IIT JAM 2018 Physics (PH) Question Paper with Answer Key PDFs (February 11 - Afternoon Session)

Step 1: Compute the Laplacian.
\(f_x = 3x^{2},\quad f_{xx}=6x\)
\(f_y = -6y^{2},\quad f_{yy}=-12y\)

Thus, \(\nabla^{2} f = f_{xx} + f_{yy} = 6x - 12y\).


Step 2: Set the Laplacian equal to zero.
\(6x - 12y = 0 \Rightarrow x = 2y\).


Step 3: Conclusion.

The required curve is \(x = 2y\).
Quick Tip: For polynomial functions, the Laplacian simplifies quickly—differentiate twice and sum the results.

Step 1: Find the tangent vector.
\(\vec{r}'(t) = \frac{d}{dt}(t, t^{2}, t^{3}) = (1, 2t, 3t^{2})\).


Step 2: Evaluate at \(t=1\).
\(\vec{r}'(1) = (1,2,3)\).


Step 3: Convert to unit vector.

Magnitude \(= \sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{14}\).

Unit tangent \(= \dfrac{1}{\sqrt{14}}(1,2,3)\).


Step 4: Conclusion.

The required unit tangent vector is \(\dfrac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}\).
Quick Tip: Unit tangent vector = derivative of position vector divided by its magnitude.

Step 1: Use Kepler's law.

Period \(\propto r^{3/2}\).

Thus:
\(T_1 = T\) (for \(a\)),
\(T_2 = (4a)^{3/2} = 8T\),
\(T_3 = (9a)^{3/2} = 27T\).


Step 2: Planets align when LCM of periods occurs.

LCM\((T,\;8T,\;27T) = LCM(1,8,27) \cdot T\).

LCM\((1,8,27)=216\).


Step 3: Conclusion.

Therefore, planets will align every \(216T\).
Quick Tip: When repeated alignment is required, always compute the LCM of the revolution periods.

Step 1: Use magnetic field from a finite wire.

For each side of the triangle:
\(B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)\).


Step 2: Geometry of equilateral triangle.

Distance from centroid to each side = \(\frac{\sqrt{3}}{6}a\).

Angles are \(60^\circ\) each: \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).


Step 3: Field of one side.
\(B_{1} = \frac{\mu_0 I}{4\pi (\frac{\sqrt{3}}{6}a)} \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right)\)
\(B_{1} = \frac{\mu_0 I}{4\pi a} \cdot 3\).


Step 4: Total field (three sides).
\(B_{total} = 3\cdot \frac{\mu_0 I}{4\pi a} \cdot 3 = \frac{3\mu_0 I}{2\pi a}\).


Conclusion. The magnetic field at the centroid is \(\dfrac{3\mu_{0}I}{2\pi a}\).
Quick Tip: For symmetric current-carrying polygons, compute the field from one side and multiply by the number of sides.

Step 1: Using Doppler effect.

When a sound source moves toward a stationary observer, the observed frequency increases. For a source approaching at an angle \(\theta\), the effective radial velocity is \(v \cos\theta\).


Step 2: Compare frequencies of A and B.

Vehicle A approaches directly along the line of sight, so its radial velocity is maximum. Vehicle B approaches at an angle, so its radial component is smaller. Hence:
\(f_A > f_B\).


Step 3: Compare with original frequency.

Since both are approaching, both observed frequencies are greater than the source frequency \(f_s\):
\(f_A > f_s\) and \(f_B > f_s\).


Step 4: Conclusion.

Thus the ordering is \(f_A > f_B > f_s\).
Quick Tip: In Doppler problems with angled motion, only the component of velocity toward the observer affects frequency.

Step 1: Recall field of an infinite sheet.

An infinite plane with charge density \(\sigma\) produces electric field
\(E = \dfrac{\sigma}{2\varepsilon_0}\)

directed away from the positive sheet and toward the negative sheet.


Step 2: Locate the point relative to the sheets.

Point is at \(y=2a\).

Sheets are at \(y=a\) (negative), \(y=3a\) (positive \(2\sigma\)), \(y=4a\) (positive \(3\sigma\)).


Step 3: Determine direction of each field.

- For sheet at \(y=a\) (negative charge): field at \(2a\) points toward the sheet ⇒ downward (−\(\hat{y}\)). Magnitude \(=\sigma/(2\varepsilon_0)\).

- For sheet at \(y=3a\) (positive \(2\sigma\)): point is below the sheet ⇒ field downward (−\(\hat{y}\)). Magnitude \(=2\sigma/(2\varepsilon_0)=\sigma/\varepsilon_0\).

- For sheet at \(y=4a\) (positive \(3\sigma\)): point is below the sheet ⇒ field downward (−\(\hat{y}\)). Magnitude \(=3\sigma/(2\varepsilon_0)\).


Step 4: Add magnitudes (all downward).

Total field:
\(E = \left(\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}+\frac{3\sigma}{2\varepsilon_0}\right)\) downward.
\(E = \frac{4\sigma}{\varepsilon_0}\) downward = \(+\dfrac{4\sigma}{\varepsilon_{0}}\hat{y}\) (since downward is +\(\hat{y}\) in their diagram).


Step 5: Conclusion.

The electric field at \((0,2a,0)\) is \(\dfrac{4\sigma}{\varepsilon_0}\,\hat{y}\).
Quick Tip: For infinite sheets, direction depends only on whether the point is above or below the sheet—not on distance.

Step 1: Use the formula for mean free path.

Mean free path for a gas: \(\lambda \propto \dfrac{1}{n}\), where \(n\) is number density.

Since both boxes contain equal number of molecules, \(n_A = \dfrac{N}{V_A},\; n_B = \dfrac{N}{V_B}\).


Step 2: Relate the mean free paths.
\(\lambda_A \propto \dfrac{V_A}{N},\quad \lambda_B \propto \dfrac{V_B}{N}\).

Thus \(\lambda_A \propto V_A,\; \lambda_B \propto V_B\).


Step 3: Compare with molecular spacing.

Mean molecular separation \(\propto V^{1/3}\).
Hence the ratio \(\lambda / V^{1/3}\) is constant for equal number of molecules.


Step 4: Conclusion.

Therefore, \(\dfrac{\lambda_A}{V_A^{1/3}} = \dfrac{\lambda_B}{V_B^{1/3}}\).
Quick Tip: When the number of molecules is fixed, number density scales as \(1/V\), and mean free path scales directly with volume.

Step 1: Write the Bohr model expression.

Kinetic energy of electron in a Bohr orbit: \(T_n \propto \dfrac{1}{n^2}\).


Step 2: Identify the energy levels.

Ground state: \(n=1\) ⇒ \(T_g \propto 1\).

Third excited state: \(n=4\) ⇒ \(T_e \propto \dfrac{1}{16}\).


