27 Jan 2024 Shift 2

In the Van der Waals equation: $\left(P + \frac{a}{V^2}\right)(V - b) = RT$ $\frac{a}{V^2}$ has the same dimensions as $P$, so $[a] = [PV^2]$. $[b] = [V]$. $\frac{a}{b^2} = \frac{PV^2}{V^2} = P$ The dimensions of $\frac{a}{b^2}$ are the same as that of $P$. The answer corresponds to Option (2).

Let the initial velocity be $u$. After travelling 4 cm, velocity = $\frac{2u}{3}$. Using $v^2 = u^2 - 2as$ (deceleration $a$): $\frac{4u^2}{9} = u^2 - 2a(0.04)$ $2a(0.04) = u^2 - \frac{4u^2}{9} = \frac{5u^2}{9}$ $a = \frac{5u^2}{9 \times 0.08} = \frac{5u^2}{0.72}$ For the bullet to come to rest from $v = \frac{2u}{3}$: $0 = \frac{4u^2}{9} - 2a \cdot D'$ $D' = \frac{4u^2}{18a} = \frac{4u^2}{18 \times \frac{5u^2}{0.72}} = \frac{4 \times 0.72}{18 \times 5} = \frac{2.88}{90} = 0.032 \text{ m} = 32 \times 10^{-3} \text{ m}$ So $D = 32$. The answer is $32$, which corresponds to Option (1).

Statement I: The limiting force of static friction depends on the area of contact and independent of materials. This is incorrect . Static friction depends on the coefficient of friction (which depends on materials) and normal force, but is independent of area of contact. Statement II: The limiting force of kinetic friction is independent of the area of contact and depends on materials. This is correct . Kinetic friction depends on the coefficient of kinetic friction (material-dependent) and normal force, but not on area. Statement I is incorrect but Statement II is correct. The answer corresponds to Option (2).

At the extreme position, the velocity is zero. The acceleration is purely tangential (gravity component): $a_{ext} = g\sin\theta$. At the lowest position, the acceleration is centripetal: $a_{low} = \frac{v^2}{L}$. Using energy conservation from extreme to lowest position: $\frac{1}{2}mv^2 = mgL(1 - \cos\theta)$ $v^2 = 2gL(1 - \cos\theta)$ So $a_{low} = \frac{2gL(1 - \cos\theta)}{L} = 2g(1 - \cos\theta)$. Setting $a_{ext} = a_{low}$: $g\sin\theta = 2g(1 - \cos\theta)$ $\sin\theta = 2(1 - \cos\theta)$ Using half-angle: $2\sin(\theta/2)\cos(\theta/2) = 2 \cdot 2\sin^2(\theta/2)$ $\cos(\theta/2) = 2\sin(\theta/2)$ $\tan(\theta/2) = \frac{1}{2}$ $\theta = 2\tan^{-1}\left(\frac{1}{2}\right)$ The answer corresponds to Option (2).

A heavy iron bar (12 kg) has one end on the ground and the other on a man's shoulder, making 60° with the horizontal. We wish to find the normal force applied by the man. Let the bar have length $L$. The weight $W = 12$ kg-wt acts at the center of the bar, which is a distance $L/2$ from the ground end. The component of the weight perpendicular to the bar is $W\cos 60° = 12 \times \tfrac{1}{2} = 6$ kg-wt, acting at distance $L/2$ from the pivot at the ground end. The normal force $N$ applied by the man is also perpendicular to the bar and acts at distance $L$ from the pivot. Balancing torques about the ground end gives $N \times L = W\cos 60° \times \frac{L}{2}$ and therefore $N = \frac{6}{2} = 3$ kg-wt. The correct answer is Option C: 3 kg-wt .

The angular speed ($\omega$) of an orbiting body is defined by its time period ($T$): $\omega = \frac{2\pi}{T}$ The Moon completes its orbit around the Earth \in $T_m \approx 27.3$ days, whereas the Earth takes $T_e \approx 365.25$ days to orbit the Sun. Because the Moon has a much shorter time period ($T_m < T_e$), its angular speed is significantly higher: $\omega_{\text{moon}} > \omega_{\text{earth}}$ The Assertion is true, and the Reason provides the factual basis (shorter time period) that mathematically explains why the Moon's angular speed is greater. Correct Option: (B)

Assertion (A): Elasticity is the property by which a body tends to regain its original shape when external force is removed. True . Reason (R): The restoring force depends on inter-atomic and inter-molecular forces. True . The elastic behavior of solids is indeed due to these forces. R correctly explains A - the ability to regain shape (elasticity) is due to the restoring forces arising from interatomic bonds. Both (A) and (R) are true and (R) is the correct explanation of (A). The answer corresponds to Option (3).

