Step 1: Understanding the Concept:

The rule for significant figures in addition and subtraction is that the result should be rounded to the same number of decimal places as the number with the least number of decimal places in the calculation.


Step 2: Detailed Explanation:

First, we perform the standard addition of the given numbers.
\[ \begin{array}{rc} & 12.1100
& 18.0
+ & 1.012
\hline & 31.1220
\end{array} \]
Now, we identify the number of decimal places in each of the numbers being added.


12.1100 has 4 decimal places.
18.0 has 1 decimal place.
1.012 has 3 decimal places.

According to the rule, the final answer must be rounded to the least number of decimal places, which is 1.

So, we need to round the sum, 31.1220, to one decimal place.

The digit after the first decimal place is 2. Since 2 is less than 5, we do not round up the preceding digit (1).

Therefore, the rounded sum is 31.1.


Step 3: Final Answer:

The value of the sum according to the rules of significant figures is 31.1. This corresponds to option (D).
Quick Tip: For addition and subtraction, always look at the number of decimal places. For multiplication and division, look at the total number of significant figures in the least precise measurement.

Step 1: Understanding the Concept:

The atomic spectrum of hydrogen consists of several series of spectral lines, named after their discoverers. Each series corresponds to electronic transitions from higher energy levels to a specific lower energy level.


Step 2: Detailed Explanation:

The different spectral series for the hydrogen atom are defined by the principal quantum number (\(n_1\)) of the final energy level to which the electron transitions.


Lyman Series: Transitions from \(n_2 = 2, 3, 4, ...\) to \(n_1 = 1\). These lines fall in the Ultraviolet (UV) region.

Balmer Series: Transitions from \(n_2 = 3, 4, 5, ...\) to \(n_1 = 2\). These lines fall in the Visible region of the electromagnetic spectrum. This is why the lines of the hydrogen spectrum visible to the naked eye belong to this series.

Paschen Series: Transitions from \(n_2 = 4, 5, 6, ...\) to \(n_1 = 3\). These lines fall in the Infrared (IR) region.

Brackett Series: Transitions from \(n_2 = 5, 6, 7, ...\) to \(n_1 = 4\). These lines fall in the Infrared (IR) region.

Pfund Series: Transitions from \(n_2 = 6, 7, 8, ...\) to \(n_1 = 5\). These lines fall in the Infrared (IR) region.


The question specifically asks about the Balmer series, which corresponds to transitions to the n=2 level. These emissions produce photons with energies that correspond to visible light.


Step 3: Final Answer:

The Balmer series lines for atomic hydrogen are observed in the Visible spectral region. This corresponds to option (A).
Quick Tip: Create a mnemonic to remember the series and their regions. For example: "Loud Ultraviolet, Bright Visible, Powerful Infrared" for Lyman (UV), Balmer (Visible), and Paschen/Brackett/Pfund (IR).

Step 1: Understanding the Concept:

This question deals with the units derived from the quantities in Heisenberg's Uncertainty Principle. The principle relates the uncertainty in a particle's position to the uncertainty in its momentum. We need to find the SI units for the product of uncertainty in position and uncertainty in velocity.


Step 2: Key Formula or Approach:

The product in question is \( (\Delta x) \times (\Delta v) \).

We need to determine the SI units for each term and then multiply them.


The SI unit for position (and thus uncertainty in position, \(\Delta x\)) is the meter (m).

The SI unit for velocity (and thus uncertainty in velocity, \(\Delta v\)) is meters per second (m/s or ms\(^{-1}\)).



Step 3: Detailed Explanation:

We multiply the units of the two quantities:
\[ Unit of (\Delta x \cdot \Delta v) = (Unit of \Delta x) \times (Unit of \Delta v) \] \[ = (m) \times (m \cdot s^{-1}) \] \[ = m^{1+1} \cdot s^{-1} \] \[ = m^2 s^{-1} \]
The correct unit for the product of uncertainty in position and uncertainty in velocity is m\(^2\)s\(^{-1}\).

Upon reviewing the given options, none of them match the correctly derived unit of m\(^2\)s\(^{-1}\). There appears to be a typographical error in the question's options or the provided answer key. Heisenberg's Uncertainty Principle states \( \Delta x \cdot \Delta p \ge \frac{h}{4\pi} \), which implies \( \Delta x \cdot m\Delta v \ge \frac{h}{4\pi} \), and so the unit of \( \Delta x \cdot \Delta v \) is \( \frac{[h]}{[m]} = \frac{J \cdot s}{kg} = \frac{kg \cdot m^2 \cdot s^{-2} \cdot s}{kg} = m^2s^{-1} \).

However, as per the provided answer key, the correct option is (D). This indicates a significant error in the source material. We are noting this discrepancy for clarity.


Step 4: Final Answer:

Based on the provided answer key, the answer is option (D). However, based on fundamental principles of physics, the correct unit is m\(^2\)s\(^{-1}\), which is not listed.
Quick Tip: Always use dimensional analysis to check your answers in physics and chemistry problems. Derive the units from the formula to ensure they are correct. Be aware that questions in exams can sometimes contain errors.

Step 1: Understanding the Concept:

Electron gain enthalpy (\(\Delta_{eg}H\)) is the enthalpy change when a neutral gaseous atom accepts an electron to form a gaseous anion. A positive electron gain enthalpy means that energy is absorbed (an endothermic process) when an electron is added, indicating that the formation of the anion is unfavorable. Inert gases have stable, completely filled valence shells (ns\(^2\)np\(^6\)), making it very difficult to add an extra electron.


Step 2: Detailed Explanation:

For an inert gas, an incoming electron must occupy a new, higher-energy principal shell (n+1). For example:


For He (1s\(^2\)), the electron enters the 2s orbital.
For Ne (2s\(^2\)2p\(^6\)), the electron enters the 3s orbital.
For Ar (3s\(^2\)3p\(^6\)), the electron enters the 4s orbital.

This process requires a significant amount of energy, resulting in positive electron gain enthalpies for all noble gases.

The question asks for the largest positive value, which means the process is the most energetically unfavorable.

The trend down the group (He to Rn) is that the atomic size increases. The incoming electron enters an orbital that is progressively farther from the nucleus and is more effectively shielded by the inner electrons. This makes the addition of an electron slightly less unfavorable as we go down the group.

Therefore, we expect the electron gain enthalpy to become less positive down the group.


He has a value of +48 kJ/mol.
Ne has a value of +116 kJ/mol.
Ar has a value of +96 kJ/mol.
Kr has a value of +96 kJ/mol.

Neon (Ne) has an anomalously high positive value. This is because the incoming electron must be added to the n=3 shell, which is relatively high in energy, and the compact size and stable 2p\(^6\) configuration of Neon strongly resists the addition of a new electron. The electron-electron repulsion is significant.


Step 3: Final Answer:

Among the given inert gases, Neon (Ne) has the largest positive electron gain enthalpy. This corresponds to option (B).
Quick Tip: Remember that noble gases are an exception to the general trend of electron gain enthalpy. Due to their stable electronic configurations, they have positive values, meaning the process of adding an electron is endothermic.

Step 1: Understanding the Concept:

The International Union of Pure and Applied Chemistry (IUPAC) provides official names for chemical elements. For super-heavy elements (atomic number > 100), there is also a systematic naming scheme used temporarily until an official name is decided. The question asks for the official IUPAC name for the element with Z = 105.


Step 2: Detailed Explanation:

Let's analyze the options provided:


Mendelevium (Md) is the element with atomic number 101.
Nobelium (No) is the element with atomic number 102.
Lawrencium (Lr) is the element with atomic number 103.
Rutherfordium (Rf) is the element with atomic number 104.
Dubnium (Db) is the element with atomic number 105.

The systematic name for element 105 would be derived from the Latin roots for the digits: 1 (un), 0 (nil), 5 (pent), with the suffix -ium. So, the systematic name is Unnilpentium (Unp). However, the question asks for the official IUPAC name. In 1997, IUPAC officially named element 105 Dubnium (Db), in honor of the Joint Institute for Nuclear Research (JINR) in Dubna, Russia.


Step 3: Final Answer:

The official IUPAC name for the element with atomic number 105 is Dubnium. This corresponds to option (E).
Quick Tip: It is helpful to memorize the names and symbols of elements up to at least Z=118. For elements beyond 103, remember both their systematic names (e.g., Unnilquadium for 104) and their official names (e.g., Rutherfordium).

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, giving them the same electronic configuration as a noble gas. We need to examine the Lewis structure of the central atom in each compound to see if it is surrounded by eight valence electrons.


Step 2: Detailed Explanation:

Let's analyze each option:


(A) BF\(_3\): Boron (B) is the central atom. It is in Group 13 and has 3 valence electrons. It forms three single covalent bonds with three fluorine atoms. The total number of electrons around Boron is \(3 \times 2 = 6\). This is an "electron-deficient" molecule with an incomplete octet.

(B) BeH\(_2\): Beryllium (Be) is the central atom. It is in Group 2 and has 2 valence electrons. It forms two single covalent bonds with two hydrogen atoms. The total number of electrons around Beryllium is \(2 \times 2 = 4\). This is another example of an incomplete octet.

(C) SCl\(_2\): Sulfur (S) is the central atom. It is in Group 16 and has 6 valence electrons. It forms two single covalent bonds with two chlorine atoms, using 2 of its valence electrons. The remaining 4 valence electrons exist as two lone pairs on the sulfur atom. The total number of electrons around Sulfur is (2 bonding pairs \(\times\) 2 electrons/pair) + (2 lone pairs \(\times\) 2 electrons/pair) = 4 + 4 = 8. Sulfur has a complete octet.

(D) AlCl\(_3\): Aluminum (Al) is the central atom. It is in Group 13 and has 3 valence electrons. It forms three single covalent bonds with three chlorine atoms. The total number of electrons around Aluminum is \(3 \times 2 = 6\). It has an incomplete octet.

(E) LiCl: Lithium chloride is an ionic compound. Li loses an electron to form Li\(^+\) (duet configuration) and Cl gains an electron to form Cl\(^-\) (octet configuration). The concept of a central atom with a shared octet is not applicable here in the same way as for covalent molecules.



Step 3: Final Answer:

Among the given options, only in SCl\(_2\) does the central atom (Sulfur) have a complete octet of eight electrons. This corresponds to option (C).
Quick Tip: Remember the common exceptions to the octet rule: Incomplete octets (H, Be, B, Al), expanded octets (elements from Period 3 onwards like P, S, Cl, Xe), and odd-electron molecules (like NO, NO\(_2\)).

Step 1: Understanding the Concept:

The shape of a molecule is determined by the arrangement of its electron pairs (both bonding and non-bonding) around the central atom, according to the Valence Shell Electron Pair Repulsion (VSEPR) theory. A square planar shape arises from an octahedral electron geometry where two positions are occupied by lone pairs. This corresponds to the VSEPR notation AX\(_4\)E\(_2\).


Step 2: Detailed Explanation:

Let's determine the shape for each option by finding the number of valence electrons, bonding pairs, and lone pairs on the central atom.


(A) SF\(_4\): Central atom is Sulfur (S, Group 16, 6 valence e\(^-\)). It is bonded to 4 Fluorine atoms.

Bonding pairs = 4
Lone pairs = (6 - 4) / 2 = 1
VSEPR type: AX\(_4\)E\(_1\). The shape is a see-saw.

(B) NH\(_4^+\): Central atom is Nitrogen (N, Group 15, 5 valence e\(^-\)). It has a +1 charge, so we subtract 1 electron. Total valence e\(^-\) = 5 - 1 = 4. It is bonded to 4 Hydrogen atoms.

Bonding pairs = 4
Lone pairs = (4 - 4) / 2 = 0
VSEPR type: AX\(_4\). The shape is tetrahedral.

(C) CH\(_2\)Cl\(_2\): Central atom is Carbon (C, Group 14, 4 valence e\(^-\)). It is bonded to 2 H and 2 Cl atoms.

Bonding pairs = 4
Lone pairs = (4 - 4) / 2 = 0
VSEPR type: AX\(_4\). The shape is tetrahedral.

(D) CH\(_4\): Central atom is Carbon (C, Group 14, 4 valence e\(^-\)). It is bonded to 4 H atoms.

Bonding pairs = 4
Lone pairs = (4 - 4) / 2 = 0
VSEPR type: AX\(_4\). The shape is tetrahedral.

(E) XeF\(_4\): Central atom is Xenon (Xe, Group 18, 8 valence e\(^-\)). It is bonded to 4 Fluorine atoms.

Bonding pairs = 4
Lone pairs = (8 - 4) / 2 = 2
VSEPR type: AX\(_4\)E\(_2\). The electron geometry is octahedral. The two lone pairs position themselves on opposite sides (axial positions) to minimize repulsion, forcing the four fluorine atoms into a plane around the central xenon atom. This results in a square planar shape.



Step 3: Final Answer:

The molecule with a square planar shape is XeF\(_4\). This corresponds to option (E).
Quick Tip: To quickly find a square planar molecule, look for a central atom with 6 electron domains (4 bonding pairs and 2 lone pairs). This is common for noble gas compounds (like XeF\(_4\)) or transition metal complexes.

Step 1: Understanding the Concept:

This question asks for the correct thermodynamic relationship between the molar heat capacity at constant pressure (C\(_p\)) and the molar heat capacity at constant volume (C\(_v\)) for an ideal gas. This relationship is known as Mayer's relation.


Step 2: Detailed Explanation:


C\(_v\): Molar heat capacity at constant volume. When heat is supplied to a gas at constant volume, all the energy goes into increasing its internal energy (\(\Delta U\)), as no work is done (\(w = -P\Delta V = 0\)). So, \(q_v = \Delta U = nC_v\Delta T\).

C\(_p\): Molar heat capacity at constant pressure. When heat is supplied at constant pressure, the gas expands and performs work on the surroundings. Therefore, the supplied heat must not only increase the internal energy but also provide the energy for this work. Thus, more heat is required to raise the temperature by the same amount compared to the constant volume case, meaning C\(_p >\) C\(_v\).

The relationship is derived from the first law of thermodynamics (\(\Delta U = q + w\)) and the definition of enthalpy (\(H = U + PV\)). For one mole of an ideal gas, it can be shown that the difference between these two heat capacities is equal to the ideal gas constant, R.


The established relationship is: \[ C_p - C_v = R \]
This is Mayer's relation.

Let's check the given options:

(A) C\(_p\) + C\(_v\) = R is incorrect.

(B) C\(_p\) / C\(_v\) = \(\gamma\), the heat capacity ratio (adiabatic index), not R.

(C) C\(_p\) - C\(_v\) = R is correct.

(D) C\(_v\) / C\(_p\) = 1/\(\gamma\), not R.

(E) C\(_v\) - C\(_p\) = R is incorrect; the difference is -R.


Step 3: Final Answer:

The correct relationship is C\(_p\) - C\(_v\) = R. This corresponds to option (C).
Quick Tip: Remember that C\(_p\) is always greater than C\(_v\) for a gas because at constant pressure, energy is needed for both increasing internal energy and for doing expansion work. The difference C\(_p\) - C\(_v\) must be positive, which helps eliminate some incorrect options.

Step 1: Understanding the Concept:

Thermodynamic properties are classified into two types:

Extensive Properties: These properties depend on the mass or size of the system. Their value is directly proportional to the amount of substance present. Examples include mass, volume, internal energy, enthalpy, and heat capacity.
Intensive Properties: These properties are independent of the mass or size of the system. Their value does not change if the system is divided into smaller parts. Examples include temperature, pressure, density, and molar heat capacity.


Step 2: Detailed Explanation:

Let's analyze each property given:


(i) Volume: If you double the amount of a substance, its volume doubles (at constant temperature and pressure). Thus, volume is an extensive property.

(ii) Pressure: If you take a container of gas and divide it in half with a partition, the pressure in each half remains the same as the original pressure. Thus, pressure is an intensive property.

(iii) Density: Density is defined as mass per unit volume (\(\rho = m/V\)). If you double the mass of a substance, its volume also doubles, so the ratio (density) remains constant. Thus, density is an intensive property. It is a ratio of two extensive properties.