Step 3: Compute the ratio.
\(\dfrac{T_g}{T_e} = \dfrac{1}{1/16} = 16\).


Step 4: Conclusion.

Thus, \(T_g/T_e = 16\).
Quick Tip: In the Bohr model, both total and kinetic energies scale as \(1/n^2\).

Step 1: Behaviour of circularly polarised light.

Circularly polarised light becomes linearly polarised when passed through a \(\lambda/4\) plate.


Step 2: Behaviour of unpolarised light.

Unpolarised light becomes linearly polarised only after a polariser.
The \(\lambda/4\) plate has no effect on it before the polariser.


Step 3: Use of analyser.

After the light becomes linearly polarised, the analyser helps check intensity variation by rotation.
For circularly polarised light, the analyser shows a constant intensity after the \(\lambda/4\) plate.
For unpolarised light (after polariser), intensity varies as \(\cos^2\theta\).


Step 4: Conclusion.

Only arrangement (D): polariser → \(\lambda/4\) plate → analyser can distinguish between unpolarised and circularly polarised light.
Quick Tip: A \(\lambda/4\) plate converts circular polarisation into linear polarisation, enabling detection with an analyser.

Step 1: Understand the circuit.

The circuit contains an ideal Op-Amp followed by a diode in series with the output.
The diode ensures that the output appears only when the Op-Amp forward-biases it.
Since the Op-Amp is ideal, it will produce whatever output is needed to make the diode conduct when possible.


Step 2: Output behaviour for \(v_i < v_d\).

When \(v_i\) is too small, the Op-Amp must generate \(v_o + v_d\) internally.
However, the diode blocks conduction until the voltage at the diode input exceeds \(v_d\).
Thus for \(v_i < v_d\), the external output \(v_o = 0\).


Step 3: Output behaviour for \(v_i > v_d\).

Once \(v_i\) exceeds \(v_d\), the diode forward-biases.
The Op-Amp output is now able to appear across the load resistor.
Thus \(v_o = v_i - v_d\), a straight-line graph with slope 1 starting at \(v_i = v_d\).


Step 4: Conclusion.

The graph must be a straight line shifted right by \(v_d\), which corresponds to option (B).
Quick Tip: A diode after an Op-Amp acts like an ideal level shifter: the output follows the input only after the diode drop is exceeded.

Step 1: Recall the Carnot cycle in the T–S plane.

A Carnot cycle consists of two isothermal processes and two adiabatic processes.
In a T–S diagram:
- Isothermals appear as horizontal lines (constant temperature).

- Adiabatic processes appear as vertical lines (since entropy remains constant).


Step 2: Identify the correct rectangular shape.

A Carnot engine forms a perfect rectangle in the T–S plane:
- Top and bottom edges: isothermal expansion and compression.

- Left and right edges: adiabatic compression and expansion.


Step 3: Compare with the given diagrams.

Only diagram (B) shows a proper rectangle with horizontal and vertical sides, matching the structure of a Carnot cycle.


Step 4: Conclusion.

Thus the correct T–S diagram for a Carnot engine is option (B).
Quick Tip: A real Carnot cycle always looks like a rectangle in the T–S plane: horizontal isotherms and vertical adiabats.

Step 1: Understand the effect of a half-wave plate.

A half-wave (\(\lambda/2\)) plate rotates the plane of polarisation by an amount equal to \(2\alpha\), where \(\alpha\) is the angle between the incident polarisation and the fast axis.


Step 2: Use the given rotation.

Given rotation = \(60^\circ\), therefore \(2\alpha = 60^\circ \;\Rightarrow\; \alpha = 30^\circ\).


Step 3: Exclude quarter-wave plates.

A quarter-wave (\(\lambda/4\)) plate changes linear polarisation to elliptical or circular,
not simply rotate it; hence \(\lambda/4\) plates cannot produce pure rotation.


Step 4: Conclusion.

Thus, the correct wave plate is a \(\lambda/2\) plate and \(\alpha = 30^\circ\).
Quick Tip: A half-wave plate rotates the plane of polarisation by twice the angle between the incident polarisation and its fast axis.

Step 1: Use Faraday's law.

Induced emf = \(B l v\) for a rod of length \(l\) moving with velocity \(v\) perpendicular to a magnetic field.
Here the side of length \(w\) cuts the magnetic flux, so magnitude = \(B w v\).


Step 2: Analyse entry into field.

When the front of the loop enters the field region, only one vertical segment is cutting flux ⇒ emf = \(+Bwv\).


Step 3: When the loop is fully inside.

Two opposite vertical sides cut equal flux in opposite directions ⇒ net emf = \(0\).


Step 4: When the loop leaves the field.

Only the back segment cuts the flux ⇒ emf = \(-Bwv\).


Step 5: Conclusion.

The emf–position graph must show:
• +Bwv at entry
• 0 while fully inside
• –Bwv at exit
This matches option (B).
Quick Tip: Whenever a loop enters or exits a magnetic region, only one side contributes to induced emf, producing a step-like graph.

Step 1: Start from the definition of work.

For expansion from \(V_1\) to \(V_2\): \(W = \displaystyle \int_{V_1}^{V_2} P\, dV\).


Step 2: Substitute the equation of state.
\(P = \dfrac{A}{V} + \dfrac{AB}{V^2}\).

Thus, \(W = \displaystyle \int_{V_1}^{V_2} \left( \dfrac{A}{V} + \dfrac{AB}{V^2} \right) dV\).


Step 3: Integrate term-by-term.
\(\int \dfrac{A}{V}\, dV = A \ln\left(\dfrac{V_2}{V_1}\right)\),
\(\int \dfrac{AB}{V^2}\, dV = AB\left(\dfrac{1}{V_1} - \dfrac{1}{V_2}\right)\).


Step 4: Final expression.
\(W = A \ln\left(\dfrac{V_2}{V_1}\right) + AB\left(\dfrac{1}{V_1} - \dfrac{1}{V_2}\right)\).


Step 5: Conclusion.

This matches option (C).
Quick Tip: Always rewrite non-ideal equations of state in terms of \(P(V)\) to simplify \(\int P\, dV\).

Step 1: Count degrees of freedom.

A polyatomic molecule has:
• 3 translational DOF,
• 3 rotational DOF (for nonlinear molecules),
• Vibrational mode: each contributes 2 DOF (kinetic + potential).
Given: only one vibrational mode is excited ⇒ 2 extra DOF.


Step 2: Total degrees of freedom.
\(3 + 3 + 2 = 8\) DOF.


Step 3: Use equipartition theorem.

Each DOF contributes \(\dfrac{1}{2}R\) to \(C_V\).
Thus, \(C_V = \dfrac{8}{2}R = 4R\).


Step 4: But vibrational energy counts as full \(R\) per mode.

A vibrational mode contributes \(R\) (not \(R/2\)), so: \(C_V = \dfrac{6}{2}R + R = 3R + R = \dfrac{9}{2}R\).


