming or aromatization. This process converts aliphatic hydrocarbons (like n-alkanes) into aromatic hydrocarbons. It involves a combination of cyclization and dehydrogenation.


Step 2: Detailed Explanation:

The starting material is n-hexane (CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_3\)), an alkane with a six-carbon chain.

The conditions are:

Catalyst: Oxides of molybdenum (Mo\(_2\)O\(_3\)), chromium (Cr\(_2\)O\(_3\)), or vanadium (V\(_2\)O\(_5\)) supported on alumina (Al\(_2\)O\(_3\)).
Temperature: High temperature (around 773 K or 500 °C).
Pressure: High pressure (10-20 atm).

Under these conditions, alkanes with six or more carbons undergo two main steps:

1. Cyclization: The linear n-hexane chain first cyclizes to form cyclohexane.
\[ n-hexane \xrightarrow{catalyst} cyclohexane + H_2 \]
2. Dehydrogenation (Aromatization): The cyclohexane formed then loses hydrogen atoms (three molecules of H\(_2\)) to become the stable aromatic compound, benzene.
\[ cyclohexane \xrightarrow{catalyst} benzene + 3H_2 \]
The overall reaction is: \[ CH_3(CH_2)_4CH_3 \xrightarrow[773K, 10-20 atm]{Mo_2O_3} C_6H_6 + 4H_2 \]
The final, stable product of this aromatization reaction is benzene.


Step 3: Final Answer:

The reaction of n-hexane under these conditions is aromatization, which yields benzene. Therefore, option (D) is correct.
Quick Tip: Remember that alkanes with 6, 7, or 8 carbon atoms are particularly suitable for aromatization. n-Hexane \(\rightarrow\) Benzene n-Heptane \(\rightarrow\) Toluene (Methylbenzene) n-Octane \(\rightarrow\) Ethylbenzene or Xylenes This is a high-yield question if you recognize the reagents and conditions for catalytic reforming.

Step 1: Understanding the Concept:

Ozonolysis is a reaction where alkenes are cleaved at the carbon-carbon double bond by reaction with ozone (O\(_3\)), followed by a work-up step (e.g., with Zn/H\(_2\)O). Each carbon atom of the original double bond becomes a carbonyl carbon (C=O) in the products. To identify the original alkene, we can reverse the process by joining the carbonyl carbons of the products.


Step 2: Detailed Explanation:

The products of the ozonolysis are:

Propan-2-one (also known as acetone): The structure is (CH\(_3\))\(_2\)C=O.
Formaldehyde (also known as methanal): The structure is H\(_2\)C=O.


To find the parent alkene, we remove the oxygen atoms from the carbonyl groups and join the two carbon atoms with a double bond: \[ (CH_3)_2C=O \quad + \quad O=CH_2 \]
Removing the oxygens and joining the carbons gives: \[ (CH_3)_2C = CH_2 \]
This structure is 2-methylpropene. The common name for 2-methylpropene is Isobutene.


Let's check the other options:

(A) 2-Butene (CH\(_3\)CH=CHCH\(_3\)) would give two moles of ethanal.
(B) 1-Butene (CH\(_3\)CH\(_2\)CH=CH\(_2\)) would give propanal and formaldehyde.
(D) 2-Methyl-2-butene ((CH\(_3\))\(_2\)C=CHCH\(_3\)) would give propan-2-one and ethanal.
(E) 2,3-Dimethyl-2-butene ((CH\(_3\))\(_2\)C=C(CH\(_3\))\(_2\)) would give two moles of propan-2-one.


Step 3: Final Answer:

The alkene that yields propan-2-one and formaldehyde upon ozonolysis is 2-methylpropene, which is commonly known as isobutene. Therefore, option (C) is correct.
Quick Tip: Working backward from ozonolysis products is a common exam question. Simply take the two carbonyl products, erase their oxygen atoms, and form a double bond between the two carbon atoms that were bonded to the oxygens. This will give you the structure of the original alkene.

Step 1: Understanding the Concept:

The acidity of alkynes is due to the C-H bond where the carbon is sp-hybridized. In an sp-hybridized orbital, there is 50% s-character. The high s-character means the electrons are held closer to the nucleus, making the carbon atom more electronegative. This high electronegativity stabilizes the negative charge on the conjugate base (the acetylide anion, R--C\(\equiv\)C\(^-\)) formed after the proton (H\(^+\)) is lost. Only terminal alkynes (which have a \(\equiv\)C--H bond) are acidic. Internal alkynes (R--C\(\equiv\)C--R') have no acidic hydrogen on the triple-bonded carbons.


Step 2: Detailed Explanation:

Let's analyze the acidity of the given compounds:


1. CH \(\equiv\) CH (Ethyne or Acetylene):

This is a terminal alkyne with two acidic hydrogens. The sp-hybridized carbons make these hydrogens acidic.


2. CH\(_3\) -- C \(\equiv\) CH (Propyne):

This is also a terminal alkyne with one acidic hydrogen (the one attached to the terminal sp-carbon). However, it has a methyl group (CH\(_3\)) attached to the triple bond. The methyl group is an electron-donating group (+I effect). This electron-donating effect destabilizes the negative charge on the acetylide anion (CH\(_3\)--C\(\equiv\)C\(^-\)) after the proton is lost, making it a weaker acid compared to ethyne.


3. CH\(_3\) -- C \(\equiv\) C -- CH\(_3\) (But-2-yne):

This is an internal alkyne. There are no hydrogen atoms directly bonded to the sp-hybridized carbons. Therefore, it is not acidic at all in this context. Its acidity is negligible compared to terminal alkynes.


Step 3: Ordering the Acidity:

Based on the analysis:

Ethyne is the most acidic because it has two acidic hydrogens and no destabilizing electron-donating groups.
Propyne is less acidic than ethyne due to the destabilizing +I effect of the methyl group.
But-2-yne is the least acidic (essentially non-acidic) because it is an internal alkyne with no terminal C-H bond.

The correct decreasing order of acidity is:

CH \(\equiv\) CH \(>\) CH\(_3\) -- C \(\equiv\) CH \(>\) CH\(_3\) -- C \(\equiv\) C -- CH\(_3\)


This matches option (A).
Quick Tip: Remember two key factors for alkyne acidity: 1. \textbf{Terminal vs. Internal}: Only terminal alkynes are acidic. Internal alkynes are not. 2. \textbf{Inductive Effects}: Electron-donating groups (like alkyl groups) attached to the C\(\equiv\)C bond decrease acidity by destabilizing the conjugate base. Electron-withdrawing groups would increase acidity.

Step 1: Understanding the Concept:

The reaction of toluene with chlorine can proceed via two different mechanisms depending on the reaction conditions. It is crucial to distinguish between them:

Electrophilic Aromatic Substitution (EAS): Occurs on the benzene ring. It requires a Lewis acid catalyst (like FeCl\(_3\), AlCl\(_3\), or Fe) and is typically done in the dark to prevent radical reactions.
Free Radical Substitution (FRS): Occurs on the alkyl side-chain (the methyl group). It requires energy input in the form of UV light or high temperature and does not use a Lewis acid catalyst.


Step 2: Detailed Explanation:

The given reaction conditions are:

Reagent: Cl\(_2\)
Catalyst: Fe (Iron metal, which reacts with Cl\(_2\) to form the actual Lewis acid catalyst, FeCl\(_3\))
Condition: In the dark.

These conditions are characteristic of Electrophilic Aromatic Substitution. The Lewis acid (FeCl\(_3\)) polarizes the Cl-Cl bond, generating a strong electrophile, "Cl\(^+\)", which then attacks the electron-rich benzene ring.


The starting material is toluene (methylbenzene). The methyl group (--CH\(_3\)) is an activating group and is ortho, para-directing. This means it directs incoming electrophiles to the positions ortho (C2, C6) and para (C4) to the methyl group.

Therefore, the chlorination of toluene under these conditions will produce a mixture of two main products:

o-Chlorotoluene (ortho-chlorotoluene)
p-Chlorotoluene (para-chlorotoluene)


Option (A), Benzyl chloride (C\(_6\)H\(_5\)CH\(_2\)Cl), would be the product of Free Radical Substitution if the reaction were carried out in the presence of UV light.


Step 3: Final Answer:

The reaction is an electrophilic aromatic substitution on the ring, directed by the methyl group to the ortho and para positions. The product is a mixture of o- and p-Chlorotoluene. Therefore, option (B) is correct.
Quick Tip: For halogenation of alkylbenzenes, the conditions are the key to the product: \textbf{Halogen + Lewis Acid (FeCl\(_3\), AlBr\(_3\)) + Dark} \(\rightarrow\) Ring substitution (ortho/para). \textbf{Halogen + UV light / Heat} \(\rightarrow\) Side-chain substitution (benzylic position). Memorize this distinction as it is a very common topic for questions.

Step 1: Understanding the Concept:

This question asks about the primary industrial method for the large-scale synthesis of phenol. While several methods exist, one particular process dominates modern industrial production due to its efficiency and the production of a valuable co-product.


Step 2: Detailed Explanation:

Let's review the major methods for phenol synthesis:

Dow Process: This older method involves the hydrolysis of chlorobenzene with NaOH at high temperature and pressure. While historically important, it is largely obsolete. So, (A) is not the current large-scale method.
From Benzene Sulfonic Acid: Benzene is sulfonated and then fused with NaOH. This is also an older method.
From Aniline (Diazotization): Aniline can be converted to a diazonium salt, which is then hydrolyzed to phenol. This is a useful laboratory synthesis but not an industrial large-scale process.
The Cumene Process: This is the dominant industrial method, accounting for about 95% of global phenol production. The process starts with benzene and propene, which are reacted to form cumene (isopropylbenzene).

The cumene is then oxidized with air to form cumene hydroperoxide.

Finally, the cumene hydroperoxide is treated with dilute acid (like H\(_2\)SO\(_4\)) to yield phenol and acetone (a valuable co-product).

\[ Benzene + Propene \xrightarrow{H_3PO_4} Cumene \xrightarrow[2. H^+/H_2O]{1. O_2} Phenol + Acetone \]

Because cumene is the key intermediate that is directly converted to phenol in the main step, it is considered the compound used for the manufacture.


Step 3: Final Answer:

The modern, large-scale industrial manufacture of phenol is predominantly done via the Cumene Process, which uses cumene as the intermediate. Therefore, option (D) is correct.
Quick Tip: When a question asks about the industrial production of phenol, the answer is almost always related to the \textbf{Cumene Process}. Remember the key reactant (Cumene) and the two products (Phenol and Acetone). This process is favored because it is economical and produces two useful chemicals.

Step 1: Understanding the Concept:

This question asks to identify a specific named reaction in organic chemistry based on its reactants and products. The reaction involves the coupling of two aryl halide molecules using sodium metal.


Step 2: Detailed Explanation:

Let's define the reactions listed in the options:

(A) Fittig reaction: This reaction involves the coupling of two molecules of an aryl halide using sodium metal in the presence of dry ether to form a biphenyl compound.
\[ 2Ar-X + 2Na \xrightarrow{Dry Ether} Ar-Ar + 2NaX \]
This exactly matches the description in the question.

(B) Wurtz-Fittig reaction: This is a modification where one molecule of an aryl halide and one molecule of an alkyl halide are reacted with sodium in dry ether to form an alkylated aromatic compound.
\[ Ar-X + R-X + 2Na \xrightarrow{Dry Ether} Ar-R + 2NaX \]

(C) Swarts reaction: This is a halogen exchange reaction used to prepare alkyl fluorides by heating an alkyl chloride or bromide with a metallic fluoride (like AgF, Hg\(_2\)F\(_2\)). It does not involve sodium or aryl halides in this context.

(D) Williamson's synthesis: This is a method for preparing ethers by reacting an alkyl halide with a sodium alkoxide or sodium phenoxide.
\[ R-O^-Na^+ + R'--X \rightarrow R-O-R' + NaX \]

(E) Kolbe's reaction: This reaction is used to prepare salicylic acid by treating sodium phenoxide with carbon dioxide under pressure.


The reaction described in the question -- an aryl halide reacting with Na to form a biphenyl -- is the definition of the Fittig reaction.


Step 3: Final Answer:

The described reaction is the Fittig reaction. Therefore, option (A) is correct.
Quick Tip: Remember the Wurtz family of reactions: \textbf{Wurtz Reaction:} Alkyl halide + Alkyl halide + Na \(\rightarrow\) Alkane \textbf{Fittig Reaction:} Aryl halide + Aryl halide + Na \(\rightarrow\) Biphenyl \textbf{Wurtz-Fittig Reaction:} Alkyl halide + Aryl halide + Na \(\rightarrow\) Alkylarene All three use sodium metal in dry ether as the reagent.