(iv) Heat capacity: Heat capacity is the amount of heat needed to raise the system's temperature by one degree. If you have twice the amount of substance, you will need twice the amount of heat to cause the same temperature change. Thus, heat capacity is an extensive property. (Note: Specific heat capacity and molar heat capacity are intensive).


The extensive properties from the list are Volume (i) and Heat capacity (iv).


Step 3: Final Answer:

The extensive properties are (i) and (iv). This corresponds to option (D).
Quick Tip: A simple test to distinguish between extensive and intensive properties is to imagine dividing the system in half. If the property's value is halved, it's extensive. If it remains the same, it's intensive.

Step 1: Understanding the Concept:

The pH scale measures the acidity or alkalinity of a solution. A high pH value (pH > 7) indicates a basic or alkaline solution, while a low pH value (pH < 7) indicates an acidic solution. The highest pH will correspond to the strongest and most concentrated basic solution among the choices.


Step 2: Key Formula or Approach:

For strong acids: pH = -log[H\(^+\)].

For strong bases: pOH = -log[OH\(^-\)], and pH = 14 - pOH.


Step 3: Detailed Explanation:

Let's calculate the approximate pH for each solution:


(A) 0.10 M HCl: HCl is a strong acid, so it dissociates completely. [H\(^+\)] = 0.10 M = 10\(^{-1}\) M.

pH = -log(10\(^{-1}\)) = 1.

(B) 0.50 M H\(_2\)SO\(_4\): H\(_2\)SO\(_4\) is a strong acid. For the first dissociation, it is complete. Assuming complete dissociation for both protons for a quick comparison: [H\(^+\)] \(\approx\) 2 \(\times\) 0.50 M = 1.0 M.

pH = -log(1.0) = 0. (This is a very acidic solution).

(C) 0.10 M NaOH: NaOH is a strong base, so it dissociates completely. [OH\(^-\)] = 0.10 M = 10\(^{-1}\) M.

pOH = -log(10\(^{-1}\)) = 1.

pH = 14 - pOH = 14 - 1 = 13.

(D) 0.5 M HCl: HCl is a strong acid. [H\(^+\)] = 0.5 M.

pH = -log(0.5) = -log(5 \(\times\) 10\(^{-1}\)) = -(log 5 + log 10\(^{-1}\)) = - (0.7 - 1) = 0.3.

(E) 0.01 M NaOH: NaOH is a strong base. [OH\(^-\)] = 0.01 M = 10\(^{-2}\) M.

pOH = -log(10\(^{-2}\)) = 2.

pH = 14 - pOH = 14 - 2 = 12.


Comparing the calculated pH values: 1, 0, 13, 0.3, and 12. The highest value is 13.


Step 4: Final Answer:

The 0.10 M NaOH solution has the highest pH value of 13. This corresponds to option (C).
Quick Tip: To find the highest pH, look for the bases. The higher the concentration of a strong base, the higher the pH. To find the lowest pH, look for the acids. The higher the concentration of a strong acid, the lower the pH.

Step 1: Understanding the Concept:

Oxidation state (or oxidation number) is a number assigned to an element in a chemical combination which represents the number of electrons lost or gained by an atom of that element. We need to apply the rules for assigning oxidation states to magnesium on both the reactant and product sides of the equation.


Step 2: Detailed Explanation:


Reactant side: The reactant is Magnesium in its solid elemental form, Mg\(_{(s)}\). According to the rules of oxidation states, any element in its free or uncombined state has an oxidation state of 0.

Product side: The product is magnesium nitride, Mg\(_3\)N\(_2\). This is an ionic compound.

Magnesium (Mg) is an alkaline earth metal (Group 2). In its compounds, it almost always exhibits an oxidation state of +2.
Nitrogen (N) is in Group 15. When it forms a binary compound with a less electronegative element like a metal, it typically takes on an oxidation state of -3 (as the nitride ion, N\(^{3-}\)).
We can verify this by checking if the compound is neutral: (3 \(\times\) Mg's charge) + (2 \(\times\) N's charge) = (3 \(\times\) (+2)) + (2 \(\times\) (-3)) = +6 - 6 = 0. The assignments are correct.


Therefore, the oxidation state of Magnesium (Mg) changes from 0 in Mg\(_{(s)}\) to +2 in Mg\(_3\)N\(_2\).


Step 3: Final Answer:

The change in the oxidation state of Magnesium is from 0 to +2. This corresponds to option (A).
Quick Tip: Memorize the key rules for oxidation states: an element by itself is 0, Group 1 metals are +1, Group 2 metals are +2, and fluorine is -1 in compounds. Use these as a starting point to solve for other elements in a compound.

Step 1: Understanding the Concept:

The question relates three important parameters in electrochemistry: resistance (R), conductivity (\(\kappa\)), and the cell constant (G or l/A). The cell constant is a characteristic of the geometry of the conductivity cell and is independent of the solution measured in it.


Step 2: Key Formula or Approach:

The relationship between conductivity, resistance, and cell constant is given by the formula: \[ Conductivity (\kappa) = \frac{1}{Resistance (R)} \times Cell Constant (G^) \]
Rearranging this formula to solve for the cell constant: \[ Cell Constant (G^) = Conductivity (\kappa) \times Resistance (R) \]

Step 3: Detailed Explanation:

We are given the following values:


Resistance (R) = 100 \(\Omega\)
Conductivity (\(\kappa\)) = 1.29 S m\(^{-1}\)

Now, we substitute these values into the rearranged formula. Note that the unit Siemens (S) is equivalent to \(\Omega^{-1}\). \[ G^ = (1.29 S m^{-1}) \times (100 \, \Omega) \] \[ G^ = (1.29 \, \Omega^{-1} m^{-1}) \times (100 \, \Omega) \] \[ G^ = 129 m^{-1} \]
The calculated cell constant is 129 m\(^{-1}\). Now we must check the units of the options. Options (C) and (D) are in cm\(^{-1}\). We need to convert our answer to cm\(^{-1}\) to compare.

We know that 1 m = 100 cm.

Therefore, 1 m\(^{-1}\) = (100 cm)\(^{-1}\) = \(\frac{1}{100}\) cm\(^{-1}\).
\[ G^ = 129 \times \left(\frac{1}{100} cm^{-1}\right) \] \[ G^ = 1.29 cm^{-1} \]

Step 4: Final Answer:

The value of the cell constant is 1.29 cm\(^{-1}\). This corresponds to option (D).
Quick Tip: Pay close attention to units in electrochemistry problems. Conductivity is often given in S m\(^{-1}\) or S cm\(^{-1}\). Ensure all your units are consistent before and after the calculation. Remember that Cell Constant = Conductivity \(\times\) Resistance.

Step 1: Understanding the Concept:

This problem applies Henry's Law, which states that the partial pressure of a gas in the vapor phase (p) is proportional to the mole fraction of the gas (x) in the solution. The proportionality constant is Henry's law constant (K\(_H\)).


Step 2: Key Formula or Approach:

Henry's Law is expressed as: \[ p = K_H \cdot x \]
We need to find the mole fraction (x), so we can rearrange the formula: \[ x = \frac{p}{K_H} \]

Step 3: Detailed Explanation:

We are given the following values:

Partial pressure of N\(_2\) (p) = 7.648 bar
Henry's law constant (K\(_H\)) = 76.4 k bar

Before we can use the formula, we must ensure that the units of pressure for p and K\(_H\) are the same. Let's convert K\(_H\) from kilobars (k bar) to bars.

1 k bar = 1000 bar
\[ K_H = 76.4 k bar = 76.4 \times 1000 bar = 76400 bar \]
Now, we can substitute the values into the rearranged formula to find the mole fraction (x) of N\(_2\). \[ x_{N_2} = \frac{p}{K_H} = \frac{7.648 bar}{76400 bar} \]
To simplify the calculation, we can write 76400 as 7.64 \(\times\) 10\(^4\). \[ x_{N_2} = \frac{7.648}{76400} = \frac{7.648}{7.648 \times 10^4} \] \[ x_{N_2} = \frac{1}{10^4} = 10^{-4} \]

Step 4: Final Answer:

The mole fraction of N\(_2\) in the solution is 10\(^{-4}\). This corresponds to option (C).
Quick Tip: The most common source of error in Henry's Law problems is unit mismatch. Always convert the partial pressure and Henry's constant to the same unit (e.g., bar, Pa, or atm) before dividing.

N/A Quick Tip: Remember the classic examples for different reaction orders. Hydrolysis of esters and inversion of cane sugar are textbook examples of pseudo first-order reactions because water is the solvent and its concentration is effectively constant.

N/A Quick Tip: Pay close attention to the axes of the plot and whether the logarithm is natural (ln) or base-10 (log). For a first-order reaction: - A plot of \(\ln[R]\) vs. \(t\) gives a slope of \(-k\). - A plot of \(\log[R]\) vs. \(t\) gives a slope of \(-k/2.303\). - A plot of \(\ln([R]_0/[R])\) vs. \(t\) gives a slope of \(+k\). - A plot of \(\log([R]_0/[R])\) vs. \(t\) gives a slope of \(+k/2.303\).

N/A Quick Tip: Memorize the two major exceptions to the Aufbau principle in the first transition series (3d series): Chromium (Cr, Z=24): \([Ar] 3d^5 4s^1\) (not \(3d^4 4s^2\)) Copper (Cu, Z=29): \([Ar] 3d^{10} 4s^1\) (not \(3d^9 4s^2\)) This is due to the extra stability of half-filled and fully-filled d-orbitals.

N/A Quick Tip: The trend for melting points in the 3d series generally peaks around Group 6 (Cr) due to the maximum number of unpaired d-electrons and strong interatomic bonding. Remember that Manganese (Mn) is a notable exception with a surprisingly low melting point.

N/A Quick Tip: Zinc, Cadmium, and Mercury (Group 12) are often considered d-block elements but not true transition metals because they do not have an incomplete d-subshell in their elemental or common ionic states. Their chemistry is dominated by the +2 oxidation state.

N/A Quick Tip: This reaction is a common laboratory method for the preparation of oxygen gas. Memorizing the products of thermal decomposition of common salts like permanganates, nitrates, and carbonates is very helpful for competitive exams.

N/A Quick Tip: Remember the key exceptions in lanthanoid configurations which are driven by the stability of empty, half-filled, and fully-filled f-orbitals: Lanthanum (La): \([Xe] 5d^1 6s^2\) Cerium (Ce): \([Xe] 4f^1 5d^1 6s^2\) Europium (Eu): \([Xe] 4f^7 6s^2\) (half-filled stability) Gadolinium (Gd): \([Xe] 4f^7 5d^1 6s^2\) (maintains half-filled f-shell) Ytterbium (Yb): \([Xe] 4f^{14} 6s^2\) (fully-filled stability) Lutetium (Lu): \([Xe] 4f^{14} 5d^1 6s^2\) (maintains fully-filled f-shell)

N/A Quick Tip: A key difference between lanthanoids and actinoids is reactivity and the range of oxidation states. Actinoids are generally more reactive and show a greater variety of oxidation states than lanthanoids. This is due to the smaller energy gap between the 5f, 6d, and 7s orbitals compared to the 4f, 5d, and 6s orbitals.

N/A Quick Tip: In problems involving coordination compounds and AgNO₃, the number of moles of AgCl precipitated per mole of complex directly tells you the number of chloride ions acting as counter ions (i.e., outside the coordination sphere).

N/A Quick Tip: When naming coordination compounds, remember the key rules: 1. Cation is named before the anion. 2. Ligands are named first in alphabetical order, then the metal. 3. Use prefixes (di, tri, tetra) for simple ligands and (bis, tris, tetrakis) for complex ligands. 4. If the complex ion is an anion, the metal's name ends in "-ate" (e.g., ferrate, cuprate). 5. The oxidation state of the metal is written in Roman numerals.

N/A Quick Tip: Always look for ambidentate ligands like NO₂⁻/ONO⁻ (nitro/nitrito), SCN⁻/NCS⁻ (thiocyanato/isothiocyanato), and CN⁻/NC⁻ (cyano/isocyano). Their presence is a strong indicator of linkage isomerism. Be cautious of potential errors in answer keys and rely on fundamental concepts.

N/A Quick Tip: To quickly solve such problems, memorize the spectrochemical series (at least the common strong and weak field ligands). For d⁶ ions like Co³⁺, a strong field ligand will cause pairing, leading to a diamagnetic \(t_{2g}^6\) configuration.

Step 1: Understanding the Concept:

The question asks to identify a heterocyclic aromatic compound. This requires understanding two terms:

- Heterocyclic: A cyclic compound where at least one atom in the ring structure is an element other than carbon. This other atom is called a heteroatom (e.g., oxygen, nitrogen, sulfur).

- Aromatic: A cyclic, planar compound with a continuous ring of p-orbitals that follows Hückel's rule, having (4n+2) \(\pi\) electrons, where n is a non-negative integer.


Step 2: Analysis of Options:

Let's analyze each option:

(A) Phenol, (B) Aniline, and (C) Toluene are all derivatives of benzene. Benzene is a carbocyclic (or homocyclic) compound, meaning its ring is made up entirely of carbon atoms. Thus, these are not heterocyclic.

(D) Naphthalene consists of two fused benzene rings. It is aromatic but also carbocyclic/homocyclic.

(E) Furan has a five-membered ring structure containing four carbon atoms and one oxygen atom. Since the ring contains a heteroatom (oxygen), it is a heterocyclic compound. Furan is also planar, cyclic, has a conjugated system, and possesses 6 \(\pi\) electrons (4 from the two double bonds and 2 from one of the lone pairs on the oxygen atom), which satisfies Hückel's rule for n=1. Therefore, furan is a heterocyclic aromatic compound.


Step 3: Final Answer:

Furan is the only compound in the list that is both heterocyclic (contains oxygen in the ring) and aromatic. This corresponds to option (E).
Quick Tip: Remember the common examples of five and six-membered heterocyclic aromatic compounds: \textbf{Five-membered rings:} Furan (O), Thiophene (S), Pyrrole (NH). \textbf{Six-membered rings:} Pyridine (N). These are frequently asked in competitive exams.

Step 1: Understanding the Concept:

The resonance effect (R effect) describes the polarity produced in a molecule by the interaction of lone pairs of electrons with \(\pi\)-bonds or the interaction of two \(\pi\)-bonds in adjacent atoms. It is transmitted along a conjugated system.

- +R effect (electron-donating): A group donates electrons to the conjugated system.
- -R effect (electron-withdrawing): A group withdraws electrons from the conjugated system. This is also known as the -M (mesomeric) effect.


Step 2: Analysis of Options:

A group shows a -R effect if it is attached to a conjugated system and has a multiple bond to a more electronegative atom. This allows it to pull \(\pi\)-electron density from the system towards itself. Let's analyze the groups:

(A) -OH, (B) -OCH\(_{3}\), and (C) -NH\(_{2}\): In these groups, the atom directly attached to the conjugated system (O or N) has lone pairs of electrons. These lone pairs can be delocalized into the conjugated system, thus donating electron density. They all exhibit a +R effect.

(E) -NHCOCH\(_{3}\): The nitrogen atom has a lone pair, which it can donate to the aromatic ring (+R effect). Although the C=O group is withdrawing, the overall effect of the group on an aromatic ring is electron-donating via resonance.

(D) -NO\(_{2}\) (nitro group): The structure is -N(=O)(-O). The nitrogen atom is attached to the conjugated system and is also double-bonded to a more electronegative oxygen atom. This arrangement allows the nitro group to strongly withdraw electron density from the conjugated system through resonance. Therefore, the -NO\(_{2}\) group shows a strong -R effect.


Step 3: Final Answer:

The -NO\(_{2}\) group is a powerful electron-withdrawing group that exhibits the -R effect. This corresponds to option (D).
Quick Tip: A simple way to identify -R groups is to look at the atom (Y) in a group -X=Y attached to a conjugated system. If Y is more electronegative than X, the group is typically electron-withdrawing (-R). Examples: -NO\(_{2}\), -CN, -CHO, -COOH. For +R groups, the atom attached to the system usually has a lone pair (e.g., -OH, -NH\(_{2}\), -Cl).