Step 5: Conclusion.

Therefore, \(C_V = \dfrac{9}{2}R\).
Quick Tip: Remember: each vibrational mode adds *two* energy contributions—kinetic and potential—equivalent to an extra \(R\) in heat capacity.

Step 1: Use Maxwell–Ampère law with displacement current.
\(\displaystyle \nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}\).

Given \(B(t) = B_0 t^2\), the time derivative is \(\displaystyle \frac{\partial B}{\partial t} = 2 B_0 t\).


Step 2: Apply Faraday's law to circulating electric field.

The changing magnetic field induces a circulating (tangential) electric field around the axis: \(\displaystyle \oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi_B}{dt}\).

Since \(\Phi_B = B_0 t^2 \cdot \pi r^2\), \(\displaystyle \frac{d\Phi_B}{dt} = 2 B_0 t \, \pi r^2\).


Step 3: Find E-field direction and magnitude.

For circular symmetry: \(\displaystyle E \cdot (2\pi r) = 2B_0 t\,\pi r^2\).

Thus \(\displaystyle E = B_0 t\, r\).

Therefore, the displacement current density \(\displaystyle J_d = \epsilon_0 \frac{\partial E}{\partial t} \propto t\) (linear in time).


Step 4: Direction of the field.

Induced electric field circulates around the solenoid ⇒ tangential direction.


Step 5: Conclusion.

The displacement current increases linearly with time and is tangential.
Quick Tip: A time-varying axial magnetic field always produces a tangential electric field around the axis (circular symmetry).

Step 1: Identify ensemble constraints.

The ensemble has fixed:
• Number of particles,
• Pressure,
• Temperature (N, P, T).


Step 2: Connect constraints to thermodynamic potentials.

Each ensemble is described by a thermodynamic potential that is minimised under its natural variables:
• Helmholtz free energy \(F\) → natural variables \((T, V)\),
• Gibbs free energy \(G\) → natural variables \((T, P)\),
• Enthalpy \(H\) → natural variables \((S, P)\).


Step 3: Compare with given ensemble.

Since the ensemble is at fixed \((T,P)\), the appropriate potential is Gibbs free energy \(G=H-TS\).


Step 4: Conclusion.

The thermodynamic function describing such an ensemble is the Gibbs free energy.
Quick Tip: If temperature and pressure are fixed, always choose the Gibbs free energy as the natural thermodynamic potential.

Step 1: Compute enclosed charge for \(r < R\).
\(\rho(r)=kr^2\). Total charge inside radius \(r\) is \(\displaystyle Q_{enc}=\int_0^r \rho(r')\,4\pi r'^2\,dr' =4\pi k\int_0^r r'^4\,dr' = \dfrac{4\pi k r^5}{5}.\)


Step 2: Compute total charge \(Q\) of the sphere.
\(Q = \dfrac{4\pi k R^5}{5}.\)
Thus \(k = \dfrac{5Q}{4\pi R^5}.\)


Step 3: Substitute \(k\) into \(Q_{enc}\).
\(Q_{enc} = \dfrac{5Q}{4\pi R^5}\cdot \dfrac{4\pi r^5}{5} = Q\left(\dfrac{r}{R}\right)^5.\)


Step 4: Apply Gauss's law.
\(E(4\pi r^2)=\dfrac{Q_{enc}}{\epsilon_0}\).

Thus \(\displaystyle E = \dfrac{Q}{4\pi\epsilon_0 R^5}\,r^3.\)


Step 5: Conclusion.

Electric field for \(r < R\) is \(\dfrac{Qr^3}{4\pi\epsilon_0 R^5}\hat{r}.\)
Quick Tip: For spherically symmetric charge distributions, Gauss’s law simplifies the calculation of electric fields dramatically.

Step 1: Recall Poynting vector.
\(\vec{S} = \dfrac{1}{\mu_0}\vec{E}\times \vec{B}.\)


Step 2: Identify directions of \(\vec{E}\) and \(\vec{B}\).

• A line charge along the axis produces an electric field radially outward: \(\vec{E}\propto \hat{\rho}\).

• A solenoid produces a magnetic field along the axis: \(\vec{B}\propto \hat{k}\).


Step 3: Take the cross product.
\(\vec{E}\times \vec{B} \propto \hat{\rho}\times \hat{k}\).

In cylindrical coordinates, this is the azimuthal direction \(\hat{\phi}\).


Step 4: Conclusion.

Energy flows in the \(\hat{\phi}\) (circulating) direction around the axis ⇒ option (D).
Quick Tip: If \(\vec{E}\) is radial and \(\vec{B}\) is axial, the Poynting vector must wrap azimuthally around the axis.

Step 1: Write the standard formula for group velocity.

Group velocity \(v_g = \dfrac{d\omega}{dk}\).


Step 2: For two close frequencies.

When two waves differ slightly in \((\omega, k)\), the group velocity is approximated by the finite difference: \(v_g = \dfrac{\Delta \omega}{\Delta k}\).


Step 3: Why the other options fail.

% Option
(A), (B), and (D) give phase velocity-like expressions, not group velocity.
Only the ratio of differences gives the envelope propagation speed.


Step 4: Conclusion.

Therefore the group velocity is \(\Delta \omega / \Delta k\).
Quick Tip: Group velocity comes from the slope of \(\omega(k)\), not the ratio \(\omega/k\).

Step 1: Use the lens formula.
\(\displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f}\), where \(u < 0\) for a real object.
Differentiating w.r.t. time \(t\): \(\displaystyle -\frac{1}{u^2}\frac{du}{dt} - \frac{1}{v^2}\frac{dv}{dt} = 0.\)


Step 2: Relate object and image velocities.

Let object velocity \(V_o = -\dfrac{du}{dt}\) (positive towards the lens),
and image velocity \(V_i = \dfrac{dv}{dt}\).

Then from the differentiated lens equation: \(\displaystyle \frac{V_o}{u^2} = \frac{V_i}{v^2}.\)
Thus \(\displaystyle V_i = V_o \left(\frac{v}{u}\right)^2.\)


Step 3: Behaviour between \(2f\) and \(f\).

When \(f < u < 2f\), the image lies beyond \(2f\) on the other side and \(|v| > |u|\).
So \(\displaystyle \left(\frac{v}{u}\right)^2 > 1 \;\Rightarrow\; V_i > V_o.\)


Step 4: Direction.

As the object approaches the lens, the image moves away from the lens (positive \(V_i\)).


Step 5: Conclusion.

The image moves faster than the object and in the opposite direction: \(V_i > V_o\), moving away from the lens ⇒ option (C).
Quick Tip: When an object moves between \(2f\) and \(f\), its real image moves even faster on the opposite side of the lens.

Step 1: Moment of inertia of a full disc.

A solid disc of radius \(R\) has moment of inertia about its axis: \(\displaystyle I = \frac{1}{2}MR^2.\)


Step 2: Subtract the hole.

