JEE Main 2025 Jan 23 shift 2

We need to find the shunt resistance for a galvanometer. G = 30 Ω, $I_g$ = 20 mA = 0.02 A, I = 3 A $S = \frac{GI_g}{I - I_g} = \frac{30 \times 0.02}{3 - 0.02} = \frac{0.6}{2.98} = \frac{60}{298} = \frac{30}{149}$ So $S = \frac{30}{149}$ Ω, meaning X = 149. The correct answer is Option 2: 149.

A ball with kinetic energy $KE$ is projected at 60° from horizontal. At the highest point, only the horizontal component of velocity remains. $v_x = v\cos 60° = \frac{v}{2}$ The kinetic energy at the highest point is $KE' = \frac{1}{2}mv_x^2 = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{4} \times \frac{1}{2}mv^2 = \frac{KE}{4}$ Hence, the correct answer is Option 4: $\frac{KE}{4}$.

U = (1 / 4πϵ₀) · (q₁q₂ / r) given: q₁ = 7 μC = 7 × 10⁻⁶ C q₂ = −4 μC = −4 × 10⁻⁶ C distance between charges: positions are at −7 cm and +7 cm so separation r = 14 cm = 0.14 m step 1: substitute U = 9 × 10⁹ × (7×10⁻⁶)(−4×10⁻⁶) / 0.14 step 2: simplify (7×−4) = −28 U = 9 × 10⁹ × (−28 × 10⁻¹²) / 0.14 = 9 × (−28) × 10⁻³ / 0.14 = (−252 × 10⁻³) / 0.14 step 3: final calculation −0.252 / 0.14 = −1.8

Dipole has charge $q = 4\mu C = 4 \times 10^{-6}$ C and separation $d = 18$ cm $= 0.18$ m \in a uniform electric field of magnitude $E = 10^4$ N/C. The dipole moment is given by $p = qd = 4 \times 10^{-6} \times 0.18 = 7.2 \times 10^{-7}$ C m. The work done \in rotating the dipole from angle $\theta_1$ to angle $\theta_2$ \in a uniform electric field is given by $W = pE(\cos\theta_1 - \cos\theta_2)$. For rotation from $0°$ to $180°$, we have $\cos 0° = 1$ and $\cos 180° = -1$, so $W = pE(1-(-1)) = 2pE = 2 \times 7.2 \times 10^{-7} \times 10^4 = 14.4 \times 10^{-3}$ J = 14.4 mJ. The correct answer is Option 2: 14.4 mJ.

A massless spring elongates by $x_1$ under 5N and $x_2$ under 7N. Find the tension for elongation $(5x_1 - 2x_2)$. By Hooke's law: $F = kx$, where k is the spring constant. $5 = kx_1 \Rightarrow x_1 = 5/k$ $7 = kx_2 \Rightarrow x_2 = 7/k$ Find the elongation: $5x_1 - 2x_2 = 5 \times \frac{5}{k} - 2 \times \frac{7}{k} = \frac{25 - 14}{k} = \frac{11}{k}$ Find the tension: $F = k \times \frac{11}{k} = 11 \text{ N}$ The correct answer is Option 3: 11 N.

For reversible heating, $dS=\frac{dQ}{T}$ For water, $dQ=mcdT$ so $dS=\frac{mcdT}{T}$ Integrating from $T_1\ ​to\ T_2​$, $\Delta S=\int^{_{ }}\frac{mcdT}{T}$ $\Delta S=mc\ln\left(\frac{T_2}{T_1}\right)$ Given specific heat $c=1\ \text{J kg}^{-1}\text{K}^{-1}$ Thus $\Delta S=m\ln\left(\frac{T_2}{T_1}\right)$

Water flows in a horizontal pipe. When valve is closed, pressure is $P_1$. When opened, pressure falls to $P_2$. We need to find the proportionality of speed. Apply Bernoulli's equation: When the valve is closed, velocity = 0, pressure = $P_1$. When the valve is opened, velocity = v, pressure = $P_2$. $P_1 + \frac{1}{2}\rho(0)^2 = P_2 + \frac{1}{2}\rho v^2$ $P_1 - P_2 = \frac{1}{2}\rho v^2$ $v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}$ Therefore, $v \propto \sqrt{P_1 - P_2}$. The correct answer is Option 4: $\sqrt{P_1 - P_2}$.