Step 1: Understanding the Concept:

The question describes a specific name reaction in organic chemistry used for the synthesis of alkanes. We need to identify the reaction based on the reactants and products.


Step 2: Identifying the Reaction Type:

The general form of the reaction is the treatment of an alkyl halide (R-X) with sodium metal in the presence of dry ether to form a symmetrical alkane (R-R) with double the number of carbon atoms.
\[ 2R-X + 2Na \xrightarrow{dry ether} R-R + 2NaX \]

Step 3: Detailed Explanation:

The given reaction is:

Reactants: Bromoethane (CH\(_{3}\)CH\(_{2}\)-Br) and metallic Sodium (Na).

Solvent: Dry ethereal solution.

Product: n-butane (CH\(_{3}\)CH\(_{2}\)CH\(_{2}\)CH\(_{3}\)).

The reaction can be written as:
\[ 2CH_3CH_2-Br + 2Na \xrightarrow{dry ether} CH_3CH_2-CH_2CH_3 + 2NaBr \]
This specific reaction, where two molecules of an alkyl halide are coupled by sodium metal in dry ether to form a higher alkane, is known as the Wurtz reaction.

Let's briefly consider the other options:

- Kolbe's reaction: Electrolysis of an aqueous solution of a sodium or potassium salt of a carboxylic acid to produce an alkane.

- Williamson reaction: Synthesis of ethers from an alkoxide and an alkyl halide.

- Fittig reaction: Similar to Wurtz, but uses aryl halides to form biaryls.

- Friedel-Crafts reaction: Alkylation or acylation of an aromatic ring using a Lewis acid catalyst.


Step 4: Final Answer:

The described synthesis of n-butane from bromoethane and sodium is a classic example of the Wurtz reaction. This corresponds to option (B).
Quick Tip: To easily remember these coupling reactions: \textbf{Wurtz} = Alkyl + Alkyl \textbf{Fittig} = Aryl + Aryl \textbf{Wurtz-Fittig} = Alkyl + Aryl All three use sodium metal in dry ether.

Step 1: Understanding the Concept of Aromaticity:

Aromaticity is a special property of certain cyclic compounds that makes them unusually stable. To be aromatic, a compound must satisfy all of Hückel's criteria:

1. It must be cyclic.

2. It must be planar.

3. It must have a continuous ring of p-orbitals (i.e., be fully conjugated).

4. It must contain (4n+2) \(\pi\) electrons, where n is an integer (0, 1, 2, ...).

A compound that fails one or more of these criteria is considered non-aromatic or anti-aromatic.


Step 2: Analysis of Options:

Let's evaluate each compound against the criteria for aromaticity:

(B) Benzene: It is cyclic, planar, fully conjugated (alternating double and single bonds), and has 6 \(\pi\) electrons (which fits 4n+2 for n=1). It is the archetypal aromatic compound.

(C) Thiophene and (D) Pyridine: These are heterocyclic compounds. Both are cyclic, planar, fully conjugated, and have 6 \(\pi\) electrons participating in the aromatic system. They are aromatic.

(E) Naphthalene: It consists of two fused benzene rings. It is cyclic, planar, fully conjugated, and has 10 \(\pi\) electrons (which fits 4n+2 for n=2). It is aromatic.

(A) Cyclohexene: It is a six-membered cyclic compound, but it is not fully conjugated. The molecule contains four sp\(^3\)-hybridized carbon atoms and only one double bond. The presence of these sp\(^3\) carbons breaks the continuous ring of p-orbitals. Since it fails the conjugation criterion, it is non-aromatic.


Step 3: Final Answer:

Cyclohexene does not have a continuous conjugated system of \(\pi\) electrons and is therefore non-aromatic. This corresponds to option (A).
Quick Tip: A quick way to spot a non-aromatic compound is to look for sp\(^3\)-hybridized carbon atoms within the ring. If you find any, the ring cannot be fully conjugated and is therefore not aromatic.

Step 1: Understanding Substituent Effects:

In electrophilic aromatic substitution, substituents on the benzene ring influence both the rate of reaction (activation/deactivation) and the position of the incoming electrophile (orientation).

- Activating groups: Increase the rate of reaction compared to benzene. They are typically ortho, para-directing.

- Deactivating groups: Decrease the rate of reaction. Most are meta-directing, but there is a key exception.

- Ortho, para-directing groups: Direct the incoming electrophile to the positions ortho and para to themselves.

- Meta-directing groups: Direct the incoming electrophile to the meta position.

The question asks for a group that is both deactivating and ortho, para-directing.


Step 2: Analysis of Groups:

Let's analyze the groups based on their electronic effects:

(A) -CH\(_{3}\) (Alkyl group): Donates electron density through the +I (inductive) effect. It is a weak activating group and is ortho, para-directing.

(B) -OH (Hydroxyl group): Strongly donates electron density through the +R (resonance) effect, which outweighs its -I effect. It is a strong activating group and is ortho, para-directing.

(D) -NO\(_{2}\) and (E) -COOH: Both are strongly electron-withdrawing through -R and -I effects. They are strong deactivating groups and are meta-directing.

(C) -Cl (Halogen): Halogens present a unique case. They are strongly electronegative, so they withdraw electron density from the ring through the -I effect, making the ring less reactive towards electrophiles (deactivating). However, they also have lone pairs of electrons that can be donated to the ring via the +R effect. This resonance effect stabilizes the carbocation intermediates formed during ortho and para attack more than meta attack. Thus, halogens are ortho, para-directing. The inductive effect determines the rate (deactivation), while the resonance effect determines the orientation (ortho, para).


Step 3: Final Answer:

Halogens, such as -Cl, are the classic examples of groups that are deactivating yet ortho, para-directing. This corresponds to option (C).
Quick Tip: Memorize this important exception: All halogens (-F, -Cl, -Br, -I) are deactivating but ortho, para-directing in electrophilic aromatic substitution. This is a very frequently tested concept.

Step 1: Understanding Acidity of Hydrocarbons:

Certain hydrocarbons can act as weak acids. Specifically, the hydrogen atom attached to an sp-hybridized carbon atom (as in a terminal alkyne) is acidic enough to be removed by a very strong base, such as sodium metal (Na) or sodamide (NaNH\(_{2}\)). This reaction forms a sodium salt (an acetylide). The question asks which hydrocarbon can form a disodium salt, implying it must have two such acidic hydrogens.


Step 2: Analysis of Options:

Let's analyze the acidity of the C-H bonds in each option:

(A) Ethane (CH\(_{3}\)-CH\(_{3}\)): All hydrogens are attached to sp\(^3\)-hybridized carbons. These are not acidic.

(B) Ethene (CH\(_{2}\)=CH\(_{2}\)) and (C) Benzene (C\(_{6}\)H\(_{6}\)): All hydrogens are attached to sp\(^2\)-hybridized carbons. While more acidic than sp\(^3\) C-H bonds, they are not acidic enough to react with sodium metal.

(D) Propyne (CH\(_{3}\)-C\(\equiv\)CH): This is a terminal alkyne. It has one acidic hydrogen attached to the sp-hybridized carbon at the end of the triple bond. It will react with sodium to form a monosodium salt (monosodium propynide).
\[ 2CH_3C\equivCH + 2Na \rightarrow 2CH_3C\equivC^-Na^+ + H_2 \]
(E) Ethyne (H-C\(\equiv\)C-H), also known as acetylene: This is also a terminal alkyne. Crucially, it has two acidic hydrogens, one on each sp-hybridized carbon. Therefore, it can react with excess sodium metal in two steps to form a disodium salt (disodium acetylide).
\[ H-C\equivC-H + Na \rightarrow H-C\equivC^-Na^+ + \frac{1}{2}H_2 \] \[ H-C\equivC^-Na^+ + Na \rightarrow ^-Na^+C\equivC^-Na^+ + \frac{1}{2}H_2 \]

Step 3: Final Answer:

Ethyne is the only hydrocarbon listed with two acidic hydrogens capable of reacting with excess sodium to form a disodium salt. This corresponds to option (E).
Quick Tip: The acidity of hydrogens on carbon atoms follows the order of hybridization: sp > sp\(^2\) > sp\(^3\). Only the hydrogen on a terminal alkyne (sp-carbon) is acidic enough to be removed by bases like Na or NaNH\(_{2}\).

Step 1: Understanding Factors Affecting Boiling Point:

The boiling point of a substance is determined by the strength of its intermolecular forces. For non-ionic compounds like haloalkanes, the primary intermolecular forces are dipole-dipole interactions and London dispersion forces (a type of van der Waals force).


Step 2: Analysis of Intermolecular Forces:

1. Molecular Mass and Size: London dispersion forces increase with the size of the molecule and the number of electrons (which correlates with molecular mass). Larger molecules have more polarizable electron clouds, leading to stronger temporary dipoles and stronger attraction.

2. Dipole Moment: While halomethanes are polar, the effect of the dipole moment on the boiling point is generally less significant than the effect of London dispersion forces, especially as the halogen atom gets larger.


Let's compare the molecular masses (approximate, in g/mol):

(A) CH\(_{3}\)F: 12 + 3 + 19 = 34

(B) CH\(_{3}\)CH\(_{2}\)F: 24 + 5 + 19 = 48

(C) CH\(_{3}\)Cl: 12 + 3 + 35.5 = 50.5

(E) CH\(_{3}\)Br: 12 + 3 + 80 = 95

(D) CH\(_{3}\)I: 12 + 3 + 127 = 142


Among the methyl halides (CH\(_{3}\)X), the molecular mass increases significantly from F to I. This leads to a dramatic increase in the strength of London dispersion forces. CH\(_{3}\)I has the largest molecular mass and the largest, most polarizable electron cloud. Therefore, it experiences the strongest London dispersion forces among all the given options. Even though CH\(_{3}\)CH\(_{2}\)F has more carbons, its molecular mass is much lower than that of CH\(_{3}\)Br and CH\(_{3}\)I. The boiling point trend for methyl halides is: CH\(_{3}\)I > CH\(_{3}\)Br > CH\(_{3}\)Cl > CH\(_{3}\)F.


Note: The provided answer key states (E) CH\(_{3}\)Br is the correct answer. This is factually incorrect based on established chemical principles and experimental data (Boiling point of CH\(_{3}\)I is 42.4 \(^\circ\)C, while for CH\(_{3}\)Br it is 3.6 \(^\circ\)C). The correct answer should be CH\(_{3}\)I due to its significantly higher molecular mass leading to stronger intermolecular dispersion forces. The solution follows the chemically correct reasoning.


Step 3: Final Answer:

CH\(_{3}\)I has the highest molecular mass and size, leading to the strongest London dispersion forces and thus the highest boiling point. This corresponds to option (D).
Quick Tip: For haloalkanes with the same alkyl group, the boiling point increases as you go down the halogen group (F < Cl < Br < I). This is because the increasing size and polarizability of the halogen atom lead to much stronger London dispersion forces, which override the effect of the decreasing dipole moment.

Step 1: Understanding the Reaction:

The reaction involves an aromatic ring (chlorobenzene) reacting with an acyl chloride (acetyl chloride) in the presence of a Lewis acid catalyst (anhydrous AlCl\(_{3}\)). This is a characteristic reaction of aromatic compounds where a hydrogen atom on the ring is replaced by another group.


Step 2: Identifying the Reaction Mechanism:

The reaction described is a Friedel-Crafts Acylation. The general mechanism involves three steps:

1. Generation of an electrophile (an acylium ion) by the reaction of the acyl chloride with the Lewis acid catalyst.
\[ CH_3COCl + AlCl_3 \rightarrow [CH_3C=O]^+ + [AlCl_4]^- \]
2. Attack of the \(\pi\)-electron system of the aromatic ring on the electrophile to form a resonance-stabilized carbocation (sigma complex).

3. Loss of a proton (H\(^+\)) from the sigma complex to restore the aromaticity of the ring.


Step 3: Classifying the Reaction:

In this reaction, the aromatic hydrogen of chlorobenzene is substituted by an acetyl group (\(-COCH_{3}\)).

- The species attacking the electron-rich benzene ring is an electrophile (the acylium ion, CH\(_{3}\)CO\(^+\)).

- A hydrogen atom on the ring is replaced, or substituted, by this electrophile.

Therefore, the reaction is an electrophilic substitution reaction.

The other options are incorrect:

- Nucleophilic substitution involves a nucleophile attacking an electron-deficient center. Aromatic rings are electron-rich and do not undergo nucleophilic substitution easily.

- Free radical substitution is characteristic of alkanes reacting with halogens in the presence of UV light.

- Addition reactions (nucleophilic or electrophilic) would result in the loss of aromaticity, which does not happen here. Addition reactions are characteristic of alkenes and alkynes.


Step 4: Final Answer:

The Friedel-Crafts acylation of chlorobenzene is a classic example of an electrophilic aromatic substitution reaction. This corresponds to option (B).
Quick Tip: Recognize the key reagents for electrophilic aromatic substitution: \textbf{Halogenation:} X\(_{2}\)/FeX\(_{3}\) \textbf{Nitration:} Conc. HNO\(_{3}\)/Conc. H\(_{2}\)SO\(_{4}\) \textbf{Sulfonation:} Fuming H\(_{2}\)SO\(_{4}\) (SO\(_{3}\)) \textbf{Friedel-Crafts Alkylation:} R-X/AlCl\(_{3}\) \textbf{Friedel-Crafts Acylation:} R-CO-X/AlCl\(_{3}\)

Step 1: Understanding Optical Activity:

A compound is optically active if it is chiral. Chirality is the property of a molecule that is non-superimposable on its mirror image. The most common cause of chirality in organic molecules is the presence of a chiral center (or asymmetric carbon atom). A chiral center is a carbon atom that is bonded to four different atoms or groups.


Step 2: Analysis of Structures:

Let's draw the structure of each compound and look for a chiral center.

(A) 1-Chlorobutane: CH\(_{2}\)(Cl)-CH\(_{2}\)-CH\(_{2}\)-CH\(_{3}\). The C1 carbon is bonded to two H atoms, so it's not chiral. No other carbon is bonded to four different groups. It is achiral.

(B) neo-Pentyl chloride: (CH\(_{3}\))\(_{3}\)C-CH\(_{2}\)Cl. The C1 carbon has two H atoms. The C2 carbon has three identical methyl groups. It is achiral.

(C) Isobutyl chloride: (CH\(_{3}\))\(_{2}\)CH-CH\(_{2}\)Cl. The C1 carbon has two H atoms. The C2 carbon has two identical methyl groups. It is achiral.

(D) tert-Butyl chloride: (CH\(_{3}\))\(_{3}\)C-Cl. The central carbon is bonded to three identical methyl groups and a chlorine atom. It is achiral.

(E) 2-Chlorobutane: CH\(_{3}\)-CH(Cl)-CH\(_{2}\)-CH\(_{3}\). Let's examine the C2 carbon. It is bonded to:

1. A hydrogen atom (-H)

2. A chlorine atom (-Cl)

3. A methyl group (-CH\(_{3}\))

4. An ethyl group (-CH\(_{2}\)CH\(_{3}\))

Since the C2 carbon is attached to four different groups, it is a chiral center. The molecule is chiral and therefore optically active.


Step 3: Final Answer:

2-Chlorobutane is the only compound among the options that has a chiral center, making it optically active. This corresponds to option (E).
Quick Tip: To quickly find a chiral center, scan the molecule for any carbon atom with four single bonds. Then, check if the four groups attached to it are all different from each other. Groups like -CH\(_{3}\), -CH\(_{2}\)-, and -CH- can all be part of a larger group.

Step 1: Understanding the Nomenclature:

The question asks for the systematic (IUPAC) or common name corresponding to "Phenetole". This is a test of knowledge of common names for specific organic compounds, particularly ethers.


Step 2: Identifying the Structure:

Phenetole is the common name for an ether consisting of an ethyl group (-CH\(_{2}\)CH\(_{3}\)) and a phenyl group (-C\(_{6}\)H\(_{5}\)) joined by an oxygen atom.

The structure is C\(_{6}\)H\(_{5}\)-O-CH\(_{2}\)CH\(_{3}\).