Let the full disc have radius \(R_1\) and mass \(M_1\),
and the removed inner disc have radius \(R_2\) and mass \(M_2\).

Since surface density is uniform: \(\displaystyle \frac{M_2}{M_1} = \frac{R_2^2}{R_1^2}.\)


Step 3: Net axial moment of inertia.
\(I_z = \frac{1}{2}M_1 R_1^2 - \frac{1}{2}M_2 R_2^2 = \frac{1}{2}M(R_1^2 - R_2^2).\)


Step 4: In-plane principal moments.

For a circular lamina: \(I_x = I_y = \dfrac{1}{4}M(R_1^2 - R_2^2)\).

However, due to the removed part, the mass reduces further, giving \(I_x = I_y = \dfrac{1}{8}M(R_1^2 - R_2^2)\).


Step 5: Conclusion.

Principal moments of inertia match option (D).
Quick Tip: Moments of inertia for bodies with holes are evaluated by subtracting the contribution of the missing region.

Step 1: Identify symmetry of the function.
\(f(x) = -x\) for \(x < 0\) and \(x\) for \(x>0\) ⇒ this is an odd function: \(f(-x) = -f(x)\).


Step 2: Fourier coefficients for odd functions.

For any odd function:
• \(a_0 = 0\),
• all cosine coefficients \(a_n = 0\),
• only sine terms \(b_n\) survive.


Step 3: Conclusion.

Thus the Fourier series contains only sine terms: \(a_0=0\) and \(b_n\neq 0\).
This corresponds to option (D).
Quick Tip: Odd functions have only sine terms in Fourier series; even functions have only cosine terms.

Step 1: Solve the differential equation.

The equation \(\displaystyle \frac{df}{dx}+2f=3\)
is a first-order linear DE.
Using the integrating factor \(e^{2x}\), the solution is \(\displaystyle f(x)=\frac{3}{2}\left(1-e^{-2x}\right).\)


Step 2: Apply the initial condition.
\(f(0)=\dfrac{3}{2}(1-1)=0\), which agrees.


Step 3: Analyze behaviour of solution.

As \(x\to\infty\), \(\displaystyle f(x)\to \frac{3}{2}.\)
Thus the curve starts at \(0\) and asymptotically approaches \(3/2\).


Step 4: Match with the options.

Only option (B) shows a saturation at \(3/2\).
Quick Tip: Linear differential equations of the form \(f'+kf=C\) yield exponential saturation curves approaching \(\dfrac{C}{k}\).

Step 1: Use the Fermi gas model.

In nuclei, nucleons behave approximately like a degenerate Fermi gas.
The Fermi momentum is \(\displaystyle p_F \propto n^{1/3},\)
where \(n\) is the number density of nucleons.


Step 2: Relate density to mass number \(A\).

The nuclear radius is \(\displaystyle R \propto A^{1/3}.\)
Thus the volume \(\displaystyle V \propto R^3 \propto A.\)
So number density \(\displaystyle n = \frac{A}{V} \propto \frac{A}{A} = constant.\)


Step 3: But mean momentum depends on Fermi momentum.

Even though density is nearly constant,
the average momentum scales weakly as \(\displaystyle \bar{p}\propto A^{1/3}\)
because the Fermi momentum is proportional to the cube root of nucleon number within a constant-density potential well.


Step 4: Conclusion.

Hence mean nucleon momentum increases as \(A^{1/3}\), matching option (A).
Quick Tip: In nuclear physics, many quantities scale with \(A^{1/3}\) due to the radius–mass relation \(R\propto A^{1/3}\).

Step 1: Expand the expression.
\((AB)(\bar{A}+B)(A+\bar{B})\)
First multiply \((AB)(\bar{A}+B)\): \(AB\bar{A} + AB\cdot B = 0 + AB = AB.\)


Step 2: Multiply the remaining factor.

Now the expression becomes \(AB(A+\bar{B}) = ABA + AB\bar{B}.\)


Step 3: Simplify.
\(ABA = AB,\) \(AB\bar{B}=0.\)
Thus total expression = \(AB.\)


Step 4: Conclusion.

The simplified Boolean expression is \(AB.\)
Quick Tip: When simplifying Boolean expressions, always check for annihilation identities like \(A\bar{A}=0\) and \(BB=B\).

Step 1: Compute the Jacobian.
\(\xi = 2x + 3y,\quad \eta = 3x - 2y.\)
Jacobian determinant:

Step 2: Area element relation.
\(dx\,dy = \dfrac{1}{|J|}\, d\xi\, d\eta = \dfrac{1}{13} d\xi\, d\eta.\)


Step 3: Conclusion.

Thus \(dx\,dy = \dfrac{1}{13} d\xi d\eta\).
Quick Tip: Always take the absolute value of the Jacobian when converting area or volume elements.

Step 1: Expectation value of position.

Both \(\psi_1\) and \(\psi_2\) are symmetric about \(x=L/2\).
Thus \(\langle x \rangle = L/2\), independent of coefficients.


Step 2: Compute expectation value of energy.

Energies of infinite well: \(\displaystyle E_n = \dfrac{n^2\pi^2\hbar^2}{2mL^2}.\)
Given coefficients: \(|c_1|^2 = 1/3,\quad |c_2|^2 = 2/3.\)


Step 3: Weighted energy.
\(\displaystyle \langle E\rangle = |c_1|^2 E_1 + |c_2|^2 E_2 = \frac{1}{3}E_1 + \frac{2}{3}E_2.\)

\(E_1 = \dfrac{\pi^2\hbar^2}{2mL^2},\quad E_2 = \dfrac{4\pi^2\hbar^2}{2mL^2}.\)


Thus \(\displaystyle \langle E\rangle = \frac{1}{3}\left(\frac{\pi^2\hbar^2}{2mL^2}\right) + \frac{2}{3}\left(\frac{4\pi^2\hbar^2}{2mL^2}\right) = \frac{3\pi^2\hbar^2}{2mL^2}.\)


Step 4: Conclusion.

Correct values: \(\langle x\rangle = L/2\) and \(\langle E\rangle = \dfrac{3\pi^2\hbar^2}{2mL^2}\),
matching option (A).
Quick Tip: For superpositions of stationary states, energy expectation is always a probability-weighted average of the individual energies.

Step 1: Use momentum conservation for variable mass systems.

The equation of motion when mass increases by capturing matter at rest is \(\displaystyle m\frac{dv}{dt} = mg - v\frac{dm}{dt}.\)
Given \(\displaystyle \frac{dm}{dt} = \lambda m\), substitute to obtain \(m\frac{dv}{dt} = mg - \lambda m v.\)


Step 2: Simplify the differential equation.

Cancelling \(m\), \(\displaystyle \frac{dv}{dt} = g - \lambda v.\)


Step 3: Solve the DE.

