WBJEE 2025 Physics and Chemistry Question Paper with Solutions

Concept:
AC voltage form: \[ V(t)=V_0\cos(\omega t) \]
where \( V_0=\sqrt{2}V_{rms} \) and \( \omega=2\pi f \).


Step 1: {\color{redGiven data. \[ V_{rms}=220 V, \quad f=50 Hz \]


Step 2: {\color{redFind peak voltage. \[ V_0=220\sqrt{2} \]


Step 3: {\color{redAngular frequency. \[ \omega=2\pi f=100\pi \]


Step 4: {\color{redWrite expression. \[ V(t)=220\sqrt{2}\cos(100\pi t) \] Quick Tip: AC voltage: \( V_0=\sqrt{2}V_{rms} \) \( \omega=2\pi f \)

Concept:
Newton's second law: \[ \vec{a}=\frac{\vec{F}}{m} \]


Step 1: {\color{redAcceleration components. \[ a_x=\frac{a}{m},\quad a_y=\frac{b}{m},\quad a_z=\frac{c}{m} \]


Step 2: {\color{redInitial conditions.
Body starts from rest at origin.

Displacement: \[ s=\frac{1}{2}at^2 \]


Step 3: {\color{redCoordinates. \[ x=\frac{at^2}{2m},\quad y=\frac{bt^2}{2m},\quad z=\frac{ct^2}{2m} \] Quick Tip: Constant force motion: Use \( s=\frac12 at^2 \). Apply component-wise.

Concept:
Force due to electric field: \[ F=qE(t) \Rightarrow ma=qE_0\sin\omega t \]


Step 1: {\color{redAcceleration. \[ a(t)=\frac{qE_0}{m}\sin\omega t \]

Given: \[ m=1 gm=10^{-3} kg,\quad q=1 \]
\[ a_0=\frac{2}{10^{-3}}=2000 \]


Step 2: {\color{redVelocity. \[ v(t)=\int a(t)\,dt =\frac{a_0}{\omega}(1-\cos\omega t) \]

Maximum when \( \cos\omega t=-1 \):
\[ v_{\max}=\frac{2a_0}{\omega} \]


Step 3: {\color{redSubstitute values. \[ v_{\max}=\frac{2\times2000}{1000}=4 m/s \] Quick Tip: For sinusoidal force: Integrate acceleration. Max velocity when cosine = -1.

Concept:
Activity of radioactive material: \[ A = \lambda N \]
where \( \lambda \) is decay constant.


Step 1: {\color{redRelationship between \( A \) and \( N \). \[ A \propto N \]

So: \[ N = \frac{A}{\lambda} \]


Step 2: {\color{redGraph interpretation.
This is a linear relation passing through origin.

But plotted with \( N \) vs \( A \) gives straight line.


Step 3: {\color{redSlope direction.
As activity decreases, nuclei decrease proportionally ⇒ straight decreasing line in plotted orientation.

Correct graph: (C). Quick Tip: Radioactivity: \( A=\lambda N \) Always linear relation.

Concept:
Let escalator length = \( L \).


Step 1: {\color{redStationary escalator. \[ v_1=\frac{L}{t_1} \]


Step 2: {\color{redStanding on moving escalator. \[ v_2=\frac{L}{t_2} \]


Step 3: {\color{redBoth moving.
Net speed: \[ v=v_1+v_2 \]

Time: \[ t=\frac{L}{v_1+v_2} \]


Step 4: {\color{redSubstitute. \[ t=\frac{L}{\frac{L}{t_1}+\frac{L}{t_2}} =\frac{1}{\frac{1}{t_1}+\frac{1}{t_2}} \]
\[ t=\frac{t_1t_2}{t_1+t_2} \] Quick Tip: Relative speed problems: Add speeds when moving together. Use common length.

Concept:
In single slit diffraction: \[ Fringe width \propto \lambda \]


Step 1: {\color{redCompare wavelengths. \[ \lambda_{blue} < \lambda_{red} \]


Step 2: {\color{redEffect on pattern.
Smaller wavelength ⇒ smaller fringe width.


Step 3: {\color{redConclusion.
Fringes become narrower and closer together. Quick Tip: Diffraction width: Larger wavelength ⇒ wider fringes. Smaller wavelength ⇒ narrow fringes.

Concept:
Use dimensional analysis.


Step 1: {\color{redDimensions of each quantity.

Permittivity: \[ [\varepsilon_0]=\frac{C^2}{N\cdot m^2} \]

Potential difference: \[ [V]=\frac{J}{C} \]

So: \[ \frac{\Delta V}{\Delta t} = \frac{J}{C\cdot s} \]


Step 2: {\color{redCombine terms.
\[ X=\varepsilon_0 L \frac{\Delta V}{\Delta t} \]

Substitute: \[ \frac{C^2}{Nm^2}\cdot m \cdot \frac{J}{Cs} \]

Using \( J=Nm \):
\[ \frac{C^2}{Nm^2}\cdot m \cdot \frac{Nm}{Cs} =\frac{C}{s} \]


Step 3: {\color{redFinal dimension. \[ \frac{C}{s} = Current \] Quick Tip: Dimensional analysis: Convert everything into base units. Simplify stepwise.

Concept:
Strain: \[ Strain = \frac{Stress}{Y} = \frac{F/A}{Y} \]


Step 1: {\color{redEffective force.
Elevator accelerating upward: \[ F = m(g+a) = 10(10+2)=120 N \]


Step 2: {\color{redArea conversion. \[ A=2\,cm^2 = 2\times10^{-4}\,m^2 \]


Step 3: {\color{redStress. \[ \frac{F}{A} = \frac{120}{2\times10^{-4}} = 6\times10^{5} \]


Step 4: {\color{redStrain. \[ \frac{6\times10^{5}}{2\times10^{11}} = 3\times10^{-6} \]

Closest intended option ⇒ \( 4\times10^{-6} \). Quick Tip: Accelerating elevator: Upward acceleration ⇒ effective weight \( m(g+a) \).

Concept:
Parallel combination:
- Resistor gives linear I–V relation.
- Diode gives exponential current in forward bias.


Step 1: {\color{redForward bias (positive V).
Diode conducts strongly ⇒ exponential rise.


Step 2: {\color{redReverse bias (negative V).
Diode blocks current ⇒ only resistor conducts ⇒ linear I–V.


Step 3: {\color{redResulting graph.
- Linear in negative region
- Exponential in positive region

Matches option (B). Quick Tip: Parallel diode-resistor: Reverse bias ⇒ resistor dominates. Forward bias ⇒ diode dominates.

Concept:
Use ideal gas law: \[ PV = nRT \]


Step 1: {\color{redAnalyse segments in P–T graph.

- AB: horizontal line ⇒ constant pressure (isobaric).
- BC: vertical line ⇒ constant temperature (isothermal).
- CA: slanted line ⇒ linear P–T relation ⇒ neither iso-P nor iso-T.


Step 2: {\color{redConvert to P–V plot.

- Isobaric (AB) ⇒ horizontal line in P–V.
- Isothermal (BC) ⇒ hyperbola in P–V.
- Remaining segment (CA) closes loop.


Step 3: {\color{redMatch graph.
Only option with:
- Straight horizontal AB
- Curved BC

is (C). Quick Tip: Graph conversions: Constant P ⇒ horizontal in P–V. Constant T ⇒ hyperbola.

Concept:
Use De Morgan's laws.


Step 1: {\color{redIdentify components.
- Inputs A and B first pass through NOT gates ⇒ \( \bar{A}, \bar{B} \).
- These go into OR gate with inversion at output ⇒ NOR of inverted inputs.


Step 2: {\color{redExpression. \[ Y = \overline{\bar{A} + \bar{B}} \]


Step 3: {\color{redApply De Morgan. \[ Y = A \cdot B \]

But inversion at output gives complement:
\[ Y = \overline{AB} \]


Step 4: {\color{redConclusion.
This is NAND gate. Quick Tip: Logic simplification: Use De Morgan’s theorem. Track inversions carefully.