In the IUPAC system, ethers are named as alkoxyalkanes (or alkoxyarenes). The larger group is considered the parent chain/ring, and the smaller group with the oxygen is named as an alkoxy substituent.

In this case, the benzene ring is the larger group (parent). The -O-CH\(_{2}\)CH\(_{3}\) group is named "ethoxy".

Therefore, the IUPAC name for Phenetole is Ethoxybenzene.

Let's check the other options:

- (C) Methoxybenzene (C\(_{6}\)H\(_{5}\)-O-CH\(_{3}\)) is commonly known as Anisole.

- The other options are aliphatic ethers.


Step 3: Final Answer:

Phenetole is the common name for ethoxybenzene. This corresponds to option (A).
Quick Tip: It is highly beneficial to memorize the common names of simple aromatic compounds as they are frequently used in exams. Key examples include: \textbf{Toluene:} Methylbenzene \textbf{Aniline:} Aminobenzene \textbf{Phenol:} Hydroxybenzene \textbf{Anisole:} Methoxybenzene \textbf{Phenetole:} Ethoxybenzene

Step 1: Understanding the Concept:

The question asks for the synthesis of a tertiary alcohol (2-methylpropan-2-ol) from a ketone (acetone). This transformation involves adding an alkyl group to the carbonyl carbon and converting the carbonyl group into a hydroxyl group. This is characteristic of a Grignard reaction.


Step 2: Detailed Explanation:

Let's analyze the reaction and the reagents.

Starting Material: Acetone (propan-2-one), a ketone with the structure CH\(_3\)-CO-CH\(_3\).

Product: 2-methylpropan-2-ol, a tertiary alcohol with the structure (CH\(_3\))\(_3\)COH.

The transformation involves changing a C=O bond to a C-OH bond and adding one methyl (-CH\(_3\)) group.


Let's evaluate the given reagents:


(A) Pd / H\(_2\): This is a catalytic hydrogenation reagent. It reduces ketones to secondary alcohols. Acetone would be reduced to propan-2-ol, not 2-methylpropan-2-ol.

(B) B\(_2\)H\(_6\) / H\(_2\)O\(_2\), NaOH: This is the hydroboration-oxidation reagent. It reduces ketones to secondary alcohols, similar to NaBH\(_4\) or LiAlH\(_4\). It would also produce propan-2-ol.

(C) CH\(_3\)MgI / H\(_2\)O: This is a Grignard reagent (methylmagnesium iodide) followed by an acidic workup (hydrolysis). The Grignard reagent acts as a nucleophile (CH\(_3^-\)).

Step 1: The nucleophilic methyl group from CH\(_3\)MgI attacks the electrophilic carbonyl carbon of acetone. The \(\pi\)-bond of the C=O group breaks, and the electrons move to the oxygen atom, forming an alkoxide intermediate.

\[ CH_3COCH_3 + CH_3MgI \rightarrow (CH_3)_3COMgI \]
Step 2: Hydrolysis (addition of H\(_2\)O or H\(_3\)O\(^+\)) protonates the alkoxide to yield the final tertiary alcohol product, 2-methylpropan-2-ol.

\[ (CH_3)_3COMgI + H_2O \rightarrow (CH_3)_3COH + Mg(OH)I \]

This sequence correctly produces the desired product.

(D) LiAlH\(_4\) and (E) NaBH\(_4\): These are strong and mild reducing agents, respectively. They reduce ketones by adding a hydride ion (H\(^-\)), converting them to secondary alcohols. They would also convert acetone to propan-2-ol.



Step 3: Final Answer:

The correct reagent to convert acetone to 2-methylpropan-2-ol is the Grignard reagent CH\(_3\)MgI followed by hydrolysis. This corresponds to option (C).
Quick Tip: To synthesize alcohols using Grignard reagents: Formaldehyde + Grignard \(\rightarrow\) Primary alcohol Other aldehydes + Grignard \(\rightarrow\) Secondary alcohol Ketones + Grignard \(\rightarrow\) Tertiary alcohol

Step 1: Understanding the Concept:

The acidity of a compound is determined by its ability to donate a proton (H\(^+\)). The strength of an acid is related to the stability of its conjugate base formed after donating the proton. A more stable conjugate base corresponds to a stronger acid.


Step 2: Detailed Explanation:

Let's compare the acidity of the given compounds. Options (A), (B), (C), and (E) are phenols or substituted phenols. Option (D) is an aliphatic alcohol.


Phenols vs. Alcohols: Phenols are generally more acidic than aliphatic alcohols. When a phenol loses a proton, it forms a phenoxide ion. This phenoxide ion is stabilized by resonance, as the negative charge is delocalized over the benzene ring. When an alcohol like ethanol loses a proton, it forms an ethoxide ion (CH\(_3\)CH\(_2\)O\(^-\)). The negative charge is localized on the oxygen atom. Furthermore, the ethyl group (-C\(_2\)H\(_5\)) is an electron-donating group (+I effect), which destabilizes the ethoxide ion by intensifying the negative charge. Because the phenoxide ion is much more stable than the ethoxide ion, phenol is a much stronger acid than ethanol. Therefore, ethanol is the weakest acid among all choices.

Comparing the Phenols:

Phenol (C\(_6\)H\(_5\)OH): This is our reference.
p-Nitrophenol: The nitro group (-NO\(_2\)) is a strong electron-withdrawing group (-R and -I effects). It delocalizes the negative charge of the phenoxide ion very effectively, greatly stabilizing it. Thus, p-nitrophenol is a much stronger acid than phenol.
p-Cresol and m-Cresol: The methyl group (-CH\(_3\)) is an electron-donating group (+I and hyperconjugation effects). It destabilizes the phenoxide ion by pushing electron density onto the ring, making it less stable than the simple phenoxide ion. Therefore, cresols are weaker acids than phenol.


The order of acidity is: p-Nitrophenol \(>\) Phenol \(>\) Cresols \(>\) Ethanol.


Step 3: Final Answer:

Ethanol is an aliphatic alcohol whose conjugate base is not resonance-stabilized, making it significantly less acidic than phenols. It is the weakest acid in the list. This corresponds to option (D).
Quick Tip: Acidity of phenols increases with electron-withdrawing groups (EWG) like -NO\(_2\), -CN, -X and decreases with electron-donating groups (EDG) like -CH\(_3\), -OCH\(_3\), -NH\(_2\). Aliphatic alcohols are generally weaker acids than water and phenols.

Step 1: Understanding the Concept:

This is a direct question about the composition of a specific named reagent used in organic chemistry. The Lucas reagent is used to distinguish between primary, secondary, and tertiary alcohols.


Step 2: Detailed Explanation:

The Lucas test involves the reaction of an alcohol with the Lucas reagent. The reaction is an S\(_N\)1 reaction where the hydroxyl group of the alcohol is replaced by a chloride ion. The anhydrous ZnCl\(_2\) acts as a Lewis acid, which coordinates with the oxygen atom of the hydroxyl group, making it a better leaving group (H\(_2\)O). This facilitates the formation of a carbocation intermediate.

The composition of the Lucas reagent is a solution of anhydrous zinc chloride (ZnCl\(_2\)) in concentrated hydrochloric acid (HCl).

The rate of reaction depends on the stability of the carbocation formed:

Tertiary alcohols react almost instantaneously because they form a stable tertiary carbocation, resulting in immediate cloudiness or turbidity due to the formation of the insoluble alkyl chloride.
Secondary alcohols react more slowly, typically within 5-10 minutes, as they form a less stable secondary carbocation.
Primary alcohols do not react at room temperature because the primary carbocation is highly unstable.


Step 3: Final Answer:

The Lucas reagent is a mixture of concentrated HCl and anhydrous ZnCl\(_2\). This corresponds to option (B).
Quick Tip: Remember the Lucas test results: Tertiary gives Turbidity Instantly (TTI), Secondary gives Turbidity Slowly (TTS), and Primary gives No Turbidity (PNT) at room temperature. The reagent is "concentrated HCl + anhydrous ZnCl\(_2\)".

Step 1: Understanding the Concept:

Nylon-6,6 is a type of polyamide, a condensation polymer. The name "6,6" indicates that it is formed from two monomers, each containing six carbon atoms. We need to identify the dicarboxylic acid monomer.


Step 2: Detailed Explanation:

The two monomers used in the synthesis of Nylon-6,6 are:

Hexamethylenediamine: A diamine with six carbon atoms. Its structure is H\(_2\)N-(CH\(_2\))\(_6\)-NH\(_2\).
Adipic acid: A dicarboxylic acid with six carbon atoms.

We need to find the IUPAC name for adipic acid. Let's name the given dicarboxylic acids:

(A) Ethanedioic acid: HOOC-COOH (2 carbons). Common name: Oxalic acid.
(B) Propanedioic acid: HOOC-CH\(_2\)-COOH (3 carbons). Common name: Malonic acid.
(C) Butanedioic acid: HOOC-(CH\(_2\))\(_2\)-COOH (4 carbons). Common name: Succinic acid.
(D) Pentanedioic acid: HOOC-(CH\(_2\))\(_3\)-COOH (5 carbons). Common name: Glutaric acid.
(E) Hexanedioic acid: HOOC-(CH\(_2\))\(_4\)-COOH (6 carbons). Common name: Adipic acid.

The polymerization reaction involves the formation of an amide bond between the amine group of one monomer and the carboxylic acid group of the other monomer, with the elimination of a water molecule. \[ n H_2N(CH_2)_6NH_2 + n HOOC(CH_2)_4COOH \rightarrow \left[ -NH(CH_2)_6NHCO(CH_2)_4CO- \right]_n + 2n H_2O \]
Thus, the six-carbon dicarboxylic acid required is hexanedioic acid (adipic acid).


Step 3: Final Answer:

The carboxylic acid used in the manufacture of nylon-6,6 is hexanedioic acid. This corresponds to option (E).
Quick Tip: For polymers like Nylon-X,Y, 'X' refers to the number of carbons in the diamine monomer and 'Y' refers to the number of carbons in the dicarboxylic acid monomer. For Nylon-6,6, both monomers have 6 carbons.

Step 1: Understanding the Concept:

This is a question about a specific named reaction in organic chemistry. The Etard reaction is a method for the direct oxidation of an aromatic or heterocyclic bound methyl group to an aldehyde.


Step 2: Detailed Explanation:

The Etard reaction specifically refers to the oxidation of toluene (or its derivatives) to benzaldehyde using chromyl chloride (CrO\(_2\)Cl\(_2\)) as the oxidizing agent. The reaction is typically carried out in an inert, non-polar solvent such as carbon disulfide (CS\(_2\)) or carbon tetrachloride (CCl\(_4\)). This is followed by a hydrolysis step to yield the aldehyde. The reaction proceeds via the formation of a brown chromium complex, which prevents further oxidation of the aldehyde to a carboxylic acid.


Let's look at the other options:


(A) Cl\(_2\) / hv, H\(_2\)O\(^+\): This is free radical chlorination of the side chain of toluene, followed by hydrolysis. It can produce benzaldehyde, but it's not the Etard reaction.
(C) CO, HCl, anhydrous AlCl\(_3\) / CuCl: These are the reagents for the Gattermann-Koch reaction, used to formylate benzene.
(D) SnCl\(_2\), HCl: This reagent pair is used in the Stephen reduction, which converts nitriles to aldehydes.
(E) DIBAL-H: Diisobutylaluminium hydride is a reducing agent used, for example, to reduce esters or nitriles to aldehydes.


Step 3: Final Answer:

The reagents employed in the Etard reaction are chromyl chloride (CrO\(_2\)Cl\(_2\)) in an inert solvent like CS\(_2\), followed by hydrolysis. This corresponds to option (B).
Quick Tip: Associate key named reactions with their specific reagents. For aldehyde synthesis: \textbf{Etard Reaction:} Toluene \(\rightarrow\) Benzaldehyde (Reagent: CrO\(_2\)Cl\(_2\)) \textbf{Stephen Reduction:} Nitrile \(\rightarrow\) Aldehyde (Reagent: SnCl\(_2\)/HCl) \textbf{Rosenmund Reduction:} Acid Chloride \(\rightarrow\) Aldehyde (Reagent: H\(_2\)/Pd-BaSO\(_4\)) \textbf{Gattermann-Koch:} Benzene \(\rightarrow\) Benzaldehyde (Reagent: CO, HCl/AlCl\(_3\))

Step 1: Understanding the Concept:

This is a multi-step synthesis problem involving common reactions of aromatic compounds, specifically the reduction of a nitro group, diazotization of an amine, and a Sandmeyer-type reaction.


Step 2: Detailed Explanation:

Let's trace the reaction sequence step by step.

Step 1: Nitrobenzene to Compound (X)

Nitrobenzene (C\(_6\)H\(_5\)NO\(_2\)) is treated with Sn / HCl. This is a standard method for the reduction of a nitro group to a primary amino group. \[ C_6H_5NO_2 \xrightarrow{Sn / HCl} C_6H_5NH_2 \]
So, compound (X) is Aniline.


Step 2: Compound (X) to Compound (Y)

Aniline (X) is treated with NaNO\(_2\) and HCl at a low temperature (273-278 K or 0-5 \(^\circ\)C). This reaction is called diazotization. It converts the primary aromatic amine into a diazonium salt. \[ C_6H_5NH_2 \xrightarrow{NaNO_2 / HCl, 278 K} C_6H_5N_2^+Cl^- \]
So, compound (Y) is Benzenediazonium chloride.


Step 3: Compound (Y) to Compound (Z)

Benzenediazonium chloride (Y) is treated with Cu / HBr. This is a variation of the Sandmeyer reaction (sometimes called the Gattermann reaction when copper powder is used instead of a cuprous salt). This reaction replaces the diazonium group (-N\(_2^+\)Cl\(^-\)) with a bromine atom from HBr. \[ C_6H_5N_2^+Cl^- \xrightarrow{Cu / HBr} C_6H_5Br + N_2 + CuCl \]
So, the final compound (Z) is Bromobenzene.


Step 3: Final Answer:

The final compound 'Z' is Bromobenzene. This corresponds to option (D).
Quick Tip: The sequence Nitrobenzene \(\rightarrow\) Aniline \(\rightarrow\) Benzenediazonium salt is a very common and important pathway in aromatic chemistry. The diazonium salt is a versatile intermediate that can be converted into many other functional groups (e.g., -Cl, -Br, -CN, -OH, -H).

Step 1: Understanding the Concept:

The question asks about the synthesis of a carbylamine (also known as an isocyanide or isonitrile). This synthesis is a specific named reaction known as the Carbylamine reaction or Hofmann's isocyanide test.


Step 2: Detailed Explanation:

The Carbylamine reaction is a chemical test for the detection of primary amines. In this reaction, a primary amine (aliphatic or aromatic) is heated with chloroform (CHCl\(_3\)) and a strong base, typically ethanolic or alcoholic potassium hydroxide (KOH). This reaction produces an isocyanide, which is characterized by an extremely unpleasant or foul smell.

The reaction for aniline (a primary aromatic amine) is: \[ C_6H_5NH_2 + CHCl_3 + 3KOH (alcoholic) \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O \]
The product, C\(_6\)H\(_5\)NC, is phenyl isocyanide, which is a carbylamine. The reaction proceeds via the formation of a dichlorocarbene (:CCl\(_2\)) intermediate from the reaction of CHCl\(_3\) and KOH.

Secondary and tertiary amines do not give this reaction, making it a reliable test for the primary amino group.


Step 3: Final Answer:

Carbylamine (phenyl isocyanide) is obtained when aniline is heated with chloroform and ethanolic KOH. This corresponds to option (C).
Quick Tip: Remember the Carbylamine test as the "foul smell test" for primary amines. The key reagents are always Chloroform (CHCl\(_3\)) and a strong alcoholic base (like KOH).

Step 1: Understanding the Concept:

The boiling point of a substance depends on the strength of the intermolecular forces between its molecules. For amines of comparable molecular mass, the main intermolecular forces to consider are hydrogen bonding and van der Waals forces. The extent of hydrogen bonding is the dominant factor.