Step 1: Understanding the Concept:

This is a question of recalling the chemical name for a very common pharmaceutical compound, Aspirin.


Step 2: Detailed Explanation:

Aspirin is one of the most widely used medications for pain relief (analgesic), fever reduction (antipyretic), and as an anti-inflammatory drug. Its chemical structure is derived from salicylic acid.


Salicylic acid is 2-hydroxybenzoic acid. It has a hydroxyl group (--OH) and a carboxylic acid group (--COOH) attached to a benzene ring at adjacent positions.


Aspirin is synthesized by the acetylation of the hydroxyl group of salicylic acid, typically using acetic anhydride. The "acetyl" group (--COCH\(_3\)) replaces the hydrogen of the hydroxyl group.


The resulting compound is acetylsalicylic acid.


Let's look at the other options:

(A) Methyl salicylate: This is the "oil of wintergreen," an ester formed by reacting the carboxylic acid group of salicylic acid with methanol.
(B) Sodium benzoate: A common food preservative, the sodium salt of benzoic acid.
(D) 2-methyl salicylic acid: A derivative of salicylic acid with an additional methyl group on the ring.
(E) Ethyl salicylate: An ester formed by reacting salicylic acid with ethanol.


Step 3: Final Answer:

The chemical name for aspirin is acetylsalicylic acid. Therefore, option (C) is correct.
Quick Tip: It's useful to remember the structures and names of common aromatic compounds related to salicylic acid: \textbf{Salicylic Acid:} The parent compound (has -OH and -COOH). \textbf{Aspirin (Acetylsalicylic Acid):} The -OH group is acetylated. \textbf{Oil of Wintergreen (Methyl Salicylate):} The -COOH group is esterified with methanol.

Step 1: Understanding the Concept:

This question asks for the specific reagent used to remove the hydroxyl (--OH) group from a phenol molecule, effectively reducing it to benzene. This is a deoxygenation reaction.


Step 2: Detailed Explanation:

The conversion of phenol to benzene is a classic reduction reaction in organic chemistry. \[ C_6H_5OH \xrightarrow{Reagent} C_6H_6 \]
The reagent required for this transformation is Zinc dust (Zn dust). When phenol is distilled or heated with zinc dust, the zinc atom removes the oxygen atom from the hydroxyl group, forming zinc oxide (ZnO) and leaving benzene as the organic product. \[ C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO \]

Let's analyze the other reagents:

(A) Na / Hg (Sodium amalgam): A reducing agent, but not typically used for this transformation. It's used in reductions like the Clemmensen reduction (with HCl).
(C) Cr\(_2\)O\(_3\): An oxidizing agent or a catalyst for dehydrogenation (as seen in aromatization), not for reducing phenol.
(D) LiAlH\(_4\) (Lithium aluminium hydride) and (E) NaBH\(_4\) (Sodium borohydride): These are powerful hydride-based reducing agents. However, they are used for reducing carbonyl compounds (aldehydes, ketones, acids, esters) to alcohols. They do not reduce the C--O bond of a phenol.


Step 3: Final Answer:

The standard reagent for converting phenol to benzene is heating with zinc dust. Therefore, option (B) is correct.
Quick Tip: The reduction of phenol to benzene with zinc dust is a very specific and frequently tested reaction. It's one of the few ways to directly remove the phenolic -OH group. Memorize this reagent-reaction pair: Phenol + Zn dust \(\rightarrow\) Benzene.

Step 1: Understanding the Concept:

Hydrogen bonding is a strong type of dipole-dipole attraction between a hydrogen atom bonded to a highly electronegative atom (like N, O, or F) and another nearby electronegative atom.

Intermolecular H-bonding: Occurs \textit{between two or more different molecules. This generally leads to higher boiling points and water solubility.
Intramolecular H-bonding: Occurs \textit{within a single molecule. This is only possible when the hydrogen donor and acceptor groups are close to each other, often forming a stable five- or six-membered ring. This type of bonding can lower the boiling point and water solubility compared to isomers that only form intermolecular H-bonds.


Step 2: Detailed Explanation:

Let's analyze the options:

(A) Water (H\(_2\)O), (B) Methanol (CH\(_3\)OH), and (C) Phenol (C\(_6\)H\(_5\)OH): All these molecules have --OH groups and will form extensive intermolecular hydrogen bonds with each other. There is no possibility for intramolecular H-bonding.
(D) o-Nitrophenol: In this molecule, a hydroxyl (--OH) group and a nitro (--NO\(_2\)) group are on adjacent carbons (ortho positions) of the benzene ring. The proximity of these two groups allows a hydrogen bond to form between the hydrogen of the --OH group and one of the oxygen atoms of the --NO\(_2\) group. This forms a stable six-membered ring structure \textit{within the molecule. This is intramolecular hydrogen bonding.
(E) p-Nitrophenol: Here, the --OH and --NO\(_2\) groups are on opposite ends of the benzene ring (para positions). They are too far apart to form a hydrogen bond within the same molecule. Instead, the --OH group of one molecule will form a hydrogen bond with the --NO\(_2\) group of a \textit{different molecule. This is intermolecular hydrogen bonding.


Step 3: Final Answer:

o-Nitrophenol is the classic example of a molecule exhibiting intramolecular hydrogen bonding due to the ortho positioning of the --OH and --NO\(_2\) groups. Therefore, option (D) is correct.
Quick Tip: Look for molecules with two functional groups capable of H-bonding (like -OH, -NH\(_2\), -NO\(_2\), -CHO) in ortho positions on a benzene ring. These are prime candidates for intramolecular H-bonding. A key consequence is that o-nitrophenol is steam volatile while p-nitrophenol is not, a property used for their separation.

Step 1: Understanding the Concept:

The acidity of a carboxylic acid depends on the stability of its conjugate base, the carboxylate anion (R--COO\(^-\)). The presence of electron-withdrawing groups (EWGs) on the R group stabilizes the negative charge of the anion through the negative inductive effect (--I effect), thereby increasing the acidity. The strength of the acid is directly proportional to the strength and number of the EWGs.


Step 2: Detailed Explanation:

All the given compounds are derivatives of acetic acid, with electron-withdrawing groups attached to the \(\alpha\)-carbon. We need to compare the strength of the --I effect of these groups. A stronger --I effect leads to a stronger acid. The question asks for the weakest acid, which will be the one with the weakest electron-withdrawing effect.


Let's establish the order of the --I effect for the relevant groups: \[ -NO_2 > -CN > -F > -Cl > -Br > -I \]
Also, the effect is additive. The more EWGs present, the stronger the effect. \[ -CCl_3 > -CHCl_2 > -CH_2Cl \]

Now let's compare the acids based on their EWGs:

(D) O\(_2\)N--CH\(_2\)COOH: Has the --NO\(_2\) group, which has the strongest --I effect among the single substituent groups listed. This will be the strongest acid among (A), (B), and (D).
(B) NC--CH\(_2\)COOH: Has the --CN group. Its --I effect is stronger than halogens but weaker than --NO\(_2\). This is a very strong acid.
(A) FCH\(_2\)COOH: Has the --F group. Fluorine is the most electronegative halogen, so its --I effect is stronger than chlorine.
(E) Cl\(_2\)CHCOOH: Has two chlorine atoms. The additive effect of two chlorines is stronger than that of a single fluorine.
(C) Cl\(_3\)C--COOH: Has three chlorine atoms. The additive effect of three chlorines makes it a very strong acid, even stronger than O\(_2\)N--CH\(_2\)COOH.


Let's order the acids from strongest to weakest based on the EWG effect:
1. Cl\(_3\)C--COOH (Strongest --I effect due to three Cl atoms)
2. O\(_2\)N--CH\(_2\)COOH (Strongest --I effect from a single substituent)
3. NC--CH\(_2\)COOH
4. Cl\(_2\)CHCOOH (Two Cl atoms have a stronger effect than one F atom)
5. FCH\(_2\)COOH (Weakest overall --I effect among the options)

Therefore, fluoroacetic acid is the weakest acid in this list.


Step 3: Final Answer:

Comparing the electron-withdrawing capabilities of the substituents, the single fluorine atom in FCH\(_2\)COOH provides the least stabilization to the conjugate base compared to the other options. Thus, it is the weakest acid. Therefore, option (A) is correct.
Quick Tip: To compare the acidity of substituted carboxylic acids, memorize the order of the inductive effect (--I) of common groups: --NO\(_2\) > --CN > --F > --Cl > --Br > --I. Also, remember that the effect is additive (more groups = stronger effect) and distance-dependent (the closer the group to --COOH, the stronger the effect).

Step 1: Understanding the Concept:

The question describes a base-catalyzed reaction between an aldehyde and a ketone. We need to identify the specific type of named reaction this represents. The key features are the nature of the carbonyl compounds (aldehyde/ketone, presence/absence of \(\alpha\)-hydrogens) and the catalyst (base).


Step 2: Detailed Explanation:

Let's analyze the reactants and the reaction type:

Reactants:

Benzaldehyde (C\(_6\)H\(_5\)CHO): An aldehyde with no \(\alpha\)-hydrogens.
Acetophenone (C\(_6\)H\(_5\)COCH\(_3\)): A ketone with three acidic \(\alpha\)-hydrogens on its methyl group.

Catalyst: NaOH, a strong base.
Reaction: The base (OH\(^-\)) abstracts an \(\alpha\)-hydrogen from acetophenone to form an enolate ion. This enolate then acts as a nucleophile and attacks the carbonyl carbon of benzaldehyde. The initial aldol addition product then undergoes dehydration (loses water) upon gentle heating to form an \(\alpha\), \(\beta\)-unsaturated ketone, benzalacetophenone.

This reaction fits the definition of an aldol condensation. Because the condensation occurs between two different carbonyl compounds (an aldehyde and a ketone), it is specifically called a cross aldol condensation.


When a cross aldol reaction occurs between a carbonyl compound with \(\alpha\)-hydrogens and one without, it is often called a Claisen-Schmidt condensation. This is a specific type of cross aldol condensation.


Let's look at the other options:

(A) Cannizzaro reaction: Involves the self-oxidation and reduction of an aldehyde with no \(\alpha\)-hydrogens in the presence of a concentrated base. This is not happening here.
(B) Aldol condensation: This is the general term. "Cross aldol condensation" is more specific and therefore a better answer.
(C) Wolf-Kishner reduction and (D) Clemmensen reduction: These are reactions that reduce a carbonyl group (C=O) to a methylene group (CH\(_2\)), which is not the product here.


Step 3: Final Answer:

The reaction between two different carbonyl compounds, benzaldehyde and acetophenone, in the presence of a base is a cross aldol condensation. Therefore, option (E) is correct.
Quick Tip: To differentiate between aldol and Cannizzaro reactions, check the reactants: \textbf{Aldol Condensation: Requires at least one reactant with \(\alpha\)-hydrogens and usually a dilute base. \textbf{Cross Aldol Condensation}: Involves two different carbonyl reactants. \textbf{Cannizzaro Reaction}: Requires an aldehyde with NO \(\alpha\)-hydrogens and a concentrated base.

Step 1: Understanding the Concept:

This is a fact-based question about the industrial applications of common carboxylic acids. We need to identify the acid that has widespread use in the specific industries mentioned.


Step 2: Detailed Explanation:

Let's review the primary uses of the listed acids:

(A) Methanoic acid (Formic acid, HCOOH): This is a simple but versatile acid.

Rubber Industry: Used as a coagulant to process latex into raw rubber.
Leather Industry: Used in tanning and fixing dyes to leather.
Textile Industry: Used as a mordant in dyeing and for finishing textiles.
It also has uses as a preservative, an antibacterial agent, and in chemical synthesis. Its properties make it suitable for all the industries listed.

(B) Ethanoic acid (Acetic acid, CH\(_3\)COOH): The main component of vinegar. It is a major industrial chemical used to produce vinyl acetate monomer (for paints and adhesives), cellulose acetate (for photographic film), and as a solvent. While used in textiles, its application profile does not match all the listed industries as well as formic acid.
(C) Benzoic acid (C\(_6\)H\(_5\)COOH): Primarily used as a food preservative (or in the form of its salts like sodium benzoate) and as a precursor for the synthesis of other chemicals like phenol.
(D) Salicylic acid: Used primarily in medicine and cosmetics, especially in anti-acne treatments and for the synthesis of aspirin.
(E) Butanoic acid (Butyric acid): Known for its unpleasant odor. It is mainly used in the synthesis of esters for perfumes and flavorings.


Comparing the applications, methanoic acid (formic acid) is the one with established, significant uses across all the mentioned sectors: rubber, textile, dyeing, leather, and electroplating.


