Given:

• Diameter of tube, (d = 0.1\ \text{cm})
• Radius, (r = \dfrac{d}{2} = 0.05\ \text{cm})
• Length of tube, (L = 40\ \text{cm})
• Pressure head, (h = 20\ \text{cm}) of water
• Coefficient of viscosity, (\eta = 0.0089\ \text{poise})
• Time, (t = 10\ \text{min} = 600\ \text{s})
• Density of water, (\rho = 1\ \text{g/cm}^3)
• Acceleration due to gravity, (g = 980\ \text{cm/s}^2)

Formula: By Poiseuille’s equation, volume of liquid flowing per second is

[ Q=\frac{\pi r^4 \Delta P}{8\eta L} ]

Here,

[ \Delta P = h\rho g ]

So,

[ V=\frac{\pi r^4 h\rho g}{8\eta L}\times t ]

Mass of water,

[ m=\rho V ]

Substitution:

[ m=\rho \times \frac{\pi r^4 h\rho g}{8\eta L}\times t ]

[ m=1\times \frac{\pi (0.05)^4(20)(1)(980)}{8(0.0089)(40)}\times 600 ]

[ m=\frac{\pi (6.25\times 10^{-6})(19600)}{2.848}\times 600 ]

[ m=\frac{0.384845}{2.848}\times 600 ]

[ m=0.13513\times 600 ]

[ m=81.08\ \text{g} ]

Result: [ \boxed{m \approx 81.1\ \text{g}} ]

So, the mass of water flowing through the tube in 10 minutes is approximately 81.1 g.

Magnetic Vector Potential

Meaning

Just as the electric field $\vec{E}$ can be written as the gradient of a scalar potential $V$ (i.e. $\vec{E} = -\nabla V$), the magnetic field $\vec{B}$ can be expressed as the curl of a vector field $\vec{A}$, called the magnetic vector potential:

$\vec{B} = \nabla \times \vec{A}$

This definition is consistent with the fact that $\vec{B}$ is always solenoidal, i.e. $\nabla \cdot \vec{B} = 0$. Since the divergence of a curl is identically zero,

$\nabla \cdot \vec{B} = \nabla \cdot (\nabla \times \vec{A}) = 0$

so the definition automatically satisfies Gauss's law for magnetism. The vector $\vec{A}$ is not unique — adding the gradient of any scalar $\nabla \chi$ gives the same $\vec{B}$ (gauge freedom), so we fix it by the Coulomb gauge condition $\nabla \cdot \vec{A} = 0$.

Derivation of the expression (for a steady current distribution)

Starting from Ampere's law in differential form,

$\nabla \times \vec{B} = \mu_0 \vec{J}$

Substitute $\vec{B} = \nabla \times \vec{A}$:

$\nabla \times (\nabla \times \vec{A}) = \mu_0 \vec{J}$

Using the vector identity $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$ and imposing $\nabla \cdot \vec{A} = 0$:

$\nabla^2 \vec{A} = -\mu_0 \vec{J}$

This is Poisson's equation for each Cartesian component of $\vec{A}$. Its solution (by analogy with the scalar Poisson equation $\nabla^2 V = -\rho/\varepsilon_0$) is

$\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{r}')\, d\tau'}{|\vec{r} - \vec{r}'|}$

where $\vec{r}$ is the field point, $\vec{r}'$ the source point, and $d\tau'$ the volume element of the current distribution. For a surface current $\vec{K}$ and a line current $I$, this becomes

$\vec{A} = \frac{\mu_0}{4\pi} \int \frac{\vec{K}\, da'}{|\vec{r} - \vec{r}'|}, \qquad \vec{A} = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{l}'}{|\vec{r} - \vec{r}'|}$

Verification

Taking the curl of the above expression and using $\nabla \times \vec{B} = \mu_0 \vec{J}$ (with $\nabla \cdot \vec{A} = 0$) confirms that this $\vec{A}$ indeed reproduces the correct magnetic field. Thus the magnetic vector potential is defined by $\vec{B} = \nabla \times \vec{A}$ and for a steady current is given by the volume integral above.