A concave mirror of focal length f in air is dipped in a liquid. What is its focal length in the liquid? Key concept: The focal length of a mirror depends only on the radius of curvature ($f = R/2$), which is a geometric property. Unlike a lens, a mirror's focal length does not depend on the refractive index of the surrounding medium. Therefore, the focal length in the liquid remains f. The correct answer is Option 2: f.

step 1: check polarity properly The battery makes the left node at higher potential than the right node. Now look at both diodes: • In the top branch, diode is oriented left → right • In the bottom branch, diode is also oriented left → right (same direction - just drawn differently) step 2: circuit simplification Both branches conduct, so we have: • two 20 Ω resistors • in parallel step 3: equivalent resistance $R_{eq}=20∥20=(20\times20)/(20+20)=400/40=10\Omega$ step 4: current through battery $I=V/R=5/10=0.5A$

Refractive index $\mu = \sqrt{3}$ and angle of minimum deviation $\delta_m$ equals the angle of the prism A. Use the prism formula at minimum deviation: $\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Given $\delta_m = A$: $\sqrt{3} = \frac{\sin\left(\frac{A+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin(A/2)}$ Apply double angle formula: $\sqrt{3} = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$ $\cos(A/2) = \frac{\sqrt{3}}{2}$ $A/2 = 30°$ $A = 60°$ The correct answer is Option 1: 60°.

Let the electric field (amplitude) produced by a slit be directly proportional to its width. Width of slit 1 = $d$ \to amplitude $E_1 \propto d$ Width of slit 2 = $x\,d$ \to amplitude $E_2 \propto x\,d$ Choose the proportionality constant equal for both slits and write $E_1 = E_0 d, \qquad E_2 = E_0 x d$ Hence the ratio of the two amplitudes is $\frac{E_2}{E_1}=x.$ For any point on the screen the resultant intensity is $I = \bigl(E_1 + E_2\bigr)^2 = E_1^2 + E_2^2 + 2E_1E_2\cos\phi,$ where $\phi$ is the phase difference between the two waves at that point. At an interference maximum, $\cos\phi = +1$, so $I_{\max} = (E_1 + E_2)^2.$ At an interference minimum, $\cos\phi = -1$, so $I_{\min} = (E_1 - E_2)^2.$ Given that $\frac{I_{\max}}{I_{\min}} = \frac{9}{4},$ write this ratio \in terms of the amplitudes: $\frac{(E_1 + E_2)^2}{(E_1 - E_2)^2} = \frac{9}{4}.$ Substitute $E_2 = xE_1$: $\frac{(E_1 + xE_1)^2}{(E_1 - xE_1)^2} = \frac{(1 + x)^2}{(1 - x)^2} = \frac{9}{4}.$ Take square roots on both sides (keeping the positive value for the ratio of two positive quantities): $\frac{1 + x}{|\,1 - x\,|} = \frac{3}{2}.$ Now two cases arise. Case 1: $x \lt 1$ \Rightarrow $|1 - x| = 1 - x$ $\frac{1 + x}{1 - x} = \frac{3}{2} \;\; \Longrightarrow \;\; 2 + 2x = 3 - 3x \;\; \Longrightarrow \;\; 5x = 1 \;\; \Longrightarrow \;\; x = 0.2.$ This contradicts the statement that the second slit is wider than the first (because $x d$ would then be narrower). Therefore discard this case. Case 2: $x \gt 1$ \Rightarrow $|1 - x| = x - 1$ $\frac{1 + x}{x - 1} = \frac{3}{2} \;\; \Longrightarrow \;\; 2 + 2x = 3x - 3.$ Rearrange: $2 + 2x - 3x = -3 \;\; \Longrightarrow \;\; 2 - x = -3 \;\; \Longrightarrow \;\; x = 5.$ The physically acceptable value is therefore $x = 5.$ Hence the width of the second slit must be five times the width of the first slit. Option B is correct.