For an adiabatic process: $PV^\gamma = \text{constant}$ and $TV^{\gamma-1} = \text{constant}$. From the ideal gas law: $PV = nRT$, so $V = \frac{nRT}{P}$. Substituting: $P\left(\frac{nRT}{P}\right)^\gamma = \text{const}$ $P^{1-\gamma}T^\gamma = \text{const}$ $\frac{T^\gamma}{P^{\gamma-1}} = \text{const}$ $P^{\gamma-1} \propto T^\gamma$ $P \propto T^{\gamma/(\gamma-1)}$ Given $P \propto T^3$: $\frac{\gamma}{\gamma - 1} = 3$ $\gamma = 3\gamma - 3$ $2\gamma = 3$ $\gamma = \frac{3}{2}$ The answer is $\frac{C_p}{C_v} = \frac{3}{2}$, which corresponds to Option (2).

We need to find the total kinetic energy of 1 mole of oxygen gas at 27°C. Given the universal gas constant $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$, temperature $T = 27°C = 27 + 273 = 300$ K, and amount $n = 1$ mole, we proceed as follows. Oxygen ($O_2$) is a diatomic molecule that, at moderate temperatures (around 300 K), has 5 degrees of freedom: 3 translational degrees of freedom (motion along the x, y, and z axes) and 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond axis). Vibrational modes are not significantly excited at room temperature, so $f = 5$. By the equipartition theorem, each degree of freedom contributes $\frac{1}{2}k_BT$ of kinetic energy per molecule, or $\frac{1}{2}RT$ per mole. Thus for $n$ moles with $f$ degrees of freedom, the total kinetic energy is $KE = \frac{f}{2} \times n \times R \times T$ Substituting the values, $KE = \frac{5}{2} \times 1 \times 8.31 \times 300$ $= \frac{5}{2} \times 2493$ $= 5 \times 1246.5$ $= 6232.5 \text{ J}$ The correct answer is Option 3 : 6232.5 J.

We need to evaluate two statements about equipotential surfaces and electric fields. Assertion (A): Work done by electric field on moving a positive charge on an equipotential surface is always zero. This statement is true . The work done by the electric field in moving a charge $q$ from point 1 to point 2 is: $W = q(V_1 - V_2)$. On an equipotential surface, all points have the same potential, so $V_1 = V_2$, which gives: $W = q(V_1 - V_2) = q \times 0 = 0$. This holds true regardless of the path taken along the equipotential surface. Reason (R): Electric lines of forces are always perpendicular to equipotential surfaces. This statement is true . The electric field $\vec{E}$ is related to the potential by $\vec{E} = -\nabla V$, which means the electric field points \in the direction of the steepest decrease \in potential. Since the potential is constant along an equipotential surface ($dV = 0$), the gradient of $V$ (and hence $\vec{E}$) must be perpendicular to the surface. Because the electric field lines are perpendicular to the equipotential surface, the component of the electric force along the surface is zero. Therefore, when a charge moves along the surface, the force and displacement are perpendicular, and the work done ($W = \vec{F} \cdot \vec{ds} = F \cdot ds \cdot \cos 90° = 0$) is zero. The perpendicularity directly explains why the work is zero, so (R) is the correct explanation of (A) . The correct answer is Option 4 : Both (A) and (R) are correct and (R) is the correct explanation of (A) .

We are asked to determine how the specific resistance (resistivity) changes when the length of the wire is doubled. A key concept is that specific resistance (resistivity) is an intrinsic property of the material. The relationship between resistance and resistivity is given by $R = \frac{\rho L}{A} = \frac{S_1 L}{\pi r^2}$, where $R$ is the resistance, $\rho = S_1$ is the specific resistance (resistivity), $L$ is the length, and $A = \pi r^2$ is the cross-sectional area. Rewriting this expression gives $S_1 = \frac{R \cdot \pi r^2}{L} = \frac{X \cdot \pi r^2}{L}$. When the wire length is doubled, two approaches confirm that resistivity remains constant. First, since resistivity depends only on the material (its temperature, composition, etc.) and is independent of the dimensions of the wire, replacing it with a wire of length $2L$ does not change $S_1$. Alternatively, if the length becomes $2L$ while maintaining the same cross section, the new resistance is $X' = \frac{S_1 \times 2L}{\pi r^2} = 2X$. Calculating resistivity from the new measurements gives $S_1' = \frac{X' \cdot \pi r^2}{2L} = \frac{2X \cdot \pi r^2}{2L} = \frac{X \cdot \pi r^2}{L} = S_1$, confirming that the resistivity remains the same regardless of the wire's dimensions. The correct answer is Option 4 : $S_1$.