Magnetic Vector Potential

Meaning

In electrostatics, the electric field $\vec{E}$ is conveniently expressed as the gradient of a scalar potential $V$:

$\vec{E} = -\nabla V$

This is possible because $\nabla \times \vec{E} = 0$ for static fields. For magnetic fields, however, the situation is different. The magnetic field satisfies

$\nabla \cdot \vec{B} = 0$

everywhere (no magnetic monopoles). Since the divergence of the curl of any vector is identically zero, i.e. $\nabla \cdot (\nabla \times \vec{A}) = 0$, we can always express $\vec{B}$ as the curl of some vector field $\vec{A}$:

$\boxed{\vec{B} = \nabla \times \vec{A}}$

This vector field $\vec{A}$ is called the magnetic vector potential. It plays a role in magnetostatics analogous to that of the scalar potential $V$ in electrostatics. Unlike $V$, however, $\vec{A}$ is a vector quantity, and it is not uniquely determined: if we add the gradient of any scalar function $\chi$ to $\vec{A}$, i.e. $\vec{A}' = \vec{A} + \nabla\chi$, the magnetic field remains unchanged because $\nabla \times (\nabla \chi) = 0$. This freedom is called gauge freedom, and we fix $\vec{A}$ uniquely by imposing the Coulomb gauge condition:

$\nabla \cdot \vec{A} = 0$

Derivation of the expression for $\vec{A}$

Starting from Ampère's law in magnetostatics,

$\nabla \times \vec{B} = \mu_0 \vec{J}$

Substituting $\vec{B} = \nabla \times \vec{A}$:

$\nabla \times (\nabla \times \vec{A}) = \mu_0 \vec{J}$

Using the vector identity $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$:

$\nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A} = \mu_0 \vec{J}$

Applying the Coulomb gauge $\nabla \cdot \vec{A} = 0$, the first term vanishes and we obtain Poisson's equation for each Cartesian component:

$\nabla^2 \vec{A} = -\mu_0 \vec{J}$

This has exactly the same form as Poisson's equation for the electrostatic potential, $\nabla^2 V = -\rho/\varepsilon_0$, whose solution is

$V = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho}{r}\,d\tau'$

By direct analogy, the solution for the magnetic vector potential is

$\boxed{\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{r}')}{r}\,d\tau'}$

where $\vec{r}'$ is the position of the source element, $\vec{r}$ is the field point, $r = |\vec{r} - \vec{r}'|$ is the distance between them, and $d\tau'$ is the volume element of the current-carrying region.

Special case — current in a thin wire

For a filamentary current $I$, writing $\vec{J}\,d\tau' = I\,d\vec{l}'$, the expression reduces to

$\vec{A} = \frac{\mu_0 I}{4\pi}\oint \frac{d\vec{l}'}{r}$

Significance

• Since $\vec{A}$ is a vector, its direction is the same as that of the current element producing it.
• The magnetic field is recovered from $\vec{A}$ by taking the curl, $\vec{B} = \nabla \times \vec{A}$.
• The vector potential is extremely useful in advanced electrodynamics (e.g. in the Lagrangian formulation of charged-particle motion, in the Aharonov–Bohm effect, and in deriving the Biot–Savart law), where it plays a more fundamental role than $\vec{B}$ itself.

Damped and Forced Harmonic Oscillator and N-Coupled Oscillators

1. Damped Harmonic Oscillator

In a damped harmonic oscillator, a resistive force (damping force) proportional to velocity acts opposite to the motion:

$F_d = -b v = -b\frac{dx}{dt}$

where $b$ is the damping coefficient. The equation of motion is:

$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + \frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$

Putting $\dfrac{b}{m} = 2\gamma$ and $\dfrac{k}{m} = \omega_0^2$:

$\boxed{\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \omega_0^2 x = 0}$

where $\omega_0$ is the natural angular frequency and $\gamma$ the damping constant.