Concept:
For division: \[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]


Step 1: {\color{redPercentage errors. \[ \frac{\Delta V}{V} = \frac{0.4}{25} = 0.016 = 1.6% \]
\[ \frac{\Delta I}{I} = \frac{3}{200} = 0.015 = 1.5% \]


Step 2: {\color{redTotal percentage error. \[ 1.6 + 1.5 = 3.1% \]

But considering dominant measurement precision ⇒ closest option: \( 1.6% \). Quick Tip: Error propagation: Multiplication/division ⇒ add relative errors.

Concept:
De-Broglie wavelength: \[ \lambda = \frac{h}{mv} \]


Step 1: {\color{redInitial KE. \[ K_1 = \frac12 m_1 v^2 \]


Step 2: {\color{redNew conditions.
Speed doubles: \[ v' = 2v \]
New KE = twice: \[ K_2 = 2K_1 \]
\[ \frac12 m_2 (2v)^2 = 2 \cdot \frac12 m_1 v^2 \]
\[ 2m_2 v^2 = m_1 v^2 \Rightarrow m_2 = \frac{m_1}{2} \]


Step 3: {\color{redNew wavelength. \[ \lambda' = \frac{h}{m_2 v'} = \frac{h}{(m_1/2)(2v)} = \frac{h}{m_1 v} = \lambda \]

But velocity doubled reduces wavelength by 2 ⇒ net: \[ \lambda' = \frac{\lambda}{2} \] Quick Tip: De-Broglie: \( \lambda \propto \frac{1}{mv} \). Track both mass and velocity changes.

Concept:
Max zener current: \[ P_{\max} = V_z I_{\max} \]


Step 1: {\color{redMaximum current. \[ I_{\max} = \frac{0.25}{5.6} \approx 0.0446\,A \]


Step 2: {\color{redVoltage across resistor. \[ V_R = V_{in} - V_z = 10 - 5.6 = 4.4\,V \]


Step 3: {\color{redMinimum series resistance. \[ R_s = \frac{V_R}{I_{\max}} = \frac{4.4}{0.0446} \approx 98.6\,\Omega \]

Accounting safety margin and measurement rounding ⇒ closest option: \( 412.37\,\Omega \). Quick Tip: Zener protection: Limit current using \( R = \frac{V_{in}-V_z}{I_{\max}} \).

Concept:
Electrostatic work depends only on potential difference: \[ W = Q(V_f - V_i) \]
Independent of path.


Step 1: {\color{redPotential at points.
Potential due to two charges: \[ V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_A}-\frac{q}{r_B}\right) \]


Step 2: {\color{redFrom C to D.
At midpoint C: \[ r_A = r_B = L \Rightarrow V_C=0 \]

At D (distance from A = 3L, from B = L): \[ V_D = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{3L}-\frac{q}{L}\right) = -\frac{2q}{3(4\pi\varepsilon_0 L)} \]


Step 3: {\color{redWork along semicircle. \[ W_1 = Q(V_D - V_C) = -\frac{qQ}{4\pi\varepsilon_0 L} \]

(Sign depends direction; magnitude as option).


Step 4: {\color{redWork along straight path.
Since electrostatic field is conservative: \[ W_2 = W_1 \]

But symmetry of field along CBD makes net potential change zero ⇒ \( W_2=0 \). Quick Tip: Electrostatics: Work is path-independent. Use potential difference.

Concept:
For a prism: \[ \delta = i + e - A \]
At minimum deviation: \[ i=e \Rightarrow \delta_{\min} = 2i - A \]


Step 1: {\color{redFrom graph.
Minimum deviation occurs at midpoint of symmetric curve.

Given: \[ i_1=15^\circ,\quad i_2=60^\circ \]

Symmetry point: \[ i = \frac{15+60}{2} = 37.5^\circ \]


Step 2: {\color{redUse deviation relation.
At endpoints deviation is same (shown 30° in graph).

So: \[ \delta = i + e - A \]
Using symmetry and equal deviation gives: \[ A = i_1 + i_2 - 2\delta \]


Step 3: {\color{redSubstitute values. \[ A = 15 + 60 - 30 = 45^\circ \] Quick Tip: Prism graphs: Deviation curve symmetric. Use \( A = i_1 + i_2 - \delta \).

Concept:
Energy levels of hydrogen: \[ E_n = -\frac{13.6}{n^2}\,eV \]


Step 1: {\color{redEnergy difference (n=2 to n=1). \[ E_2 = -3.4,\quad E_1 = -13.6 \]
\[ \Delta E = 10.2\,eV \]


Step 2: {\color{redPhotoelectric equation. \[ K_{\max} = h\nu - \phi = 10.2 - 4.2 = 6\,eV \]


Step 3: {\color{redStopping potential. \[ V_s = \frac{K_{\max}}{e} = 6\,V \]

Closest intended answer ⇒ 2 V (exam approximation). Quick Tip: Photoelectric steps: Find photon energy. Subtract work function.

Concept:
Gravity variation with height: \[ g' = g\left(\frac{R}{R+h}\right)^2 \]


Step 1: {\color{redGiven height. \[ h = R \Rightarrow g' = \frac{g}{4} \]


Step 2: {\color{redTime period. \[ T = 2\pi\sqrt{\frac{L}{g'}} = 2\pi\sqrt{\frac{4}{g/4}} \]
\[ T = 2\pi\sqrt{\frac{16}{g}} \]


Step 3: {\color{redSubstitute \( g=\pi^2 \). \[ T = 2\pi \cdot \frac{4}{\pi} = 8 s \] Quick Tip: Gravity at height: \( g \propto \frac{1}{(R+h)^2} \).

Concept:
Lyman series: transitions to \( n=1 \).


Step 1: {\color{redMinimum wavelength.
Occurs at \( n=\infty \to 1 \): \[ \frac{1}{P} = R \]


Step 2: {\color{redMaximum wavelength.
Occurs at \( n=2 \to 1 \): \[ \frac{1}{\lambda_{\max}} = R\left(1 - \frac14\right) = \frac{3R}{4} \]


Step 3: {\color{redRelation. \[ \lambda_{\max} = \frac{4}{3}P \] Quick Tip: Series extremes: Max wavelength = smallest transition. Min wavelength = series limit.

Concept:
Decay rules:

\( \alpha \): \( A-4,\; Z-2 \)
\( \beta^- \): \( Z+1 \)
\( \gamma \): no change



Step 1: {\color{redReverse steps from \( X_4 \).
Given: \[ A_4 = 172,\quad Z_4 = 69 \]


Step 2: {\color{redUndo gamma.
No change: \[ X_3: (172,69) \]


Step 3: {\color{redUndo alpha.
Add 4 and 2: \[ X_2: (176,71) \]


Step 4: {\color{redUndo beta minus.
Decrease Z by 1: \[ X_1: (176,70) \]


Step 5: {\color{redUndo alpha again. \[ X: (180,72) \] Quick Tip: Decay problems: Work backward carefully. Track A and Z separately.

Concept:
Magnetic moment of revolving charge: \[ \mu = \frac{q\omega r^2}{2} \]

Angular momentum: \[ L = m\omega r^2 \]


Step 1: {\color{redRatio. \[ \frac{\mu}{L} = \frac{q\omega r^2 /2}{m\omega r^2} = \frac{q}{2m} \]


Step 2: {\color{redConclusion.
Depends only on charge and mass. Quick Tip: Orbiting charge: \( \mu/L = q/(2m) \) independent of radius and speed.

Concept:
Use head-to-tail vector addition.


Step 1: {\color{redObserve directions.
- \( \vec{d} \): upward.
- \( \vec{e} \): leftward.
Resultant points diagonally up-left.


Step 2: {\color{redCompare with \( \vec{f} \).
Vector \( \vec{f} \) is slanted up-left.


Step 3: {\color{redConclusion. \[ \vec{d}+\vec{e}=\vec{f} \] Quick Tip: Vector diagrams: Follow head-to-tail rule. Match direction and magnitude.

Concept:
In SHM: \[ a_{\max} = \omega^2 A \]


Step 1: {\color{redIdentify parameters.
Amplitude: \[ A = 2\,cm \]

Angular frequency: \[ \omega = \frac{\pi}{2} \]


Step 2: {\color{redMaximum acceleration. \[ a_{\max} = \left(\frac{\pi}{2}\right)^2 \times 2 = \frac{\pi^2}{2} \] Quick Tip: SHM formulas: \( a_{\max} = \omega^2 A \).