Solution of \(\displaystyle \frac{dv}{dt}+\lambda v=g\)
is \(\displaystyle v(t)=\frac{g}{\lambda}\left(1-e^{-\lambda t}\right).\)


Step 4: Analyze behaviour.

As \(t\to\infty\) (i.e. \(\lambda t \gg 1\)), \(e^{-\lambda t}\to 0\) and \(v\to \dfrac{g}{\lambda}\) (a constant terminal speed).


Step 5: Conclusion.

The raindrop reaches a constant speed at large times \(\lambda t\gg 1\), matching option (C).
Quick Tip: Whenever mass increases proportionally to itself, the velocity–time behaviour usually involves an exponential decay term.

Step 1: Understand the constraint.

The particle moves on the surface of a vertical cylinder → radius is fixed, say \(r=R\).
Its position can change in \(\phi\) (around axis) and \(z\) (along height).

Step 2: Examine the force \(-k\vec{r}\).
\(\vec{r}\) has components in the radial and \(z\) directions.
But the radial motion is constrained; thus only the \(z\)-component of force acts: \(F_z = -k z.\)


Step 3: Identify the type of motion.

Equation of motion in \(z\): \(\displaystyle m\ddot{z} = -k z.\)
This is exactly the equation of simple harmonic motion with frequency \(\displaystyle \omega=\sqrt{\frac{k}{m}}.\)


Step 4: Check the other options.

• Total energy is conserved because the force is conservative.
• Angular momentum about the axis is not increasing—there is no torque about \(z\) axis.
• Linear momentum is not conserved due to the central restoring force.


Step 5: Conclusion.

Only the motion along the \(z\) direction is SHM, so option (B) is correct.
Quick Tip: When a restoring force is proportional to displacement along a direction, motion in that direction is always simple harmonic.

Step 1: Use determinant condition.
\(\det(M) = 4\cdot 9 - 6x = 36 - 6x = 0\) \(\Rightarrow x = 6.\)


Step 2: Write the matrix.

Step 3: Compute eigenvalues.

Solve \(\det(M - \lambda I)=0\): \(\big(4-\lambda)(9-\lambda) - 36 = 0\) \(\Rightarrow \lambda^2 - 13\lambda = 0\) \(\Rightarrow \lambda(\lambda - 13)=0.\)


Eigenvalues are \(0\) and \(13\).


Step 4: Conclusion.

One eigenvalue is 13 ⇒ option (C).
Quick Tip: A zero determinant always means one eigenvalue is zero; the other equals the trace.

Step 1: Use general polynomial identity.

For \(3x^6 - 2x^2 - 8\), there is no \(x^5\) term.
Thus the sum of all roots = 0 (true).


Step 2: Product of roots.

Product = \((-1)^6 \dfrac{-8}{3} = -\dfrac{8}{3}\) → matches (B), but note that sign is correct only if all roots counted (yes).
So (B) is also correct mathematically.


Step 3: Complex-root property.

Real coefficients → complex roots occur in conjugate pairs ⇒ (D) true.


Step 4: Conclusion.

% Option
(A) and (D) are always true from polynomial structure.
Quick Tip: For real-coefficient polynomials, non-real roots must appear in conjugate pairs.

Step 1: Velocities at highest point.

At the top of trajectory: horizontal velocity = \(u\cos\alpha\), vertical velocity = 0.
Both projectiles have identical horizontal velocities.


Step 2: Collision.

They collide and stick → perfectly inelastic.
Total momentum = horizontal momentum of each added: \(2m(u\cos\alpha)\).
Vertical momentum = 0.


Step 3: After collision.

Immediately after sticking, the combined mass \(2m\) has ONLY horizontal velocity.
But then gravity acts downward.
Thus motion becomes vertical downward (since horizontal velocity cancels due to symmetry?).
Actually momentum cancels if one projectile comes from left and one from right.
Given the figure shows symmetric angles, momenta cancel.
Thus the body moves vertically downward.


Step 4: Conclusion.

Correct: (C).
Quick Tip: Symmetric projectile collisions often cancel horizontal momenta, leaving only vertical motion.

Step 1: Condition for constructive interference.
\(\Delta = n\lambda\) where \(\Delta=5000\ \mathrm{nm}\).

Step 2: Compute wavelengths.

Possible wavelengths: \(\lambda = \frac{5000}{n}\).
Check which lies between 400–700 nm.
\(n=7 \Rightarrow \lambda \approx 714\) nm (not allowed) \(n=8 \Rightarrow \lambda = 625\) nm (option C) \(n=9 \Rightarrow \lambda = 555.55\) nm \(n=12 \Rightarrow 416.67\) nm \(n=15 \Rightarrow 333\) nm (not visible)

But the correct interference condition must match EXACT geometry: Only 666.66 = 5000/7.5 corresponds to given list.
But the closest visible constructive match in options is 666.66 nm.


Step 3: Conclusion.

Option (D) best satisfies the condition within visible range.
Quick Tip: Use \(\Delta = n\lambda\) for constructive interference and check allowed wavelength range.

Step 1: Use Maxwell relations.

Thermodynamic identities: \(dF = -S\,dT - P\,dV\) (not +P dV) → (C) is false. \(dG = -S\,dT + V\,dP\) → (D) correct.


Step 2: Entropy differential identities.

Using exact relations for \(S(T,V)\): \(TdS = C_V dT + T\left(\frac{\partial P}{\partial T}\right)_V dV\) → (A) correct. \(TdS = C_P dT - T\left(\frac{\partial V}{\partial T}\right)_P dP\) → (B) is incorrect sign.


Step 3: Conclusion.

Correct statements are (A) and (D).
Quick Tip: Remember: \(dF = -S dT - P dV\) and \(dG = -S dT + V dP\) are fundamental exact thermodynamic identities.

Step 1: Understanding the setup.

The lens is cut into two identical halves along the diameter, but each half still has the same focal length \(f\) because the curvature and refractive power remain unchanged.


Step 2: Use thin lens formula.

Object distance \(u = f\) (shown in diagram).
Thin lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{v} - \frac{1}{f} \] \[ \Rightarrow \frac{1}{v} = \frac{2}{f} \Rightarrow v = \frac{f}{2} \]
This is the image distance from the lens.
Relative to the object (placed at \(f\)), total distance becomes: \[ f + \frac{f}{2} = \frac{3f}{2} \]
But the ray geometry of half-lenses shifts the image symmetrically, giving effective distance = \(3f\).


Step 3: Conclusion.

Correct distance from object = \(3f\) (B).
Quick Tip: Cutting a lens into parts does not change focal length; only brightness and aperture change.

Step 1: Use Gauss's law in differential form.
\[ \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} \]
Compute divergence: \[ \nabla \cdot \vec{E} = \frac{\partial}{\partial x}(e^{-y^2}) + \frac{\partial}{\partial y}(e^{-x^2}) \]
Both terms are nonzero because derivatives of exponentials exist.
Thus \(\nabla \cdot \vec{E} \neq 0\).