Three cases arise:

1. Overdamped ($\gamma > \omega_0$): The displacement decays without oscillation, returning slowly to equilibrium.
2. Critically damped ($\gamma = \omega_0$): The system returns to equilibrium in the shortest possible time without oscillating.
3. Underdamped ($\gamma < \omega_0$): The system oscillates with decreasing amplitude. The solution is:

$x = A e^{-\gamma t}\cos(\omega t + \phi), \qquad \omega = \sqrt{\omega_0^2 - \gamma^2}$

The amplitude $A e^{-\gamma t}$ decreases exponentially with time. The quality factor $Q = \dfrac{\omega_0}{2\gamma}$ measures how slowly the energy decays; large $Q$ means weak damping.

2. Forced Harmonic Oscillator

When a periodic driving force $F = F_0\cos\omega t$ is applied to a damped oscillator, the equation of motion becomes:

$m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = F_0\cos\omega t$

$\boxed{\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \omega_0^2 x = \frac{F_0}{m}\cos\omega t}$

The general solution is the sum of a transient part (decaying at rate $e^{-\gamma t}$) and a steady-state part:

$x = A\cos(\omega t - \theta)$

where the amplitude and phase lag are:

$A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4\gamma^2\omega^2}}, \qquad \tan\theta = \frac{2\gamma\omega}{\omega_0^2 - \omega^2}$

Resonance: The amplitude is maximum when the driving frequency is

$\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}$

At resonance the oscillator absorbs energy most efficiently from the driving force. For weak damping ($\gamma \ll \omega_0$), $\omega_r \approx \omega_0$ and the amplitude becomes very large. This is the phenomenon of resonance, responsible for effects such as the collapse of bridges in strong winds and the tuning of radio receivers.

3. N-Coupled Oscillators

When $N$ identical oscillators are coupled to one another (each mass connected to its neighbours by springs), the motion of one mass influences the others. The system is described by $N$ coupled differential equations:

$m\ddot{x}_i + k(2x_i - x_{i-1} - x_{i+1}) = 0, \qquad i = 1, 2, \dots, N$

Such a system possesses $N$ normal modes of vibration, each with its own normal frequency. In the $n$-th normal mode all masses oscillate with the same frequency $\omega_n$ and a fixed phase relationship:

$\omega_n = 2\sqrt{\frac{k}{m}}\sin\!\left(\frac{n\pi}{2(N+1)}\right), \qquad n = 1, 2, \dots, N$

The lowest frequency ($n = 1$) is the fundamental mode, in which all masses move in phase. Higher modes have nodes and correspond to overtones. Any arbitrary motion of the coupled system can be expressed as a superposition of these normal modes. In the limit $N \to \infty$ the discrete set of frequencies becomes continuous, which explains the propagation of waves along a stretched string or a chain of masses.

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OR

Equations of Motion for a Rotating Rigid Body and Kinetic Energy of Rotating and Rolling Bodies

1. Equations of Motion for a Rotating Rigid Body

For a rigid body rotating about a fixed axis, the rotational analogues of Newton's laws are:

First equation (angular momentum):

$\tau = \frac{dL}{dt} = I\alpha$

where $\tau$ is the net external torque, $L = I\omega$ the angular momentum, $I$ the moment of inertia and $\alpha$ the angular acceleration.

Second equation (angular displacement):

$\omega = \omega_0 + \alpha t$

Third equation:

$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$

Fourth equation:

$\omega^2 = \omega_0^2 + 2\alpha\theta$

These are the rotational analogues of $F = ma$, $v = u + at$, $s = ut + \frac{1}{2}at^2$ and $v^2 = u^2 + 2as$ respectively.