Step 2: Detailed Explanation:

Let's classify the given amines:

(i) n-C\(_4\)H\(_9\)NH\(_2\) (Butan-1-amine): This is a primary (1\(^\circ\)) amine. It has two hydrogen atoms attached to the nitrogen atom (-NH\(_2\)). This allows it to form extensive intermolecular hydrogen bonds, acting as both a hydrogen bond donor and acceptor.
(ii) (C\(_2\)H\(_5\))\(_2\)NH (Diethylamine): This is a secondary (2\(^\circ\)) amine. It has one hydrogen atom attached to the nitrogen atom (-NH-). It can still form hydrogen bonds, but to a lesser extent than a primary amine because it has fewer N-H bonds.
(iii) C\(_2\)H\(_5\)N(CH\(_3\))\(_2\) (N,N-Dimethylethanamine): This is a tertiary (3\(^\circ\)) amine. It has no hydrogen atoms attached to the nitrogen atom. Therefore, it cannot form hydrogen bonds with other tertiary amine molecules (it cannot act as a hydrogen bond donor). It can only act as a hydrogen bond acceptor with other molecules like water. The main intermolecular forces are dipole-dipole interactions and van der Waals forces.

Since all three amines are isomers (C\(_4\)H\(_{11}\)N, molecular mass = 73 g/mol), their van der Waals forces are comparable. The deciding factor is hydrogen bonding.

The strength of hydrogen bonding follows the order: Primary amine \(>\) Secondary amine \(>\) Tertiary amine.

Consequently, the energy required to overcome these forces (and thus the boiling point) will follow the same order.

Boiling Point Order: n-C\(_4\)H\(_9\)NH\(_2\) \(>\) (C\(_2\)H\(_5\))\(_2\)NH \(>\) C\(_2\)H\(_5\)N(CH\(_3\))\(_2\).

This corresponds to the order (i) \(>\) (ii) \(>\) (iii).


Step 3: Final Answer:

The correct order of boiling points is (i) \(>\) (ii) \(>\) (iii). This corresponds to option (A).
Quick Tip: For isomeric amines, the boiling point order is always: Primary \(>\) Secondary \(>\) Tertiary. This is due to the decreasing ability to form intermolecular hydrogen bonds as the number of hydrogen atoms on the nitrogen decreases.

Step 1: Understanding the Concept:

This is a biology/biochemistry question that requires knowledge of vitamins and the diseases or symptoms caused by their deficiency. Convulsions are sudden, violent, irregular movements of the body, caused by involuntary contraction of muscles.


Step 2: Detailed Explanation:

Let's review the deficiency symptoms for each vitamin listed:

(A) Riboflavin (Vitamin B\(_2\)): Deficiency can cause ariboflavinosis, with symptoms like cracks in the lips (cheilosis), inflammation of the tongue (glossitis), and skin disorders. It is not typically associated with convulsions.
(B) Thiamine (Vitamin B\(_1\)): Deficiency causes Beriberi, a disease affecting the nervous system (peripheral neuropathy) and cardiovascular system.
(C) Ascorbic acid (Vitamin C): Deficiency causes Scurvy, characterized by bleeding gums, weakness, and poor wound healing.
(D) Pyridoxine (Vitamin B\(_6\)): Pyridoxine is crucial for the synthesis of neurotransmitters, such as GABA (gamma-aminobutyric acid), an inhibitory neurotransmitter. A deficiency of Vitamin B\(_6\) can lead to insufficient GABA production, resulting in neurological symptoms, including irritability, depression, and convulsions, especially in infants.
(E) Vitamin D: Deficiency causes Rickets in children (softening and weakening of bones) and osteomalacia in adults.


Step 3: Final Answer:

The deficiency of Pyridoxine (Vitamin B\(_6\)) is known to cause convulsions. This corresponds to option (D).
Quick Tip: Create a table or flashcards to memorize the vitamins, their chemical names, sources, and key deficiency diseases. For Vitamin B6 (Pyridoxine), link it to neurotransmitter synthesis and neurological symptoms like convulsions.

Step 1: Understanding the Concept:

Optical activity in a molecule is the ability to rotate the plane of plane-polarized light. This property arises from chirality. A molecule is chiral if it is non-superimposable on its mirror image. For amino acids, this typically means the \(\alpha\)-carbon (the carbon atom next to the carboxyl group) must be a chiral center, i.e., bonded to four different groups.


Step 2: Detailed Explanation:

The general structure of an \(\alpha\)-amino acid is: \[ \begin{array}{c} COOH
|
H_2N - C - H
|
R \end{array} \]
The four groups attached to the \(\alpha\)-carbon are:

A hydrogen atom (-H)
An amino group (-NH\(_2\))
A carboxyl group (-COOH)
A side chain (-R)

For the \(\alpha\)-carbon to be chiral, the side chain (-R) must be different from the other three groups (-H, -NH\(_2\), -COOH).

Let's examine the side chains of the given amino acids:

(A) Glycine: The side chain (R) is a hydrogen atom (-H). This means the \(\alpha\)-carbon is bonded to two hydrogen atoms, as well as an amino group and a carboxyl group. Since it is not bonded to four different groups, the \(\alpha\)-carbon is achiral. Therefore, glycine is optically inactive.
(B) Alanine: The side chain (R) is a methyl group (-CH\(_3\)). The four different groups are -H, -NH\(_2\), -COOH, and -CH\(_3\). Alanine is chiral and optically active.
(C) Valine: The side chain (R) is an isopropyl group (-CH(CH\(_3\))\(_2\)). The four different groups are -H, -NH\(_2\), -COOH, and -CH(CH\(_3\))\(_2\). Valine is chiral and optically active.
(D) Leucine: The side chain (R) is an isobutyl group (-CH\(_2\)CH(CH\(_3\))\(_2\)). The four different groups are -H, -NH\(_2\), -COOH, and -CH\(_2\)CH(CH\(_3\))\(_2\). Leucine is chiral and optically active.


Step 3: Final Answer:

Glycine is the only standard proteinogenic amino acid that is achiral and therefore optically inactive. This corresponds to option (A).
Quick Tip: Remember that Glycine is the simplest amino acid and the unique exception to chirality among the 20 common amino acids because its side chain is just a hydrogen atom.

Step 1: Understanding the Concept:

The 'barn' is a unit of area used in nuclear and particle physics to quantify the cross-section of a scattering process. The question asks for its value in the SI base unit for area, which is the square meter (m\(^2\)).


Step 2: Detailed Explanation:

The size of an atomic nucleus is on the order of femtometers (1 fm = 10\(^{-15}\) m). The cross-sectional area of a typical nucleus is therefore on the order of (10\(^{-14}\) m)\(^2\) = 10\(^{-28}\) m\(^2\). This value was defined as one barn.

The etymology of the name comes from the phrase "as big as a barn," which was wartime slang used by physicists at Purdue University to describe the large probability of interaction for certain nuclear reactions.

The exact definition is: \[ 1 barn = 10^{-28} m^2 \]
This can also be expressed in other units: \[ 1 barn = 100 fm^2 = 10^{-24} cm^2 \]

Step 3: Final Answer:

The value of one barn in SI units is 10\(^{-28}\) m\(^2\). This corresponds to option (A).
Quick Tip: Remember that the barn is a unit of area, roughly the size of a Uranium nucleus's cross-section. The key conversion to memorize for SI units is 1 barn = 10\(^{-28}\) m\(^2\).

Step 1: Understanding the Concept:

This question asks for the definition of a fundamental kinematic quantity. We need to analyze the given expression and relate it to the standard definitions of motion.


Step 2: Detailed Explanation:

Let's break down the expression \( \frac{X_2 - X_1}{\Delta t} \):

Numerator (X\(_2\) - X\(_1\)): This term represents the change in position of the body. The change in position is also known as displacement (\(\Delta\)X).
Denominator (\(\Delta\)t): This term represents the time interval over which the change in position occurred.

The expression is therefore the total displacement divided by the total time interval.

By definition:

Average velocity (\(\bar{v}\)) is defined as the total displacement divided by the total time interval.
\[ \bar{v} = \frac{Total Displacement}{Total Time Interval} = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{\Delta t} \]
Average acceleration (\(\bar{a}\)) is the change in velocity (\(\Delta\)v) divided by the time interval (\(\Delta\)t).
Instantaneous velocity (v) is the limit of the average velocity as the time interval approaches zero, i.e., the derivative of position with respect to time (\(v = dX/dt\)).
Instantaneous acceleration (a) is the limit of the average acceleration as the time interval approaches zero (\(a = dv/dt\)).
Average displacement is not a standard term; displacement itself is the change in position.

The given expression exactly matches the definition of average velocity.


Step 3: Final Answer:

The expression \( \frac{X_2 - X_1}{\Delta t} \) is defined as average velocity. This corresponds to option (B).
Quick Tip: Remember the key kinematic definitions: Velocity is the rate of change of position. Acceleration is the rate of change of velocity. The term "average" refers to a quantity over a finite interval, while "instantaneous" refers to a quantity at a specific moment in time (involving a limit or derivative).

Step 1: Understanding the Concept:

This is a kinematics problem involving two phases of motion: constant acceleration followed by constant retardation (deceleration). We need to calculate the distance covered in each phase and add them to find the total distance.


Step 2: Key Formula or Approach:

We will use the equations of motion for constant acceleration:

\(v = u + at\)
\(s = ut + \frac{1}{2}at^2\)
\(v^2 = u^2 + 2as\)

where u = initial velocity, v = final velocity, a = acceleration, t = time, and s = distance.


Step 3: Detailed Explanation:

First, convert the time from minutes to seconds:
t = 1 minute = 60 s.


Phase 1: Acceleration

Initial velocity, \(u_1 = 0\) (starts from rest)
Acceleration, \(a_1 = 2\) m/s\(^2\)
Time, \(t_1 = 60\) s

Let's find the distance travelled in this phase (\(s_1\)) and the final velocity (\(v_1\)).
Using the second equation of motion for distance: \[ s_1 = u_1t_1 + \frac{1}{2}a_1t_1^2 \] \[ s_1 = (0)(60) + \frac{1}{2}(2)(60)^2 \] \[ s_1 = (1)(3600) = 3600 m \]
Now, let's find the velocity at the end of this phase, which will be the initial velocity for the next phase.
Using the first equation of motion: \[ v_1 = u_1 + a_1t_1 \] \[ v_1 = 0 + (2)(60) = 120 m/s \]

Phase 2: Retardation

Initial velocity, \(u_2 = v_1 = 120\) m/s
Retardation, so acceleration is negative, \(a_2 = -2\) m/s\(^2\)
Time, \(t_2 = 60\) s
Final velocity, \(v_2 = 0\) (comes to rest). We can use this to check our calculation. \(v = u + at = 120 + (-2)(60) = 120 - 120 = 0\). This confirms the time is correct.

Let's find the distance travelled in this phase (\(s_2\)).
Using the second equation of motion: \[ s_2 = u_2t_2 + \frac{1}{2}a_2t_2^2 \] \[ s_2 = (120)(60) + \frac{1}{2}(-2)(60)^2 \] \[ s_2 = 7200 - (1)(3600) = 3600 m \]

Total Distance
The total distance travelled is the sum of the distances from both phases. \[ S_{total} = s_1 + s_2 = 3600 m + 3600 m = 7200 m \]

Step 4: Final Answer:

The total distance travelled by the car is 7200 m. This corresponds to option (E).
Quick Tip: For a symmetrical motion problem like this (same magnitude of acceleration and retardation over the same time), the distance covered in both phases will be equal. You can calculate the distance for the first phase and simply double it.

Step 1: Understanding the Concept:

This problem applies Newton's Second Law of Motion, which relates force (F), mass (m), and acceleration (a).


Step 2: Key Formula or Approach:

Newton's Second Law is given by the formula: \[ F = ma \]
We are given information about two bodies, let's denote them by subscripts 1 and 2.

\(m_1 = 2\) kg
\(m_2 = 3\) kg

We are told that the forces acting on them are the same: \[ F_1 = F_2 \]

Step 3: Detailed Explanation:

Using Newton's Second Law for each body: \[ F_1 = m_1 a_1 \] \[ F_2 = m_2 a_2 \]
Since \(F_1 = F_2\), we can set the expressions equal to each other: \[ m_1 a_1 = m_2 a_2 \]
We need to find the ratio of their respective accelerations, which is \(a_1 : a_2\) or \( \frac{a_1}{a_2} \).
To get this ratio, we rearrange the equation: \[ \frac{a_1}{a_2} = \frac{m_2}{m_1} \]
Now, substitute the given masses: \[ \frac{a_1}{a_2} = \frac{3 kg}{2 kg} = \frac{3}{2} \]
So, the ratio of their accelerations is 3 : 2.


Step 4: Final Answer:

The ratio of the respective accelerations (\(a_1 : a_2\)) is 3 : 2. This corresponds to option (D).
Quick Tip: From \(F=ma\), if the force (F) is constant, acceleration (a) is inversely proportional to mass (m). This means the heavier object will have less acceleration, and the lighter object will have more. The ratio of accelerations will be the inverse of the ratio of their masses.

Step 1: Understanding the Concept:

This question asks for the physical quantity represented by the area under a force-time (F-t) graph. This is a fundamental concept in mechanics related to Newton's second law.


Step 2: Key Formula or Approach:

From Newton's second law, force is the rate of change of momentum: \[ F = \frac{dp}{dt} \]
where \(p\) is the momentum.

To find the total change in momentum over a time interval from \(t_1\) to \(t_2\), we can rearrange and integrate this equation: \[ dp = F \, dt \] \[ \int_{p_1}^{p_2} dp = \int_{t_1}^{t_2} F \, dt \] \[ p_2 - p_1 = \Delta p = \int_{t_1}^{t_2} F \, dt \]
The quantity \(\Delta p\) (change in momentum) is defined as the impulse (J). The integral on the right-hand side represents the area under the F-t curve.


Step 3: Detailed Explanation:

The area under a curve of a function y(x) plotted against x is given by the definite integral \(\int y \, dx\).

In this case, the curve is force (F) plotted against time (t). So, the area under the curve is given by the integral \(\int F \, dt\).

As shown in Step 2, this integral is defined as the impulse imparted to the object. Therefore, the area under the force-time graph represents impulse.

Let's analyze the other options:

Work done is the area under a force-displacement (F-x) graph, \(W = \int F \, dx\).
Torque is a rotational force (\(\tau = r \times F\)).
Power is the rate of doing work (\(P = dW/dt = F \cdot v\)).
Kinetic energy is the energy of motion (\(K = \frac{1}{2}mv^2\)).


Step 4: Final Answer:

The area under the force-time graph is the impulse. This corresponds to option (B).
Quick Tip: Remember the two important graphical areas in mechanics: Area under Force-Time graph = Impulse = Change in Momentum. Area under Force-Displacement graph = Work Done = Change in Kinetic Energy.

Step 1: Understanding the Concept:

This problem involves the principle of conservation of energy, taking into account the work done by a non-conservative force (air resistance). The initial potential energy of the raindrop is converted into kinetic energy and heat due to air resistance.


Step 2: Key Formula or Approach:

The work-energy theorem states that the change in mechanical energy is equal to the work done by non-conservative forces. \[ E_{initial} = E_{final} + E_{lost} \]
Here, the initial energy is purely gravitational potential energy (PE), the final energy is purely kinetic energy (KE), and the energy lost is due to air resistance. \[ PE_{initial} = KE_{final} + E_{lost} \] \[ mgh = \frac{1}{2}mv^2 + E_{lost} \]

Step 3: Detailed Explanation:

First, we list the given values and convert them to SI units.