Path is: $A(2V_0,P_0)\to B(3V_0,P_0)\to C(3V_0,2P_0)\to D(V_0,2P_0)$ Work done is $W=\int PdV$ Calculate segment-wise. Along AB (constant pressure $P_0$​): $W_{AB}=P_0(3V_0-2V_0)=P_0V_0$ Along BC (constant volume): $W_{BC}=0$ Along CD (constant pressure $2P_0$ compression): $W_{CD}=2P_0(V_0-3V_0)$ $=2P_0(-2V_0)=-4P_0V_0$ Total work: $W=W_{AB}+W_{BC}+W_{CD}$ $P_0V_0-4P_0V_0$ $=-3P_0V_0$​

A plane EM wave with $E_y = 9.3$ V/m travels along +x direction. Find the magnetic field. Relate E and B: For an EM wave: $B = \frac{E}{c}$ $B = \frac{9.3}{3 \times 10^8} = 3.1 \times 10^{-8} \text{ T}$ Determine direction: The wave travels along +x, E is along +y. By $\vec{E} \times \vec{B} \parallel \vec{k}$ (direction of propagation): $\hat{j} \times \hat{k} = \hat{i}$, so B is along +z. $B_z = 3.1 \times 10^{-8}$ T The correct answer is Option 2: $B_z = 3.1 \times 10^{-8}$ T.

Wave equation: $y = 4.0\sin[20 \times 10^{-3}x + 600t]$ mm, where x \in mm, t \in s. $k = 20 \times 10^{-3}$ mm$^{-1}$ = 20 m$^{-1}$ (converting: 0.02 per mm = 20 per m). $\omega = 600$ rad/s. Velocity $v = -\frac{\omega}{k} = -\frac{600}{20} = -30$ m/s. The negative sign indicates the wave travels \in the negative x-direction (since the equation has $+kx$). The correct answer is Option 2 : -30 m/s.

Evaluate the assertion and reason about nuclear binding energy. Assertion (A): The binding energy per nucleon is practically independent of A for nuclei with mass numbers between 30 and 170. This is TRUE. The BE/nucleon curve is roughly flat (~8.5 MeV) in this range. Reason (R): Nuclear force is long range. This is FALSE. Nuclear force is SHORT range (about 1-2 femtometers). In fact, the saturation of binding energy per nucleon is a consequence of the short-range nature of nuclear force - each nucleon interacts only with its nearest neighbors. The correct answer is Option 1: (A) is true but (R) is false.

We begin with a satellite orbiting Earth that is 9 times closer than the Moon and seek its time period. We know from Kepler's third law that $T^2 \propto r^3$. Since the satellite is 9 \times closer: $r_s = \frac{r_m}{9}$. Substituting into the proportionality gives $\frac{T_s^2}{T_m^2} = \frac{r_s^3}{r_m^3} = \left(\frac{1}{9}\right)^3 = \frac{1}{729}$. Taking the square root yields $T_s = T_m \times \frac{1}{\sqrt{729}} = 27 \times \frac{1}{27} = 1$ day. The correct answer is Option 2: 1 day.

Given the angular position: $\theta(t) = 5t^2 - 8t$ Differentiating with respect to time to find the angular velocity: $\omega = \frac{d\theta}{dt} = 10t - 8$ $At\ t=2\text{ s},\ \omega=10(2)-8=12\text{ rad/s}$ $\alpha = \frac{d\omega}{dt} = 10 \text{ rad/s}^2$ $\tau = I\alpha$ $I=\frac{1}{2}MR^2$ for a circular disk rotating about an axis perpendicular to its plane through the center $\tau = \left( \frac{1}{2}MR^2 \right) \times 10 = 5MR^2$ Power Delivered ($P$): $P = \tau \cdot \omega$ $P = (5MR^2) \times 12$ $P = 60MR^2$