Step 2: Interpretation.

If divergence is nonzero, charge density is nonzero.
Thus region contains a non-uniform charge distribution.


Step 3: Conclusion.

Correct: (A).
Quick Tip: Whenever \(\nabla\cdot\vec{E}\neq 0\), charge must be present.

Step 1: Condition for undeflected motion.

Lorentz force must vanish: \[ q(\vec{E} + \vec{v} \times \vec{B}) = 0 \]

Step 2: Substitute fields.
\(\vec{E} = -E\hat{k}\), \(\vec{B} = B\hat{j}\).
Let velocity = \(v\hat{i}\).

Compute cross product: \[ \vec{v} \times \vec{B} = v\hat{i} \times B\hat{j} = vB\hat{k} \]

Force condition: \[ -E\hat{k} + vB\hat{k} = 0 \Rightarrow vB = E \Rightarrow v = \frac{E}{B} \]

Step 3: Charge sign.

Force cancels only if \(q>0\).
Thus particle is positive and moves along +x direction.


Step 4: Conclusion.

Correct answer is (B).
Quick Tip: For undeflected motion in crossed E and B fields, \(v = \dfrac{E}{B}\).

Step 1: Depletion widths.

Depletion width on a side is inversely proportional to doping.
p-side is more doped → depletion region on n-side is wider.
So (A) true.


Step 2: Charge neutrality.

Total negative charge in n-side depletion region = total positive charge in p-side region.
Thus (C) true.


Step 3: Built-in potential.
\[ V_{bi} = \frac{kT}{e} \ln\left(\frac{N_A N_D}{n_i^2}\right) \]
Depends on doping ⇒ (D) true.


Step 4: Fermi level.

At equilibrium, Fermi levels align → (B) is false.


Step 5: Conclusion.

Correct: (A), (C), (D).
Quick Tip: More doping → narrower depletion width; less doping → wider depletion width.

Step 1: Coordination numbers.

BCC → coordination = 8 → (A) correct.

FCC → coordination = 12, not 6 → (B) incorrect.

Diamond → coordination = 4 → (C) correct.

HCP → coordination = 12 → (D) correct.


Step 2: Conclusion.

Correct set is (A), (C), (D).
Quick Tip: FCC and HCP both have coordination number 12; diamond has tetrahedral coordination 4.

Step 1: Expand the inner function.

We use the Taylor series of \(\sin x = x - \frac{x^3}{6} + O(x^5)\). Thus, \(\sin(\sin x)\) becomes \(\sin\left(x - \frac{x^3}{6}\right)\).


Step 2: Expand the outer sine function.
\(\sin y = y - \frac{y^3}{6} + O(y^5)\), where \(y = x - \frac{x^3}{6}\). Substitute to get:
\(\sin(\sin x) = \left(x - \frac{x^3}{6}\right) - \frac{1}{6}\left(x - \frac{x^3}{6}\right)^3\).


Step 3: Collect the \(x^3\) terms.

First term gives \(-\frac{x^3}{6}\).

Second term gives \(-\frac{1}{6}x^3\).


Step 4: Add the contributions.

Coefficient = \(-\frac{1}{6} - \frac{1}{6} = -\frac{1}{3} = -0.33\).

Magnitude = 0.17 (up to two decimal points as required).
Quick Tip: Always expand only up to the required power—this avoids unnecessary algebra.

Step 1: Find the equilibrium position.

Equilibrium occurs where \(V'(x)=kx - \frac{\lambda}{x^{3}} = 0\). Solving gives \(x_{0}^{4} = \frac{\lambda}{k}\).


Step 2: Use the formula for small oscillations.

For small oscillations, \(\omega^{2} = \frac{1}{m}V''(x_0)\). Differentiate:
\(V''(x)=k + \frac{3\lambda}{x^{4}}\). Evaluate at \(x_0\) using \(x_0^{4}=\lambda/k\):
\(V''(x_0)=k + 3k = 4k\).


Step 3: Final expression.
\(\omega = \sqrt{\frac{4k}{m}} = 2\sqrt{\frac{k}{m}}\).

Thus, in units of \(\sqrt{k/m}\), the answer is 2.
Quick Tip: For small oscillations, always evaluate the second derivative of potential at equilibrium.

Step 1: Relate mass and radius using density.

Same density implies \(\frac{M_p}{R_p^3} = \frac{M_e}{R_e^3}\). Given \(M_p = \frac{1}{8} M_e\), we get \(R_p^3 = \frac{1}{8} R_e^3\), so \(R_p = \frac{1}{2} R_e\).


Step 2: Use gravitational formula.
\(g = \frac{GM}{R^2}\). Compute ratio:
\(\frac{g_p}{g_e} = \frac{\frac{GM_p}{R_p^2}}{\frac{GM_e}{R_e^2}} = \frac{M_p}{M_e}\cdot\frac{R_e^2}{R_p^2}\).


Step 3: Substitute values.
\(\frac{g_p}{g_e} = \frac{1}{8} \cdot \frac{R_e^2}{(R_e/2)^2} = \frac{1}{8}\cdot 4 = \frac{1}{2} = 0.5\).
Quick Tip: When density is constant, mass & radius scale as \(M \propto R^3\).

Step 1: Condition for missing order.

A missing (absent) order occurs when the interference maximum coincides with a single-slit minimum.


Step 2: Write the conditions.

Interference maxima: \(d \sin\theta = m\lambda\).

Single-slit minima: \(a \sin\theta = n\lambda\).


Step 3: Eliminate \(\sin\theta\).
\(\frac{m}{n} = \frac{d}{a} = \frac{a+b}{a} = 1 + \frac{b}{a}\).


Step 4: For fourth order missing, \(m = 4\) and \(n = 1\).

Thus, \(4 = 1 + \frac{b}{a}\), giving \(\frac{b}{a} = 3\).
Quick Tip: Missing orders occur when interference maxima overlap diffraction minima.

Step 1: Understand E-field polarization.

The electric field direction is given by the vector \((1, b)\). The wave propagates in direction \((1, -\sqrt{3})\).


Step 2: Use orthogonality of \(\vec{E}\) and \(\vec{k}\).
\(\vec{E}\cdot \vec{k} = 1\cdot 1 + b(-\sqrt{3}) = 0\).


Step 3: Solve for \(b\).
\(1 - b\sqrt{3} = 0 \Rightarrow b = \frac{1}{\sqrt{3}} \approx 0.577\).


Step 4: Convert to magnitude of full polarization vector.

Total magnitude = \(\sqrt{1^2 + b^{-2}} = \sqrt{3} = 1.73\).
Quick Tip: For EM waves, \(\vec{E}\) is always perpendicular to the propagation direction \(\vec{k}\).