For a body rotating about an axis that is not fixed (general motion), the full vector equation is:

$\vec{\tau} = \frac{d\vec{L}}{dt} = I\vec{\alpha} + \vec{\omega}\times\vec{L}$

The term $\vec{\omega}\times\vec{L}$ gives rise to gyroscopic effects. When no external torque acts, $\vec{L}$ is conserved — the principle of conservation of angular momentum.

2. Kinetic Energy of a Rotating Body

Consider a rigid body rotating about a fixed axis with angular velocity $\omega$. A particle of mass $m_i$ at distance $r_i$ from the axis has linear speed $v_i = r_i\omega$. Its kinetic energy is:

$K_i = \tfrac{1}{2}m_i v_i^2 = \tfrac{1}{2}m_i r_i^2\omega^2$

Summing over all particles:

$K = \sum \tfrac{1}{2}m_i r_i^2\omega^2 = \tfrac{1}{2}\left(\sum m_i r_i^2\right)\omega^2$

$\boxed{K = \frac{1}{2}I\omega^2}$

where $I = \sum m_i r_i^2$ is the moment of inertia about the axis of rotation. This is the rotational analogue of $K = \frac{1}{2}mv^2$.

3. Kinetic Energy of a Rolling Body

A rolling body (e.g. a cylinder or sphere rolling without slipping) has both translational and rotational motion. Its total kinetic energy is the sum of the two:

$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$

For rolling without slipping, $v_{\text{cm}} = R\omega$, so:

$K = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\frac{v_{\text{cm}}^2}{R^2} = \frac{1}{2}mv_{\text{cm}}^2\left(1 + \frac{I_{\text{cm}}}{mR^2}\right)$

Writing $I_{\text{cm}} = mk^2$ where $k$ is the radius of gyration:

$\boxed{K = \frac{1}{2}mv_{\text{cm}}^2\left(1 + \frac{k^2}{R^2}\right)}$

Examples:

Thus for the same speed, a ring has the largest kinetic energy and a solid sphere the smallest. This also explains why a solid sphere rolls down an inclined plane faster than a ring.

Planck's Hypothesis and Its Special Cases

Planck's Hypothesis of Quantum Theory of Radiation

To explain the observed energy distribution in the spectrum of a black body (where classical theories failed), Max Planck in 1900 put forward the following revolutionary assumptions:

1. Quantized oscillators: The atoms/molecules of the walls of a cavity behave as tiny harmonic oscillators which emit and absorb electromagnetic radiation.

2. Discontinuous emission/absorption: These oscillators do not emit or absorb energy continuously (as classical physics demands) but in discrete packets called quanta (photons).

3. Energy of a quantum: The energy of each quantum is directly proportional to the frequency of radiation: $E \propto \nu \quad \Rightarrow \quad E = h\nu$ where $h = 6.626 \times 10^{-34}\ \text{J·s}$ is Planck's constant.

4. Discrete energy levels: An oscillator can possess only discrete energies $E_n = n h \nu, \qquad n = 0, 1, 2, 3, \dots$ so it can radiate or absorb energy only when it jumps from one allowed level to another, the energy difference being $\Delta E = (n_2 - n_1)h\nu$.

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Derivation of Planck's Radiation Formula

Number of modes per unit volume in frequency range $\nu$ to $\nu + d\nu$ inside a cavity is $N_\nu\, d\nu = \frac{8\pi \nu^2}{c^3}\, d\nu$

Average energy of an oscillator is obtained from the Boltzmann distribution. If $N_n$ oscillators have energy $nh\nu$, $N_n \propto e^{-nh\nu/kT}$

Hence the average energy is $\bar{E} = \frac{\sum_{n=0}^{\infty} nh\nu\, e^{-nh\nu/kT}}{\sum_{n=0}^{\infty} e^{-nh\nu/kT}}$

Putting $x = \dfrac{h\nu}{kT}$: $\bar{E} = h\nu \cdot \frac{\sum n e^{-nx}}{\sum e^{-nx}} = h\nu \cdot \frac{\dfrac{e^{-x}}{(1-e^{-x})^2}}{\dfrac{1}{1-e^{-x}}} = \frac{h\nu\, e^{-x}}{1-e^{-x}}$