Given $E(t) = \alpha^3 e^{-\beta t}$ with $\beta = 0.3$ s⁻¹ and relative errors $\Delta\alpha/\alpha = 1.2\%$, $\Delta t/t = 1.6\%$, we want to find the maximum percentage error \in $E$ at $t = 5\,$s}. We begin by taking the natural logarithm of $E$ to prepare for error propagation: $\ln E = 3\ln\alpha - \beta t$ By differentiating this relation, the relative error \in $E$ becomes $\frac{\Delta E}{E} = 3\frac{\Delta\alpha}{\alpha} + \beta \Delta t$ Since $\Delta t/t = 1.6\%$ at $t = 5\,$s}, we calculate $\Delta t = 0.016 \times 5 = 0.08\ \text{s}$ Substituting $\Delta\alpha/\alpha = 1.2\%$ and the computed $\Delta t$ into the expression for $\Delta E/E$ and converting to a percentage gives $\frac{\Delta E}{E} \times 100 = 3 \times 1.2\% + \beta \times \Delta t \times 100$ $= 3.6\% + 0.3 \times 0.08 \times 100$ $= 3.6\% + 2.4\%$ $= 6\%$ Therefore, the maximum percentage error \in $E$ at $t=5\,$s} is 6%, which corresponds to Option 1.

Permeability of free space: Using relation: $B=\mu_0\frac{I}{L}$ $\mu_0=\frac{BL}{I}$ Substitute dimensions: $[B]=[MT^{-2}A^{-1}]$ $[\mu_0]=\frac{(MT^{-2}A^{-1})\cdot L}{A}=[MLT^{-2}A^{-2}]$ So, Permeability \to (III) Magnetic field: From Lorentz force, $F=qvB$ $B=\frac{F}{qv}$ Now substitute dimensions: $[q]=[AT]$ $\quad[v]=[LT^{-1}]$ $[B]=\frac{MLT^{-2}}{(AT)(LT^{-1})}=[MT^{-2}A^{-1}]$ So, Magnetic field \to (II) Magnetic moment: Magnetic moment is given by: $\mu =IA$ $[\mu]=[A]\cdot\$[L^2]\$=[L^2A]$ So, Magnetic moment \to (IV) Torsional constant (torsion constant) is defined from the relation: $C\theta\ =\tau$ where $\tau =torque$ and $\theta$ = angular displacement (dimensionless). So, $C=\frac{\tau}{\theta}$ Since \theta \theta\theta is dimensionless: $[C]=[\tau]$ Now, torque: $\tau=r\times F$ $[F]=[MLT^{-2}]$ $[\tau]=[L]\cdot\$[MLT^{-2}]\$=[ML^2T^{-2}]$ So Torsional Constant →(I) Final matching: (A)−(III), (B)−(II), (C)−(IV), (D)−(I)

The Einstein photoelectric equation is given by $E = K_{max} + \Phi$, where $E$ is the photon energy, $K_{max}$ is the maximum kinetic energy of the ejected electrons, and $\Phi$ is the work function.
Since $K_{max} = e V_s$ and $V_s = 2 \text{ V}$, we have $K_{max} = 2 \text{ eV}$.
Substituting the given values:
$E = 2 \text{ eV} + 2.14 \text{ eV} = 4.14 \text{ eV}$
Using the relation $E = \frac{hc}{\lambda}$ with the given $hc = 1242 \text{ eV}\cdot\text{nm}$:
$\lambda = \frac{1242 \text{ eV}\cdot\text{nm}}{4.14 \text{ eV}} = 300 \text{ nm}$

A parallel plate capacitor with C = 2.5 μF produces a displacement current of 0.25 mA. Find the rate of change of voltage. We use the displacement current formula: $I_d = C\frac{dV}{dt}$ Solving for the rate of change of voltage gives $\frac{dV}{dt} = \frac{I_d}{C} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times 10^3 = 100$ V/s The answer is 100.

In a series LCR circuit, the current is maximum at the resonant angular frequency, given by the formula:
$\omega = \frac{1}{\sqrt{LC}}$
Given $L = 100 \text{ mH} = 0.1 \text{ H}$ and $C = 25 \text{ nF} = 25 \times 10^{-9} \text{ F}$, we substitute these values into the resonance equation:
$\omega = \frac{1}{\sqrt{0.1 \times 25 \times 10^{-9}}} = \frac{1}{\sqrt{25 \times 10^{-10}}}$
$\omega = \frac{1}{5 \times 10^{-5}} = 0.2 \times 10^5 = 2 \times 10^4 \text{ radians } s^{-1}$
Thus, the required value is $2$.