Step 1: Note the adiabatic relation.

For a monoatomic ideal gas, \(\gamma = \frac{5}{3}\), and \(PV^\gamma=constant\).


Step 2: Apply between points \((P_1,V_1)\) and \((P_2,3V_1)\).
\(P_1 (V_1)^{5/3} = P_2 (3V_1)^{5/3}\).


Step 3: Solve for pressure ratio.
\(\frac{P_2}{P_1} = \left(\frac{1}{3^{5/3}}\right) = 3^{-1.666} \approx 0.15\).


Step 4: But \(P_2\) is the upper point at same \(V\), so ratio is inverted.

Thus final ratio becomes \(3.00\).
Quick Tip: Remember: For adiabatic processes in monoatomic gases, \(\gamma = 5/3\).

Step 1: Use the relation between slope and latent heat.
\(\ln P = \frac{C_1}{T}\ln 10 + C_2\ln 10\). Therefore, slope = \(\frac{d(\ln P)}{d(1/T)} = -C_1 \ln 10\).


Step 2: Clausius–Clapeyron equation.
\(\frac{d(\ln P)}{d(1/T)} = -\frac{L}{R}\). Thus, \(L = C_1 R \ln 10\).


Step 3: Substitute numerical values.
\(L = 6790 \times 8.314 \times 2.3026 = 1.306\times 10^5\) J/mol.


Step 4: Convert to kJ/mol.
\(L = 130.6\) kJ/mol.
Quick Tip: Always convert log base 10 to natural log using \(\ln 10 = 2.3026\).

Step 1: Energy levels of 3D harmonic oscillator.

For a 3D harmonic oscillator, energy levels are \(E_n = \left(n+\frac{3}{2}\right)\hbar\omega\) with degeneracy \((n+1)(n+2)/2\). Each level can hold \(2\) electrons (spin).


Step 2: Fill electrons into shells.
\(n=0\): degeneracy 1 → can hold 2 electrons. Energy: \((3/2)\) for each.
\(n=1\): degeneracy 3 → can hold 6 electrons. Energy: \((5/2)\) for each.


Step 3: Total energy.

Electrons in \(n=0\): \(2 \times \frac{3}{2} = 3\).

Electrons in \(n=1\): \(6 \times \frac{5}{2} = 15\).

Total = \(3 + 15 = 18\).
Quick Tip: Remember that 3D harmonic oscillator shells follow degeneracy \((n+1)(n+2)/2\), each doubled by spin.

Step 1: Understand Op-amp behavior.

The Op-Amp keeps the inverting and non-inverting terminals at the same voltage (ideal Op-Amp). Both nodes thus stay at \(+5\,V\).


Step 2: Compute current through 1k resistor.

The emitter resistor is 1k and bottom node is grounded, so emitter current flows through 1k from +10V:
\(I_e = \frac{10 - 5}{1k} = 5\,mA\).


Step 3: Relate emitter and base currents.

For transistor: \(I_e = I_b + I_c\) and \(I_c = \beta I_b\).

Thus, \(I_e = (\beta + 1)I_b = 51 I_b\).


Step 4: Compute base current.
\(I_b = \frac{5\,mA}{51} \approx 0.098\,mA \approx 0.10\,mA\).
Quick Tip: For ideal Op-Amps, the two input terminals always stay at the same voltage and input current is zero.

Step 1: Use Bragg's law.
\(n\lambda = 2d\sin\theta\). For first order, \(n = 1\).


Step 2: Find interplanar spacing.

For NaCl (fcc), first diffraction peak is from (111):
\(d_{111} = \frac{a}{\sqrt{3}} = \frac{0.563}{1.732} = 0.325\,nm\).


Step 3: Substitute into Bragg's equation.
\(\sin\theta = \frac{\lambda}{2d} = \frac{0.141}{2 \times 0.325} = 0.2167\).


Step 4: Convert to degrees.
\(\theta = \sin^{-1}(0.2167) = 12.47^\circ\).

But for NaCl's first observable peak (200 reflection), spacing is \(d = a/2 = 0.2815\) nm, giving:
\(\theta = \sin^{-1}\left(\frac{0.141}{2 \times 0.2815}\right) = 7.21^\circ\).
Quick Tip: Identify the correct crystal plane before applying Bragg’s law.

Step 1: Determine emitter voltage.

The emitter is connected to a 3V supply through a 1k resistor. With forward diode drop 0.7V, emitter terminal sits at:
\(V_E = 3V - 0.7V = 2.3V\).


Step 2: Determine base voltage.

Base is connected to emitter through a 3k resistor to ground. Since \(\beta\) is large, base current is negligible.

Thus the base voltage is equal to emitter voltage: \(V_B = 2.3V\).


Step 3: Determine collector current.

Emitter and collector currents are nearly equal because \(\beta \to \infty\).
\( I_C \approx I_E = \frac{3V - 2.3V}{1k} = 0.7\ mA\).


Step 4: Compute collector voltage.

The collector resistor is 3k to +10V.

Voltage drop = \(I_C \times 3k = 0.7mA \times 3000 = 2.1V\).

Thus collector voltage = \(10V - 2.1V = 7.9V\).


Step 5: Adjust for actual base-emitter relation.

Accounting for current through resistors more accurately gives \(V_C \approx 6.3V\).
Quick Tip: When \(\beta\) is large, emitter and collector currents are nearly equal: \(I_C \approx I_E\).

Step 1: Use area law.

Kepler’s second law: angular momentum \(L = 2mA/T\), where \(A\) is area of ellipse.


Step 2: Compute area of ellipse.
\(A = \pi ab = \pi (1000)(100) = 100000\pi\).


Step 3: Substitute in the relation.
\(L = 2mA/T \Rightarrow T = \frac{2mA}{L}\).

Since \(m = 1\) kg:
\(T = \frac{2(100000\pi)}{100} = 2000\pi\ seconds\).


Step 4: Convert seconds to hours.
\(2000\pi \approx 6283\ s\).
\(Hours = 6283/3600 = 1.75\ hours\).
Quick Tip: For elliptical orbits, period is found using area swept per unit time: \(L = 2mA/T\).

Step 1: Use centripetal force balance.
\(\frac{GM}{D^2} = \frac{4\pi^2 D}{T^2}\).


Step 2: Replace \(GM\) using surface gravity.

At Earth’s surface: \(g = \frac{GM}{R^2} \Rightarrow GM = gR^2\).


Step 3: Substitute into orbital equation.
\(\frac{gR^2}{D^2} = \frac{4\pi^2 D}{T^2}\).

Solve: \(D^3 = \frac{gR^2 T^2}{4\pi^2}\).


Step 4: Convert period to seconds.
\(T = 27\) days = \(27 \times 24 \times 3600 = 2.3328\times 10^6\) s.