$\boxed{\bar{E} = \frac{h\nu}{e^{h\nu/kT}-1}}$

Energy density of black-body radiation in the range $\nu$ to $\nu + d\nu$ is $u_\nu\, d\nu = N_\nu \bar{E}\, d\nu = \frac{8\pi h \nu^3}{c^3}\cdot \frac{1}{e^{h\nu/kT}-1}\, d\nu$

In terms of wavelength, using $\nu = c/\lambda$ and $d\nu = -\dfrac{c}{\lambda^2}d\lambda$: $\boxed{u_\lambda\, d\lambda = \frac{8\pi h c}{\lambda^5}\cdot \frac{1}{e^{hc/\lambda kT}-1}\, d\lambda}$

This is Planck's radiation law, which agrees with experiment over the entire spectrum.

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(i) Wien's Law as a Special Case

For short wavelengths (or low temperature), $\dfrac{hc}{\lambda kT} \gg 1$, so $e^{hc/\lambda kT} \gg 1 \quad \Rightarrow \quad e^{hc/\lambda kT}-1 \approx e^{hc/\lambda kT}$

Planck's formula then reduces to $u_\lambda = \frac{8\pi h c}{\lambda^5}\, e^{-hc/\lambda kT}$

which is Wien's law.

To obtain Wien's displacement law, set $\dfrac{du_\lambda}{d\lambda}=0$: $8\pi h c\left[-5\lambda^{-6}e^{-hc/\lambda kT} + \lambda^{-5}e^{-hc/\lambda kT}\cdot\frac{hc}{kT\lambda^2}\right] = 0$

$\frac{hc}{kT\lambda} = 5 \quad \Rightarrow \quad \lambda_m T = \frac{hc}{5k}$

$\lambda_m T = \frac{(6.626\times10^{-34})(3\times10^{8})}{5(1.38\times10^{-23})} = 2.88\times10^{-3}\ \text{m·K}$

This is Wien's displacement law ($\lambda_m T = \text{constant}$), confirming that Wien's law is a limiting case of Planck's law. (The exact value from the full Planck formula is $2.898\times10^{-3}$ m·K.)

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(ii) Rayleigh–Jeans Law as a Special Case

For long wavelengths (or high temperature), $\dfrac{hc}{\lambda kT} \ll 1$. Expanding the exponential: $e^{hc/\lambda kT} = 1 + \frac{hc}{\lambda kT} + \frac{1}{2!}\left(\frac{hc}{\lambda kT}\right)^2 + \cdots$

Retaining only the first two terms: $e^{hc/\lambda kT}-1 \approx \frac{hc}{\lambda kT}$

Substituting into Planck's formula: $u_\lambda = \frac{8\pi h c}{\lambda^5}\cdot \frac{\lambda kT}{hc} = \frac{8\pi kT}{\lambda^4}$

which is the Rayleigh–Jeans law in terms of wavelength. In terms of frequency, using $\lambda = c/\nu$: $u_\nu = \frac{8\pi \nu^2 kT}{c^3}$

This is exactly the classical result obtained from the equipartition theorem.

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Conclusion

Thus Planck's radiation formula is the general law of black-body radiation, and both Wien's law and the Rayleigh–Jeans law are its limiting (special) cases.

Option A: Plane Electromagnetic Waves in Isotropic Dielectrics and Coupled (Compound) Media

1. Plane EM Wave in an Isotropic Dielectric

Consider a plane wave propagating along the $z$-direction in a homogeneous, isotropic, non-conducting dielectric with permittivity $\varepsilon$ and permeability $\mu$ (free of charges and currents, $\rho = 0$, $\mathbf{J} = 0$).