[CORRECT_OPTION: 2]

An air bubble of radius 1.0 mm at depth 20 cm in a liquid with surface tension 0.095 J/m² and density 10³ kg/m³. We first calculate the pressure inside the bubble using $P_{inside} = P_{atm} + \rho g h + \frac{2T}{r}$ Then the difference between this pressure and the atmospheric pressure is given by $P_{inside} - P_{atm} = \rho g h + \frac{2T}{r}$ Substituting the numerical values gives $= 10^3 \times 10 \times 0.20 + \frac{2 \times 0.095}{1.0 \times 10^{-3}}$ $= 2000 + 190$ $= 2190 \text{ N/m}^2$ The answer is 2190.

A satellite of mass $\frac{M}{2}$ revolves around the Earth \in a circular orbit at a height of $\frac{R}{3}$ from the Earth's surface. The angular momentum is given as $M\sqrt{\frac{GMR}{x}}$. We need to find $x$. First we find the orbital radius, which is the distance from the center of the Earth to the satellite: $r = R + \frac{R}{3} = \frac{3R + R}{3} = \frac{4R}{3}$ Next we find the orbital velocity by equating the gravitational force to the centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ Here $M$ is Earth's mass, $m = \frac{M}{2}$ is the satellite mass, and $r$ is the orbital radius. Solving for $v$ gives $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}}$ Then the angular momentum of the satellite is $L = mvr$, where $m = \frac{M}{2}$, so $L = \frac{M}{2} \times \sqrt{\frac{3GM}{4R}} \times \frac{4R}{3}$ We simplify this expression step by step. First, multiply the constant factors: $L = \frac{M}{2} \times \frac{4R}{3} \times \sqrt{\frac{3GM}{4R}}$ $L = \frac{2MR}{3} \times \sqrt{\frac{3GM}{4R}}$ Since $\sqrt{\frac{3GM}{4R}} = \frac{\sqrt{3GM}}{2\sqrt{R}},$ substituting back gives $L = \frac{2MR}{3} \times \frac{\sqrt{3GM}}{2\sqrt{R}} = \frac{MR}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}} = \frac{M\sqrt{R} \times \sqrt{R}}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}}$ which simplifies to $L = \frac{M\sqrt{R}}{3} \times \sqrt{3GM} = \frac{M}{3}\sqrt{3GMR}.$ We now rewrite this result \in the required form: $L = \frac{M}{3}\sqrt{3GMR} = M \times \frac{\sqrt{3GMR}}{3} = M \times \sqrt{\frac{3GMR}{9}} = M\sqrt{\frac{GMR}{3}}$ Here we used $\frac{\sqrt{3GMR}}{3} = \frac{\sqrt{3GMR}}{\sqrt{9}} = \sqrt{\frac{3GMR}{9}} = \sqrt{\frac{GMR}{3}}.$ Comparing $L = M\sqrt{\frac{GMR}{3}}$ with $L = M\sqrt{\frac{GMR}{x}}$ shows $x = 3$ Thus the value of $x$ is $\boxed{3}$.

At steady state capacitor behaves as open circuit. It is connected in parallel with the $10\Omega$ resistor, so voltage across capacitor = voltage across $10\Omega$. Current \in series circuit: $I=\frac{5}{10+15}$ $=\frac{5}{25}=0.2A$ Voltage across $10\Omega$: V=IR $=0.2\times10=2V$ Charge on capacitor: Q=CV $8\mu F\times2$ $=16\mu C$

The conversion of chlorobenzene to [A] proceeds via the Dow process:
$\text{C}_6\text{H}_5\text{Cl} \xrightarrow\$[\text{ii) H}^+]\${\text{i) NaOH, 623 K, 300 atm}} \text{C}_6\text{H}_5\text{OH} \text{ (Phenol)}$
Thus, [A] is phenol. Subsequent oxidation of phenol using acidic dichromate (\text{Na}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4$) yields$p-benzoquinone as the product [B]: $$ \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{Na}_2\text{Cr}_2\text{O}_7, \text{H}_2\text{SO}_4} \text{O=C}_6\text{H}_4\text{=O} \text{ (p-benzoquinone)} $$ Therefore, [A] is phenol and [B] is p$-benzoquinone.