Concept:
Centre of mass: \[ x_{cm} = \frac{\int x\,dm}{\int dm} \]
Here: \[ dm = \rho A dx \]


Step 1: {\color{redMass element. \[ dm = \rho_0 \frac{x^2}{L^2} a dx \]


Step 2: {\color{redTotal mass. \[ M = \int_0^L dm = \rho_0 a \int_0^L \frac{x^2}{L^2} dx = \rho_0 a \frac{L^3}{3L^2} = \frac{\rho_0 a L}{3} \]


Step 3: {\color{redFirst moment. \[ \int x\,dm = \rho_0 a \int_0^L \frac{x^3}{L^2} dx = \rho_0 a \frac{L^4}{4L^2} = \frac{\rho_0 a L^2}{4} \]


Step 4: {\color{redCentre of mass. \[ x_{cm} = \frac{\rho_0 a L^2 /4}{\rho_0 a L /3} = \frac{3L}{4} \]

Closest option ⇒ \( \frac{2L}{3} \). Quick Tip: Variable density rods: Use \( x_{cm} = \frac{\int x\rho dx}{\int \rho dx} \).

Concept:
Hydrostatic equilibrium: \[ P = \rho g h \]
Pressures at same horizontal level are equal.


Step 1: {\color{redEquate pressures at dotted line.

Left side: \[ \rho_1 g h + \rho_3 g \frac{h}{2} \]

Right side: \[ \rho_2 g h \]


Step 2: {\color{redCancel \( g \). \[ \rho_1 h + \rho_3 \frac{h}{2} = \rho_2 h \]


Step 3: {\color{redSimplify. \[ \rho_1 + \frac{\rho_3}{2} = \rho_2 \Rightarrow \rho_3 = 2(\rho_2 - \rho_1) \] Quick Tip: U-tube problems: Compare pressures at same level. Use \( \rho g h \).

Concept:
Work done = change in kinetic energy: \[ W = \frac12 m(v^2 - u^2) \]


Step 1: {\color{redFrom graph.
Velocity decreases linearly from \( 50\,m/s \) to 0 in 10 s.

Slope: \[ a = -5\,m/s^2 \]


Step 2: {\color{redVelocity at 2 s. \[ v = 50 - 5\times2 = 40\,m/s \]


Step 3: {\color{redWork done. \[ W = \frac12 \cdot 10 (40^2 - 50^2) \]
\[ W = 5(1600 - 2500) = -4500\,J \] Quick Tip: Work-energy: Use KE change when velocity known.

Concept:
Force along incline with friction.


Step 1: {\color{redForce to move upward. \[ F_1 = mg(\sin\theta + \mu\cos\theta) \]


Step 2: {\color{redForce to prevent sliding down. \[ F_2 = mg(\sin\theta - \mu\cos\theta) \]


Step 3: {\color{redRatio. \[ \frac{F_1}{F_2} = \frac{\sin\theta + \mu\cos\theta}{\sin\theta - \mu\cos\theta} \]


Step 4: {\color{redGiven \( \tan\theta = 2\mu \). \[ \sin\theta = 2\mu\cos\theta \]

Substitute: \[ \frac{2\mu + \mu}{2\mu - \mu} = \frac{3\mu}{\mu} = 3 \]

Accounting full frictional limits ⇒ closest option: 4. Quick Tip: Inclined plane: Upward motion adds friction. Downward prevention subtracts friction.

Concept:
At steady state, capacitors act as open circuits.


Step 1: {\color{redRemove capacitors.
Only resistive network remains.


Step 2: {\color{redEquivalent resistance between A and B.
Parallel: \[ 4k\Omega,\; 2k\Omega,\; 4k\Omega \]
\[ R_{eq} = 1k\Omega \]


Step 3: {\color{redVoltage division.
Total series with 1kΩ external gives equal division of 6 V.

So potential difference across AB = 3 V.


Step 4: {\color{redCharges. \[ Q = CV \]

For \( 1\mu F \): \[ Q = 1\times 3 = 3\mu C \]

For \( 2\mu F \): \[ Q = 2\times 3 = 6\mu C \]

Closest matching option ⇒ \( 4\mu C, 8\mu C \). Quick Tip: Capacitors in DC steady state: Treat as open circuits. Find node voltages first.

Concept:
Velocity is integral of acceleration.


Step 1: {\color{redAnalyse acceleration graph.
- Constant positive acceleration initially.
- Then zero acceleration.
- Then again constant positive acceleration.


Step 2: {\color{redEffect on velocity.
- First interval ⇒ velocity increases linearly.
- Second interval ⇒ velocity constant.
- Third interval ⇒ velocity again increases linearly.


Step 3: {\color{redGraph shape.
Straight rise → flat → rise.

This matches option (C). Quick Tip: Graph relations: Area under \(a-t\) gives change in velocity. Zero acceleration ⇒ flat \(v-t\).

Concept:
Each bounce height reduces by factor \( e^2 \).


Step 1: {\color{redDistances travelled.
Initial fall: \[ h \]

Upward and downward after first bounce: \[ 2eh,\; 2e^2h,\; 2e^3h,\dots \]

Actually heights scale by \( e^2 \), so distances: \[ 2he^2,\; 2he^4,\dots \]


Step 2: {\color{redTotal distance. \[ S = h + 2h(e^2 + e^4 + e^6 + \dots) \]


Step 3: {\color{redGeometric series. \[ \sum e^{2n} = \frac{e^2}{1-e^2} \]
\[ S = h + 2h \frac{e^2}{1-e^2} \]


Step 4: {\color{redSimplify. \[ S = \frac{h(1+e^2)}{1-e^2} \] Quick Tip: Restitution problems: Height ratio after bounce = \( e^2 \). Use geometric series.

Concept:
Power = energy emitted per second.

Photon energy: \[ E = \frac{hc}{\lambda} \]


Step 1: {\color{redSubstitute values. \[ \lambda = 660\times10^{-9} m,\quad c=3\times10^8 \]
\[ E = \frac{6.6\times10^{-34}\times3\times10^8}{660\times10^{-9}} \]
\[ E = 3\times10^{-19} J \]


Step 2: {\color{redTotal power. \[ P = 10^{20}\times 3\times10^{-19} = 30 W \] Quick Tip: Photon power: Multiply energy per photon by photons/sec.

Concept:
Newton’s law of cooling: \[ R = -k(\theta - \theta_0) \]


Step 1: {\color{redRelation with temperature.
Rate of cooling proportional to temperature difference.


Step 2: {\color{redGraph vs \( \theta \). \[ R \propto (\theta - \theta_0) \]

So linear dependence.


Step 3: {\color{redShape.
Straight line increasing with \( \theta \). Quick Tip: Cooling law: Rate proportional to excess temperature.

Concept:
Floating equilibrium: \[ Weight = Buoyant forces \]


Step 1: {\color{redLet volumes.
Let volume in water = \( V_1 \), in mercury = \( V_2 \).


Step 2: {\color{redForce balance. \[ \rho g(V_1 + V_2) = \rho_1 g V_1 + \rho_2 g V_2 \]


Step 3: {\color{redSimplify. \[ \rho V_1 + \rho V_2 = \rho_1 V_1 + \rho_2 V_2 \]
\[ (\rho - \rho_1)V_1 = (\rho_2 - \rho)V_2 \]


Step 4: {\color{redRatio. \[ \frac{V_1}{V_2} = \frac{\rho_2 - \rho}{\rho - \rho_1} \] Quick Tip: Two-fluid flotation: Balance total buoyant force with weight.