Step 3: Final Answer:

Methanoic acid has diverse industrial applications that match the description in the question. Therefore, option (A) is correct.
Quick Tip: Associate common carboxylic acids with their primary uses: \textbf{Methanoic acid (Formic):} Rubber, leather, textiles. \textbf{Ethanoic acid (Acetic):} Vinegar, solvents, plastics (vinyl acetate). \textbf{Benzoic acid:} Food preservation. \textbf{Salicylic acid:} Medicine (Aspirin), skincare.

Step 1: Understanding the Concept:

The question asks for the starting material (an amine) required to synthesize a specific amide, acetanilide. The reaction described is the acylation of an amine.

Acetanilide: The name indicates an acetyl group (--COCH\(_3\)) attached to the nitrogen of an aniline molecule. Aniline is the common name for benzenamine (C\(_6\)H\(_5\)NH\(_2\)). So, acetanilide is N-phenylacetamide (C\(_6\)H\(_5\)NHCOCH\(_3\)).
Reagents: Acetic anhydride ((CH\(_3\)CO)\(_2\)O) is the acylating agent that provides the acetyl group. Pyridine is a base used to catalyze the reaction and neutralize the acetic acid byproduct.


Step 2: Detailed Explanation:

The reaction is the nucleophilic acyl substitution where the lone pair on the nitrogen of the amine attacks one of the carbonyl carbons of acetic anhydride.
To form acetanilide (C\(_6\)H\(_5\)NHCOCH\(_3\)), the starting amine must provide the C\(_6\)H\(_5\)NH-- part. This amine is C\(_6\)H\(_5\)NH\(_2\).


The IUPAC name for C\(_6\)H\(_5\)NH\(_2\) is Benzenamine. The common name is aniline.


Let's check the other options:

(A) Ethanamine (CH\(_3\)CH\(_2\)NH\(_2\)) would produce N-ethylacetamide.
(B) Methanamine (CH\(_3\)NH\(_2\)) would produce N-methylacetamide.
(D) N-Methylaniline (C\(_6\)H\(_5\)NHCH\(_3\)) is a secondary amine and would produce N-methyl-N-phenylacetamide.
(E) N-Methylethanamine (CH\(_3\)CH\(_2\)NHCH\(_3\)) would produce N-ethyl-N-methylacetamide.


The reaction is: \[ \underset{Benzenamine (Aniline)}{C_6H_5NH_2} + \underset{Acetic Anhydride}{(CH_3CO)_2O} \xrightarrow{Pyridine} \underset{Acetanilide}{C_6H_5NHCOCH_3} + \underset{Acetic Acid}{CH_3COOH} \]

Step 3: Final Answer:

The amine required to prepare acetanilide is benzenamine (aniline). Therefore, option (C) is correct.
Quick Tip: To solve synthesis questions, it's often easiest to analyze the structure of the product. Break down the name "Acetanilide": \textbf{Acet-}: Refers to the acetyl group (CH\(_3\)CO--), which comes from the acetic anhydride. \textbf{-anilide}: Refers to the aniline group (C\(_6\)H\(_5\)NH--), which comes from the starting amine, aniline (benzenamine). This deconstruction quickly points to the correct reactants.

Step 1: Understanding the Concept:

The Hinsberg test is a chemical test used to distinguish between primary (1°), secondary (2°), and tertiary (3°) amines. The reagent used is Hinsberg's reagent, which is benzenesulfonyl chloride (C\(_6\)H\(_5\)SO\(_2\)Cl). The reactivity depends on the presence of a hydrogen atom attached to the nitrogen of the amine.


Primary amines (R-NH\(_2\)): Have two H atoms on the nitrogen. They react with Hinsberg's reagent to form an N-alkylbenzenesulfonamide, which is acidic and soluble in alkali (like KOH or NaOH) because it still has one acidic hydrogen on the nitrogen.
Secondary amines (R\(_2\)NH): Have one H atom on the nitrogen. They react to form an N,N-dialkylbenzenesulfonamide. This product has no acidic hydrogen on the nitrogen, so it is insoluble in alkali.
Tertiary amines (R\(_3\)N): Have no H atoms on the nitrogen. They do not have the necessary hydrogen to be replaced, so they do not react with benzenesulfonyl chloride to form a stable sulfonamide.


Step 2: Detailed Explanation:

We need to classify each of the given amines:


(A) Methanamine (CH\(_3\)NH\(_2\)): A primary amine. It will react.
(B) N-Methylethanamine (CH\(_3\)CH\(_2\)NHCH\(_3\)): A secondary amine. It will react.
(C) N,N-Dimethylethanamine (CH\(_3\)CH\(_2\)N(CH\(_3\))\(_2\)): A tertiary amine. There are no hydrogen atoms directly bonded to the nitrogen. It will not react with Hinsberg's reagent.
(D) 1-Propanamine (CH\(_3\)CH\(_2\)CH\(_2\)NH\(_2\)): A primary amine. It will react.
(E) 2-Propanamine ((CH\(_3\))\(_2\)CHNH\(_2\)): A primary amine. It will react.


Step 3: Final Answer:

N,N-Dimethylethanamine is a tertiary amine and lacks a hydrogen atom on the nitrogen, so it does not react with Hinsberg's reagent. Therefore, option (C) is correct.
Quick Tip: The key to the Hinsberg test is the hydrogen on the nitrogen atom. Primary (2 H's): Reacts, product soluble in base. Secondary (1 H): Reacts, product insoluble in base. Tertiary (0 H's): No reaction. Just classify the amine to predict the result.

Step 1: Understanding the Concept:

This question asks about a specific named reaction for the synthesis of aryl fluorides, which are difficult to prepare by direct fluorination. The synthesis from diazonium salts is a very important route.


Step 2: Detailed Explanation:

The preparation of aryl fluorides from diazonium salts is a two-step process known as the Balz-Schiemann reaction.


Step 1: Formation of Diazonium Tetrafluoroborate. A primary aromatic amine is first diazotized with nitrous acid (NaNO\(_2\) + HCl) at low temperatures (0-5 °C) to form a diazonium salt (e.g., ArN\(_2^+\)Cl\(^-\)). This salt is then treated with fluoroboric acid (HBF\(_4\)). This causes the precipitation of the relatively stable diazonium tetrafluoroborate salt.
\[ ArN_2^+Cl^- + HBF_4 \rightarrow \underset{(precipitate)}{ArN_2^+BF_4^-} + HCl \]

Step 2: Thermal Decomposition. The isolated diazonium tetrafluoroborate is then gently heated. It decomposes to give the desired aryl fluoride, along with nitrogen gas and boron trifluoride.
\[ ArN_2^+BF_4^- \xrightarrow{\Delta} Ar-F + N_2\uparrow + BF_3\uparrow \]

The key reagent that introduces the fluorine in a controlled manner is fluoroboric acid (HBF\(_4\)). Simple fluoride salts like NaF or KF (Sandmeyer-like reaction) do not work well for preparing aryl fluorides.


Step 3: Final Answer:

The Balz-Schiemann reaction, which uses HBF\(_4\) to convert a diazonium salt to an aryl fluoride, is the standard method. Therefore, option (E) is correct.
Quick Tip: Associate named reactions with their specific reagents for diazonium salt conversions: \textbf{Sandmeyer Reaction:} CuCl/HCl (for ArCl), CuBr/HBr (for ArBr), CuCN/KCN (for ArCN). \textbf{Gattermann Reaction:} Cu powder/HCl (for ArCl), Cu powder/HBr (for ArBr). \textbf{Balz-Schiemann Reaction:} HBF\(_4\), then heat (for ArF).

Step 1: Understanding the Concept:

Amino acids are the building blocks of proteins. They are classified based on whether the human body can synthesize them.

Essential amino acids: Cannot be synthesized by the body and must be obtained from the diet. There are 9 essential amino acids.
Non-essential amino acids: Can be synthesized by the body, usually from other molecules. There are 11 non-essential amino acids.
Conditionally essential amino acids: Are normally non-essential but can become essential under certain conditions, such as illness or in infants (e.g., Arginine).


Step 2: Detailed Explanation:

Let's classify the amino acids given in the options:

(A) Proline: This is a non-essential amino acid. The body can synthesize it from glutamate.
(B) Leucine: This is an essential amino acid. It must be obtained from the diet.
(C) Valine: This is an essential amino acid.
(D) Arginine: This is generally considered non-essential or conditionally essential. Adults can synthesize it, but the rate of synthesis may not be sufficient during periods of rapid growth (infancy) or stress.
(E) Histidine: This is an essential amino acid.

The question asks which amino acid can be synthesized in the body. Both Proline and Arginine fit this description for a healthy adult. However, Proline is always considered non-essential, while Arginine's classification is sometimes debated (conditionally essential). In the context of a multiple-choice question, Proline is the most unambiguously correct answer representing a non-essential amino acid. Given the provided answer key, we select Proline.


Step 3: Final Answer:

Proline is a non-essential amino acid, meaning it can be synthesized by the human body. The other options are either essential (Leucine, Valine, Histidine) or conditionally essential (Arginine). Therefore, Proline is the best answer.
Quick Tip: A useful mnemonic to remember the 9 essential amino acids is "PVT TIM HALL": \textbf{Phenylalanine \textbf{V}aline \textbf{T}hreonine \textbf{T}ryptophan \textbf{I}soleucine \textbf{M}ethionine \textbf{H}istidine \textbf{A}rginine (sometimes included) \textbf{L}ysine \textbf{L}eucine Any amino acid not on this list is non-essential (can be synthesized).

Step 1: Understanding the Concept:

This question requires matching different types of carbohydrates with specific examples. We need to know the classification of common sugars based on their functional group (aldose/ketose), number of carbons (hexose), number of monomer units (disaccharide/polysaccharide), and their ability to act as a reducing agent.


Step 2: Detailed Explanation:

Let's classify each example:

i) Maltose: A disaccharide made of two glucose units. It has a free hemiacetal group, so it is a reducing disaccharide.
ii) Glycogen: A polymer of glucose, used for energy storage in animals. It is a polysaccharide.
iii) Glucose: A monosaccharide with an aldehyde group and six carbon atoms. It is an aldohexose.
iv) Sucrose: A disaccharide made of glucose and fructose. The anomeric carbons of both units are involved in the glycosidic bond, so it has no free hemiacetal group. It is a non-reducing disaccharide.
v) Fructose: A monosaccharide with a ketone group and six carbon atoms. It is a ketohexose.


Now we can match the columns:

a) Aldohexose matches with (iii) Glucose.
b) Ketohexose matches with (v) Fructose.
c) Non-reducing disaccharide matches with (iv) Sucrose.
d) Reducing disaccharide matches with (i) Maltose.
e) Polysaccharide matches with (ii) Glycogen.


The correct matching is: a)-(iii); b)-(v); c)-(iv); d)-(i); e)-(ii).


Step 3: Final Answer:

The correct set of matches is a)-(iii), b)-(v), c)-(iv), d)-(i), e)-(ii), which corresponds to option (A).
Quick Tip: Remember the key examples for each carbohydrate class: \textbf{Monosaccharides:} Glucose (aldohexose), Fructose (ketohexose). \textbf{Disaccharides:} Sucrose (non-reducing), Lactose/Maltose (reducing). \textbf{Polysaccharides:} Starch/Glycogen/Cellulose (storage/structural). A sugar is reducing if it has a free hemiacetal group. In sucrose, the anomeric carbons are locked in the bond, making it non-reducing.

Step 1: Understanding the Concept:

This question deals with the relationship between fundamental quantities in radioactive decay: the decay constant (\(\lambda\)) and the mean life (\(\tau\)). We need to find the dimensions of their product.


Step 2: Key Formula or Approach:

The mean life (\(\tau\)) of a radioactive substance is defined as the reciprocal of its decay constant (\(\lambda\)). \[ \tau = \frac{1}{\lambda} \]
Alternatively, the decay constant is the reciprocal of the mean life: \[ \lambda = \frac{1}{\tau} \]

Step 3: Detailed Explanation:

We are asked to find the dimensional formula for the product \(\lambda \times \tau\).
Using the relationship from Step 2, we can substitute one variable in terms of the other.
Let's substitute \(\lambda = \frac{1}{\tau}\): \[ Product = \lambda \times \tau = \left(\frac{1}{\tau}\right) \times \tau = 1 \]
The product of the decay constant and the mean life is the number 1, which is a pure number. A pure, dimensionless number has no physical dimensions of Mass (M), Length (L), or Time (T).


Therefore, the dimensional formula is M\(^0\)L\(^0\)T\(^0\).


Alternatively, let's consider the units.
The decay constant \(\lambda\) represents the probability of decay per unit time. Its unit is s\(^{-1}\), so its dimension is [T\(^{-1}\)].