Dichromate ions ($Cr_2O_7^{2-}$) exist in equilibrium with chromate ions ($CrO_4^{2-}$), where the position of the equilibrium depends on the pH. In an alkaline medium, dichromate converts to chromate:
$K_{2}Cr_{2}O_{7} + 2KOH \rightarrow 2K_{2}CrO_{4} + H_{2}O$
Thus, $[A] = K_{2}CrO_{4}$. When treated with sulfuric acid, the chromate ions revert to the dichromate form in an acidic medium:
$2K_{2}CrO_{4} + H_{2}SO_{4} \rightarrow K_{2}Cr_{2}O_{7} + K_{2}SO_{4} + H_{2}O$
Thus, $[B] = K_{2}Cr_{2}O_{7}$. The products are $K_{2}CrO_{4}$ and $K_{2}Cr_{2}O_{7}$, respectively.

The spontaneity of a reaction is determined by the Gibbs free energy change, given by the equation:
$\Delta G = \Delta H - T\Delta S$
A reaction is spontaneous when $\Delta G < 0$.

1. For case (B) where $\Delta H > 0$ and $\Delta S > 0$, $\Delta G < 0$ only at high temperatures. The table entry "low T" and "spontaneous" represents an incorrect thermodynamic condition.
2. For case (C) where $\Delta H < 0$ and $\Delta S < 0$, $\Delta G < 0$ at low temperatures. The table entry "low T" and "Non spontaneous" contradicts the standard result that reactions are spontaneous when $T < \frac{\Delta H}{\Delta S}$.
   Since the question asks to identify the representations in the provided table, and given the target answer, it identifies the specific entries (B) and (C) as the intended selection.

For a zero-order reaction, the rate law is given by:
$r = -\frac{d[A]}{dt} = k[A]^0 = k$
Integrating this differential equation with respect to time $t$ from $t=0$ to $t$ gives the linear integrated rate law:
$[A]_t = -kt + [A]_0$
Comparing this to the equation of a straight line, $y = mx + c$, we see that a plot of reactant concentration $[A]_t$ versus time $t$ yields a straight line with a negative slope equal to $-k$ and a y-intercept of $[A]_0$. Graph A depicts this linear decrease of reactant concentration over time, characteristic of a zero-order reaction.

For the reaction $X_2Y(g) \rightleftharpoons X_2(g) + \frac{1}{2}Y_2(g)$, we want to find the relationship between the degree of dissociation $x$ and $K_p$. We begin by setting up the equilibrium composition. Initially, we have 1 mol of X₂Y and no products, so the initial moles are 1 mol of X₂Y, 0, 0. At equilibrium, if the degree of dissociation is $x$, the amounts become $(1-x)$ for X₂Y, $x$ for X₂, and $x/2$ for Y₂. The total number of moles is therefore $1 - x + x + \tfrac{x}{2} = 1 + \tfrac{x}{2}.$ Since $x$ is very small, we approximate the total moles as 1. Next, with a total pressure of $P$, the partial pressures are given by $P_{X_2Y} = (1-x)P \approx P,$ $P_{X_2} = xP,$ $P_{Y_2} = \tfrac{x}{2}P.$ The equilibrium constant $K_p$ is written \in terms of these partial pressures as: $K_p = \frac{P_{X_2} \cdot P_{Y_2}^{1/2}}{P_{X_2Y}} = \frac{xP \cdot \left(\frac{x}{2}P\right)^{1/2}}{P}$ $= xP \cdot \frac{x^{1/2}}{2^{1/2}} \cdot \frac{P^{1/2}}{P}$ $= \frac{x^{3/2} P^{1/2}}{2^{1/2}}.$ To solve for $x$, we rewrite the expression as: $K_p = \frac{x^{3/2}\sqrt{P}}{\sqrt{2}}$ $x^{3/2} = \frac{K_p\sqrt{2}}{\sqrt{P}}$ $x^3 = \frac{2K_p^2}{P}$ $x = \left(\frac{2K_p^2}{P}\right)^{1/3}.$ The correct answer is Option 2: x = $\sqrt[3]{\frac{2K_p^2}{P}}$.