Mass, \(m = 10 g = 10 \times 10^{-3} kg = 0.01 kg\)
Height, \(h = 50 m\)
Acceleration due to gravity, \(g = 10 m/s^2\)
Energy lost due to air resistance, \(E_{lost} = 3 J\)

Next, we calculate the initial potential energy (PE) of the raindrop. \[ PE_{initial} = mgh = (0.01 kg)(10 m/s^2)(50 m) = 5 J \]
Now, we use the energy conservation equation to find the final kinetic energy (KE) just before it hits the ground. \[ KE_{final} = PE_{initial} - E_{lost} \] \[ KE_{final} = 5 J - 3 J = 2 J \]
Finally, we use the formula for kinetic energy to find the final velocity (v). \[ KE_{final} = \frac{1}{2}mv^2 \] \[ 2 = \frac{1}{2}(0.01)v^2 \] \[ 4 = 0.01 \times v^2 \] \[ v^2 = \frac{4}{0.01} = 400 \] \[ v = \sqrt{400} = 20 m/s \]

Step 4: Final Answer:

The velocity of the drop on striking the ground is 20 ms\(^{-1}\). This corresponds to option (E).
Quick Tip: In energy conservation problems, always account for all forms of energy. If non-conservative forces like friction or air resistance are present, the initial mechanical energy will not equal the final mechanical energy. The difference is the energy "lost" or converted into heat.

Step 1: Understanding the Concept:

This problem relates power, force, and velocity. The power delivered by the motor is used to overcome the downward forces (gravity and friction) and move the lift at a certain speed. Assuming the lift moves at a constant speed, the upward force provided by the motor must balance the total downward force.


Step 2: Key Formula or Approach:

The relationship between power (P), force (F), and velocity (v) is: \[ P = F \cdot v \]
To find the speed (v), we can rearrange this to: \[ v = \frac{P}{F} \]
Here, F is the total upward force the motor must exert.


Step 3: Detailed Explanation:

First, identify all the forces acting on the lift.

Downward force due to gravity (Weight): \(W = mg\)
Downward frictional force: \(F_{friction} = 2000 N\)
Upward force exerted by the motor: \(F_{motor}\)

Calculate the weight of the lift and load: \[ W = (1000 kg) \times (10 m/s^2) = 10000 N \]
The total downward force is the sum of the weight and the frictional force. \[ F_{downward} = W + F_{friction} = 10000 N + 2000 N = 12000 N \]
Since the power delivered results in the upward motion, the motor must provide an upward force equal to the total downward force to move the lift (assuming constant velocity). \[ F_{motor} = F_{upward} = F_{downward} = 12000 N \]
Now, we can use the power formula to find the speed (v).
We are given:

Power, \(P = 36000 W\)
Total upward force, \(F = 12000 N\)
\[ v = \frac{P}{F} = \frac{36000 W}{12000 N} = 3 m/s \]

Step 4: Final Answer:

The speed of the lift is 3 ms\(^{-1}\). This corresponds to option (C).
Quick Tip: When calculating the force needed to lift an object at a constant speed, always sum up all the forces that oppose the motion. In this case, it's both gravity and friction. Then apply the formula \(P = F_{total} \times v\).

Step 1: Understanding the Concept:

This question tests the fundamental properties of rotational motion of a rigid body. A rigid body is one where the distance between any two constituent particles remains constant. When it rotates about a fixed axis, all particles move in circles.


Step 2: Detailed Explanation:

Let's analyze each statement:

(A) \(\omega\) is directed perpendicular to the axis of rotation: This is incorrect. The angular velocity \(\omega\) is an axial vector. Its direction is along the axis of rotation, determined by the right-hand grip rule.
(B) all the particles move with same speed: This is incorrect. All particles of the rigid body have the same angular speed (\(\omega\)). However, their linear speed (v) depends on their perpendicular distance (r) from the axis of rotation, according to the formula \(v = r\omega\). Particles farther from the axis move faster.
(C) \(\omega\) is a scalar quantity: This is incorrect. Angular velocity is a vector quantity, having both magnitude (angular speed) and direction.
(D) \(\omega\) has no direction: This is incorrect, as explained in (A) and (C). It has a well-defined direction along the axis of rotation.
(E) different particles move in different circles: This is correct. Each particle in the rigid body moves in a circular path. The center of each circle lies on the axis of rotation. Particles at different perpendicular distances (r) from the axis will trace out circles of different radii. Only particles at the same perpendicular distance from the axis will move in the same circle.


Step 3: Final Answer:

The correct statement is that different particles move in different circles. This corresponds to option (E).
Quick Tip: For a rigid body rotating about a fixed axis, remember that angular quantities (\(\omega, \alpha\)) are the same for every particle in the body. In contrast, linear quantities (v, a) vary with the particle's distance from the axis of rotation.

Step 1: Understanding the Concept:

This problem requires the use of the Perpendicular Axis Theorem to relate the moment of inertia of a planar object about an axis perpendicular to its plane to the moments of inertia about two perpendicular axes lying in its plane.


Step 2: Key Formula or Approach:

The Perpendicular Axis Theorem states that for a planar lamina, the moment of inertia (\(I_z\)) about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia (\(I_x\) and \(I_y\)) about two mutually perpendicular axes lying in the plane of the lamina and intersecting at the point where the perpendicular axis passes through it. \[ I_z = I_x + I_y \]

Step 3: Detailed Explanation:

Let the circular disc lie in the x-y plane.

The "central axis" is the axis perpendicular to the plane of the disc and passing through its center. Let's call this the z-axis. The moment of inertia about this axis is given as \(I_z = I\).
A "diameter" of the disc is an axis lying in the plane of the disc and passing through its center. We can choose any two perpendicular diameters as our x-axis and y-axis.
Due to the symmetry of the circular disc, the moment of inertia about any diameter is the same. Therefore, the moment of inertia about the x-axis (\(I_x\)) is equal to the moment of inertia about the y-axis (\(I_y\)). Let's call this value \(I_{diameter}\).
\[ I_x = I_y = I_{diameter} \]

Now, we apply the Perpendicular Axis Theorem: \[ I_z = I_x + I_y \]
Substituting the known terms: \[ I = I_{diameter} + I_{diameter} \] \[ I = 2 \times I_{diameter} \]
Solving for the moment of inertia about the diameter (\(I_{diameter}\)): \[ I_{diameter} = \frac{I}{2} \]

Step 4: Final Answer:

The moment of inertia of the disc about its diameter is I/2. This corresponds to option (D).
Quick Tip: Memorize the Perpendicular Axis Theorem (\(I_z = I_x + I_y\)) and recognize that it applies only to 2D (planar) objects. For a symmetric object like a disc or a square, the moments of inertia about the in-plane axes (\(I_x, I_y\)) are often equal.

Step 1: Understanding the Concept:

This question asks to identify a fundamental law of planetary motion based on its statement. The statement describes the relationship between the area swept by a planet's position vector and time.


Step 2: Detailed Explanation:

Let's review Kepler's three laws of planetary motion:

Kepler's First Law (Law of Orbits): Every planet moves in an elliptical orbit with the Sun situated at one of the two foci of the ellipse. This describes the shape of the orbit.
Kepler's Second Law (Law of Areas): The radius vector drawn from the Sun to a planet sweeps out equal areas in equal intervals of time. This means the areal velocity (\(\frac{dA}{dt}\)) of a planet is constant. This law is a consequence of the conservation of angular momentum. It implies that a planet moves faster when it is closer to the Sun (perihelion) and slower when it is farther away (aphelion).
Kepler's Third Law (Law of Periods): The square of the orbital period (\(T\)) of a planet is directly proportional to the cube of the semi-major axis (\(a\)) of its orbit. Mathematically, \(T^2 \propto a^3\).

The statement given in the question, "The line that joins any planet to the sun sweeps out equal areas in equal intervals of time," is the exact definition of Kepler's second law.


Step 3: Final Answer:

The statement is Kepler's second law. This corresponds to option (D).
Quick Tip: Create simple mnemonics for Kepler's laws: 1st Law: Law of \textbf{E}llipses 2nd Law: Law of \textbf{A}reas (\textbf{E}qual \textbf{A}reas) 3rd Law: Law of \textbf{P}eriods (\(T^2 \propto R^3\))

Step 1: Understanding the Concept:

This question is about the mechanical properties of different states of matter. Moduli of elasticity describe a material's resistance to deformation. We need to identify which states of matter can resist the types of deformation associated with Young's modulus and shear modulus.


Step 2: Detailed Explanation:

Let's define the moduli:

Young's Modulus (Y): It is the ratio of longitudinal stress to longitudinal strain. It measures a material's resistance to a change in length when stretched or compressed. For a material to have a well-defined length, it must have a definite shape.
Shear Modulus (G) or Modulus of Rigidity: It is the ratio of shearing stress to shearing strain. It measures a material's resistance to a change in shape (a deformation where parallel planes slide past one another).

Now let's consider the states of matter:

Solids: Solids have a definite shape and volume. They resist changes in both length and shape. Therefore, Young's modulus and shear modulus are well-defined for solids.
Liquids and Gases (Fluids): Fluids do not have a definite shape. They take the shape of their container. They cannot sustain a shearing stress; they will simply flow. Because they do not resist a change in shape, their shear modulus is effectively zero. Similarly, because they do not have a fixed initial length, Young's modulus is not a meaningful concept for fluids. Fluids do, however, resist changes in volume, so they have a Bulk Modulus.

Therefore, both Young's modulus and shear modulus are properties that are only defined for solids.


Step 3: Final Answer:

Young's modulus and shear modulus can be defined only in solids. This corresponds to option (E).
Quick Tip: Remember the defining characteristic of a fluid (liquid or gas) is its inability to support shear stress. This is why the shear modulus is zero for fluids, and consequently, only solids possess rigidity and defined Young's and shear moduli.

Step 1: Understanding the Concept:

This question explores how two key properties of liquids, surface tension and viscosity, are affected by changes in temperature. Both properties are related to the intermolecular forces within the liquid.


Step 2: Detailed Explanation:


Effect of Temperature on Surface Tension (T): Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. It arises from the cohesive intermolecular forces among the liquid molecules. When the temperature of a liquid is increased, the average kinetic energy of its molecules increases. This increased movement of molecules tends to weaken the cohesive forces between them. As the intermolecular forces decrease, the surface tension also decreases. At the boiling point, the surface tension of a liquid becomes zero.
Effect of Temperature on Viscosity (\(\eta\)) of a Liquid: Viscosity is a measure of a fluid's resistance to flow. In liquids, viscosity is primarily caused by the cohesive forces between molecules that create an internal friction. Similar to surface tension, when the temperature of a liquid increases, the kinetic energy of the molecules increases, which overcomes the intermolecular cohesive forces. This allows the liquid layers to slide past one another more easily, resulting in a decrease in viscosity.

Therefore, for a liquid, both surface tension and the coefficient of viscosity decrease as the temperature increases.

(Note: The original question used 'n' for viscosity, which has been corrected to the standard symbol \(\eta\)).


Step 3: Final Answer:

With an increase in temperature, both surface tension (T) and viscosity (\(\eta\)) of a liquid decrease. This corresponds to option (B).
Quick Tip: A practical example is heating cooking oil or honey. When heated, they become "thinner" and flow more easily, which is a clear demonstration that their viscosity decreases with temperature. This weakening of intermolecular forces also reduces surface tension.

Step 1: Understanding the Concept:

This phenomenon is described by Torricelli's Law, which is a specific application of Bernoulli's principle to fluid flowing out of an orifice (a small opening). The law relates the speed of the exiting fluid (efflux speed) to the height of the fluid above the opening.


Step 2: Key Formula or Approach:

We can apply Bernoulli's equation between two points: point 1 at the free surface of the water in the tank and point 2 at the opening.
Bernoulli's equation is: \[ P_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2 \]
where P is pressure, \(\rho\) is fluid density, g is acceleration due to gravity, h is height, and v is speed.


Step 3: Detailed Explanation:

Let's set up the variables for our two points.

Point 1 (at the surface of the water):

The pressure \(P_1\) is the atmospheric pressure, \(P_{atm}\).
Let the height of the opening be our reference level, so the height of the surface is \(h_1 = h\).
Since the tank is large, the speed at which the water level at the surface drops is negligible, so we can approximate \(v_1 \approx 0\).

Point 2 (at the small opening):

The water flows out into the atmosphere, so the pressure \(P_2\) is also atmospheric pressure, \(P_{atm}\).
The height is at our reference level, so \(h_2 = 0\).
The speed of the water flowing out is \(v_2 = v\), which is what we need to find.


Now, substitute these into Bernoulli's equation: \[ P_{atm} + \rho g h + \frac{1}{2}\rho (0)^2 = P_{atm} + \rho g (0) + \frac{1}{2}\rho v^2 \]
Simplifying the equation: \[ P_{atm} + \rho g h = P_{atm} + \frac{1}{2}\rho v^2 \]
The \(P_{atm}\) terms cancel out: \[ \rho g h = \frac{1}{2}\rho v^2 \]
The density \(\rho\) also cancels out: \[ g h = \frac{1}{2}v^2 \]
Now, solve for the speed v: \[ v^2 = 2gh \] \[ v = \sqrt{2gh} \]
This result is known as Torricelli's Law. It shows that the speed of efflux is the same as the speed an object would acquire by free-falling from a height h.


Step 4: Final Answer:

The speed of the water flowing out is \(\sqrt{2gh}\). This corresponds to option (E).
Quick Tip: Remembering Torricelli's Law (\(v = \sqrt{2gh}\)) can save you time. The derivation from Bernoulli's principle is straightforward, but knowing the final formula for the speed of efflux is very useful for competitive exams.

Step 1: Understanding the Concept:

The question asks for the molar specific heat at constant volume (\(C_v\)) for a diatomic gas. According to the law of equipartition of energy, the total internal energy of a gas is distributed equally among its degrees of freedom. Each degree of freedom contributes \(\frac{1}{2}kT\) to the average energy per molecule, or \(\frac{1}{2}RT\) to the molar internal energy.


Step 2: Key Formula or Approach:

The molar specific heat at constant volume is defined as the rate of change of molar internal energy (\(U\)) with respect to temperature (\(T\)).
\[ C_v = \frac{dU}{dT} \]
For a gas with \(f\) degrees of freedom, the molar internal energy is given by:
\[ U = \frac{f}{2}RT \]
Therefore, the formula for \(C_v\) becomes:
\[ C_v = \frac{d}{dT}\left(\frac{f}{2}RT\right) = \frac{f}{2}R \]

Step 3: Detailed Explanation:

A diatomic gas molecule (like O\(_2\), N\(_2\)) at normal temperatures has 5 degrees of freedom:

- 3 translational degrees of freedom (movement along x, y, and z axes).

- 2 rotational degrees of freedom (rotation about two axes perpendicular to the line connecting the atoms).

Vibrational modes are generally not active at normal temperatures.

So, for a diatomic gas, \(f = 5\).

Now, we can substitute the values of \(f\) and R into the formula for \(C_v\).

Given:

- Degrees of freedom, \(f = 5\).

- Universal gas constant, \(R = 8.2\) Jmol\(^{-1}\)K\(^{-1}\).
\[ C_v = \frac{5}{2}R \] \[ C_v = \frac{5}{2} \times 8.2 \] \[ C_v = 2.5 \times 8.2 \] \[ C_v = 20.5 \, Jmol^{-1}K^{-1} \]

Step 4: Final Answer:

The value of \(C_v\) for the diatomic gas is 20.5 Jmol\(^{-1}\)K\(^{-1}\), which corresponds to option (A).
Quick Tip: For competitive exams, quickly recall the degrees of freedom (\(f\)) for different types of gases: Monatomic gas (e.g., He, Ne, Ar): \(f = 3\) (translational only). So, \(C_v = \frac{3}{2}R\). Diatomic gas (e.g., O\(_2\), H\(_2\)): \(f = 5\) (3 translational + 2 rotational). So, \(C_v = \frac{5}{2}R\). Non-linear polyatomic gas (e.g., H\(_2\)O): \(f = 6\) (3 translational + 3 rotational). So, \(C_v = 3R\). Memorizing these helps solve problems much faster.

Step 1: Understanding the Concept:

The question describes a standing wave in a steel rod. When a rod is clamped at its middle, this point becomes a node (a point of zero displacement). The ends of the rod are free to vibrate, so they become antinodes (points of maximum displacement). The fundamental frequency is the lowest frequency at which the rod can vibrate in this configuration.


Step 2: Key Formula or Approach:

The relationship between wave speed (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) is given by:
\[ v = f \lambda \]
We need to determine the wavelength (\(\lambda\)) for the fundamental mode of vibration based on the boundary conditions.