The mean life \(\tau\) is a measure of time. Its unit is s, so its dimension is [T].

The product of their dimensions is: \[ [\lambda] \times [\tau] = [T^{-1}] \times [T] = [T^{-1+1}] = [T^0] \]
Since there are no Mass or Length dimensions involved, the full dimensional formula is [M\(^0\)L\(^0\)T\(^0\)].


Step 4: Final Answer:

The product of the decay constant and the mean life is a dimensionless quantity. Its dimensional formula is L\(^0\)M\(^0\)T\(^0\). This corresponds to option (D).
Quick Tip: Remember the key relationships in radioactivity: Mean life \(\tau = 1/\lambda\) Half-life \(t_{1/2} = \ln(2)/\lambda = 0.693/\lambda\) \(t_{1/2} = 0.693 \tau\) From the first relation, it is immediately clear that \(\lambda\tau = 1\), which is dimensionless.

Step 1: Understanding the Concept:

This problem involves the kinematics of an object in free fall under constant gravitational acceleration, g. We will use the equations of motion for uniformly accelerated motion. The object is dropped, so its initial velocity is zero.


Step 2: Key Formula or Approach:

The relevant equation of motion is: \[ s = ut + \frac{1}{2}at^2 \]
where:

\(s\) is the distance traveled
\(u\) is the initial velocity
\(a\) is the acceleration (here, \(a=g\))
\(t\) is the time

Since the object is dropped, \(u = 0\), and the formula simplifies to \(s = \frac{1}{2}gt^2\). From this, we can see that \(t = \sqrt{\frac{2s}{g}}\), which means time is proportional to the square root of the distance fallen, \(t \propto \sqrt{s}\).


Step 3: Detailed Explanation:

Case 1: Falling the full height h

The object is dropped from height h and reaches the ground. The distance fallen is \(s_1 = h\).
The time taken is given as \(t\).
Using the formula: \[ h = \frac{1}{2}gt^2 \quad \quad (Equation \ 1) \]

Case 2: Passing through height h/2

The object is at a height of h/2 from the ground. This means it has fallen a distance from the top of \(s_2 = h - h/2 = h/2\).
Let the time taken to fall this distance be \(t'\).
Using the formula: \[ \frac{h}{2} = \frac{1}{2}gt'^2 \quad \quad (Equation \ 2) \]

Finding the relationship between t' and t:

We have two equations. Let's divide Equation 2 by Equation 1: \[ \frac{h/2}{h} = \frac{\frac{1}{2}gt'^2}{\frac{1}{2}gt^2} \]
Cancel out the common terms (\(h\), \(1/2\), \(g\)): \[ \frac{1}{2} = \frac{t'^2}{t^2} \]
Now, solve for \(t'\): \[ t'^2 = \frac{t^2}{2} \] \[ t' = \sqrt{\frac{t^2}{2}} = \frac{t}{\sqrt{2}} \]

Step 4: Final Answer:

The time after which the object was passing through a point at a height h/2 (i.e., had fallen a distance h/2) is \(t/\sqrt{2}\). This corresponds to option (B).
Quick Tip: For any object in free fall from rest, the distance fallen is proportional to the square of the time (\(s \propto t^2\)), or the time taken is proportional to the square root of the distance fallen (\(t \propto \sqrt{s}\)). So, if \(t_1\) is the time to fall distance \(s_1\) and \(t_2\) is the time to fall distance \(s_2\), then \(\frac{t_2}{t_1} = \sqrt{\frac{s_2}{s_1}}\). Here, \(s_1=h\) and \(s_2=h/2\). So, \(\frac{t'}{t} = \sqrt{\frac{h/2}{h}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\).

Step 1: Understanding the Concept:

We need to find the magnitude of the difference between two unit vectors. The magnitude of a vector difference \(|\vec{A} - \vec{B}|\) can be found using the law of cosines for vector subtraction or by using the dot product.


Step 2: Key Formula or Approach:

The formula for the magnitude of the difference between two vectors \(\vec{A}\) and \(\vec{B}\) is given by: \[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta} \]
where \(\theta\) is the angle between the two vectors.


Alternatively, we can use the dot product property: \(|\vec{V}|^2 = \vec{V} \cdot \vec{V}\).
So, \(|\hat{A} - \hat{B}|^2 = (\hat{A} - \hat{B}) \cdot (\hat{A} - \hat{B})\).


Step 3: Detailed Explanation:

Method 1: Using the magnitude formula

We are given:

\(\hat{A}\) and \(\hat{B}\) are unit vectors, so their magnitudes are \(|\hat{A}| = 1\) and \(|\hat{B}| = 1\).
The angle between them is \(\theta = 60^\circ\).

Substitute these values into the formula: \[ |\hat{A} - \hat{B}| = \sqrt{(1)^2 + (1)^2 - 2(1)(1)\cos(60^\circ)} \]
We know that \(\cos(60^\circ) = 1/2\). \[ |\hat{A} - \hat{B}| = \sqrt{1 + 1 - 2(1/2)} \] \[ |\hat{A} - \hat{B}| = \sqrt{2 - 1} \] \[ |\hat{A} - \hat{B}| = \sqrt{1} = 1 \]

Method 2: Using the dot product
\[ |\hat{A} - \hat{B}|^2 = (\hat{A} - \hat{B}) \cdot (\hat{A} - \hat{B}) \] \[ |\hat{A} - \hat{B}|^2 = \hat{A} \cdot \hat{A} - \hat{A} \cdot \hat{B} - \hat{B} \cdot \hat{A} + \hat{B} \cdot \hat{B} \]
Since \(\hat{A} \cdot \hat{B} = \hat{B} \cdot \hat{A}\), this simplifies to: \[ |\hat{A} - \hat{B}|^2 = |\hat{A}|^2 - 2(\hat{A} \cdot \hat{B}) + |\hat{B}|^2 \]
Recall that \(\hat{A} \cdot \hat{B} = |\hat{A}||\hat{B}|\cos\theta\). \[ |\hat{A} - \hat{B}|^2 = (1)^2 - 2(1)(1)\cos(60^\circ) + (1)^2 \] \[ |\hat{A} - \hat{B}|^2 = 1 - 2(1/2) + 1 = 1 - 1 + 1 = 1 \] \[ |\hat{A} - \hat{B}| = \sqrt{1} = 1 \]

Step 4: Final Answer:

The value of \(|\hat{A} - \hat{B}|\) is 1. This corresponds to option (D).
Quick Tip: For two unit vectors, the formulas for the magnitude of their sum and difference simplify nicely: \(|\hat{A} + \hat{B}| = \sqrt{1+1+2\cos\theta} = \sqrt{2(1+\cos\theta)} = 2\cos(\theta/2)\) \(|\hat{A} - \hat{B}| = \sqrt{1+1-2\cos\theta} = \sqrt{2(1-\cos\theta)} = 2\sin(\theta/2)\) Using the second formula here: \(|\hat{A} - \hat{B}| = 2\sin(60^\circ/2) = 2\sin(30^\circ) = 2(1/2) = 1\). This is a very fast way to solve the problem.

Step 1: Understanding the Concept:

The weight a person experiences (apparent weight) is the normal reaction force (N) exerted on them by the surface they are standing on (the lift floor). The actual weight of the person is \(W = mg\), which is the force of gravity acting on them. Apparent weight can differ from actual weight when the reference frame (the lift) is accelerating.


Step 2: Key Formula or Approach:

We can apply Newton's second law (\(F_{net} = ma\)) to the person in the lift. The forces acting on the person are the gravitational force \(mg\) (downward) and the normal force \(N\) (upward). Let's take the upward direction as positive.
The net force is \(F_{net} = N - mg\).
So, \(N - mg = ma\), where \(a\) is the acceleration of the lift.
The apparent weight is \(N\). \[ N = mg + ma = m(g + a) \]
Here, \(a\) is positive for upward acceleration and negative for downward acceleration.


Step 3: Detailed Explanation:

We are looking for the condition for "weight loss", which means the apparent weight (N) is less than the actual weight (\(mg\)). \[ N < mg \] \[ m(g + a) < mg \] \[ g + a < g \] \[ a < 0 \]
An acceleration \(a < 0\) means the acceleration is in the downward direction. So, the lift must be accelerating downwards.


Let's analyze the given options based on the formula \(N = m(g+a)\):

(A) downward with uniform velocity: Uniform velocity means acceleration \(a = 0\). So, \(N = m(g+0) = mg\). The apparent weight equals the actual weight. No weight loss.
(B) upward with constant acceleration: The acceleration is upward, so \(a\) is positive (\(a > 0\)). Then \(N = m(g+a) > mg\). The person experiences weight gain.
(C) downward with constant acceleration: The acceleration is downward, so \(a\) is negative (\(a < 0\)). Then \(N = m(g-|a|) < mg\). The person experiences weight loss.
(D) upward with uniform velocity: Same as (A), \(a = 0\), so \(N = mg\). No weight loss.
(E) upward with variable acceleration: If the acceleration is upward, there will be weight gain, not loss.


Step 4: Final Answer:

A person experiences weight loss when the lift accelerates downwards. Therefore, option (C) is correct.
Quick Tip: Remember this simple rule for lifts (elevators): \textbf{Accelerating Up} or \textbf{Decelerating Down}: You feel heavier (\(a > 0\)). \textbf{Accelerating Down} or \textbf{Decelerating Up}: You feel lighter (\(a < 0\)). \textbf{Constant Velocity (Up or Down)}: You feel your normal weight (\(a = 0\)). \textbf{Free Fall}: You feel weightless (\(a = -g\)).

Step 1: Understanding the Concept:

The problem states that initially, the body is stationary under the action of three forces. This means the body is in equilibrium. When one force is removed, the equilibrium is disturbed, and the remaining forces will cause an acceleration according to Newton's second law.


Step 2: Key Formula or Approach:

1. Condition for equilibrium: If the body is stationary, the net force is zero. \(\vec{F}_{net} = \sum \vec{F}_i = 0\).
2. Newton's Second Law: When a force is removed, the new net force will be \(\vec{F}_{net}'\). The acceleration will be \(\vec{a} = \vec{F}_{net}' / m\).


Step 3: Detailed Explanation:

Initial State (Equilibrium):

The body is stationary, so the vector sum of the three forces is zero. \[ \vec{F_1} + \vec{F_2} + \vec{F_3} = 0 \]
This equation implies that the resultant of any two forces is equal in magnitude and opposite in direction to the third force.
Specifically, the resultant of \(\vec{F_1}\) and \(\vec{F_2}\) balances \(\vec{F_3}\): \[ \vec{F_1} + \vec{F_2} = -\vec{F_3} \]

Final State (Force \(\vec{F_3}\) removed):

When the force \(\vec{F_3}\) is removed, the new net force acting on the body, \(\vec{F}_{net}'\), is just the vector sum of the remaining forces, \(\vec{F_1}\) and \(\vec{F_2}\). \[ \vec{F}_{net}' = \vec{F_1} + \vec{F_2} \]
From the equilibrium condition, we know that \(\vec{F_1} + \vec{F_2} = -\vec{F_3}\).
So, the new net force is: \[ \vec{F}_{net}' = -\vec{F_3} \]
This means the net force on the body after \(\vec{F_3}\) is removed is a force with the same magnitude as \(\vec{F_3}\) but in the opposite direction.


Calculating the Acceleration:

According to Newton's second law, \(\vec{a} = \vec{F}_{net}' / m\).
The magnitude of the acceleration is: \[ |\vec{a}| = \frac{|\vec{F}_{net}'|}{m} = \frac{|-\vec{F_3}|}{m} = \frac{F_3}{m} \]
The information that \(\vec{F_1}\) and \(\vec{F_2}\) are mutually perpendicular is extra information that allows you to calculate the magnitude of \(\vec{F_3}\) as \(F_3 = \sqrt{F_1^2 + F_2^2}\), but it is not needed to find the acceleration in terms of \(F_3\).


Step 4: Final Answer:

The magnitude of the acceleration of the body when \(\vec{F_3}\) is removed is \(F_3 / m\). This corresponds to option (A).
Quick Tip: For any system in equilibrium with multiple forces, if one force is removed, the net resulting force is equal in magnitude and opposite in direction to the force that was removed. This is a direct consequence of the equilibrium condition \(\sum \vec{F}_i = 0\).

Step 1: Understanding the Concept:

Power is the rate at which work is done or energy is transferred. For an object moving under the influence of a force, the instantaneous power delivered by that force is the product of the force and the object's instantaneous velocity. We need to find how this power depends on time for an object starting from rest and moving with constant acceleration.