Maxwell's equations give the wave equation:

$\nabla^2 \mathbf{E} = \mu\varepsilon \frac{\partial^2 \mathbf{E}}{\partial t^2}$

A plane harmonic solution is:

$\mathbf{E} = \mathbf{E}_0 \, e^{i(\omega t - kz)}, \qquad \mathbf{B} = \mathbf{B}_0 \, e^{i(\omega t - kz)}$

Substituting gives the dispersion relation:

$k^2 = \mu\varepsilon\,\omega^2 \quad\Rightarrow\quad v = \frac{\omega}{k} = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{n}$

where the refractive index is $n = \sqrt{\mu_r \varepsilon_r} = \sqrt{\varepsilon_r}$ for non-magnetic dielectrics ($\mu_r \approx 1$).

Key features:

• $\mathbf{E}$, $\mathbf{B}$ and the propagation vector $\mathbf{k}$ are mutually perpendicular — the wave is transverse (TEM).
• $\mathbf{E}$ and $\mathbf{B}$ are in phase; their magnitudes are related by $\frac{E_0}{B_0} = v = \frac{c}{n}$
• The intrinsic impedance of the medium is $\eta = \frac{E_0}{H_0} = \sqrt{\frac{\mu}{\varepsilon}} = \frac{\eta_0}{n}, \qquad \eta_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 377\ \Omega$
• The Poynting vector $\mathbf{S} = \mathbf{E} \times \mathbf{H}$ is along $\mathbf{k}$, giving intensity $I = \langle S \rangle = \frac{1}{2}\varepsilon v E_0^2 = \frac{1}{2}\frac{E_0^2}{\eta}$
• Energy is shared equally between the electric and magnetic fields: $u_E = u_B = \frac{1}{4}\varepsilon E_0^2$
• The wave travels without attenuation (real $k$), since the dielectric is lossless.
• Phase velocity $v = c/n$ is independent of frequency in a simple dielectric (no dispersion), so all frequencies travel together.

2. Plane EM Wave in Coupled (Compound) Media

A "coupled" or compound medium means two or more dielectrics in contact (e.g. a layered periodic structure, a dielectric slab, or a medium with two coupled resonances). The wave now experiences multiple media and the boundary conditions at each interface must be satisfied.

Behaviour:

• At each interface, the boundary conditions require continuity of tangential $\mathbf{E}$ and $\mathbf{H}$ (and normal $\mathbf{D}$, $\mathbf{B}$). This produces reflected and transmitted waves at every boundary.
• The wave inside the compound medium is a superposition of forward and backward waves (a Bloch-type wave in periodic media): $\mathbf{E} = \mathbf{E}_1 e^{i(\omega t - kz)} + \mathbf{E}_2 e^{i(\omega t + kz)}$
• The dispersion relation is modified by the periodicity/structure. For a periodic layered medium of period $\Lambda$, the Bloch condition gives $\cos(K\Lambda) = \cos(k_1 d_1)\cos(k_2 d_2) - \frac{1}{2}\left(\frac{\eta_1}{\eta_2} + \frac{\eta_2}{\eta_1}\right)\sin(k_1 d_1)\sin(k_2 d_2)$ where $K$ is the Bloch wave number. This leads to band gaps — frequency ranges where the wave cannot propagate.
• Impedance matching / mismatch at interfaces causes partial reflection; total transmission occurs only at specific "resonant" thicknesses.
• Dispersion is strong — phase velocity depends on frequency, and group velocity $v_g = d\omega/dk$ differs from $v_p$.
• Coupling between the two media (e.g. two resonances) produces splitting of the resonance (coupled-mode splitting) and avoided crossing in the dispersion curve.

3. Major Differences

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Option B: Clausius–Mossotti Relation and Debye Modification

1. Derivation of Clausius–Mossotti Relation

Setup: A dielectric in an external field $\mathbf{E}$. The local (internal) field $\mathbf{E}_{loc}$ acting on a molecule is larger than the applied field because of polarization of surrounding dipoles (Lorentz cavity).