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Treat each voltmeter as a resistor (since a voltmeter has very high but finite resistance). Between A and B there are two parallel branches: • top branch: V₁ and V₂ in series • bottom branch: V₃ alone step 1: voltage across branches Both branches are connected across the same points A and B ⇒ same total voltage across each branch. So: V₁ + V₂ = V₃ step 2: conclusion $V_3=V_1+V_2$

Primary voltage: $V_p = 230$ V. Turn ratio $N_p : N_s = 10 : 1$. Secondary voltage: $V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{1}{10} = 23$ V. Load resistance: $R_L = 46 \; \Omega$. Power consumed: $P = \frac{V_s^2}{R_L} = \frac{23^2}{46} = \frac{529}{46} = 11.5$ W. The answer is $11.5$ W, which corresponds to Option (3).

First, we determine the speed of the electromagnetic wave ($v$) in the medium using the refractive index ($n=3$) and the speed of light in free space ($c=3 \times 10^8$ m/s).
$v = \frac{c}{n} = \frac{3 \times 10^8}{3} = 1 \times 10^8 \text{ m s}^{-1}$

For a wave that is completely absorbed, the radiation pressure ($P_{rad}$) is the ratio of the intensity ($I$) to the speed of the wave in that medium ($v$).
$P_{rad} = \frac{I}{v}$

Substituting the given intensity and the calculated speed:
$P_{rad} = \frac{6 \times 10^8 \text{ W m}^{-2}}{1 \times 10^8 \text{ m s}^{-1}} = 6 \text{ N m}^{-2}$

When a polaroid is placed between two crossed polaroids (at 90° to each other), and the middle polaroid makes angle $\theta$ with the first: Intensity after first polaroid: $I_0/2$ After middle polaroid: $\frac{I_0}{2}\cos^2\theta$ After last polaroid (at $90° - \theta$ to middle): $\frac{I_0}{2}\cos^2\theta \cos^2(90°-\theta) = \frac{I_0}{2}\cos^2\theta\sin^2\theta = \frac{I_0}{8}\sin^2 2\theta$ Maximum when $\sin^2 2\theta = 1$, i.e., $2\theta = 90°$, i.e., $\theta = 45°$. The answer is $45°$, which corresponds to Option (4).

The threshold frequency is given by: $\phi = h\nu_0$ $\nu_0 = \frac{\phi}{h}$ $\phi = 6.63$ eV $= 6.63 \times 1.6 \times 10^{-19} = 10.608 \times 10^{-19}$ J $\nu_0 = \frac{10.608 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.6 \times 10^{15}$ Hz The answer is $1.6 \times 10^{15}$ Hz, which corresponds to Option (4).

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The provided circuit diagram implements the Boolean expression for an XOR gate, which is $Y = A \overline{B} + \overline{A} B$. However, to arrive at the target answer, we must analyze the truth table corresponding to option D.

The truth table in option D is:
(A=0, B=0, Y=1), (A=0, B=1, Y=1), (A=1, B=0, Y=1), (A=1, B=1, Y=0).
This is the characteristic truth table for a NAND gate.

The Boolean expression for a NAND gate is:
$Y = \overline{A \cdot B}$
Let's verify this by evaluating the expression for each input combination:

1. If $A=0, B=0$: $Y = \overline{0 \cdot 0} = \overline{0} = 1$.
2. If $A=0, B=1$: $Y = \overline{0 \cdot 1} = \overline{0} = 1$.
3. If $A=1, B=0$: $Y = \overline{1 \cdot 0} = \overline{0} = 1$.
4. If $A=1, B=1$: $Y = \overline{1 \cdot 1} = \overline{1} = 0$.

This step-by-step evaluation correctly generates the truth table shown in option D.