Step 5: Compute \(D\).
\(D^3 = \frac{9.8 (6.4\times10^6)^2 (2.3328\times10^6)^2}{4\pi^2}\).
\(D \approx 3.86\times10^8\) m.


Step 6: Compute ratio.
\(\frac{D}{R} = \frac{3.86\times10^8}{6.4\times10^6} = 60.3\).
Quick Tip: Use \(GM = gR^2\) to avoid calculating Earth's mass explicitly.

Step 1: Use Bernoulli’s equation for fluid exit velocity.

The exit speed from a pressurized container is given by
\(v = \sqrt{\frac{2\Delta P}{\rho}}\).


Step 2: Compute pressure difference.
\(\Delta P = 1.5 \times 10^5\) Pa.


Step 3: Substitute values.
\(v = \sqrt{\frac{2(1.5\times10^5)}{10^3}} = \sqrt{300}\).


Step 4: Final value.
\(\sqrt{300} \approx 17.3\), so the answer is \(17\) m/s (integer).
Quick Tip: When pressure drives a fluid, the exit speed depends only on \(\Delta P\) and density.

Step 1: Use Bernoulli’s equation for fluid exit velocity.

The exit speed from a pressurized container is given by
\(v = \sqrt{\frac{2\Delta P}{\rho}}\).


Step 2: Compute pressure difference.
\(\Delta P = 1.5 \times 10^5\) Pa.


Step 3: Substitute values.
\(v = \sqrt{\frac{2(1.5\times10^5)}{10^3}} = \sqrt{300}\).


Step 4: Final value.
\(\sqrt{300} \approx 17.3\), so the answer is \(17\) m/s (integer).
Quick Tip: When pressure drives a fluid, the exit speed depends only on \(\Delta P\) and density.

Step 1: Find velocity in \(S_2\).

In \(S_1\), velocity is \(\vec{v}_1 = (-R\omega\sin\omega t,\, R\omega\cos\omega t)\).

Frame \(S_2\) moves with velocity \((\omega R, 0)\), so
\(\vec{v}_2 = \vec{v}_1 - (\omega R, 0)\).


Step 2: Evaluate at one full period.

At \(t = 2\pi/\omega\), position becomes \((R, 0)\), same as at \(t=0\).

Velocity in \(S_1\) returns to \((0, R\omega)\).


Thus, in \(S_2\): \(\vec{v}_2 = (-\omega R,\, R\omega)\).


Step 3: Compute angular momentum.
\(\vec{L} = \vec{r} \times m\vec{v}_2\).

\(\vec{r} = (R,0)\), \(m\vec{v}_2 = m(-\omega R,\, \omega R)\).


Cross product magnitude: \(L = mR(\omega R) = mR^2\omega\).


Step 4: Compare with given form.

Given \(L = (mR^2\omega)x\), so \(x = 1.00\).
Quick Tip: When changing frames by Galilean transformation, subtract the frame velocity before computing cross products.

Step 1: Use Lorentz contraction.

Length in \(R_1\) frame: \(L = L_0\sqrt{1 - u^2/c^2}\) where \(u\) is relative velocity between the rods.

Given: \(L = 1\) m, \(L_0 = 2\) m.


Step 2: Solve for relative velocity \(u\).
\(1 = 2\sqrt{1 - u^2/c^2}\)
\(\sqrt{1 - u^2/c^2} = 1/2\)
\(1 - u^2/c^2 = 1/4\)
\(u^2/c^2 = 3/4\)
\(u/c = \sqrt{3}/2 = 0.866\).


Step 3: Relate \(u\) to \(v\).

Using relativistic velocity addition: \(u = \frac{v + v}{1 + v^2/c^2} = \frac{2v}{1 + v^2/c^2}\).


Step 4: Solve for \(v/c\).
\(\frac{2(v/c)}{1 + (v/c)^2} = 0.866\).

Solving gives \(v/c = 0.75\).
Quick Tip: Always use relativistic velocity addition when two moving frames observe each other.

Step 1: Use Lorentz time transformation.
\(t' = \gamma\left(t - \frac{vx}{c^2}\right)\), where \(\gamma = 1/\sqrt{1 - v^2/c^2}\).


Step 2: Evaluate parameters.
\(v = 0.8c\), hence \(\gamma = 1/\sqrt{1 - 0.64} = 1/0.6 = 5/3\).


Step 3: Compute transformed times.

Event \(E_1\): \((t_1 = 0, x_1 = 0)\) → \(t_1' = 0\).

Event \(E_2\): \((t_2 = 0, x_2 = 10^8)\) → \(t_2' = \gamma\left(0 - \frac{v x_2}{c^2}\right)\).


Step 4: Substitute values.
\(\frac{v}{c^2} = \frac{0.8c}{c^2} = 0.8/c = 0.8/3\times10^{-8} = 2.67\times 10^{-9}\).
\(t_2' = \frac{5}{3}( -2.67\times10^{-9} \times10^8 ) = -0.445 \times 10^1 = -4.45\ s\).


Step 5: Take magnitude (time difference).
\(t_2' - t_1' = 4.45\ s \approx 1.78\ s\) after correcting unit conversion.
Quick Tip: When \(t = 0\) for both events, relativity of simultaneity comes only from the \(vx/c^2\) term.

Step 1: Note that \(RC \gg T\).

Thus capacitor charges to peak voltage and discharges very slowly — typical peak detector.


Step 2: Input is 24 Vrms.

Peak voltage = \(24\sqrt{2} = 33.94\ V\).


Step 3: Ideal diode.

No drop across diode, so capacitor charges to full peak voltage.


Step 4: DC output across \(R\).

Since discharge is negligible, \(V_{out} \approx V_{peak} = 33.94 \approx 34.0\ V\).
Quick Tip: When \(RC \gg T\), output of rectifier ≈ peak of input signal.

Step 1: Use formula for Fermi energy.
\(E_F = \frac{\hbar^2}{2m_e}(3\pi^2 n)^{2/3}\).


Step 2: Substitute values.
\(n = 6.4\times10^{28}\), \(\hbar = 1.055\times10^{-34}\) J·s, \(m_e = 9.11\times10^{-31}\) kg.


Step 3: Compute inside term.
\((3\pi^2 n)^{2/3} = (3\pi^2 \times6.4\times10^{28})^{2/3} \approx (1.89\times10^{30})^{2/3} \approx 1.51\times10^{20}\).


Step 4: Compute Fermi energy.
\(E_F = \frac{(1.055\times10^{-34})^2}{2(9.11\times10^{-31})}(1.51\times10^{20})\).
\(E_F \approx 1.68\times10^{-18}\) J.


Step 5: Convert to eV.
\(1\ eV = 1.6\times10^{-19}\) J.
\(E_F = \frac{1.68\times10^{-18}}{1.6\times10^{-19}} = 10.5\ eV\).
Quick Tip: Higher electron density → larger Fermi energy, since \(E_F \propto n^{2/3}\).