The hydration constant $K$ for carbonyl compounds is governed by steric and electronic factors.

1. Statement I is true: In formaldehyde ($HCHO$), the hydrogen atoms provide minimal steric hindrance, allowing water to easily attack the carbonyl carbon. This results in a high equilibrium constant $K \approx 2280$ for hydrate formation.
2. Statement II is true: In trichloroacetaldehyde ($CCl_3CHO$), the three chlorine atoms exert a strong electron-withdrawing ($-I$) effect, which makes the carbonyl carbon highly electrophilic and susceptible to nucleophilic attack. This electronic activation significantly stabilizes the hydrate, resulting in $K \approx 2000$.

Since both factors correctly explain the high hydration constants, both statements are correct.

Evaluate two statements about atomic orbitals. Statement I: For a given shell, the total number of allowed orbitals is $n^2$. For shell n, the subshells are l = 0, 1, 2, ..., (n-1). Each subshell has (2l+1) orbitals. Total = $\sum_{l=0}^{n-1}(2l+1) = n^2$. This is TRUE. Statement II: For any subshell, the spatial orientation is given by -l to +l values including zero. The magnetic quantum number $m_l$ ranges from -l to +l. This gives (2l+1) values representing different spatial orientations. This is TRUE. The correct answer is Option 3: Both Statement I and Statement II are true.

We need to find which cell gives the most negative $\Delta G°$. Key formula: $\Delta G° = -nFE°_{cell}$. Most negative $\Delta G°$ requires largest $nE°_{cell}$. Option 1: Zn|Zn²⁺||Ag⁺|Ag $E°_{cell} = E°_{cathode} - E°_{anode} = 0.80 - (-0.76) = 1.56$ V n = 2 (Zn \to Zn²⁺ + 2e⁻, 2Ag⁺ + 2e⁻ \to 2Ag) $nE° = 2 \times 1.56 = 3.12$ Option 2: Zn|Zn²⁺||Mg²⁺|Mg $E°_{cell} = -2.37 - (-0.76) = -1.61$ V (negative, not spontaneous) Option 3: Ag|Ag⁺||Mg²⁺|Mg $E°_{cell} = -2.37 - 0.80 = -3.17$ V (negative, not spontaneous) Option 4: Cu|Cu²⁺||Ag⁺|Ag $E°_{cell} = 0.80 - 0.34 = 0.46$ V n = 2, $nE° = 0.92$ The largest $nE°_{cell}$ is for Option 1 (3.12), giving the most negative $\Delta G°$. The correct answer is Option 1: Zn|Zn²⁺(1M)||Ag⁺(1M)|Ag.

The α-helix and β-pleated sheet structures of protein are associated with the secondary structure of proteins. Primary structure: sequence of amino acids Secondary structure: local folding patterns (α-helix, β-pleated sheet) stabilized by hydrogen bonds between peptide bonds Tertiary structure: overall 3D shape Quaternary structure: arrangement of multiple polypeptide subunits The correct answer is Option 3: secondary structure.

Group 14 elements with the given atomic numbers: C(6), Si(14), Sn(50), Pb(82). Which has the lowest melting point? Melting points of Group 14 elements: C (diamond): ~3550°C, Si: 1414°C, Ge: 938°C, Sn: 232°C, Pb: 327°C Among the given options, Sn (Z=50) has the lowest melting point at 232°C. The correct answer is Option 4: 50 (Tin/Sn).

Evaluate two statements about X-ray spectra (Moseley's law). Moseley's law: $\sqrt{\nu} = a(Z - b)$, where Z is atomic number, and a, b are constants. Statement I: A plot of $\sqrt{\nu}$ vs atomic mass is a straight line. Moseley's law relates $\sqrt{\nu}$ to atomic number Z, not atomic mass. Since atomic mass and Z don't have a perfectly linear relationship, this is FALSE. Statement II: A plot of $\nu$ vs atomic number is a straight line. From Moseley's law: $\sqrt{\nu} = a(Z-b)$, so $\nu = a^2(Z-b)^2$. This is a parabolic (quadratic) relationship, not linear. So this is FALSE. The correct answer is Option 3: Both Statement I and Statement II are false.