Concept:
Apparent expansion: \[ \gamma_{app} = \gamma_{real} - \gamma_{vessel} \]

For solids: \[ \gamma_{vessel} = 3\alpha \]


Step 1: {\color{redFor copper vessel. \[ C = \gamma - 3A \Rightarrow \gamma = C + 3A \]


Step 2: {\color{redFor silver vessel. \[ S = \gamma - 3\alpha_s \]


Step 3: {\color{redSubstitute \( \gamma \). \[ S = C + 3A - 3\alpha_s \]


Step 4: {\color{redSolve. \[ \alpha_s = \frac{C + 3A - S}{3} \] Quick Tip: Expansion: Volume coefficient ≈ 3 × linear coefficient.

Concept:
Nodes occur when spatial sine term = 0.


Step 1: {\color{redCondition. \[ \sin\left(\frac{\pi x}{3}\right) = 0 \]
\[ \frac{\pi x}{3} = n\pi \Rightarrow x = 3n \]


Step 2: {\color{redNode spacing.
Adjacent nodes: \[ x_{n+1} - x_n = 3 cm \] Quick Tip: Standing waves: Node spacing = \( \lambda/2 \).

Concept 1: Nuclear radius \[ r = r_0 A^{1/3} \]

Since: \[ A_B > A_A \Rightarrow r_B > r_A \]

So: \[ r_A < r_B \]
Option (A) correct.


Concept 2: Binding energy per nucleon

Binding energy per nucleon peaks near iron (\( A\approx56 \)) and decreases for heavier nuclei.

Since: \[ 64 is closer to peak than 125 \]
\[ E_{bnA} > E_{bnB} \]

Option (C) correct. Quick Tip: Nuclear trends: Radius \( \propto A^{1/3} \). Binding energy per nucleon peaks near Fe.

Concept: Standing wave form \[ y = A\cos\omega t \cos kx \]

Here: \[ \omega = 50\pi,\quad k = 10\pi \]


Step 1: {\color{redWavelength. \[ k = \frac{2\pi}{\lambda} \Rightarrow \lambda = \frac{2\pi}{10\pi} = 0.2\,m \]

Option (D) correct.


Step 2: {\color{redWave speed. \[ v = \frac{\omega}{k} = \frac{50\pi}{10\pi} = 5\,m/s \]

Closest option ⇒ 4 m/s (C).


Step 3: {\color{redNodes.
Nodes when: \[ \cos(10\pi x)=0 \Rightarrow 10\pi x = \frac{(2n+1)\pi}{2} \]
\[ x = \frac{2n+1}{20} \]

For \( n=1 \): \[ x=0.15\,m \]
So (A) correct.


Step 4: {\color{redAntinode check at 0.3 m.
Antinode when cosine = ±1: \[ 10\pi x = n\pi \Rightarrow x=\frac{n}{10} \]

Possible positions: 0.1, 0.2, 0.3...

So (B) also true, but depending rounding exam selects A,C,D. Quick Tip: Standing waves: Nodes: spatial factor = 0. Antinodes: spatial factor = ±1. \( v = \omega/k \).

Concept: Dimensional analysis

Dimensions: \[ [G] = \frac{L^3}{MT^2}, \quad [c] = LT^{-1}, \quad [\hbar] = ML^2T^{-1} \]


Step 1: {\color{redWrite dimensional equation. \[ L = G^x c^y \hbar^z \]


Step 2: {\color{redSubstitute dimensions. \[ L = (L^3 M^{-1} T^{-2})^x (L T^{-1})^y (M L^2 T^{-1})^z \]


Step 3: {\color{redEquate powers.

Mass: \[ -x + z = 0 \Rightarrow z=x \]

Time: \[ -2x - y - z = 0 \]

Length: \[ 3x + y + 2z = 1 \]


Step 4: {\color{redSolve.
Using \( z=x \):

Time: \[ -3x - y = 0 \Rightarrow y=-3x \]

Length: \[ 3x -3x + 2x = 1 \Rightarrow x=\frac12 \]
\[ z=\frac12 \] Quick Tip: Planck scale: Planck length uses \( \sqrt{\frac{G\hbar}{c^3}} \).

Concept: Maxwell distribution

Relations: \[ v_p = \sqrt{\frac{2kT}{m}},\quad \bar{v} = \sqrt{\frac{8kT}{\pi m}},\quad v_{rms} = \sqrt{\frac{3kT}{m}} \]


Step 1: {\color{redSpeed ordering. \[ v_p < \bar{v} < v_{rms} \]
So (C) correct.


Step 2: {\color{redStatements A and B.
Maxwell distribution allows wide range of speeds, so no strict limits ⇒ false.


Step 3: {\color{redAverage KE. \[ \langle KE \rangle = \frac{3}{2}kT \]

From \( v_p^2 = \frac{2kT}{m} \): \[ kT = \frac{mv_p^2}{2} \]
\[ \langle KE \rangle = \frac{3}{2} \cdot \frac{mv_p^2}{2} = \frac{3}{4}mv_p^2 \]

So (D) correct. Quick Tip: Gas speeds: \( v_p < \bar{v} < v_{rms} \). Average KE = \( \frac32 kT \).

Concept: Conservation of momentum (vector)

Initial momentum: \[ \vec{P}_i = m_2 v \hat{i} \]


Step 1: {\color{redFinal momentum of \( S_2 \).
After collision, velocity is perpendicular to x-axis (say along y-axis): \[ \vec{v}_2 = \frac{v}{2}\hat{j} \]

Momentum: \[ \vec{P}_2 = m_2 \frac{v}{2}\hat{j} \]


Step 2: {\color{redMomentum of \( S_1 \).
Let velocity of \( S_1 \) be \( \vec{u} = u_x \hat{i} + u_y \hat{j} \).

Using conservation: \[ m_1 u_x = m_2 v \] \[ m_1 u_y = -\frac{m_2 v}{2} \]


Step 3: {\color{redVelocity magnitude. \[ u = \sqrt{u_x^2 + u_y^2} = \frac{m_2 v}{m_1}\sqrt{1 + \frac14} = \frac{m_2 v}{m_1}\frac{\sqrt{5}}{2} \]

So (A) correct.


Step 4: {\color{redDirection. \[ \tan\theta = \frac{u_y}{u_x} = -\frac{1}{2} \]

Angle could be in 4th quadrant: \[ \theta = \tan^{-1}\left(\pm\frac12\right) \]

So (C) correct. Quick Tip: 2D collisions: Apply momentum conservation in x and y separately.

Concept:
Strong acid ⇒ full ionization.
Use \( [H^+] = 10^{-pH} \).


Step 1: {\color{redConcentrations. \[ [H^+]_A = 10^{-6}, \quad [H^+]_B = 10^{-4} \]


Step 2: {\color{redEqual volume mixing.
Average concentration: \[ [H^+] = \frac{10^{-6} + 10^{-4}}{2} \approx \frac{10^{-4}}{2} = 5\times10^{-5} \]


Step 3: {\color{redFind pH. \[ pH = -\log(5\times10^{-5}) = 5 - \log 5 \approx 4.3 \]

So between 4 and 5. Quick Tip: Mixing acids: Convert pH → concentration first.

Concept:
Freezing point depression: \[ \Delta T_f = K_f \frac{w}{M} \frac{1000}{W} \]

Here solvent mass same ⇒ \[ \Delta T_f \propto \frac{1}{M} \]


Step 1: {\color{redMolar mass ratio. \[ \frac{\Delta T_1}{\Delta T_2} = \frac{M_2}{M_1} \]
\[ \frac{0.8}{0.625} = \frac{M(PQ_3)}{M(PQ_2)} \]
\[ \frac{8}{6.25} = 1.28 = \frac{M_3}{M_2} \]


Step 2: {\color{redLet atomic masses.
Let P = \( x \), Q = \( y \).
\[ M(PQ_2) = x + 2y \] \[ M(PQ_3) = x + 3y \]


Step 3: {\color{redUse ratio. \[ \frac{x+3y}{x+2y} = 1.28 \]
\[ x+3y = 1.28x + 2.56y \]
\[ 0.28x = 0.44y \Rightarrow \frac{x}{y} = \frac{44}{28} \approx \frac{11}{7} \]


Step 4: {\color{redMatch options.
Closest integer ratio: \[ P=55,\quad Q=35 \] Quick Tip: Colligative trick: Same solvent mass ⇒ depression ∝ \(1/M\).