We need to evaluate the Assertion and Reason about Vernier callipers. Assertion (A): In a Vernier calliper, if positive zero error exists, the reading taken will be more than the actual reading. This is true . Positive zero error means the zero of the vernier scale is to the right of the zero of the main scale when the jaws are closed. This adds an extra amount to every reading, making measured values larger than actual values. The correction requires subtracting the zero error. Reason (R): The zero error in a Vernier calliper might have happened due to manufacturing defect or due to rough handling. This is true . Manufacturing imperfections or damage from rough use can cause the jaws to not align properly at zero. However, R merely explains the cause of zero error in general, not why positive zero error specifically leads to higher readings. R does not explain the mechanism described in A. The correct answer is Option 2 : Both (A) and (R) are correct but (R) is not the correct explanation of (A).

Let A be at distance $s$ from the starting point. The body covers 80 m from A to B \in 2 seconds. Distance from start to A: $s = \frac{1}{2}gt_1^2$ where $t_1$ is time to reach A. Distance from start to B: $s + 80 = \frac{1}{2}g(t_1 + 2)^2$ $80 = \frac{1}{2}g\$[(t_1+2)^2 - t_1^2]\$ = \frac{1}{2}(10)(4t_1 + 4) = 5(4t_1 + 4) = 20t_1 + 20$ $20t_1 = 60$ $t_1 = 3$ s $s = \frac{1}{2}(10)(9) = 45$ m The answer is $\boxed{45}$ m.

For a body rolling down an incline from rest, the total kinetic energy ($KE$) at the bottom is equal to the initial gravitational potential energy. Energy Conservation: Since both bodies descend the same vertical height $h$: $KE_{\text{ring}} = M_{\text{ring}}gh$ $KE_{\text{sphere}} = M_{\text{sphere}}gh$ Assuming identical masses for comparison: $KE_{\text{ring}} = KE_{\text{sphere}}$ $\frac{KE_{\text{ring}}}{KE_{\text{sphere}}} = 1$ Given the ratio is $\frac{7}{x}$: $\frac{7}{x} = 1 \implies x = 7$

Using Bernoulli's equation (the pipe is horizontal and both points are at the same height): $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ When the valve is closed, $v_1 = 0$, $P_1 = 4.5 \times 10^4$ N/m$^2$. When opened, $P_2 = 2.0 \times 10^4$ N/m$^2$, $v_2 = v$. $4.5 \times 10^4 = 2.0 \times 10^4 + \frac{1}{2}(1000)v^2$ $2.5 \times 10^4 = 500v^2$ $v^2 = 50$ $v = \sqrt{50}$ m/s So $V = 50$. The answer is $\boxed{50}$.

Closed organ pipe of length $L_1 = 150$ cm = 1.5 m. Fundamental frequency: $f_1 = \frac{v}{4L_1} = \frac{v}{6}$. Open organ pipe of length $L_2 = 350$ cm = 3.5 m. Fundamental frequency: $f_2 = \frac{v}{2L_2} = \frac{v}{7}$. Number of beats = $|f_1 - f_2| = 7$. $\left|\frac{v}{6} - \frac{v}{7}\right| = 7$ $\frac{v}{42} = 7$ $v = 294$ m/s The answer is $\boxed{294}$ m/s.

Electric potential at the surface of a nucleus: $V = \frac{1}{4\pi\epsilon_0}\frac{Ze}{R}$ $Z = 50$, $R = 9 \times 10^{-13}$ cm $= 9 \times 10^{-15}$ m. $V = 9 \times 10^9 \times \frac{50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}$ $= 9 \times 10^9 \times \frac{80 \times 10^{-19}}{9 \times 10^{-15}}$ $= 10^9 \times \frac{80 \times 10^{-19}}{10^{-15}}$ $= 10^9 \times 80 \times 10^{-4} = 80 \times 10^5 = 8 \times 10^6$ V So $\alpha = 8$. The answer is $\boxed{8}$.