Step 3: Detailed Explanation:

For the fundamental mode of vibration in a rod clamped at the middle:

- The center is a Node (N).

- The two ends are Antinodes (A).

The simplest standing wave pattern that fits this condition is A-N-A.

The distance between an antinode and the next node is always \(\frac{\lambda}{4}\).

The total length of the rod, L, covers the distance from one antinode (end) to the node (middle) and then to the other antinode (other end).

So, \(L = \frac{\lambda}{4} + \frac{\lambda}{4} = \frac{\lambda}{2}\).

This means the wavelength of the fundamental mode is twice the length of the rod:
\[ \lambda = 2L \]
Given:

- Length of the rod, \(L = 1\) m.

- Fundamental frequency, \(f = 3\) kHz = \(3 \times 10^3\) Hz.

First, calculate the wavelength:
\[ \lambda = 2 \times 1 \, m = 2 \, m \]
Now, calculate the speed of sound (\(v\)) using the wave equation:
\[ v = f \lambda \] \[ v = (3 \times 10^3 \, Hz) \times (2 \, m) \] \[ v = 6000 \, m/s \]

Step 4: Final Answer:

The speed of sound in the steel rod is 6000 ms\(^{-1}\), which corresponds to option (B).
Quick Tip: Visualize the standing wave patterns for different boundary conditions. For a rod clamped at the center, the fundamental mode has a node in the middle and antinodes at the ends. The length of the rod is half a wavelength (\(L=\lambda/2\)). If the rod were clamped at one end, the fundamental mode would have a node at the clamped end and an antinode at the free end, making the length a quarter of a wavelength (\(L=\lambda/4\)).

Step 1: Understanding the Concept:

The question asks for the frequency of a periodic wave shown in a graph. Frequency is the number of complete cycles of a wave that occur in one unit of time. It is the reciprocal of the time period (\(T\)), which is the time taken to complete one full cycle.


Step 2: Key Formula or Approach:

The relationship between frequency (\(f\)) and time period (\(T\)) is:
\[ f = \frac{1}{T} \]

Step 3: Detailed Explanation:

First, we need to determine the time period (\(T\)) from the given graph.

The graph shows the wave's amplitude (\(y\)) as a function of time (\(t\)). The x-axis represents time in microseconds (\(\mu\)s).

A complete cycle of the wave is the smallest repeating pattern. Looking at the graph:

- The wave starts at \(t=0\), goes up, and stays high.

- At \(t=2 \, \mu\)s, it goes down and stays low.

- At \(t=4 \, \mu\)s, it goes up again, starting the next cycle.

Therefore, the time taken for one complete cycle (the time period) is \(T = 4 \, \mu\)s.

Convert the time period to seconds:
\[ T = 4 \, \mus = 4 \times 10^{-6} \, s \]
Now, calculate the frequency using the formula:
\[ f = \frac{1}{T} = \frac{1}{4 \times 10^{-6} \, s} \] \[ f = 0.25 \times 10^{6} \, Hz \]
The unit \(10^6\) Hz is equivalent to Megahertz (MHz).
\[ f = 0.25 \, MHz \]

Step 4: Final Answer:

The frequency of the periodic wave is 0.25 MHz, which corresponds to option (B).
Quick Tip: Always pay close attention to the units on the axes of a graph. In this problem, the time axis is in microseconds (\(\mu\)s), not seconds. A common mistake is to forget this conversion, which would lead to an incorrect answer.

Step 1: Understanding the Concept:

This problem deals with Coulomb's Law, which describes the electrostatic force between two stationary point charges. The law states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.


Step 2: Key Formula or Approach:

Coulomb's Law is given by the formula:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \(F\) is the electrostatic force, \(q_1\) and \(q_2\) are the magnitudes of the charges, \(r\) is the distance between them, and \(k\) is Coulomb's constant.


Step 3: Detailed Explanation:

Let the two point charges be \(q_1\) and \(q_2\).

Initial situation:

The distance between the charges is \(d\).

The force \(F\) is given by:
\[ F = k \frac{q_1 q_2}{d^2} \quad --- (1) \]
Final situation:

One of the charges is "moved away by a distance d/2". This means the initial distance \(d\) is increased by \(d/2\).

The new distance of separation, \(d'\), is:
\[ d' = d + \frac{d}{2} = \frac{2d + d}{2} = \frac{3d}{2} \]
Now, we calculate the new force, \(F'\), using this new distance:
\[ F' = k \frac{q_1 q_2}{(d')^2} = k \frac{q_1 q_2}{\left(\frac{3d}{2}\right)^2} \] \[ F' = k \frac{q_1 q_2}{\frac{9d^2}{4}} \] \[ F' = \frac{4}{9} \left( k \frac{q_1 q_2}{d^2} \right) \quad --- (2) \]
Now, substitute equation (1) into equation (2):
\[ F' = \frac{4}{9} F \]

Step 4: Final Answer:

The new force between the charges is \(\frac{4}{9}F\), which corresponds to option (C). (Note: Options B and C are identical).
Quick Tip: Read the wording of distance changes carefully. "Moved away by a distance x" means the new distance is \(d+x\). "Moved to a distance x" means the new distance is \(x\). This distinction is crucial for setting up the problem correctly.

Step 1: Understanding the Concept:

A molecule is considered nonpolar if it has a net dipole moment of zero. This can occur in two ways: either all the bonds in the molecule are nonpolar, or the molecule has polar bonds but its symmetrical geometry causes the individual bond dipoles to cancel each other out.


Step 3: Detailed Explanation:

Let's analyze each molecule:

(A) CO\(_2\) (Carbon Dioxide):

The molecule has a linear structure: O=C=O.

Oxygen is more electronegative than carbon, so each C=O bond is polar, with the dipole moment pointing from C to O.

However, the two bond dipoles are equal in magnitude and point in exactly opposite directions. As a result, they cancel each other out, and the net dipole moment of the molecule is zero. Thus, CO\(_2\) is a nonpolar molecule.


(B) H\(_2\)O (Water):

The molecule has a bent (V-shaped) structure due to the two lone pairs on the oxygen atom.

The O-H bonds are polar because oxygen is highly electronegative. The bond dipoles point from H to O.

Because of the bent geometry, the two bond dipoles do not cancel out. They add up vectorially to give a significant net dipole moment. Thus, H\(_2\)O is a polar molecule.


(C) CH\(_3\)OH (Methanol):

This molecule has a tetrahedral geometry around the carbon and a bent geometry around the oxygen. It is asymmetrical.

It contains polar C-O and O-H bonds. Due to the lack of symmetry, these bond dipoles do not cancel, resulting in a net dipole moment. Thus, Methanol is a polar molecule.


(D) HCl (Hydrogen Chloride):

This is a simple diatomic molecule. Chlorine is more electronegative than hydrogen, creating a polar covalent bond. Since there is only one bond, there is a net dipole moment. Thus, HCl is a polar molecule.


(E) NaCl (Sodium Chloride):

This is an ionic compound, formed by the complete transfer of an electron from Na to Cl, creating Na\(^+\) and Cl\(^-\) ions. The electrostatic attraction between these ions results in a very large dipole moment. It is extremely polar.


Step 4: Final Answer:

Based on the analysis, CO\(_2\) is the only nonpolar molecule among the given options because of its linear symmetry. This corresponds to option (A).
Quick Tip: Molecular geometry is key to determining polarity. Symmetrical shapes like linear (CO\(_2\)), trigonal planar (BF\(_3\)), and tetrahedral (CCl\(_4\)) often lead to nonpolar molecules, provided all surrounding atoms are identical. Asymmetrical shapes like bent (H\(_2\)O) or trigonal pyramidal (NH\(_3\)) almost always result in polar molecules.

Step 1: Understanding the Concept:

The resistance of a wire depends on its intrinsic properties (resistivity) and its physical dimensions (length and cross-sectional area). The question asks how the resistance changes when these dimensions are altered.


Step 2: Key Formula or Approach:

The formula for the resistance (\(R\)) of a wire is:
\[ R = \rho \frac{l}{A} \]
where \(\rho\) (rho) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is its cross-sectional area.


Step 3: Detailed Explanation:

Initial Wire:

Length = \(l\)

Cross-sectional area = \(A\)

The resistance is given as \(R\). So, we have:
\[ R = \rho \frac{l}{A} \quad --- (1) \]
(The resistivity \(\rho\) is the same for both wires because they are made of the "same material".)


New Wire:

New length, \(l' = 3l\)

New cross-sectional area, \(A' = \frac{A}{3}\)

Let the new resistance be \(R'\). Using the resistance formula for the new wire:
\[ R' = \rho \frac{l'}{A'} \]
Substitute the new dimensions into this formula:
\[ R' = \rho \frac{3l}{\frac{A}{3}} \] \[ R' = \rho \frac{3l \times 3}{A} \] \[ R' = \rho \frac{9l}{A} \] \[ R' = 9 \left( \rho \frac{l}{A} \right) \quad --- (2) \]
Now, substitute the expression for \(R\) from equation (1) into equation (2):
\[ R' = 9R \]

Step 4: Final Answer:

The resistance of the new wire is 9R, which corresponds to option (C).
Quick Tip: Resistance is directly proportional to length (\(R \propto l\)) and inversely proportional to the cross-sectional area (\(R \propto \frac{1}{A}\)). If length is multiplied by a factor of 'x' and area is divided by a factor of 'y', the new resistance will be \(x \times y\) times the old resistance. Here, \(x=3\) and \(y=3\), so the new resistance is \(3 \times 3 = 9\) times the original.

Step 1: Understanding the Concept:

"Energy dissipated per unit time" is the definition of electric power (\(P\)). The question asks for the power dissipated by a resistor when connected to a voltage source.


Step 2: Key Formula or Approach:

There are three common formulas for electric power (\(P\)):

1. \( P = VI_{source} \)

2. \( P = I^2 R_{wire} \)

3. \( P = \frac{V_{source}^2}{R_{wire}} \)

Since the problem provides the voltage of the battery and the resistance of the wire, the most direct formula to use is the third one.


Step 3: Detailed Explanation:

We are given the following information:

- Resistance of the wire, \(R_{wire} = 2R\).

- Voltage of the battery, \(V_{source} = 2V\).

We need to find the power, \(P\).

Using the formula \( P = \frac{V_{source}^2}{R_{wire}} \):
\[ P = \frac{(2V)^2}{2R} \]
Now, let's simplify the expression:
\[ P = \frac{4V^2}{2R} \] \[ P = \frac{2V^2}{R} \]

Step 4: Final Answer:

The energy dissipated per unit time (power) is \(\frac{2V^2}{R}\), which corresponds to option (C).
Quick Tip: Be very careful with the variable names. The problem defines the resistance as '2R' and voltage as '2V', while the options use 'R' and 'V' as base variables. Don't get confused. Substitute the given values (2R and 2V) into the standard power formula \( P = V^2/R \).

Step 1: Understanding the Concept:

This question asks for the standard formula for the magnetic field (\(B\)) produced at the very center of a circular loop of wire carrying a current (\(I\)). This formula is a direct result of applying the Biot-Savart Law.


Step 2: Key Formula or Approach:

The Biot-Savart Law gives the magnetic field \(d\vec{B}\) produced by a small current element \(I d\vec{l}\):
\[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \]
To find the total magnetic field at the center of a circular loop, we need to integrate this expression over the entire loop.


Step 3: Detailed Explanation:

For a circular loop of radius \(r\), the current element \(d\vec{l}\) is always perpendicular to the vector \(\vec{r}\) pointing from the element to the center of the loop. Therefore, the angle between \(d\vec{l}\) and \(\hat{r}\) is 90°, and \(\sin(90^\circ) = 1\).

The magnitude of the magnetic field from a small element \(dl\) is:
\[ dB = \frac{\mu_0}{4\pi} \frac{I dl}{r^2} \]
By the right-hand rule, the direction of the magnetic field from every element \(dl\) on the loop points along the axis, perpendicular to the plane of the loop. So, we can simply sum up the magnitudes.

To find the total magnetic field \(B\), we integrate \(dB\) around the entire loop:
\[ B = \int_{loop} dB = \int_0^{2\pi r} \frac{\mu_0}{4\pi} \frac{I}{r^2} dl \]
Since \(\frac{\mu_0 I}{4\pi r^2}\) is constant, we can take it out of the integral:
\[ B = \frac{\mu_0 I}{4\pi r^2} \int_0^{2\pi r} dl \]
The integral of \(dl\) over the entire loop is just the circumference of the loop, which is \(2\pi r\).
\[ B = \frac{\mu_0 I}{4\pi r^2} (2\pi r) \]
Simplifying the expression:
\[ B = \frac{\mu_0 I}{2r} \]

Step 4: Final Answer:

This is a standard result. The magnetic field at the centre of a current loop of radius \(r\) carrying a current \(I\) is \(\frac{\mu_0 I}{2r}\), which corresponds to option (A).
Quick Tip: Do not confuse the formula for the magnetic field at the center of a circular loop with the formula for the magnetic field at a distance \(r\) from a long, straight wire. \textbf{Center of a loop:} \(B = \frac{\mu_0 I}{2r}\) (no \(\pi\)) \textbf{Long straight wire:} \(B = \frac{\mu_0 I}{2\pi r}\) (has \(\pi\)) Remembering this distinction can prevent common errors.

Step 1: Understanding the Concept:

The question asks for the magnetic field inside a long solenoid. A solenoid is a coil of wire, and when current flows through it, it produces a nearly uniform magnetic field inside, especially near the center.


Step 2: Key Formula or Approach:

The formula for the magnetic field (\(B\)) inside a long solenoid is:
\[ B = \mu_0 n I \]
where:

- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7}\) T·m/A).

- \(n\) is the number of turns per unit length.

- \(I\) is the current flowing through the wire.

The number of turns per unit length (\(n\)) is calculated as \(n = \frac{N}{L}\), where \(N\) is the total number of turns and \(L\) is the length of the solenoid.


Step 3: Detailed Explanation:

First, let's list the given values:

- Current, \(I = 1\) A.

- Length of the solenoid, \(L = 1\) m.

- Total number of turns, \(N = 7000\).

- Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}\) T·m/A.


Next, calculate the number of turns per unit length (\(n\)):
\[ n = \frac{N}{L} = \frac{7000 turns}{1 m} = 7000 turns/m \]

Now, substitute these values into the formula for the magnetic field:
\[ B = \mu_0 n I \] \[ B = (4\pi \times 10^{-7} \, T·m/A) \times (7000 \, m^{-1}) \times (1 \, A) \] \[ B = 4\pi \times 7000 \times 10^{-7} \, T \] \[ B = 28000\pi \times 10^{-7} \, T \] \[ B = 2.8\pi \times 10^4 \times 10^{-7} \, T \] \[ B = 2.8\pi \times 10^{-3} \, T \]
To get a numerical value, use the approximation \(\pi \approx 3.14159\):
\[ B \approx 2.8 \times 3.14159 \times 10^{-3} \, T \] \[ B \approx 8.796 \times 10^{-3} \, T \]
This value is approximately \(8.8 \times 10^{-3}\) T.


Step 4: Final Answer:

The magnetic field produced at the middle of the solenoid is approximately \(8.8 \times 10^{-3}\) T, which corresponds to option (D).
Quick Tip: For solenoid problems, always check if you are given the total turns \(N\) or the turns per unit length \(n\). If you are given \(N\) and length \(L\), you must first calculate \(n = N/L\) before using the formula \(B = \mu_0 n I\).

Step 1: Understanding the Concept:

Electrical resonance is a phenomenon that occurs in an AC circuit when the inductive reactance and capacitive reactance are equal in magnitude. This causes the circuit's impedance to be at a minimum (in a series circuit) or maximum (in a parallel circuit), leading to a significant increase in current or voltage at a specific frequency known as the resonant frequency.


Step 2: Detailed Explanation:

Let's analyze the roles of the circuit elements in an AC circuit:

- Resistor (R): Its impedance (\(Z_R = R\)) is independent of frequency. It only dissipates energy.