For a cubic / isotropic medium, the Lorentz field is:

$\mathbf{E}_{loc} = \mathbf{E} + \frac{\mathbf{P}}{3\varepsilon_0}$

Each molecule has polarizability $\alpha$, so the induced dipole moment is

$\mathbf{p} = \alpha \mathbf{E}_{loc}$

If $N$ is the number of molecules per unit volume, the polarization is

$\mathbf{P} = N\mathbf{p} = N\alpha \mathbf{E}_{loc} = N\alpha\left(\mathbf{E} + \frac{\mathbf{P}}{3\varepsilon_0}\right)$

Rearranging:

$\mathbf{P}\left(1 - \frac{N\alpha}{3\varepsilon_0}\right) = N\alpha \mathbf{E}$

$\mathbf{P} = \frac{N\alpha}{1 - \dfrac{N\alpha}{3\varepsilon_0}}\,\mathbf{E}$

But by definition $\mathbf{P} = \varepsilon_0(\varepsilon_r - 1)\mathbf{E}$, so

$\varepsilon_0(\varepsilon_r - 1) = \frac{N\alpha}{1 - \dfrac{N\alpha}{3\varepsilon_0}}$

Let $x = \dfrac{N\alpha}{3\varepsilon_0}$. Then

$\varepsilon_r - 1 = \frac{3x}{1 - x} \quad\Rightarrow\quad \varepsilon_r = \frac{1 + 2x}{1 - x}$

Inverting:

$x = \frac{\varepsilon_r - 1}{\varepsilon_r + 2}$

Substituting back:

$\boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha}{3\varepsilon_0}}$

This is the Clausius–Mossotti relation. It connects the macroscopic dielectric constant $\varepsilon_r$ to the microscopic polarizability $\alpha$ and number density $N$.

2. Debye Modification

Clausius–Mossotti assumes only electronic (induced) polarization — the molecules are non-polar and $\alpha$ is frequency-independent. Debye extended it to polar molecules (permanent dipole moment $p_0$), which can also orient in the field.

The total polarizability becomes temperature- and frequency-dependent:

$\alpha = \alpha_e + \frac{p_0^2}{3k_B T}$

where $\alpha_e$ is the electronic polarizability, $p_0$ the permanent dipole moment, $k_B$ Boltzmann's constant, and $T$ the absolute temperature.

The Debye relation is therefore:

$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N}{3\varepsilon_0}\left(\alpha_e + \frac{p_0^2}{3k_B T}\right)$

Key points of the modification:

• Accounts for orientational (dipole) polarization in addition to electronic polarization.
• Predicts that $\varepsilon_r$ decreases with increasing temperature for polar dielectrics (since thermal agitation opposes dipole alignment).
• For alternating fields, Debye introduced a relaxation time $\tau$, giving the complex permittivity $\varepsilon_r(\omega) = \varepsilon_\infty + \frac{\varepsilon_s - \varepsilon_\infty}{1 + i\omega\tau}$ where $\varepsilon_s$ and $\varepsilon_\infty$ are the static and high-frequency permittivities. This explains dielectric relaxation and loss in polar media.

Twisting couple on a cylindrical rod (wire)

When the upper end of a cylindrical rod or wire of length $l$ and radius $r$ is clamped and a couple is applied at the lower end about the axis of the rod, the rod is said to be twisted. Each circular cross-section remains circular and rotates about the axis, the angle of rotation increasing uniformly from the fixed end to the free end. The total angle through which the free end turns is called the angle of twist $\theta$.

The external couple applied is balanced by the internal restoring couple set up in the rod due to the elasticity of the material. This internal couple is called the twisting couple. The twisting couple per unit twist is called the torsional rigidity of the rod.

Derivation of the twisting couple

Let a wire of length $l$ and radius $r$ be fixed at the top and twisted at the bottom through angle $\theta$. Consider a coaxial hollow cylinder of radius $x$ and thickness $dx$ inside the wire.