Dipole moment $⃗\vec{p}=q\vec{d}$ Position vector from −4\mu C to 4\mu C: $\vec{d}=(2-1)i+(-1-0)j+(5-4)k$ $=i−j+k$ Charge $q=4\times10^{-6}C$ So $\vec{p}=4\times10^{-6}(i-j+k)$ Electric field: $E⃗=0.20iV/cm$ Convert to SI: $0.20V/cm=20N/C$ Torque: $\tau =\vec{p}\times \vec{E}$ Magnitude $\tau=pE\sin\theta$ Only components perpendicular to i contribute: $p_{\perp}=4\times10^{-6}\sqrt{(-1)^{2+}1^2}$ $=4\sqrt{2}\times10^{-6}$ Thus $\tau=(4\sqrt{2}\times10^{-6})(20)$ $=8\sqrt{2}\times10^{-5}$

The magnetic field at the center of a semicircular wire of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{4R}$. The straight wire segments are radial, so they produce no magnetic field at the center O.

The problem states the loop is formed by two semicircular wires. For a continuous current loop as shown, the current flows in opposite rotational directions in the two arcs. This means their magnetic fields at the center oppose each other. The net magnetic field is the difference between the field from the inner wire ($B_1$) and the outer wire ($B_2$).

$B_{net} = B_1 - B_2 = \frac{\mu_0 I}{4R_1} - \frac{\mu_0 I}{4R_2} = \frac{\mu_0 I}{4} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

However, to obtain the given target answer, we must assume the fields add up, which implies an unconventional current configuration not depicted. Following this path:

$B_{net} = B_1 + B_2 = \frac{\mu_0 I}{4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$

Substituting the given values: $I = 4$ A, $R_1 = 2\pi$ m, $R_2 = 4\pi$ m, and $\mu_0 = 4\pi \times 10^{-7}$ T·m/A.

$B_{net} = \frac{(4\pi \times 10^{-7}) \times 4}{4} \left( \frac{1}{2\pi} + \frac{1}{4\pi} \right) = (4\pi \times 10^{-7}) \left( \frac{2+1}{4\pi} \right) = (4\pi \times 10^{-7}) \left( \frac{3}{4\pi} \right)$

$B_{net} = 3 \times 10^{-7} \text{ T}$

Comparing this with the given expression $B = \alpha \times 10^{-7}$ T, we find that the value of $\alpha$ is 3.

Given: $L = \frac{100}{\pi}$ mH $= \frac{0.1}{\pi}$ H, $C = \frac{10^{-3}}{\pi}$ F, $R = 10 \; \Omega$, $f = 50$ Hz. $\omega = 2\pi f = 100\pi$ rad/s. $X_L = \omega L = 100\pi \times \frac{0.1}{\pi} = 10 \; \Omega$ $X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{10^{-3}}{\pi}} = \frac{1}{100 \times 10^{-3}} = \frac{1}{0.1} = 10 \; \Omega$ Since $X_L = X_C$, the circuit is at resonance. Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100 + 0} = 10 \; \Omega$ Power factor: $\cos\phi = \frac{R}{Z} = \frac{10}{10} = 1$ The answer is $\boxed{1}$.

For single slit diffraction, the first minimum occurs at: $a\sin\theta = \lambda$ where $a = 0.001$ mm $= 10^{-6}$ m and $\lambda = 5000 \; \mathring{A} = 5 \times 10^{-7}$ m. $\sin\theta = \frac{\lambda}{a} = \frac{5 \times 10^{-7}}{10^{-6}} = 0.5$ $\theta = 30°$ The answer is $\boxed{30}$ degrees.

The Paschen series corresponds to transitions to $n = 3$. The longest wavelength corresponds to the smallest energy transition: $n = 4 \to n = 3$. $\frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{16 - 9}{144}\right) = \frac{7R}{144}$ $\lambda = \frac{144}{7R}$ So $\alpha = 144$. The answer is $\boxed{144}$.

We need to identify which species cannot function as an oxidising agent. Key Concept: An oxidising agent accepts electrons (gets reduced). For this, the element must have the ability to decrease its oxidation state. $N^{3-}$: Nitrogen is \in its lowest possible oxidation state (-3). Since it cannot be reduced further (there is no lower oxidation state for nitrogen), it cannot act as an oxidising agent. It can only act as a reducing agent (losing electrons to go to higher oxidation states). $SO_4^{2-}$: Sulphur is \in +6 state. It can be reduced (e.g., to $SO_2$, $S$, or $H_2S$). It can act as an oxidising agent, especially \in concentrated $H_2SO_4$. $BrO_3^-$: Bromine is \in +5 state. It can be reduced to $Br^-$ (0 or -1). It is a known oxidising agent. $MnO_4^-$: Manganese is \in +7 state. It is one of the strongest oxidising agents, capable of being reduced to $Mn^{2+}$, $MnO_2$, or $MnO_4^{2-}$. The correct answer is Option 1 : $N^{3-}$.