- Inductor (L): Its impedance, called inductive reactance (\(X_L\)), is directly proportional to frequency (\(X_L = \omega L = 2\pi f L\)). It stores energy in a magnetic field.

- Capacitor (C): Its impedance, called capacitive reactance (\(X_C\)), is inversely proportional to frequency (\(X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\)). It stores energy in an electric field.


Resonance occurs when the opposing effects of the inductor and capacitor cancel each other out. This happens when their reactances are equal:
\[ X_L = X_C \] \[ \omega L = \frac{1}{\omega C} \]
For this condition to be met, the circuit must contain both an inductor (\(L\)) and a capacitor (\(C\)).

- An RL circuit or an RC circuit cannot resonate because they only have one type of reactance.

- Circuits with only an inductor or only a capacitor cannot resonate.

- A resistor is usually present in practical resonant circuits (like an RLC circuit), but it is not essential for the phenomenon of resonance itself. The core requirement is the presence of both L and C to allow for the exchange of energy between the magnetic and electric fields.


Step 3: Final Answer:

The essential circuit elements required for exhibiting resonance are an inductor and a capacitor. This corresponds to option (E).
Quick Tip: Think of resonance as a balancing act. The inductor's reactance increases with frequency, while the capacitor's reactance decreases. Resonance is the specific frequency where these two reactances are perfectly balanced (\(X_L = X_C\)). This balance can only be achieved if both components are present in the circuit.

Step 1: Understanding the Concept:

Electromagnetic (EM) radiation is produced by the acceleration of charged particles. Different types of EM radiation correspond to different energy ranges and are produced by different physical processes at the atomic and molecular level.


Step 2: Detailed Explanation:

Let's look at the origin of different regions of the EM spectrum:

- Microwaves: Primarily produced by the rotational motion of molecules and by man-made electronic devices like klystrons and magnetrons.

- Infrared (IR) Radiation: This radiation is directly associated with the thermal motion of atoms and molecules. When molecules vibrate or rotate, they can absorb or emit photons in the infrared range. Hence, IR is often called "heat radiation." The energy levels for molecular vibrations correspond to the infrared region.

- Visible and Ultraviolet (UV) Light: These are produced by electronic transitions in atoms and molecules. When an electron in an outer shell jumps to a lower energy level, it emits a photon. The energy of these photons typically falls in the visible or UV range.

- X-rays: These are higher-energy photons produced by two main mechanisms: transitions of electrons in the inner shells of heavy atoms, or the rapid deceleration of high-energy charged particles (like electrons hitting a metal target), a process known as bremsstrahlung.


The question specifically asks about the radiation from the vibrations of atoms and molecules. This process is the primary source of infrared radiation.


Step 3: Final Answer:

The vibrations of atoms and molecules produce electromagnetic radiation in the infrared region. This corresponds to option (B).
Quick Tip: Create a mental map of the electromagnetic spectrum and its sources. A simple mnemonic for the energy order is: \textbf{R}oman \textbf{M}en \textbf{I}nvented \textbf{V}ery \textbf{U}nusual \textbf{X}-ray \textbf{G}uns (Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma). Associate these with their sources: Radio (circuits), Microwave (molecular rotation), IR (molecular vibration), Visible/UV (electron transitions), X-ray (inner electron transitions), Gamma (nuclear decay).

Step 1: Understanding the Concept:

The question asks which phenomenon does NOT occur when a monochromatic light ray passes through a prism. Monochromatic light is light of a single wavelength (and thus a single color).


Step 2: Detailed Explanation:

Let's analyze each phenomenon in the context of light passing through a prism:

(A) Dispersion: Dispersion is the splitting of white light into its constituent colors (like a rainbow). This happens because the refractive index (\(n\)) of the prism material depends on the wavelength (\(\lambda\)) of light (\(n = n(\lambda)\)). Since different colors have different wavelengths, they bend by slightly different amounts, causing them to separate. However, monochromatic light consists of only one wavelength. With nothing to split into, the phenomenon of dispersion cannot occur.


(B) Refraction: Refraction is the bending of light as it passes from one medium to another (e.g., from air into glass, and from glass back into air). This will happen at both surfaces of the prism as long as the light ray is not incident perpendicular to the surface. A monochromatic ray will still bend.


(C) Deviation: Deviation is the overall change in the direction of the light ray after it has passed through the prism. This is a direct consequence of the two refractions at the prism surfaces. A monochromatic ray will be deviated from its original path.


(D) Reflection: Whenever light hits a boundary between two different media, some of it is reflected. This is known as partial reflection. So, some of the monochromatic light will reflect off both the first and second surfaces of the prism.


(E) Total Internal Reflection (TIR): If the angle of incidence of the light ray on the second surface (from inside the prism) is greater than the critical angle for the glass-air interface, the ray will be totally internally reflected instead of refracting out. This phenomenon is possible for a monochromatic ray under the right conditions of incidence angle and prism angle.


Step 3: Final Answer:

The only phenomenon listed that requires light of multiple wavelengths is dispersion. Therefore, a monochromatic ray will not produce dispersion. This corresponds to option (A).
Quick Tip: The word "monochromatic" is the key. "Mono" means one, and "chroma" means color. Monochromatic light is single-color light. Dispersion is the separation of colors. You cannot separate a single color into multiple colors. Therefore, dispersion is impossible with monochromatic light.

Step 1: Understanding the Concept:

A simple microscope (or a magnifying glass) uses a single convex lens to produce a magnified virtual image of a small object. The magnification depends on where the final image is formed. The question specifies the case where the image is formed at the least distance of distinct vision (\(D\)), which is the near point of the eye. This provides the maximum angular magnification.


Step 2: Key Formula or Approach:

The formula for the magnifying power (\(M\)) of a simple microscope when the final image is formed at the least distance of distinct vision (\(D\)) is:
\[ M = 1 + \frac{D}{f} \]
where \(f\) is the focal length of the convex lens.


Step 2: Detailed Explanation:

We are given the following values:

- Focal length of the lens, \(f = 5\) cm.

- Least distance of distinct vision, \(D = 25\) cm.


Now, we substitute these values into the formula:
\[ M = 1 + \frac{25 \, cm}{5 \, cm} \] \[ M = 1 + 5 \] \[ M = 6 \]

Step 3: Final Answer:

The magnification of the simple microscope is 6, which corresponds to option (B).
Quick Tip: Remember the two main formulas for a simple microscope's magnification: When the image is at the near point (\(D\)) for maximum magnification: \(M = 1 + \frac{D}{f}\). When the image is at infinity for relaxed viewing: \(M = \frac{D}{f}\). Always check the problem statement to see which condition applies. If not specified, the case for the image at the near point is often assumed for maximum magnification.

Step 1: Understanding the Concept:

The photoelectric effect is the emission of electrons (photoelectrons) from a material when light shines on it. For this to happen, the energy of the incident light photons must be greater than or equal to the work function of the material. The threshold wavelength (\(\lambda_0\)) is the maximum wavelength of incident light that can cause photoemission. Any light with a wavelength longer than the threshold wavelength will not have enough energy to eject electrons.


Step 2: Key Formula or Approach:

The condition for photoemission is:

Energy of incident photon \(\geq\) Work function (\(\Phi\))
\[ E \geq \Phi \]
In terms of wavelength (\(\lambda\)), since \(E = \frac{hc}{\lambda}\), the condition becomes:
\[ \frac{hc}{\lambda} \geq \frac{hc}{\lambda_0} \]
This simplifies to:
\[ \lambda \leq \lambda_0 \]
So, photoelectrons will be emitted only if the incident wavelength is less than or equal to the threshold wavelength.


Step 2: Detailed Explanation:

We need to compare the wavelengths of the given options with the threshold wavelength, which is in the green region. The order of colors in the visible spectrum from shortest to longest wavelength is given by the acronym VIBGYOR:

Violet \(<\) Indigo \(<\) Blue \(<\) Green \(<\) Yellow \(<\) Orange \(<\) Red

Ultraviolet (UV) light has an even shorter wavelength than violet light.


Given: The threshold wavelength \(\lambda_0\) is in the green region.

We are looking for light that will not emit photoelectrons, which means we need to find light with a wavelength \(\lambda > \lambda_0\) (i.e., \(\lambda > \lambda_{green}\)).


Let's check the options:

(A) Ultraviolet: \(\lambda_{UV} < \lambda_{green}\). Will cause emission.

(B) Blue: \(\lambda_{Blue} < \lambda_{green}\). Will cause emission.

(C) Violet: \(\lambda_{Violet} < \lambda_{green}\). Will cause emission.

(D) Orange: \(\lambda_{Orange} > \lambda_{green}\). Will not cause emission.

(E) Indigo: \(\lambda_{Indigo} < \lambda_{green}\). Will cause emission.


Step 3: Final Answer:

Orange light has a longer wavelength than green light, so its photons have less energy than the work function of the material. Therefore, orange light will not emit photoelectrons. This corresponds to option (D).
Quick Tip: Remember the relationship between color, wavelength, and energy. Towards the red end of the spectrum, wavelength increases and energy decreases. Towards the violet end, wavelength decreases and energy increases. For photoemission, you need "enough energy," which means the wavelength must be "short enough."

Step 1: Understanding the Concept:

The question asks to identify the nuclear process characterized by the release of free neutrons from a nucleus. We need to analyze the products of each listed nuclear reaction.


Step 2: Detailed Explanation:

Let's examine each option:

(A) \(\alpha\) - decay: A heavy, unstable nucleus emits an alpha particle, which is a helium nucleus (\({^4_2}He\)). This particle consists of two protons and two neutrons bound together. No free neutrons are released.

Example: \({^{238}_{92}}U \rightarrow {^{234}_{90}}Th + {^4_2}\alpha\)


(B) \(\beta\) - decay: In \(\beta^-\) decay, a neutron within the nucleus is converted into a proton, and an electron (\({^0_{-1}}e\)) and an antineutrino are emitted. In \(\beta^+\) decay, a proton is converted into a neutron, and a positron (\({^0_{+1}}e\)) and a neutrino are emitted. In either case, free neutrons are not released from the nucleus.

Example (\(\beta^-\)): \({^{14}_{6}}C \rightarrow {^{14}_{7}}N + {^0_{-1}}e + \bar{\nu}_e\)


(C) Nuclear fusion: Two or more light nuclei combine to form a heavier nucleus. While some fusion reactions can release a neutron (e.g., Deuterium-Tritium fusion), it is not the defining characteristic of fusion in general. Many fusion reactions release protons or just energy.

Example: \({^2_1}H + {^3_1}H \rightarrow {^4_2}He + {^1_0}n\) (releases a neutron).

Example: \({^2_1}H + {^2_1}H \rightarrow {^3_2}He + {^1_0}n\) (releases a neutron).

Example: \({^2_1}H + {^1_1}H \rightarrow {^3_2}He + \gamma\) (no neutron released).


(D) Pair production: A high-energy photon (gamma ray) passing near a nucleus transforms into an electron-positron pair. No nucleons are involved or released.


(E) Nuclear fission: A heavy nucleus, typically after absorbing a neutron, splits into two or more smaller nuclei (fission fragments). This process releases a very large amount of energy and, crucially, several free neutrons (usually 2 or 3). The release of these neutrons is a key feature, as they can go on to trigger further fission events, leading to a self-sustaining chain reaction.

Example: \({^1_0}n + {^{235}_{92}}U \rightarrow {^{236}_{92}}U^* \rightarrow {^{141}_{56}}Ba + {^{92}_{36}}Kr + 3({^1_0}n)\)


Step 3: Final Answer:

Among the given options, nuclear fission is the process fundamentally characterized by the splitting of a heavy nucleus and the release of free neutrons. This corresponds to option (E).
Quick Tip: Think about practical applications. Nuclear power plants and atomic bombs work on the principle of a chain reaction. This chain reaction is sustained by the neutrons released during each \textbf{fission} event. This connection makes it easy to remember that fission is the process that releases neutrons.

Step 1: Understanding the Concept:

The volume of an atomic nucleus is related to its mass number (\(A\)), which is the total number of protons and neutrons. Experiments show that nuclear matter has a nearly constant density, which implies a direct relationship between the volume and the number of nucleons.


Step 2: Key Formula or Approach:

The radius \(R\) of a nucleus is given by the empirical formula:
\[ R = R_0 A^{1/3} \]
where \(R_0\) is a constant (\(\approx 1.2 \times 10^{-15}\) m) and \(A\) is the mass number.

Assuming the nucleus is spherical, its volume \(V_{nucleus}\) is:
\[ V_{nucleus} = \frac{4}{3}\pi R^3 \]
By substituting the formula for \(R\), we can find the relationship between volume and mass number.


Step 2: Detailed Explanation:

Let's derive the relationship between volume (\(V_{nucleus}\)) and mass number (\(A\)):
\[ V_{nucleus} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 \] \[ V_{nucleus} = \frac{4}{3}\pi R_0^3 (A^{1/3})^3 \] \[ V_{nucleus} = \left(\frac{4}{3}\pi R_0^3\right) A \]
Since \(\frac{4}{3}\pi R_0^3\) is a constant, we can see that the volume of a nucleus is directly proportional to its mass number:
\[ V_{nucleus} \propto A \]
We can use this proportionality to solve the problem. Let \(V_1\) be the volume for mass number \(A_1\), and \(V_2\) be the volume for mass number \(A_2\).
\[ \frac{V_2}{V_1} = \frac{A_2}{A_1} \]
Given:

- For mass number \(A_1 = 2\), the volume is \(V_1 = V\).

- We need to find the volume \(V_2\) for mass number \(A_2 = 8\).


Substitute the values into the ratio:
\[ \frac{V_2}{V} = \frac{8}{2} \] \[ \frac{V_2}{V} = 4 \] \[ V_2 = 4V \]

Step 3: Final Answer:

The volume of the nucleus with mass number 8 is 4V, which corresponds to option (C).
Quick Tip: A key consequence of \(V \propto A\) is that the density of all nuclei is approximately constant. Density \(\rho = \frac{Mass}{Volume} \approx \frac{A \cdot m_p}{k \cdot A} = constant\), where \(m_p\) is the mass of a nucleon and \(k\) is the constant of proportionality for volume. This is a fundamental property of nuclear matter.

Step 1: Understanding the Concept:

A p-n junction diode is in forward bias when the positive terminal of an external voltage source is connected to the p-type semiconductor and the negative terminal is connected to the n-type semiconductor. This configuration greatly affects the diode's electrical properties.


Step 2: Detailed Explanation:

When a p-n junction is forward biased:

1. The external voltage applies an electric field that opposes the internal electric field of the depletion region.

2. This opposition lowers the potential barrier at the junction.

3. With a lower potential barrier, the width of the depletion region decreases (narrows).

4. Majority charge carriers (holes from the p-side and electrons from the n-side) now have enough energy to overcome the reduced barrier and diffuse across the junction.

5. This diffusion of majority carriers constitutes a significant forward current, typically in the milliampere (mA) range for a small applied voltage (e.g., > 0.7 V for silicon).

6. Since a large current flows for a small voltage, the effective resistance of the diode in forward bias is very low.


Now let's evaluate the given options based on this understanding:

(A) p-n junction provides very low resistance: This is correct. The large flow of current for a small voltage indicates low resistance.


(B) width of the depletion layer increases: This is incorrect. The width decreases in forward bias. It increases in reverse bias.


(C) potential barrier increases: This is incorrect. The potential barrier is lowered by the external voltage.


(D) amount of current flow is in the range of microampere: This is incorrect. Microampere (\(\mu\)A) range current is characteristic of a reverse-biased diode (due to minority carriers). Forward current is in milliamperes (mA).


(E) current flow is due to minority carriers only: This is incorrect. The large forward current is predominantly due to the flow of majority carriers. The small reverse current is due to minority carriers.


Step 3: Final Answer:

The correct statement describing a forward-biased diode is that the p-n junction provides very low resistance, which corresponds to option (A).
Quick Tip: Think of a forward-biased diode as a "closed switch" or a one-way street with the gate open: it offers low resistance and allows current to pass easily. Conversely, a reverse-biased diode is like an "open switch" or a closed gate: it offers very high resistance and blocks the flow of current (allowing only a tiny leakage current).