Step 2: Key Formula or Approach:

1. Instantaneous Power: \(P = \vec{F} \cdot \vec{v}\). For linear motion, \(P = Fv\).
2. Equation of motion for velocity: \(v = u + at\).
3. Newton's Second Law: \(F = ma\).


Step 3: Detailed Explanation:

The problem states:

Constant force, \(F = constant\).
From \(F=ma\), this implies constant acceleration, \(a = F/m = constant\).
Initially at rest, so initial velocity \(u = 0\).

First, let's find the instantaneous velocity \(v\) at time \(t\). Using the equation of motion: \[ v = u + at \]
Since \(u=0\) and \(a\) is constant, \[ v = at \]
This shows that the velocity is directly proportional to time (\(v \propto t\)).


Now, let's find the instantaneous power \(P\) delivered at time \(t\). \[ P = F \times v \]
We know \(F\) is constant and we just found \(v = at\). \[ P = F \times (at) \]
Since both \(F\) and \(a\) are constants, their product \(Fa\) is also a constant. \[ P = (Fa)t \]
This shows that the power \(P\) is directly proportional to time \(t\). \[ P \propto t \]

Step 4: Final Answer:

The power delivered to the body is directly proportional to \(t\). This corresponds to option (B).
Quick Tip: For motion under a constant force starting from rest: Acceleration is constant (\(a\)). Velocity is proportional to time (\(v \propto t\)). Displacement is proportional to time squared (\(s \propto t^2\)). Power is proportional to time (\(P = Fv \propto t\)). Kinetic Energy is proportional to time squared (\(KE = \frac{1}{2}mv^2 \propto t^2\)). Knowing these proportionality relationships can save a lot of time.

Step 1: Understanding the Concept:

This question tests the fundamental definition of work and power in physics. Work done by a constant force is defined as \(W = \vec{F} \cdot \vec{d} = Fd\cos\theta\), where \(\theta\) is the angle between the force vector \(\vec{F}\) and the displacement vector \(\vec{d}\). Power is the rate at which work is done, \(P = W/t\). We need to identify the statement that is physically incorrect.


Step 2: Detailed Explanation:

Let's analyze each statement:

(A) Work done in uniform circular motion is zero. In uniform circular motion, the centripetal force is always directed towards the center of the circle, while the instantaneous displacement is always tangential to the circle. The angle between the force and displacement is always 90°. Since \(\cos(90^\circ) = 0\), the work done by the centripetal force over any part of the path is zero. This statement is correct.
(B) When a body is in dynamic equilibrium, work done is zero. Dynamic equilibrium means the body is moving with a constant velocity (no acceleration). If the velocity is constant, the net force on the body is zero. Since the net force is zero, the net work done on the body is also zero. This statement is correct.
(C) Work done is positive for a freely falling body under gravity. For a freely falling body, the force of gravity is downwards and the displacement is also downwards. The angle between the force and displacement is 0°. Since \(\cos(0^\circ) = 1\), the work done by gravity is positive (\(W = mgd > 0\)). This statement is correct.
(D) Work done in a stretched string is positive. When a string is stretched, an external force is applied in the direction of the stretch (displacement). The angle between the applied force and the displacement is 0°. Thus, the work done by the external force to stretch the string is positive. (The work done by the restoring force of the string is negative). The statement is generally interpreted as work done on the string. This statement is correct.
(E) Work done depends on the time taken to complete the work. The definition of work, \(W = Fd\cos\theta\), does not include time. Work depends on force, displacement, and the angle between them. Power is the quantity that depends on the time taken to do the work (\(P=W/t\)). Therefore, this statement is incorrect.


Step 3: Final Answer:

The statement that work done depends on the time taken is incorrect. It is power, not work, that depends on time. Therefore, (E) is the incorrect statement.
Quick Tip: Be careful to distinguish between the concepts of work and power. \textbf{Work (W): A transfer of energy. It depends on Force and Displacement (\(W = \vec{F} \cdot \vec{d}\)). It has no explicit time dependence. \textbf{Power (P):} The rate of work. It depends on Work and Time (\(P = dW/dt\)). A common mistake is to confuse these two quantities.

Step 1: Understanding the Concept:

This question deals with the motion of the center of mass (CM) of a system. A key principle is that the motion of the center of mass is only affected by external forces acting on the system. Internal forces, such as those from an explosion, do not change the trajectory of the center of mass.


Step 2: Detailed Explanation:


Before the explosion: The shell is moving under the influence of a single external force: gravity. Its trajectory is a parabola, and its center of mass is, of course, following this parabolic path.
During the explosion: The explosion is caused by \textit{internal forces within the shell. According to Newton's third law, for every internal force exerted on one fragment, there is an equal and opposite force exerted on another fragment. The vector sum of all these internal forces is zero.
After the explosion: The individual fragments may fly off in various directions. However, the only \textit{external force acting on the system of fragments is still gravity. Since the internal forces of the explosion sum to zero, they do not produce any net force on the system and therefore cannot change the acceleration of the center of mass. The acceleration of the center of mass remains \(\vec{a_{CM} = \vec{g}\), which is exactly the same as it was before the explosion.

Because the acceleration of the center of mass remains unchanged, and its velocity at the moment of explosion is the same as the shell's velocity just before, the center of mass of the fragments must continue to follow the exact same trajectory that the shell would have followed if it had not exploded. This trajectory is the original parabolic path.


Step 3: Final Answer:

The center of mass of the fragments continues to move along the original parabolic path, as its motion is governed only by the external force of gravity. Therefore, option (B) is correct.
Quick Tip: A fundamental principle of mechanics: \textbf{Internal forces cannot change the motion of the center of mass}. The trajectory of the center of mass of a system can only be altered by an external net force. Explosions, collisions, and people walking inside a boat are all examples of internal forces that do not affect the CM's path.

Step 1: Understanding the Concept:

The radius of gyration (k) of a body about an axis is a measure of how its mass is distributed with respect to that axis. It is related to the moment of inertia (I) and the total mass (M) of the body by the formula \(I = Mk^2\). We need to find the moments of inertia for a ring and a disc about their central axes and then use them to find the ratio of their radii of gyration.


Step 2: Key Formula or Approach:

1. Moment of inertia of a circular ring of mass M and radius R about its central axis (perpendicular to its plane) is \(I_{ring} = MR^2\).
2. Moment of inertia of a circular disc of mass M and radius R about its central axis (perpendicular to its plane) is \(I_{disc} = \frac{1}{2}MR^2\).
3. The relationship between moment of inertia and radius of gyration is \(I = Mk^2\), which means \(k = \sqrt{\frac{I}{M}}\).


Step 3: Detailed Explanation:

Let M be the mass and R be the radius for both the ring and the disc.


For the circular ring:
\[ I_{ring} = MR^2 \]
The radius of gyration of the ring, \(k_{ring}\), is: \[ k_{ring} = \sqrt{\frac{I_{ring}}{M}} = \sqrt{\frac{MR^2}{M}} = \sqrt{R^2} = R \]

For the circular disc:
\[ I_{disc} = \frac{1}{2}MR^2 \]
The radius of gyration of the disc, \(k_{disc}\), is: \[ k_{disc} = \sqrt{\frac{I_{disc}}{M}} = \sqrt{\frac{\frac{1}{2}MR^2}{M}} = \sqrt{\frac{1}{2}R^2} = \frac{R}{\sqrt{2}} \]

Calculate the ratio:

We need to find the ratio \(k_{ring} : k_{disc}\). \[ \frac{k_{ring}}{k_{disc}} = \frac{R}{R/\sqrt{2}} = R \times \frac{\sqrt{2}}{R} = \sqrt{2} \]
So, the ratio is \(\sqrt{2} : 1\).


Step 4: Final Answer:

The ratio of the radius of gyration of the ring to that of the disc is \(\sqrt{2}:1\). This corresponds to option (D).
Quick Tip: It is essential to memorize the formulas for the moment of inertia of common shapes (ring, disc, solid sphere, hollow sphere, rod). The radius of gyration can always be found from \(k = \sqrt{I/M}\). For a ring, mass is at the periphery, leading to a larger I and k compared to a disc where mass is distributed throughout.

Step 1: Understanding the Concept:

The gravitational field at a point is the gravitational force experienced per unit mass at that point. It is a vector quantity. For the net gravitational field from two masses to be zero at a point, the gravitational fields produced by each mass at that point must be equal in magnitude and opposite in direction. For two positive masses, this null point will lie on the line joining them, between the two masses.


Step 2: Key Formula or Approach:

The magnitude of the gravitational field (g) produced by a mass M at a distance r is given by: \[ g = \frac{GM}{r^2} \]
Let the point where the net field is zero be at a distance \(d\) from the mass \(m\). This point will then be at a distance \((x-d)\) from the mass \(4m\).
At this point, the magnitude of the field from \(m\) must equal the magnitude of the field from \(4m\). \[ g_m = g_{4m} \] \[ \frac{Gm}{d^2} = \frac{G(4m)}{(x-d)^2} \]

Step 3: Detailed Explanation:

Let's solve the equation for \(d\): \[ \frac{Gm}{d^2} = \frac{4Gm}{(x-d)^2} \]
Cancel the common term \(Gm\) from both sides: \[ \frac{1}{d^2} = \frac{4}{(x-d)^2} \]
Take the square root of both sides: \[ \sqrt{\frac{1}{d^2}} = \sqrt{\frac{4}{(x-d)^2}} \] \[ \frac{1}{d} = \frac{2}{x-d} \]
Now, cross-multiply to solve for \(d\): \[ 1 \times (x-d) = 2 \times d \] \[ x - d = 2d \] \[ x = 3d \] \[ d = \frac{x}{3} \]

Step 4: Final Answer:

The point at which the gravitational field is zero is at a distance of x/3 from the mass m. This corresponds to option (A).
Quick Tip: For two masses \(m_1\) and \(m_2\), the null point (where the gravitational field is zero) is always closer to the smaller mass. A shortcut formula for the distance \(d_1\) from mass \(m_1\) is: \[ d_1 = \frac{\sqrt{m_1}}{\sqrt{m_1} + \sqrt{m_2}} \times x \] Here, \(m_1 = m\), \(m_2 = 4m\). \[ d = \frac{\sqrt{m}}{\sqrt{m} + \sqrt{4m}} \times x = \frac{\sqrt{m}}{\sqrt{m} + 2\sqrt{m}} \times x = \frac{\sqrt{m}}{3\sqrt{m}} \times x = \frac{x}{3} \] This formula works for both gravitational and electrostatic forces between like charges.

Step 1: Understanding the Concept:

When a wire is stretched by a load, work is done by the applied force. This work is stored in the wire as elastic potential energy. The force required to stretch the wire is not constant; it increases linearly from zero to its final value, F (the load), as the wire extends. To calculate the work done, we must consider this varying force.


Step 2: Key Formula or Approach:

Work done by a variable force is the integral of the force over the displacement, or graphically, the area under the Force-extension graph.
Let the extension be \(l\) when the load is \(F\). Assuming the wire obeys Hooke's Law, the force \(f\) is proportional to the extension \(x\): \(f = kx\).
The work done (\(W\)) in stretching the wire from 0 to a final extension \(l\) is: \[ W = \int_0^l f(x) \,dx = \int_0^l kx \,dx \]
Alternatively, since the force increases linearly from 0 to F, the average force is \(\frac{0+F}{2} = \frac{F}{2}\).
Work done = Average Force \(\times\) Total Extension.


Step 3: Detailed Explanation:

Using the average force method:

Initial force = 0
Final force = Load (F)
Average force = \(\frac{0 + Load}{2} = \frac{1}{2} \times Load\)
Total displacement = Extension (\(l\))

Work Done = (Average Force) \(\times\) (Extension) \[ W = \frac{1}{2} \times Load \times Extension \]
This matches option (D).


Let's check the other forms. We know:
Stress = Load / Area (\(A\)) \(\implies\) Load = Stress \(\times A\)
Strain = Extension / Original Length (\(L\)) \(\implies\) Extension = Strain \(\times L\)
Substituting these into the formula for work: \[ W = \frac{1}{2} \times (Stress \times A) \times (Strain \times L) = \frac{1}{2} \times Stress \times Strain \times (AL) \]
Since \(AL\) is the volume of the wire, the work done per unit volume (energy density) is \(\frac{1}{2} \times Stress \times Strain\).
Option (B) is \(\frac{1}{2} \times\) load \(\times\) strain, which is dimensionally incorrect.


The provided answer key says (E) \(\frac{1}{4} \times\) load \(\times\) extension. This is incorrect. The standard, universally accepted formula for the potential energy stored in a stretched wire (and the work done to stretch it) is \(\frac{1}{2} \times\) Load \(\times\) Extension.