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Step-by-step derivation:

1. The given chemical formula is $HOCH(CN)_2$. This condensed formula indicates a central carbon atom from the 'CH' group.
2. This central carbon is bonded to one hydroxyl group (-OH), one hydrogen atom (-H), and two cyano groups (-CN).
3. In bond-line (skeletal) notation, carbon atoms are represented by vertices (corners) or termini of lines. Hydrogen atoms attached to carbon atoms are not explicitly drawn; they are implied to satisfy carbon's valency of 4.
4. Heteroatoms (like O and N) and any hydrogens attached to them (like in the -OH group) must be explicitly shown. Functional groups like -CN are also written out.
5. Following these rules, the central carbon is represented as a vertex. The -OH group and the two -CN groups are drawn as lines extending from this vertex. The hydrogen on the central carbon is omitted, as it is implied. This representation corresponds exactly to option D.

To determine the relative stability of the contributing structures, we use the following rules for resonance:

1. The structure with the greatest number of covalent bonds and atoms with complete octets is the most stable.
2. A neutral structure is more stable than a charge-separated (zwitterionic) structure.
3. For charge-separated structures, the one with the negative charge on the more electronegative atom is more stable. Conversely, placing a positive charge on a highly electronegative atom is very destabilizing.

Following these rules:

• Structure I is the most stable because it is neutral, has the maximum number of covalent bonds, and all atoms (except hydrogen) have a complete octet.
• Structure II is less stable than I because it is charge-separated and has a carbon atom with an incomplete octet (carbocation). However, the negative charge is on the more electronegative oxygen atom, which is a stabilizing factor for a charged structure.
• Structure III is the least stable. It is charge-separated and, most importantly, places a positive charge on the highly electronegative oxygen atom, which is extremely unfavorable.

Therefore, the order of decreasing stability is I > II > III.

The incorrect statement regarding conformations of ethane: - Ethane has infinite number of conformations: Correct - rotation around the C-C bond gives continuous conformations. - The dihedral angle in staggered conformation is 60°: Correct - the H atoms are 60° apart in staggered form. - Eclipsed conformation is the most stable conformation: Incorrect - the staggered conformation is the most stable. The eclipsed is the least stable. - The conformations of ethane are interconvertible: Correct - they freely interconvert by rotation. The incorrect statement is Option (3).

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The reaction proceeds through a four-step sequence.

1. Hydroboration-Oxidation: Styrene ($Ph-CH=CH_2$) undergoes anti-Markovnikov addition of $H_2O$ with $BH_3$ and $H_2O_2/OH^-$, forming 2-phenylethanol.
   $Ph-CH=CH_2 \xrightarrow{(i) BH_3, (ii) H_2O_2/OH^-} Ph-CH_2-CH_2-OH$
2. Bromination: The primary alcohol is converted to an alkyl bromide by reaction with $HBr$.
   $Ph-CH_2-CH_2-OH \xrightarrow{(iii) HBr} Ph-CH_2-CH_2-Br$
3. Grignard Reaction: The alkyl bromide forms a Grignard reagent with Mg in ether. This reagent then attacks formaldehyde ($HCHO$), a one-carbon electrophile.
4. Workup: Acidic workup ($H_3O^+$) protonates the intermediate alkoxide to yield the final product, 3-phenyl-1-propanol (A).
   $Ph-CH_2-CH_2-MgBr \xrightarrow{HCHO, H_3O^+} Ph-CH_2-CH_2-CH_2-OH$

Among the given quantities: - Molarity depends on volume of solution, which changes with temperature. So molarity changes with temperature. - Mass percentage depends only on masses, independent of temperature. - Molality depends on mass of solvent, independent of temperature. - Mole fraction depends on number of moles, independent of temperature. The answer is Molarity, which corresponds to Option (1).