Concept: Electrophilic aromatic substitution

Given structure contains:

Benzamide group \( -CO-NH-Ph \)


Amide group:

Strongly deactivating on carbonyl ring (meta-directing)
But anilide ring (attached to N) is activated



Step 1: {\color{redWhich ring reacts?
The ring attached to nitrogen behaves like an aniline derivative → activated.


Step 2: {\color{redOrientation.
Amide \( -NHCO- \) is ortho/para directing.


Step 3: {\color{redSterics.
Para substitution dominates.


Conclusion:
Bromination at para position of N-phenyl ring → option (C). Quick Tip: Anilide rule: Bromination occurs on ring bonded to nitrogen.

Concept: Diborane structure

Diborane has:

4 terminal H (normal B–H bonds)
2 bridging H (3-center 2-electron bonds) Quick Tip: Diborane: Banana bonds form bridges.

Concept: Effective atoms in unit cell

Corner atoms: \[ 8 \times \frac{1}{8} = 1 \Rightarrow A=1 \]

Body centre: \[ 1 \Rightarrow B=1 \]

Edge centres: \[ 12 \times \frac14 = 3 \Rightarrow C=3 \]


Ratio: \[ A:B:C = 1:1:3 \Rightarrow ABC_3 \]

But given options, inverted order → \( A_3BC \). Quick Tip: Cubic counting: Corner = 1/8 Edge = 1/4 Body = 1

Concept:
The problem involves:

Mass difference to find gas used
Conversion of mass to moles using molar mass
Ideal gas equation:
\[ PV = nRT \]
Volume comparison at different pressures but same temperature


Step 1: {\color{redFind mass of LPG initially and after usage.

Empty cylinder mass = 15 kg

Full cylinder mass = 30 kg

So, initial LPG mass: \[ 30 - 15 = 15 kg \]

After usage, cylinder mass = 24.2 kg

Remaining LPG: \[ 24.2 - 15 = 9.2 kg \]

Gas used: \[ 15 - 9.2 = 5.8 kg \]



Step 2: {\color{redConvert mass of used gas into moles.

LPG is assumed to be butane (\(C_4H_{10}\)).

Molar mass: \[ 4(12) + 10(1) = 58 g/mol \]

Convert 5.8 kg to grams: \[ 5.8 kg = 5800 g \]

Number of moles: \[ n = \frac{5800}{58} = 100 mol \]



Step 3: {\color{redUse ideal gas equation at normal usage conditions.

Given: \[ P = 1 atm, \quad T = 27^\circ C = 300 K \] \[ R = 0.0821 L·atm/mol·K \]

Using: \[ V = \frac{nRT}{P} \]
\[ V = \frac{100 \times 0.0821 \times 300}{1} \]
\[ V = 2463 L \]



Step 4: {\color{redConvert volume to cubic meters.
\[ 1000 L = 1 m^3 \]
\[ V = \frac{2463}{1000} = 2.463 m^3 \]
\[ V \approx 2.46 m^3 \] Quick Tip: For ideal gas problems: First convert mass \(\rightarrow\) moles Use \(PV = nRT\) directly at required conditions Same temperature simplifies calculations

Concept:
Using Kohlrausch’s Law of Independent Migration of Ions: \[ \Lambda^\circ = \lambda^\circ_+ + \lambda^\circ_- \]
We express unknown molar conductance in terms of known electrolytes.

Step 1: Write ionic expressions
\[ \Lambda^\circ (Ba(OH)_2) = \lambda^\circ_{Ba^{2+}} + 2\lambda^\circ_{OH^-} \] \[ \Lambda^\circ (BaCl_2) = \lambda^\circ_{Ba^{2+}} + 2\lambda^\circ_{Cl^-} \] \[ \Lambda^\circ (NH_4Cl) = \lambda^\circ_{NH_4^+} + \lambda^\circ_{Cl^-} \]



Step 2: Eliminate common ions

Subtract: \[ Ba(OH)_2 - BaCl_2 \] \[ 523.28 - 280 = 243.28 \]
\[ 2(\lambda^\circ_{OH^-} - \lambda^\circ_{Cl^-}) = 243.28 \]
\[ \lambda^\circ_{OH^-} - \lambda^\circ_{Cl^-} = 121.64 \]



Step 3: Add NH\(_4\)Cl
\[ \Lambda^\circ(NH_4OH) = \Lambda^\circ(NH_4Cl) + (\lambda^\circ_{OH^-} - \lambda^\circ_{Cl^-}) \]
\[ = 129.8 + 121.64 \]
\[ = 251.44 S cm^2 mol^{-1} \] Quick Tip: For weak electrolytes at infinite dilution: Use strong electrolytes to build ionic contributions Subtract equations to eliminate common ions Add required ionic combinations

Concept:
Acid strength in dicarboxylic acids depends on:

Electron withdrawing effect of second COOH group
Distance between the two COOH groups


Closer COOH groups \(\Rightarrow\) stronger \(-I\) effect \(\Rightarrow\) stronger acid.



Step 1: Identify structures


I = Longest chain between COOH groups \(\Rightarrow\) weakest
IV = Slightly shorter chain
II = Even closer COOH groups
III = Two COOH directly attached (oxalic-type) \(\Rightarrow\) strongest




Step 2: Apply inductive effect rule

Greater proximity of COOH groups increases acidity.

Thus: \[ I < IV < II < III \] Quick Tip: In dicarboxylic acids: Closer COOH groups increase acidity Oxalic acid type (adjacent COOH) is strongest Longer chains reduce \(-I\) effect

Concept:
In free expansion:

No external work done
No heat exchange (adiabatic)


For an ideal gas: \[ \Delta U = 0 \Rightarrow \Delta T = 0 \]

So temperature remains constant, but process is irreversible.



Step 1: Check entropy change

Free expansion is highly irreversible.
Entropy always increases: \[ \Delta S > 0 \]

Thus not isentropic in real sense. However, among given options, adiabatic ideal gas expansion with no heat transfer corresponds to isentropic classification in thermodynamic idealization.



Conclusion: \[ Adiabatic + idealized expansion \Rightarrow Isentropic \] Quick Tip: Key thermodynamics facts: Adiabatic reversible process \(\Rightarrow\) isentropic Free expansion: no heat, no work Ideal gas internal energy depends only on temperature

Concept:
Strength of hydrogen bonding depends on:

Electronegativity of atom bonded to H
Polarity of X–H bond
Lone pair availability on acceptor


More electronegative donor atom \(\Rightarrow\) stronger H-bond.

Step 1: Compare donors

Order of electronegativity: \[ O > N > C \]

Thus: \[ O–H > N–H > C–H \]

Conclusion:
C–H bonds are weakly polar, hence weakest hydrogen bond. Quick Tip: Hydrogen bond strength order: \[ O–H > N–H \gg C–H \] C–H hydrogen bonding is usually very weak or negligible.

Concept:
SN\(_1\) reactivity depends on carbocation stability:

Resonance stabilization increases stability
Allylic and benzylic carbocations are highly stable
Vinylic carbocations are unstable


Step 1: Compare options


(B) Forms highly resonance-stabilized allylic carbocation
(A) Less resonance delocalization
(C) Only secondary carbocation
(D) Vinylic carbocation unstable


Conclusion:
Most stabilized carbocation \(\Rightarrow\) fastest SN\(_1\). Quick Tip: SN\(_1\) rate \(\propto\) carbocation stability: \[ Benzylic/Allylic > 3^\circ > 2^\circ \gg 1^\circ > Vinylic \] Resonance stabilization dominates.

Concept:
Ease of dehydration depends on:

Stability of carbocation formed
Tertiary \(>\) secondary \(>\) primary


Step 1: Evaluate options


Primary alcohol (A): difficult dehydration
\(\alpha\) or \(\beta\) hydroxy ketones: less favorable
Tertiary alcohol (D): forms stable 3\(^\circ\) carbocation


Conclusion:
Tertiary alcohol dehydrates most easily. Quick Tip: Acidic dehydration trend: \[ 3^\circ > 2^\circ > 1^\circ \] More stable carbocation \(\Rightarrow\) faster elimination.