Step 4: Final Answer:

The correct expression for the work done in a stretched wire is \(\frac{1}{2} \times\) Load \(\times\) extension. Therefore, option (D) is the correct answer.
Quick Tip: The formula for work done or energy stored in systems where force is proportional to displacement (like springs and stretched wires) often has a factor of 1/2. This comes from using the average force (\(F_{avg} = (F_{initial} + F_{final})/2\)). Stretched Wire: \(W = \frac{1}{2} \times Load \times Extension\) Stretched Spring: \(U = \frac{1}{2} kx^2 = \frac{1}{2} \times Force \times Extension\) Charged Capacitor: \(U = \frac{1}{2} QV = \frac{1}{2} \times Charge \times Voltage\) This pattern can be a helpful memory aid.

Step 1: Understanding the Concept:

The total pressure inside an air bubble submerged in a liquid is the sum of three separate pressures:

The atmospheric pressure (\(P_o\)) acting on the surface of the liquid.
The gauge pressure due to the liquid column above the bubble (\(P_{gauge}\)).
The excess pressure inside the bubble due to surface tension (\(P_{excess}\)).


Step 2: Key Formula or Approach:

1. Atmospheric pressure = \(P_o\).
2. Gauge pressure at a depth \(h\) in a liquid of density \(\rho\) is \(P_{gauge} = h\rho g\).
3. The excess pressure inside a spherical bubble in a liquid is given by the Young-Laplace equation for a single spherical interface: \(P_{excess} = \frac{2T}{r}\), where T is the surface tension and r is the radius.

The total pressure inside is \(P_{inside} = P_o + P_{gauge} + P_{excess}\).


Step 3: Detailed Explanation:

Let's find the pressure just outside the bubble at depth \(h\). This pressure is the sum of the atmospheric pressure on the surface and the pressure from the liquid column. \[ P_{outside\_bubble} = P_o + h\rho g \]
Now, due to the curvature of the bubble's surface, the pressure inside must be greater than the pressure outside. This excess pressure is caused by surface tension. For an air bubble inside a liquid, there is only one liquid-air interface. The formula for excess pressure is: \[ P_{excess} = P_{inside} - P_{outside\_bubble} = \frac{2T}{r} \]
Therefore, the total pressure inside the bubble is: \[ P_{inside} = P_{outside\_bubble} + P_{excess} \]
Substituting the expressions for the outside pressure and excess pressure: \[ P_{inside} = (P_o + h\rho g) + \frac{2T}{r} \] \[ P = P_o + h\rho g + \frac{2T}{r} \]

This matches option (B). Note that option (C) with \(\frac{4T}{r}\) would be for a soap bubble in air, which has two air-liquid interfaces.


Step 4: Final Answer:

The total pressure inside the air bubble is the sum of atmospheric pressure, gauge pressure, and excess pressure due to surface tension, which is \(P = P_o + h\rho g + \frac{2T}{r}\). This corresponds to option (B).
Quick Tip: Be careful to distinguish between different types of bubbles and drops: \textbf{Liquid drop in air} (e.g., raindrop): 1 interface, \(\Delta P = 2T/r\). \textbf{Air bubble in liquid}: 1 interface, \(\Delta P = 2T/r\). \textbf{Soap bubble in air}: 2 interfaces (inner and outer), \(\Delta P = 4T/r\). The question specifies an air bubble in a liquid, so the excess pressure is \(2T/r\).

Step 1: Understanding the Concept:

The equation \(PV^\gamma = constant\) describes an adiabatic process for an ideal gas, where \(\gamma\) is the adiabatic index, defined as the ratio of specific heats, \(\gamma = C_p / C_v\). The question asks what process the equation represents if the value of \(\gamma\) is taken as 1.


Step 2: Detailed Explanation:

We are given the relation for a thermodynamic process: \[ PV^\gamma = constant \]
And we are given the condition that \(\gamma = C_p / C_v = 1\).

Substituting \(\gamma = 1\) into the equation, we get: \[ PV^1 = constant \] \[ PV = constant \]
This is the equation for an isothermal process according to the ideal gas law (\(PV = nRT\)). If \(PV\) is constant, and \(n\) and \(R\) are constants, then the temperature \(T\) must also be constant.


Let's briefly review the other processes:

Adiabatic: No heat exchange (\(Q=0\)), described by \(PV^\gamma = constant\) where \(\gamma > 1\).
Isochoric: Constant volume (\(V = constant\)).
Isobaric: Constant pressure (\(P = constant\)).


Step 3: Final Answer:

When \(\gamma = 1\), the equation becomes \(PV = constant\), which is the definition of an isothermal process. Therefore, option (C) is correct.
Quick Tip: Remember the general polytropic process equation \(PV^k = constant\). Different values of \(k\) define different processes: \(k=0 \implies P = constant\) (Isobaric) \(k=1 \implies PV = constant\) (Isothermal) \(k=\gamma \implies PV^\gamma = constant\) (Adiabatic) \(k \to \infty \implies V = constant\) (Isochoric)

Step 1: Understanding the Concept:

The root-mean-square (r.m.s.) velocity of gas molecules is a measure of their average speed, related to the kinetic energy of the gas. It depends on the temperature and the molar mass of the gas.


Step 2: Key Formula or Approach:

The formula for the r.m.s. velocity (\(v_{rms}\)) is: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:

\(R\) is the ideal gas constant.
\(T\) is the absolute temperature in Kelvin.
\(M\) is the molar mass of the gas in kg/mol.


Step 3: Detailed Explanation:

We are given the condition that the r.m.s. velocity of oxygen (O\(_2\)) is equal to the r.m.s. velocity of hydrogen (H\(_2\)). \[ v_{rms}(O_2) = v_{rms}(H_2) \]
Using the formula, we can write: \[ \sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} \]
Square both sides and cancel the common term \(3R\): \[ \frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}} \]
Now, we can solve for the temperature of oxygen, \(T_{O_2}\): \[ T_{O_2} = T_{H_2} \times \frac{M_{O_2}}{M_{H_2}} \]
We are given:

Temperature of hydrogen, \(T_{H_2} = 20 K\).
Molar mass of oxygen, \(M_{O_2} \approx 32 g/mol\).
Molar mass of hydrogen, \(M_{H_2} \approx 2 g/mol\).

(Note: We can use g/mol since the units will cancel in the ratio).
Substitute the values: \[ T_{O_2} = 20 K \times \frac{32}{2} \] \[ T_{O_2} = 20 K \times 16 \] \[ T_{O_2} = 320 K \]

Step 4: Final Answer:

The temperature of the oxygen molecule must be 320 K for its r.m.s. velocity to be equal to that of a hydrogen molecule at 20 K. This corresponds to option (B).
Quick Tip: From the r.m.s. velocity formula, you can see that \(v_{rms}^2 \propto T/M\). For the velocities to be equal, the ratio \(T/M\) must be the same for both gases. This gives a quick way to set up the problem: \(\frac{T_1}{M_1} = \frac{T_2}{M_2}\).

Step 1: Understanding the Concept:

In Simple Harmonic Motion (SHM), there is a continuous conversion between kinetic energy (energy of motion) and potential energy (stored energy). The total mechanical energy of the system remains constant, assuming no damping.

Mean Position (x=0): The point of maximum speed and zero restoring force.
Extreme Positions (x = \(\pm\)A, where A is amplitude): The points of zero speed and maximum restoring force.


Step 2: Detailed Explanation:

Let's analyze the energy at different positions:

Potential Energy (P.E.): Stored due to the restoring force. It is given by \(P.E. = \frac{1}{2}kx^2\), where \(k\) is the force constant. P.E. is zero at the mean position (\(x=0\)) and maximum at the extreme positions (\(x = \pm A\)).
Kinetic Energy (K.E.): Energy of motion, given by \(K.E. = \frac{1}{2}mv^2\). Since speed is maximum at the mean position (\(x=0\)) and zero at the extremes, K.E. is maximum at \(x=0\) and zero at the extreme positions.
Total Energy (T.E.): The sum of K.E. and P.E. It is constant throughout the motion: \(T.E. = K.E. + P.E. = \frac{1}{2}kA^2\).


Now let's evaluate the given statements:

(A) K.E. is minimum when x = 0. This is incorrect. K.E. is maximum at x=0.
(B) T.E. is zero when x = 0. This is incorrect. T.E. is constant and non-zero.
(C) P.E. is maximum when x = 0. This is incorrect. P.E. is \textit{minimum (zero) at x=0.
(D) K.E. is maximum when x is maximum. This is incorrect. K.E. is \textit{minimum (zero) when x is maximum.
(E) P.E. is maximum when x is maximum. This is correct. The displacement \(x\) is maximum at the amplitude (\(x=\pm A\)), and the potential energy \(P.E. = \frac{1{2}kx^2\) is maximum at these points.


Step 3: Final Answer:

The statement that potential energy is maximum when displacement x is maximum is correct. Therefore, option (E) is the correct choice.
Quick Tip: Think of SHM like a pendulum swing or a mass on a spring. At the center (mean position, x=0): Fastest speed \(\implies\) Max K.E., Min P.E. At the ends (extreme positions, x=max): Momentarily stops \(\implies\) Min K.E., Max P.E. The total energy is always conserved.

Step 1: Understanding the Concept:

This problem compares the resonant frequencies (modes of vibration) of a closed organ pipe and an open organ pipe of the same length. The boundary conditions (open or closed end) determine which harmonics are present.

Open pipe: Open at both ends. All harmonics are present.
Closed pipe: Closed at one end, open at the other. Only odd harmonics are present.

The "nth mode of vibration" refers to the nth possible resonant frequency in the series of allowed frequencies.


Step 2: Key Formula or Approach:

Let L be the length of the pipes and v be the speed of sound.

For an open pipe, the allowed frequencies are \(f_n = \frac{nv}{2L}\), where n = 1, 2, 3, ...
For a closed pipe, the allowed frequencies are \(f_n = \frac{nv}{4L}\), where n = 1, 3, 5, ...


Step 3: Detailed Explanation:

Third mode for the closed pipe:
The allowed modes (harmonics) are the 1st, 3rd, 5th, 7th, ...

1st mode (fundamental): n = 1 \(\implies f_1 = \frac{v}{4L}\)
2nd mode (1st overtone): n = 3 \(\implies f_3 = \frac{3v}{4L}\)
3rd mode (2nd overtone): n = 5 \(\implies f_{3rd\_mode\_closed} = \frac{5v}{4L}\)


Third mode for the open pipe:
The allowed modes (harmonics) are the 1st, 2nd, 3rd, 4th, ...

1st mode (fundamental): n = 1 \(\implies f_1 = \frac{v}{2L}\)
2nd mode (1st overtone): n = 2 \(\implies f_2 = \frac{2v}{2L}\)
3rd mode (2nd overtone): n = 3 \(\implies f_{3rd\_mode\_open} = \frac{3v}{2L}\)


Calculate the ratio:
We need the ratio of the frequency of the closed pipe's third mode to the open pipe's third mode. \[ Ratio = \frac{f_{3rd\_mode\_closed}}{f_{3rd\_mode\_open}} = \frac{5v/4L}{3v/2L} \]
Cancel the common terms v and L: \[ Ratio = \frac{5/4}{3/2} = \frac{5}{4} \times \frac{2}{3} = \frac{10}{12} = \frac{5}{6} \]
The ratio is 5:6.


Step 4: Final Answer:

The ratio of the frequencies is 5:6. This corresponds to option (C).
Quick Tip: Be very careful with the term "nth mode". For an open pipe, the nth mode is the nth harmonic. For a closed pipe, the nth mode is the (2n-1)th harmonic. So, the 3rd mode of a closed pipe is the (2*3 - 1) = 5th harmonic. The 3rd mode of an open pipe is the 3rd harmonic.

Step 1: Understanding the Concept:

This problem involves the concept of electrostatic induction and shielding by a conductor. When an external charge is brought near a conductor, the free charges within the conductor redistribute themselves. An "isolated" conductor means it is electrically neutral (has no net charge).


Step 2: Detailed Explanation:


Induction: The negative point charge (-q) attracts the positive charges within the isolated conducting plane and repels the negative charges.
Charge Redistribution: As a result, a net positive charge (totaling +q) is induced on the surface of the plane closer to the point charge. To maintain overall neutrality, a net negative charge (totaling -q) is induced on the opposite (far) surface of the plane.
Electric Field inside the Conductor: The charges redistribute in such a way that the net electric field inside the conducting material is exactly zero. The field from the external charge (-q) and the induced positive charge (+q) on the near surface cancel out the field from the induced negative charge (-q) on the far surface, within the conductor.
Electric Field on the "Other Side": For a point P on the other side of the plane, it is shielded from the direct field of the point charge -q and the induced charge +q on the near side. The only field it experiences is the one created by the induced negative charge (-q) that resides on the far surface of the plane.
Direction of the Field: For a large, flat conducting plane, the induced charge (-q) on the far surface will spread out more or less uniformly. The electric field lines from a plane of negative charge are directed perpendicularly towards the plane.