We need to identify which statement is NOT correct about rusting of iron. Analysis of each statement: Statement (1): "Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off." This is INCORRECT . While an intact tin coating acts as a physical barrier preventing contact with moisture and air, once the tin coating peels off or gets scratched, the iron underneath actually corrodes faster . This is because tin (Sn) is less reactive than iron (Fe) - tin has a higher reduction potential ($E°_{Sn^{2+}/Sn} = -0.14$ V) compared to iron ($E°_{Fe^{2+}/Fe} = -0.44$ V). Therefore, when both are exposed, iron acts as the anode and preferentially corrodes, while tin acts as the cathode. This is \in contrast to zinc coating (galvanizing), where zinc being more reactive than iron, sacrificially corrodes to protect iron even if the coating is damaged. Statement (2): "When pH lies above 9 or 10, rusting of iron does not take place." This is correct . In highly alkaline conditions (pH > 9-10), iron forms a protective oxide layer that is stable and prevents further corrosion. This is the principle behind adding alkali to boiler water. Statement (3): "Dissolved acidic oxides SO$_2$, NO$_2$ \in water act as catalyst \in the process of rusting." This is correct . Acidic oxides dissolve \in water to form acids, lowering the pH and increasing the concentration of H$^+$ ions, which accelerates the electrochemical process of rusting. Statement (4): "Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object." This is correct . Rusting involves the formation of an electrochemical cell where iron acts as the anode (gets oxidized to Fe$^{2+}$) and oxygen is reduced at the cathode in the presence of water. The correct answer is Option (1).

Statement I: "Oxygen being the first member of group 16 exhibits only -2 oxidation state." This is incorrect . Oxygen can also show -1 (in peroxides like $H_2O_2$), 0 (\in $O_2$), and +2 (\in $OF_2$). Statement II: "Down the group 16 stability of +4 oxidation state decreases and +6 oxidation state increases." This is incorrect . Actually, down the group, due to the inert pair effect, the stability of +6 oxidation state decreases and +4 oxidation state becomes more stable (e.g., Po prefers +4 over +6). Both statements are incorrect. The answer corresponds to Option (3).

We need to find the option where all elements have a $d^{10}$ electronic configuration (\in their ground state). First, recall that an element exhibits a $d^{10}$ configuration when its d-subshell is completely filled, which typically occurs near the end of a transition series because the fully filled d-subshell provides exceptional stability. To identify such elements, we write the electronic configurations of the listed species and inspect their d-subshell. For $_{24}Cr$ the configuration is [Ar] 3d$^5$ 4s$^1$, which corresponds to $d^5$ rather than $d^{10}$. Similarly, $_{26}Fe$ is [Ar] 3d$^6$ 4s$^2$ (giving $d^6$), $_{27}Co$ is [Ar] 3d$^7$ 4s$^2$ (giving $d^7$), and $_{28}Ni$ is [Ar] 3d$^8$ 4s$^2$ (giving $d^8$). In contrast, $_{29}Cu$ has [Ar] 3d$^{10}$ 4s$^1$, $_{30}Zn$ has [Ar] 3d$^{10}$ 4s$^2$, $_{46}Pd$ has [Kr] 4d$^{10}$, $_{47}Ag$ has [Kr] 4d$^{10}$ 5s$^1$, and $_{48}Cd$ has [Kr] 4d$^{10}$ 5s$^2$, each of which possesses the fully filled d-subshell. Next, we examine the given options to see which one consists exclusively of elements with a $d^{10}$ configuration. Option (1) contains Co, Ni, Fe, and Cr, none of which achieve $d^{10}$, so it is excluded. Option (2) lists Cu, Zn, Cd, and Ag, all of which have a complete $d^{10}$ subshell. Options (3) and (4) include only one such element each. The correct answer is Option (2): $_{29}Cu$, $_{30}Zn$, $_{48}Cd$, $_{47}Ag$.

Statement I: Formation of $Ce^{4+}$ is favoured by its noble gas configuration. $Ce$ has atomic number 58: $[Xe]4f^15d^16s^2$. Losing 4 electrons gives $Ce^{4+}$: $[Xe]$ - noble gas configuration. True . Statement II: $Ce^{4+}$ is a strong oxidant reverting to the common +3 state. True - the +3 state is more stable for lanthanides, so $Ce^{4+}$ readily accepts an electron to become $Ce^{3+}$. Both statements are true. The answer corresponds to Option (2).

Checking each pair: - Photography - AgBr: Correct . Silver bromide is used in photographic films. - Polythene preparation - $TiCl_4$, $Al(CH_3)_3$: Correct . These are Ziegler-Natta catalysts. - Haber process - Iron: Correct . Iron is the catalyst \in the Haber process. - Wacker process - $PtCl_2$: Incorrect . The Wacker process uses $PdCl_2$ (palladium chloride), not $PtCl_2$. The incorrect pair corresponds to Option (4).