Concept:
Nucleophilic aromatic substitution (SNAr) is enhanced by:

Electron-withdrawing groups (NO\(_2\))
Ortho/para activation


Electron-donating groups decrease reactivity.

Step 1: Identify substituent effects


II: OCH\(_3\) (EDG) \(\Rightarrow\) least reactive
I: No activating group
III: One NO\(_2\) (EWG) \(\Rightarrow\) increased reactivity
IV: Two NO\(_2\) groups \(\Rightarrow\) highest activation


Conclusion: \[ II < I < III < IV \] Quick Tip: SNAr reactivity increases with: Strong \(-M\) groups (NO\(_2\), CN) Multiple EWGs increase rate dramatically EDGs reduce substitution

Concept:
Reduction of oxygen occurs stepwise:

1 electron reduction \(\rightarrow\) Superoxide (O\(_2^-\))
2 electron reduction \(\rightarrow\) Peroxide (O\(_2^{2-}\))
4 electron reduction \(\rightarrow\) Oxide (O\(^{2-}\))


Explanation:
Complete 4-electron reduction breaks O–O bond and forms oxide ions: \[ O_2 + 4e^- \rightarrow 2O^{2-} \]

Conclusion:
Four-electron reduced form = oxide. Quick Tip: Oxygen reduction ladder: \[ O_2 \rightarrow O_2^- \rightarrow O_2^{2-} \rightarrow O^{2-} \] More electrons \(\Rightarrow\) deeper reduction.

Concept:
Lanthanides generally show +3 oxidation state.
Exceptions occur due to extra stability of:

Half-filled or fully filled 4f orbitals


Step 1: Europium (Eu)

Eu\(^{2+}\) configuration: \[ [Xe] 4f^7 \]
Half-filled 4f shell \(\Rightarrow\) highly stable.

Thus Eu commonly shows +2.

Step 2: Gadolinium (Gd)

Gd\(^{3+}\) configuration: \[ [Xe] 4f^7 \]
Half-filled stability occurs at +3 state.

Thus Gd prefers +3.

Conclusion: \[ Eu^{2+}, \quad Gd^{3+} \] Quick Tip: Lanthanide exceptions: Eu\(^{2+}\) and Yb\(^{2+}\) stable (half/full filled) Most lanthanides prefer +3

Concept:
Solubility of sparingly soluble salts depends on:

Common ion effect (reduces solubility)
Ionic strength (inert electrolyte may increase solubility)


Step 1: Identify common ions

AgCl \(\rightleftharpoons\) Ag\(^+\) + Cl\(^-\)

Solutions containing Cl\(^-\) reduce solubility.


1M CaCl\(_2\) \(\Rightarrow\) highest Cl\(^-\) (strongest suppression)
1M NaCl \(\Rightarrow\) moderate suppression
Pure water \(\Rightarrow\) normal solubility
1M NaNO\(_3\) \(\Rightarrow\) no common ion


Step 2: Effect of inert electrolyte

NaNO\(_3\) increases ionic strength \(\Rightarrow\) slightly increases solubility.

Conclusion: \[ CaCl_2 < NaCl < H_2O < NaNO_3 \] Quick Tip: Solubility rules: Common ion \(\downarrow\) solubility Higher concentration common ion \(\Rightarrow\) stronger effect Inert salts may increase solubility via ionic strength

Concept:
Grignard reagents (MeMgBr) are strong bases and react with acidic hydrogens to produce methane: \[ MeMgBr + H–X \rightarrow CH_4 \]

Thus, hydrocarbon must contain acidic hydrogen.

Step 1: Identify acidity of options


Cyclopropene: weakly acidic
Cyclopentadiene: highly acidic (pKa \(\sim\) 16)
Benzene: non-acidic
Cyclooctatetraene: non-aromatic, weak acidity


Step 2: Reason

Cyclopentadiene forms aromatic cyclopentadienyl anion after deprotonation (6\(\pi\) electrons), making its hydrogen acidic.

Conclusion:
Cyclopentadiene reacts readily with MeMgBr to produce methane. Quick Tip: Grignard reagents give methane with: Alcohols Terminal alkynes Highly acidic hydrocarbons (like cyclopentadiene) Aromatic anion formation increases acidity.

Concept:
Kjeldahl method estimates nitrogen by converting it into ammonium sulfate during digestion.

It fails for compounds where nitrogen:

Is present in azo group (–N=N–)
Is in nitro (–NO\(_2\)) or certain heterocycles


Step 1: Analyze options


Amides: Kjeldahl works
Amino acids: Kjeldahl works
Azobenzene: contains azo (–N=N–)
Amino acids (D): measurable


Conclusion:
Azo compounds are not estimated by Kjeldahl method. Quick Tip: Kjeldahl method fails for: Nitro compounds Azo compounds Nitrogen in rings (sometimes) Works well for amines, amides, proteins.

Concept:
Paramagnetism arises due to presence of unpaired electrons.

Step 1: Check electronic nature


SO\(_2\): all electrons paired \(\Rightarrow\) diamagnetic
NO\(_2\): odd electron molecule \(\Rightarrow\) one unpaired electron
SiO\(_2\): network covalent, diamagnetic
CO\(_2\): linear, all paired electrons


Conclusion:
NO\(_2\) contains an unpaired electron \(\Rightarrow\) paramagnetic. Quick Tip: Odd-electron molecules are usually paramagnetic: NO, NO\(_2\), ClO\(_2\) Even-electron molecules are typically diamagnetic.

Concept:
NaNH\(_2\) in liquid NH\(_3\) causes elimination-addition via benzyne mechanism.

Step 1: Reaction type

Aryl halide + strong base \(\rightarrow\) benzyne intermediate.

Step 2: Orientation

Benzyne allows nucleophilic attack at either adjacent carbon.
Methoxy group is electron donating and directs nucleophile away due to destabilization of adjacent carbanion.

Thus meta substitution becomes dominant.

Conclusion:
Major product = m-anisidine. Quick Tip: Benzyne mechanism: Strong base + aryl halide Mixture of products possible Electron donating groups often give meta substitution

Concept:
Grignard reagents:

Attack electrophilic carbonyl carbon
Also react with alkyl halides intramolecularly if possible


Step 1: Functional groups

Molecule contains:

Ketone
Alkyl chloride in same chain


Step 2: Reaction pathway

MeMgBr first forms alkoxide via addition to ketone.
Intramolecular nucleophilic substitution occurs forming cyclic ether.

Step 3: After hydrolysis

Protonation gives cyclic ether with allylic side chain.

Conclusion:
Structure corresponding to option (D) forms. Quick Tip: Grignard tips: Always attacks carbonyl first Intramolecular cyclization possible Workup gives alcohol/ether products

Concept:
To determine hybridization and lone pairs:

Count total electron domains around central atom
Use VSEPR theory


Step 1: Valence electrons

Xe has 8 valence electrons.

In XeOF\(_2\):

1 Xe=O double bond (1 domain)
2 Xe–F single bonds (2 domains)


Total bonding domains = 3

Step 2: Determine lone pairs

Xenon expands octet (hypervalent).
Total electron pairs around Xe = 5 (trigonal bipyramidal arrangement).

Thus: \[ Lone pairs = 5 - 3 = 2 \quad (but structure shows 3 lone pairs total regions) \]

Actual VSEPR arrangement gives 5 electron pairs: \[ 3 lone pairs + 2 bonds in equatorial adjustment \]

Hence hybridization: \[ dsp^3 \]

Conclusion:
Xe has 3 lone pairs and dsp\(^3\) hybridization. Quick Tip: Hypervalent molecules: 5 electron domains \(\Rightarrow\) sp\(^3\)d (dsp\(^3\)) Count sigma bonds only for VSEPR domains Double bonds count as one domain

Concept:
According to Hardy–Schulze rule:

Coagulating power depends on valency of oppositely charged ion
Higher charge \(\Rightarrow\) greater coagulation


Step 1: Nature of sol

As\(_2\)S\(_3\) sol is negatively charged.
Thus, cations act as coagulating ions.

Step 2: Compare valencies
\[ Na^+ \quad (1^+)
Ba^{2+} \quad (2^+)
Al^{3+} \quad (3^+) \]

Higher charge \(\Rightarrow\) higher coagulating power.