Therefore, the electric field at point P is directed towards the plane perpendicularly.


Step 3: Final Answer:

Due to electrostatic shielding and the induction of a negative charge on the far surface of the isolated plane, the electric field on the other side is directed perpendicularly towards the plane. This corresponds to option (B).
Quick Tip: Remember the key properties of conductors in electrostatic equilibrium: 1. The electric field inside the conductor is zero. 2. Any net charge resides on the surface. 3. The electric field at the surface is perpendicular to the surface. 4. A conductor shields its interior and the region beyond it from external static fields (if grounded). If isolated, an external field is produced on the far side due to induced charge separation.

Step 1: Understanding the Concept:

This problem involves calculating the equivalent capacitance for capacitors connected in series and then finding the total charge stored in the combination, which is the same as the charge drawn from the battery.


Step 2: Key Formula or Approach:

1. For capacitors in series, the reciprocal of the equivalent capacitance (\(C_{eq}\)) is the sum of the reciprocals of the individual capacitances:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots \]
2. The relationship between charge (Q), capacitance (C), and voltage (V) is given by:
\[ Q = CV \]
In a series combination, the charge Q is the same on each capacitor and is equal to the total charge drawn from the battery.


Step 3: Detailed Explanation:

Calculate the equivalent capacitance (\(C_{eq}\)):
We have three capacitors, each with \(C = 12 \, \muF\), connected in series. \[ \frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \]
Therefore, the equivalent capacitance is: \[ C_{eq} = 4 \, \muF \]

Calculate the total charge (Q):
The combination is connected to a battery with a voltage \(V = 12 V\).
The total charge drawn from the battery is: \[ Q = C_{eq} \times V \] \[ Q = (4 \, \muF) \times (12 V) \] \[ Q = 48 \, \muC \]

Step 4: Final Answer:

The charge drawn from the battery is 48 \(\mu\)C. This corresponds to option (C).
Quick Tip: For \(n\) identical capacitors \(C\) connected in series, the equivalent capacitance is \(C_{eq} = C/n\). Here, \(n=3\) and \(C=12\,\mu F\), so \(C_{eq} = 12/3 = 4\,\mu F\). This is a quick way to find the equivalent capacitance. Remember that for capacitors, the series/parallel rules are opposite to those for resistors.

Step 1: Understanding the Concept:

This question tests fundamental principles and definitions in the topic of current electricity. We need to evaluate the correctness of each statement.


Step 2: Detailed Explanation:

Let's analyze each statement:

(A) Kirchhoff's junction rule is based on conservation of energy. The junction rule (or Kirchhoff's first law) states that the sum of currents entering a junction must equal the sum of currents leaving it. This is a direct consequence of the conservation of charge, not energy. Charge cannot be created or destroyed at a junction. Kirchhoff's loop rule (second law) is based on the conservation of energy. Therefore, this statement is INCORRECT.
(B) Ohm's law asserts that the plot of current I versus potential V is linear. Ohm's law is \(V=IR\). Rearranging gives \(I = (1/R)V\). For an ohmic device, R is constant, so this is an equation of a straight line (\(y=mx\)) passing through the origin. This statement is correct.
(C) Current is a scalar quantity. Although current has a direction of flow, it does not obey the laws of vector addition (e.g., parallelogram law). Currents at a junction add algebraically. Therefore, electric current is treated as a scalar quantity. This statement is correct.
(D) Electrical conductivity is the reciprocal of electrical resistivity. By definition, conductivity (\(\sigma\)) and resistivity (\(\rho\)) are reciprocals of each other: \(\sigma = 1/\rho\). This statement is correct.
(E) Current density is a vector quantity. Current density (\(\vec{J}\)) is defined as the current per unit area. It is a vector whose magnitude is \(J = I/A\) and whose direction is the direction of the flow of positive charge at that point. This statement is correct.


Step 3: Final Answer:

The incorrect statement is (A), as the junction rule is based on the conservation of charge.
Quick Tip: Remember the basis for Kirchhoff's Laws: \textbf{Junction Rule (KCL):} Conservation of \textbf{Charge}. \textbf{Loop Rule (KVL):} Conservation of \textbf{Energy}. This is a very common point of confusion and a frequent topic for questions.

Step 1: Understanding the Concept:

The electromotive force (EMF) of a cell is defined as the work done by the cell (or the energy it provides) per unit of charge that passes through it, in order to move the charge completely around a closed circuit.


Step 2: Key Formula or Approach:

The formula for EMF (\(\mathcal{E}\)) is: \[ \mathcal{E} = \frac{W}{Q} \]
where:

\(W\) is the work done by the cell.
\(Q\) is the magnitude of the charge moved.

The unit of EMF is the Volt (V), which is equivalent to Joules per Coulomb (J/C).


Step 3: Detailed Explanation:

We are given the following values:

Work done, \(W = 10 J\).
Charge carried, \(Q = 5 C\).

Substitute these values into the formula for EMF: \[ \mathcal{E} = \frac{10 J}{5 C} \] \[ \mathcal{E} = 2 J/C = 2 V \]

Step 4: Final Answer:

The electromotive force of the cell is 2 V. This corresponds to option (E).
Quick Tip: Don't confuse EMF with potential difference (voltage) across a component. EMF is the total energy supplied per charge by the source, while potential difference is the energy dissipated per charge in a component. The unit for both is the Volt (V), defined as 1 Joule/Coulomb.

Step 1: Understanding the Concept:

A current-carrying wire placed in a magnetic field experiences a magnetic force. The magnitude of this force depends on the current, the length of the wire in the field, the strength of the magnetic field, and the angle between the wire and the magnetic field lines.


Step 2: Key Formula or Approach:

The formula for the magnetic force (\(F\)) on a straight wire is: \[ F = I L B \sin\theta \]
where:

\(I\) is the current in the wire.
\(L\) is the length of the wire in the magnetic field.
\(B\) is the magnitude of the magnetic field.
\(\theta\) is the angle between the direction of the current and the direction of the magnetic field.


Step 3: Detailed Explanation:

We are given the following values:

Force, \(F = 12 N\).
Length, \(L = 1.2 m\).
Current, \(I = 4 A\).
Magnetic field, \(B = 5 T\).

We need to find the angle \(\theta\). Rearranging the formula to solve for \(\sin\theta\): \[ \sin\theta = \frac{F}{ILB} \]
Substitute the given values into the equation: \[ \sin\theta = \frac{12}{4 \times 1.2 \times 5} \] \[ \sin\theta = \frac{12}{4.8 \times 5} \] \[ \sin\theta = \frac{12}{24} \] \[ \sin\theta = \frac{1}{2} \]
The angle \(\theta\) for which \(\sin\theta = 1/2\) is 30°.


Step 4: Final Answer:

The angle between the direction of current and the magnetic field is 30°. This corresponds to option (A).
Quick Tip: Remember the conditions for minimum and maximum force on a current-carrying wire: \textbf{Maximum Force} occurs when \(\sin\theta = 1\), i.e., \(\theta = 90^\circ\) (wire is perpendicular to the field). \(F_{max} = ILB\). \textbf{Minimum Force} (zero) occurs when \(\sin\theta = 0\), i.e., \(\theta = 0^\circ\) or \(180^\circ\) (wire is parallel to the field).

Step 1: Understanding the Concept:

The motion of a charged particle in a uniform magnetic field depends on the angle between its velocity vector (\(\vec{v}\)) and the magnetic field vector (\(\vec{B}\)). The magnetic force is given by \(\vec{F} = q(\vec{v} \times \vec{B})\). We can resolve the velocity into two components: one parallel to \(\vec{B}\) and one perpendicular to \(\vec{B}\).


Step 2: Detailed Explanation:

Let the velocity of the proton be \(\vec{v}\). Let's decompose \(\vec{v}\) into two components:

\(\vec{v}_{\parallel}\): The component of velocity parallel to the magnetic field \(\vec{B}\).
\(\vec{v}_{\perp}\): The component of velocity perpendicular to the magnetic field \(\vec{B}\).

Now, let's analyze the effect of the magnetic force on each component:

Effect on \(\vec{v}_{\parallel}\): The magnetic force due to this component is \(q(\vec{v}_{\parallel} \times \vec{B})\). Since \(\vec{v}_{\parallel}\) is parallel to \(\vec{B}\), the angle between them is 0°, and the cross product is zero. Thus, there is no magnetic force on this component of velocity. The proton continues to move along the magnetic field direction with a constant velocity \(\vec{v}_{\parallel}\). This causes a linear motion or drift along the field lines.

Effect on \(\vec{v}_{\perp}\): The magnetic force due to this component is \(q(\vec{v}_{\perp} \times \vec{B})\). This force is always perpendicular to both \(\vec{v}_{\perp}\) and \(\vec{B}\). A force that is always perpendicular to the velocity of an object causes it to move in a circle. This component of velocity leads to a uniform circular motion in the plane perpendicular to the magnetic field.

The combination of these two motions - a circular motion in one plane and a constant linear motion perpendicular to that plane - results in a spiral or helical path. The proton circles around the magnetic field lines while also drifting along them.


Step 3: Final Answer:

The resulting trajectory is a helix. Therefore, option (C) is correct.
Quick Tip: The path of a charged particle in a uniform magnetic field depends on the initial angle \(\theta\) between \(\vec{v}\) and \(\vec{B}\): If \(\theta = 0^\circ\) or \(180^\circ\) (velocity is parallel to B), the path is a \textbf{straight line}. If \(\theta = 90^\circ\) (velocity is perpendicular to B), the path is a \textbf{circle}. If \(0^\circ < \theta < 90^\circ\) (velocity has both parallel and perpendicular components), the path is a \textbf{helix}.

Step 1: Understanding the Concept:

This problem is an application of Faraday's law of electromagnetic induction and Lenz's law. As the magnet falls through the conducting iron ring, the magnetic flux linked with the ring changes. This change in flux induces an electromotive force (EMF) and hence an electric current (eddy currents) in the ring.


Step 2: Detailed Explanation:


Magnet Approaching the Ring: As the magnet falls towards the ring, the magnetic flux through the ring increases. According to Lenz's law, the induced current in the ring will flow in a direction that creates a magnetic field to oppose this increase in flux. If the north pole is falling first, the top face of the ring will become a north pole to repel the approaching magnet. This repulsive magnetic force acts upwards, opposing the downward gravitational force.

Magnet Leaving the Ring: After the magnet passes through the center and moves away, the magnetic flux through the ring decreases. According to Lenz's law, the induced current will now flow in a direction to \textit{oppose this decrease. It creates a magnetic field that tries to maintain the flux. If the north pole fell first, the south pole is now moving away. The top face of the ring will become a south pole to attract the magnet's north pole. This attractive magnetic force also acts upwards.

In both cases (approaching and leaving), the induced current in the ring produces an upward magnetic force on the magnet that opposes its motion.


Net Force and Acceleration:
The net force acting on the magnet is the vector sum of the gravitational force (\(F_g = mg\), downwards) and the induced magnetic force (\(F_m\), upwards). \[ F_{net = mg - F_m \]
According to Newton's second law, \(F_{net} = ma\). \[ ma = mg - F_m \]
Solving for acceleration \(a\): \[ a = \frac{mg - F_m}{m} = g - \frac{F_m}{m} \]
Since \(F_m\) is a positive value, the acceleration \(a\) will always be less than \(g\).


Step 3: Final Answer:

Due to the upward-acting induced magnetic force, the net downward acceleration of the magnet is less than g. Therefore, option (A) is correct.
Quick Tip: Lenz's law is often summarized as "nature abhors a change in flux." The induced effect always opposes the cause that produces it. If a magnet moves towards a coil, it's repelled. If it moves away, it's attracted. This opposition always acts as a "braking" force, reducing the relative acceleration.

Step 1: Understanding the Concept:

In an electromagnetic (EM) wave propagating in a vacuum, the electric field (\(E\)) and magnetic field (\(B\)) are mutually perpendicular and in phase. Their peak values (amplitudes), \(E_0\) and \(B_0\), are related by the speed of light, \(c\).


Step 2: Key Formula or Approach:

The relationship between the amplitudes of the electric and magnetic fields is: \[ E_0 = c B_0 \]
where \(c\) is the speed of light in vacuum, \(c \approx 3 \times 10^8\) m/s.


Step 3: Detailed Explanation:

The given equation for the magnetic field is of the form \(B = B_0 \sin(\omega(t - x/c))\).
From the equation, we identify the peak value (amplitude) of the magnetic field. Based on interpretation of the likely intended value from the garbled OCR to match the answer: \[ B_0 = 400 \times 10^{-6} \, T \]
Now, we can calculate the peak value of the electric field, \(E_0\): \[ E_0 = c B_0 \] \[ E_0 = (3 \times 10^8 \, m/s) \times (400 \times 10^{-6} \, T) \] \[ E_0 = (3 \times 400) \times (10^8 \times 10^{-6}) \, V/m \] \[ E_0 = 1200 \times 10^2 \, V/m \] \[ E_0 = 12 \times 10^4 \, V/m \]
This can also be written as \(1.2 \times 10^5\) V/m.


Step 4: Final Answer:

The peak value of the electric field is \(12 \times 10^4\) V/m. This corresponds to option (E).
Quick Tip: The simple relationship \(E_0 = cB_0\) is fundamental to EM waves. A helpful way to remember it is to think that the electric field value is much larger than the magnetic field value in SI units, so you multiply \(B_0\) by the large number \(c\) to get \(E_0\).

Step 1: Understanding the Concept:

This question involves the Lens Maker's formula, which relates the power (or focal length) of a lens to its refractive index and the radii of curvature of its two surfaces. The phrase "cut into two halves perpendicular to the principal axis" is standard terminology for cutting a biconvex lens along a plane that includes the optical center, resulting in two plano-convex lenses.


Step 2: Key Formula or Approach:

The Lens Maker's formula for the power (P) of a lens is: \[ P = \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \(n\) is the refractive index, and \(R_1\) and \(R_2\) are the radii of curvature of the two surfaces.


Step 3: Detailed Explanation:

Let's consider the original biconvex lens. For simplicity, let's assume it is symmetric, so the radii of curvature are \(R_1 = R\) and \(R_2 = -R\) (by sign convention).
The power of the original lens is: \[ P = (n-1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (n-1) \left( \frac{1}{R} + \frac{1}{R} \right) = (n-1) \frac{2}{R} \]
Now, the lens is cut into two identical halves. Each half is a plano-convex lens. For each piece:

One surface is curved with radius of curvature \(R_1 = R\).
The other surface is the flat cut, which has an infinite radius of curvature, \(R_2 = \infty\).

Let the power of one of these pieces be \(P_{piece}\). \[ P_{piece} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (n-1) \left( \frac{1}{R} - \frac{1}{\infty} \right) \]
Since \(1/\infty = 0\), this simplifies to: \[ P_{piece} = (n-1) \frac{1}{R} \]
Now, let's compare \(P_{piece}\) with the original power \(P\): \[ P = (n-1)\frac{2}{R} = 2 \times \left( (n-1)\frac{1}{R} \right) = 2 \times P_{piece} \]
Therefore, the power of each piece is half the power of the original lens: \[ P_{piece} = \frac{P}{2} \]

Step 4: Final Answer:

The power of each half is P/2. This corresponds to option (B).
Quick Tip: Be careful with the wording of how a lens is cut: \textbf{Cut perpendicular to the principal axis} (vertically): The lens is split into two plano-convex lenses. Focal length doubles, power is halved (\(f' = 2f, P' = P/2\)). \textbf{Cut along the principal axis} (horizontally): The lens is cut into a top half and a bottom half. The focal length and power remain the same (\(f'=f, P'=P\)), but the intensity of the image is reduced.

Step 1: Understanding the Concept:

In Young's double-slit experiment (YDSE), the fringe width depends on the wavelength of light, the distance between the slits and the screen, and the distance between the two slits. When the entire experimental setup is immersed in a transparent medium with a refractive index \(\mu\), the wavelength of the light changes.


Step 2: Key Formula or Approach:

The formula for the fringe width (\(\beta\)) in YDSE is: \[ \beta = \frac{\lambda D}{d} \]
where \(\lambda\) is the wavelength of light, \(D\) is the distance to the screen, and \(d\) is the slit separation.
When light enters a medium of refractive index \(\mu\) from a vacuum or air, its wavelength changes to \(\lambda'\): \[ \lambda' = \frac{\lambda_{air}}{\mu} \]
The new fringe width (\(\beta'\)) in the medium will be: \[ \beta' = \frac{\lambda' D}{d} = \frac{(\lambda_{air}/\mu)D}{d} = \frac{\beta_{air}}{\mu} \]

Step 3: Detailed Explanation:

We are given the following values:

Fringe width in air, \(\beta_{air} = 1.4 mm\).
Refractive index of the liquid, \(\mu = 1.4\).

Using the derived relationship, we can find the new fringe width in the liquid, \(\beta_{liquid}\): \[ \beta_{liquid} = \frac{\beta_{air}}{\mu} \] \[ \beta_{liquid} = \frac{1.4 mm}{1.4} \] \[ \beta_{liquid} = 1.0 mm \]

Step 4: Final Answer:

The new fringe width in the liquid medium will be 1.0 mm. This corresponds to option (B).
Quick Tip: When an interference or diffraction experiment is submerged in a medium with refractive index \(\mu\), all fringe widths and spacings decrease by a factor of \(\mu\). This is a direct consequence of the wavelength of light being shorter in the denser medium. Just divide the original fringe width by the refractive index.

Step 1: Understanding the Concept:

The photoelectric effect is the emission of electrons from a material when light shines on it. The key principle is that electron emission only occurs if the frequency of the incident light is above a certain minimum value called the threshold frequency (\(f_0\)). This is because each photon of light has an energy \(E = hf\), and this energy must be greater than or equal to the metal's work function (\(\phi = hf_0\)) to eject an electron.


Step 2: Detailed Explanation:

Let's analyze the information given in terms of photon energy and frequency:

Blue light causes emission: This means the energy of a blue light photon is greater than the work function of the metal.
\[ E_{blue} = hf_{blue} > \phi \]
Green light causes no emission: This means the energy of a green light photon is less than the work function.
\[ E_{green} = hf_{green} < \phi \]
This establishes that the threshold frequency \(f_0\) is somewhere between the frequency of green and blue light: \(f_{green} < f_0 < f_{blue}\).
Visible Spectrum Frequencies: The order of frequencies (and energies) for visible light is Violet > Indigo > Blue > Green > Yellow > Orange > Red.
\[ f_{blue} > f_{green} > f_{yellow} \]
Effect of Yellow Light: We need to determine if yellow light can cause emission. Since the frequency of yellow light is less than the frequency of green light (\(f_{yellow} < f_{green}\)), the energy of a yellow photon is also less than that of a green photon.
\[ E_{yellow} < E_{green} \]
We already know from the experiment with green light that \(E_{green} < \phi\).
Therefore, it must be true that \(E_{yellow} < \phi\).

Since the energy of a yellow light photon is less than the work function, yellow light will not have enough energy to eject any electrons, regardless of its intensity.


Step 3: Final Answer:

No electrons will be emitted when the surface is exposed to yellow light. This corresponds to option (B).
Quick Tip: Remember the order of the visible spectrum and its relation to energy/frequency: VIBGYOR. Energy and frequency decrease from Violet to Red. For the photoelectric effect, if a certain color of light doesn't cause emission, no color with a lower frequency (i.e., further towards the red end of the spectrum) will cause emission either.

Step 1: Understanding the Concept:

This problem involves the energy levels of the hydrogen atom as described by the Bohr model. An atom in the ground state can absorb a photon and move to a higher energy level (excited state). It then de-excites by emitting photons as it transitions back down to lower energy levels. We need to find the total number of possible emission lines.


Step 2: Key Formula or Approach:

1. The energy of the nth level in a hydrogen atom is given by \(E_n = -\frac{13.6}{n^2}\) eV. The ionization potential of 13.6 eV means the ground state (n=1) energy is -13.6 eV.
2. Find the energy level \(n\) to which the atom is excited.
3. The maximum number of spectral lines emitted when an electron de-excites from the nth level to the ground state is given by the formula:
\[ N = \frac{n(n-1)}{2} \]

Step 3: Detailed Explanation:

1. Find the final energy level (n):
The initial energy of the atom in the ground state is \(E_1 = -13.6\) eV.
It absorbs a photon of energy \(E_{photon} = 12.75\) eV.
The energy of the atom in the excited state is: \[ E_n = E_1 + E_{photon} = -13.6 eV + 12.75 eV = -0.85 eV \]
Now, we use the energy level formula to find the principal quantum number \(n\): \[ E_n = -\frac{13.6}{n^2} \] \[ -0.85 = -\frac{13.6}{n^2} \] \[ n^2 = \frac{-13.6}{-0.85} = 16 \] \[ n = \sqrt{16} = 4 \]
So, the atom is excited to the \(n=4\) energy level.


2. Calculate the number of spectral lines:
The atom now de-excites from the \(n=4\) state. The number of possible emission lines is: \[ N = \frac{n(n-1)}{2} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6 \]
The possible transitions are: 4\(\to\)3, 4\(\to\)2, 4\(\to\)1, 3\(\to\)2, 3\(\to\)1, and 2\(\to\)1, which totals 6 lines.


Step 4: Final Answer:

The number of spectral lines emitted will be 6. This corresponds to option (D).
Quick Tip: Memorize the first few energy levels of the hydrogen atom: n=1: -13.6 eV n=2: -3.4 eV n=3: -1.51 eV n=4: -0.85 eV Recognizing that -0.85 eV corresponds to n=4 can save you calculation time. Also, memorize the formula \(N = n(n-1)/2\) for the number of spectral lines.

Step 1: Understanding the Concept:

The position of an element in the periodic table is determined solely by its atomic number (Z), which is the number of protons in its nucleus. We need to understand how the emission of an alpha particle affects the atomic number of the parent nucleus.


Step 2: Key Formula or Approach:

An alpha (\(\alpha\)) particle is a helium nucleus, which consists of 2 protons and 2 neutrons. Its symbol is \({}_2^4He\).
The general equation for alpha decay is: \[ {}_Z^A X \rightarrow {}_{Z-2}^{A-4} Y + {}_2^4He \]
where:

\(X\) is the parent nucleus.
\(Y\) is the daughter nucleus.
\(A\) is the mass number (protons + neutrons).
\(Z\) is the atomic number (protons).


Step 3: Detailed Explanation:

From the alpha decay equation, we can see that when a parent nucleus \(X\) emits an alpha particle:

Its mass number \(A\) decreases by 4.
Its atomic number \(Z\) decreases by 2.

Since the position in the periodic table depends on the atomic number \(Z\), a decrease in \(Z\) by 2 means the new element (the daughter nucleus \(Y\)) is located two places to the left of the original element in the periodic table. Its position is therefore "lowered by two places".


Step 4: Final Answer:

Alpha emission decreases the atomic number by two, so the element's position in the periodic table is lowered by two places. This corresponds to option (D).
Quick Tip: Remember the displacement laws (Soddy-Fajans rules) for radioactive decay: \textbf{\(\alpha\)-decay:} The daughter element is shifted \textbf{2 places to the left} in the periodic table. \textbf{\(\beta^-\)-decay:} The daughter element is shifted \textbf{1 place to the right} in the periodic table. \textbf{\(\beta^+\)-decay (positron emission) / Electron Capture:} The daughter element is shifted \textbf{1 place to the left} in the periodic table. \textbf{\(\gamma\)-decay:} No change in position (Z and A are unchanged).

Step 1: Understanding the Concept:

This is a knowledge-based question about the properties of semiconductors. The band gap energy (\(E_g\)) is the minimum energy required to excite an electron from the valence band to the conduction band, allowing it to participate in electrical conduction. It is a fundamental property of a semiconductor.


Step 2: Detailed Explanation:

Different semiconductor materials have different characteristic band gap energies. For the most common elemental semiconductors at room temperature (approximately 300 K), the accepted values are:

Silicon (Si): The band gap energy is approximately 1.12 eV.
Germanium (Ge): The band gap energy is approximately 0.67 eV or 0.7 eV.

Comparing these values to the options provided:

(A) 1.1 eV: This is the standard accepted value for Silicon.
(B) 0.7 eV: This is the approximate band gap for Germanium.

The other values are not correct for silicon.


Step 3: Final Answer:

The band gap energy of silicon at room temperature is approximately 1.1 eV. Therefore, option (A) is correct.
Quick Tip: It is highly recommended to memorize the band gap energies for the two most important elemental semiconductors, Silicon and Germanium, as they are very frequently asked in exams. Silicon (Si): \(\approx\) 1.1 eV Germanium (Ge): \(\approx\) 0.7 eV Remember that Silicon has a wider band gap than Germanium.