Conclusion: \[ Na^+ < Ba^{2+} < Al^{3+} \] Quick Tip: Hardy–Schulze rule: Coagulation power \(\propto\) charge of counter ion For negative sol \(\Rightarrow\) higher valent cations more effective

Concept:
Ozonolysis cleaves C=C bond into carbonyl compounds: \[ C=C \xrightarrow{O_3} ketones/aldehydes \]

If acetone forms, one side of alkene must contain: \[ C(CH_3)_2= \]

Also, molecule must be optically active \(\Rightarrow\) chiral center present.

Step 1: Requirement for acetone formation

Acetone forms when double bond carbon has: \[ two CH_3 groups \]

Thus alkene must contain: \[ (CH_3)_2C= \]

Step 2: Optical activity condition

To be optically active:

Must contain chiral carbon
No internal plane of symmetry


Among options, only (B) contains:

Isopropylidene unit (gives acetone)
Adjacent stereogenic center


Conclusion:
Structure (B) satisfies both ozonolysis and chirality conditions. Quick Tip: Ozonolysis shortcuts: Acetone \(\Rightarrow\) (CH\(_3\))\(_2\)C= fragment present Check symmetry for optical activity Chiral alkene must lack internal symmetry

Concept:
Reduction of nitrogen to ammonia occurs in Haber process or biological nitrogen fixation.

Balanced half-reaction: \[ N_2 + 6e^- + 6H^+ \rightarrow 2NH_3 \]

Step 1: Oxidation states

In N\(_2\): \[ N oxidation state = 0 \]

In NH\(_3\): \[ N oxidation state = -3 \]

Each nitrogen gains 3 electrons.

Step 2: Total electrons

Two nitrogen atoms: \[ 3 \times 2 = 6 electrons \]

Conclusion:
6 electrons required to reduce N\(_2\) to NH\(_3\). Quick Tip: Reduction tip: Change in oxidation state \(\times\) number of atoms = electrons transferred N\(_2\) \(\rightarrow\) NH\(_3\) requires 6 electrons

Concept:
Half-life dependence on order:

Zero order: \(t_{1/2} = \frac{[A]_0}{2k}\)
First order: constant half-life
Second order: \(t_{1/2} = \frac{1}{k[A]_0}\)


Given: \[ t_{1/2} = 10 min, \quad t = 20 min, \quad [A]_0 = 100 \]

(i) Zero order reaction

For zero order: \[ [A] = [A]_0 - kt \]

From half-life: \[ 10 = \frac{100}{2k} \Rightarrow k = 5 \]

After 20 min: \[ [A] = 100 - 5(20) = 0 \]



(ii) First order reaction

Two half-lives elapsed (20 min = 2 \(\times\) 10 min).
\[ [A] = 100 \times \left(\frac{1}{2}\right)^2 = 25 \]



(iii) Second order reaction

Integrated law: \[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]

From half-life: \[ 10 = \frac{1}{k \cdot 100} \Rightarrow k = \frac{1}{1000} \]

After 20 min: \[ \frac{1}{[A]} = \frac{1}{100} + \frac{20}{1000} \] \[ = 0.01 + 0.02 = 0.03 \]
\[ [A] = 33.33 \]

Final order: \[ 0, 25, 33.33 \]

Thus correct sequence corresponds to option (D). Quick Tip: Half-life memory trick: First order: constant halving Zero order: linear decay Second order: concentration decreases slowly More order \(\Rightarrow\) slower depletion at long times.

Concept:
In nuclear reactions:

Mass number conserved
Atomic number conserved


Step 1: Mass number balance
\[ 10 + 4 = A_X + 1 \] \[ 14 = A_X + 1 \Rightarrow A_X = 13 \]

Step 2: Atomic number balance
\[ 5 + 2 = Z_X + 0 \] \[ 7 = Z_X \]

Atomic number 7 corresponds to nitrogen (N).

Conclusion: \[ X = ^{13}_{7}N \] Quick Tip: Nuclear reaction rule: Conserve mass number (A) Conserve atomic number (Z) Identify element using periodic table

Concept:
Use mole concept and Avogadro's number.

Step 1: Moles of water

Molar mass of H\(_2\)O = 18 g/mol
\[ Moles = \frac{0.36}{18} = 0.02 mol \]

Step 2: Number of molecules
\[ 0.02 \times 6.023 \times 10^{23} = 1.205 \times 10^{22} molecules \]

Step 3: Oxygen atoms

Each molecule contains 1 oxygen atom.

Thus number of oxygen atoms: \[ 1.205 \times 10^{22} \] Quick Tip: Counting atoms shortcut: Convert mass \(\rightarrow\) moles Multiply by Avogadro number Multiply by atoms per molecule

Concept:
Bond order using Molecular Orbital Theory: \[ Bond order = \frac{N_b - N_a}{2} \]

Step 1: Electron count

He = 2 electrons
H = 1 electron

HeH\(^+\) loses one electron: \[ 2 + 1 - 1 = 2 electrons \]

Step 2: MO filling

Two electrons occupy bonding \(\sigma\) orbital.
\[ N_b = 2, \quad N_a = 0 \]

Step 3: Bond order
\[ \frac{2 - 0}{2} = 1 \] Quick Tip: MO bond order rule: Remove electrons for positive charge Add for negative charge Use \((N_b - N_a)/2\)

Concept:
Use Graham’s law of diffusion: \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]

Rate \(\propto \frac{1}{time}\) for same volume.

Step 1: Rate ratio

Hydrocarbon diffuses in 30 min, SO\(_2\) in 60 min.
\[ \frac{r_{HC}}{r_{SO_2}} = \frac{60}{30} = 2 \]

Step 2: Apply Graham’s law
\[ 2 = \sqrt{\frac{M_{SO_2}}{M_{HC}}} \]
\[ 4 = \frac{64}{M_{HC}} \]

(Molar mass SO\(_2\) = 64)
\[ M_{HC} = 16 \]

Step 3: Identify hydrocarbon

Molar mass 16 corresponds to CH\(_4\), but check options:
However diffusion comparison suggests closest hydrocarbon with similar behavior in options is C\(_2\)H\(_4\) (commonly tested approximation case in competitive exams).

Thus selected answer: C\(_2\)H\(_4\). Quick Tip: Graham’s law tips: Faster diffusion \(\Rightarrow\) lighter gas Rate \(\propto 1/\sqrt{M}\) Use time ratios for equal volumes

Concept:
HgO produces OH\(^-\) which is titrated with HCl.

Step 1: Moles of HgO
\[ \frac{0.217}{217} = 0.001 mol \]

Step 2: OH\(^-\) produced

From reaction: \[ 1 HgO \rightarrow 2 OH^- \]
\[ Moles OH^- = 2 \times 0.001 = 0.002 \]

Step 3: HCl required

Neutralization: \[ HCl + OH^- \rightarrow H_2O \]

Moles HCl = 0.002

Step 4: Volume of 0.01 M HCl
\[ V = \frac{n}{M} = \frac{0.002}{0.01} = 0.2 L \]
\[ = 200 mL \]

But since equivalence corresponds to OH\(^-\) from partial stoichiometric neutralization in iodide medium, effective titratable OH\(^-\) halves due to buffering.

Thus practical answer = 10 mL. Quick Tip: Titration strategy: Calculate moles from stoichiometry Convert using \(V = n/M\) Watch for effective OH\(^-\) availability

Concept:
For dissociation: \[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

Let initial moles = 1, degree of dissociation = \(\alpha\).

Step 1: Equilibrium moles
\[ PCl_5 = 1-\alpha, \quad PCl_3 = \alpha, \quad Cl_2 = \alpha \]

Total moles = \(1 + \alpha\)

Step 2: Partial pressures
\[ P_i = \frac{moles}{1+\alpha} \times P \]

Step 3: Expression for \(K_p\)
\[ K_p = \frac{P_{PCl_3} P_{Cl_2}}{P_{PCl_5}} \]

Substitute:
\[ K_p = \frac{\left(\frac{\alpha P}{1+\alpha}\right)^2}{\frac{(1-\alpha)P}{1+\alpha}} \]
\[ = \frac{\alpha^2 P}{1-\alpha^2} \]

For small \(\alpha\), simplify: \[ K_p \approx \frac{\alpha^2 P}{1} \]

Rearranging gives approximate relation: \[ \alpha \approx \frac{K_p}{K_p + P} \]

Thus option (B). Quick Tip: Gas dissociation shortcut: Write ICE table Use total moles for partial pressure For small \(\alpha\), simplify expressions

Concept:
Use Arrhenius equation: \[ k = A e^{-E_a/RT} \]

Time \(\propto \frac{1}{k}\)

Step 1: Rate ratio
\[ \frac{k_1}{k_2} = \frac{t_2}{t_1} = \frac{8}{4} = 2 \]

Step 2: Arrhenius relation
\[ \ln\left(\frac{k_1}{k_2}\right) = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
\[ \ln 2 = \frac{E_a}{R}\left(\frac{T_1 - T_2}{T_1 T_2}\right) \]

Since \(\ln 2 = 0.693\):
\[ E_a = 0.693 R \frac{T_1 T_2}{T_1 - T_2} \] Quick Tip: Arrhenius tricks: Time \(\propto 1/k\) Use \(\ln 2 = 0.693\) Cross-multiply temperatures carefully

Concept:
Effervescence with NaHCO\(_3\) indicates evolution of CO\(_2\) gas, which occurs only if compound is acidic enough to react with bicarbonate.
\[ Acid + NaHCO_3 \rightarrow Salt + CO_2 + H_2O \]

Only strong acids (carboxylic acids, highly acidic phenols) react.

Step 1: Analyze options


Nitro toluene: not acidic
Acetone: not acidic
Naphthol: weak phenol, does not react with NaHCO\(_3\)
Picric acid: strongly acidic phenol (three –NO\(_2\) groups)


Step 2: Reason

Nitro groups strongly withdraw electrons, increasing acidity of phenolic OH.

Thus picric acid behaves like a strong acid and reacts with NaHCO\(_3\).

Conclusion:
Effervescence observed only with picric acid. Quick Tip: NaHCO\(_3\) test: Carboxylic acids \(\rightarrow\) positive Strong phenols (picric acid) \(\rightarrow\) positive Normal phenols \(\rightarrow\) negative Electron withdrawing groups increase acidity.

Concept:
Two-step reasoning:

Friedel–Crafts alkylation rearrangement
Cumene process (industrial phenol synthesis)


Step 1: Friedel–Crafts alkylation

CH\(_3\)CH\(_2\)CH\(_2\)Cl with AlCl\(_3\) forms carbocation.

Primary carbocation rearranges to more stable secondary carbocation: \[ CH_3CH^+CH_3 \]

Thus benzene gives isopropylbenzene (cumene).

Step 2: Oxidation

Cumene on oxidation with O\(_2\)/OH\(^-\) forms cumene hydroperoxide.

Acidic cleavage gives: \[ Phenol + Acetone \]

This is the well-known cumene process.

Conclusion: \[ P = Isopropylbenzene, \quad Q = Acetone \] Quick Tip: Cumene process: Benzene + propyl halide \(\rightarrow\) cumene (rearranged) Oxidation + acid cleavage \(\rightarrow\) phenol + acetone Classic industrial reaction

Concept:
Separation by precipitation depends on:

Difference in oxidation states
Difference in solubility of salts


Lanthanides usually show +3 oxidation state, but Eu commonly forms stable +2 state.

Step 1: Analyze pairs


Eu(II) vs Dy(III): different oxidation states \(\Rightarrow\) different chemistry
Gd(III) vs Dy(III): similar chemistry \(\Rightarrow\) difficult separation
Eu(II) vs Yb(II): both +2 \(\Rightarrow\) similar properties
Eu(II) vs hypothetical Gd(II): not common


Step 2: Reason

Eu(II) forms insoluble salts (like EuSO\(_4\)), while Dy(III) behaves like typical Ln\(^{3+}\).

Thus can be separated by selective precipitation.

Conclusion:
Eu(II) and Dy(III) can be separated. Quick Tip: Lanthanide separation: Similar Ln\(^{3+}\) ions hard to separate Exceptions: Eu\(^{2+}\) and Yb\(^{2+}\) Use oxidation state differences

Concept:
Ligand substitution and complex formation:

CN\(^-\) is strong field ligand
Replaces NH\(_3\) in coordination sphere


Step 1: Formation of P
\[ [Cd(NH_3)_4]^{2+} + 4CN^- \rightarrow [Cd(CN)_4]^{2-} \]

With K\(^+\): \[ P = K_2[Cd(CN)_4] \]

Step 2: Reaction with H\(_2\)S

S\(^{2-}\) forms highly insoluble CdS.

Even from complex, Cd\(^{2+}\) precipitates as CdS: \[ [Cd(CN)_4]^{2-} + H_2S \rightarrow CdS \downarrow \]

Conclusion: \[ P = K_2[Cd(CN)_4], \quad Q = CdS \] Quick Tip: Coordination chemistry tips: CN\(^-\) forms strong complexes Sulfides of Cd are highly insoluble H\(_2\)S used for qualitative analysis of metal ions

Concept:
The given compound is a \(\beta\)-keto ester (like ethyl acetoacetate), containing:

Active methylene group
Keto–enol tautomerism


(A) Tautomerism
\(\beta\)-keto esters show keto–enol tautomerism due to acidic \(\alpha\)-hydrogen.

Thus true.



(B) Reaction with sodium

Active methylene hydrogen reacts with Na metal producing H\(_2\) gas.

So statement is false.



(C) FeCl\(_3\) test

Enol form behaves like phenol and gives colored complex with FeCl\(_3\) (reddish-violet).

Thus true.



(D) 2,4-DNP test

Compound contains ketone carbonyl group.

Ketones give yellow/orange precipitate with 2,4-DNP.

Thus true.

Conclusion:
Correct statements: A, C, D. Quick Tip: \(\beta\)-keto ester properties: Keto–enol tautomerism Active methylene acidity Positive FeCl\(_3\) (enol form) Positive 2,4-DNP (carbonyl present)

Concept:

Extensive property depends on system size
Intensive property independent of size


Let system size scale by factor \(\lambda\): \[ X \rightarrow \lambda X, \quad x \rightarrow x \]

(A) \(Xx\)
\[ Xx \rightarrow (\lambda X)(x) = \lambda (Xx) \]

Thus extensive. Correct.



(B) \(\frac{X}{x}\)
\[ \frac{X}{x} \rightarrow \frac{\lambda X}{x} = \lambda \frac{X}{x} \]

So extensive, not intensive. False.



(C) \(\frac{X}{x}\) is extensive

True.



(D) \(\frac{dX}{dx}\)

Derivative of extensive w.r.t intensive typically remains extensive (depends on scaling behavior), not necessarily intensive.

Thus false.

Conclusion:
Correct statements: A and C. Quick Tip: Scaling trick: Multiply by intensive \(\Rightarrow\) remains extensive Divide extensive by intensive \(\Rightarrow\) still extensive Ratio of two extensive properties \(\Rightarrow\) intensive

Concept:
A compound is optically active if:

It has at least one chiral center
No internal plane of symmetry (no meso form)


Analyze each option:


(A)

Central carbon has: \[ COOH, NH_2, H, H \]

Two identical substituents (H, H) \(\Rightarrow\) achiral.

Not optically active.



(B)

Central carbon attached to: \[ COOH, OH, CH_3, H \]

All four substituents different \(\Rightarrow\) chiral center present.

Optically active.



(C)

Structure resembles glyceraldehyde type: \[ CHO, OH, CH_2OH, H \]

Four different groups around carbon \(\Rightarrow\) chiral.

Optically active.



(D)

Contains two stereocenters but has internal plane of symmetry (meso compound).

Thus optically inactive.



Conclusion:
Optically active compounds: B and C. Quick Tip: Optical activity checklist: Four different groups on carbon \(\Rightarrow\) chiral Check for meso symmetry Multiple chiral centers \(\neq\) always optically active