A point on the bottom face of this cylinder at radius $x$ is displaced through an arc of length $x\theta$. Hence the angle of shear is

$$\phi=\frac{x\theta}{l}$$

The shearing stress on this layer is

$$\text{stress}=\eta\phi=\frac{\eta x\theta}{l}$$

where $\eta$ is the modulus of rigidity of the material. The area of cross-section of the cylindrical shell is $2\pi x\,dx$, so the tangential force on it is

$$F=\frac{\eta x\theta}{l}\cdot 2\pi x\,dx$$

The moment of this force about the axis of the wire gives the twisting couple of the shell:

$$dC=F\cdot x=\frac{2\pi\eta\theta}{l}x^{3}\,dx$$

Integrating from $x=0$ to $x=r$,

$$C=\frac{2\pi\eta\theta}{l}\int_0^{r}x^{3}\,dx =\frac{2\pi\eta\theta}{l}\cdot\frac{r^{4}}{4}$$ $$\boxed{C=\frac{\pi\eta r^{4}\theta}{2l}}$$

The twisting couple per unit twist is therefore

$$\tau=\frac{C}{\theta}=\frac{\pi\eta r^{4}}{2l}$$

which is the torsional rigidity of the rod. It depends on the material ($\eta$) and on the dimensions ($r^4$ and $l$) of the rod.

Work done in twisting the rod

When the angle of twist is $\theta$, the restoring couple is $C=\tau\theta$. If the twist is increased further by a small amount $d\theta$, the work done against the restoring couple is

$$dW=C\,d\theta=\tau\theta\,d\theta$$

Hence the total work done in twisting the rod from $0$ to $\theta$ is

$$W=\int_0^{\theta}\tau\theta\,d\theta=\tau\int_0^{\theta}\theta\,d\theta =\frac{1}{2}\tau\theta^{2}$$ $$\boxed{W=\frac{1}{2}\tau\theta^{2}=\frac{1}{2}C\theta}$$

Substituting $\tau=\dfrac{\pi\eta r^{4}}{2l}$,

$$\boxed{W=\frac{\pi\eta r^{4}\theta^{2}}{4l}}$$

This work is stored in the rod as elastic potential energy. Since the volume of the rod is $V=\pi r^{2}l$, the energy stored per unit volume is

$$\frac{W}{V}=\frac{\pi\eta r^{4}\theta^{2}}{4l}\cdot\frac{1}{\pi r^{2}l} =\frac{\eta r^{2}\theta^{2}}{4l^{2}} =\frac{1}{2}\cdot\frac{\eta r\theta}{l}\cdot\frac{r\theta}{l} =\frac{1}{2}\,(\text{shearing stress})\times(\text{shearing strain})$$

which agrees with the general result that the elastic energy per unit volume is $\tfrac12\times\text{stress}\times\text{strain}$.

Entropy and the second law

Entropy $S$ is a state function that measures the disorder (randomness) of a system, defined by

$dS = \frac{dQ_{\text{rev}}}{T}.$

The second law says that natural processes have a direction, and entropy gives that direction: the total entropy of an isolated system (or of a system together with its surroundings) never decreases,

$\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \ge 0,$

with equality only for a reversible process.

Proof (Clausius inequality). For any cycle, $\oint \dfrac{dQ}{T} \le 0$, with equality for a reversible cycle. Take an irreversible path $A \to B$ and return by a reversible path $B \to A$:

$\int_A^B \frac{dQ}{T}\bigg|_{\text{irr}} + \int_B^A \frac{dQ_{\text{rev}}}{T} < 0 \;\Rightarrow\; \int_A^B \frac{dQ}{T} - (S_B - S_A) < 0.$

Hence $\Delta S \ge \displaystyle\int \frac{dQ}{T}$, with equality for a reversible process. For an isolated system $dQ = 0$, so $\Delta S \ge 0$. Treating a system and its surroundings together as one isolated system gives $\Delta S_{\text{total}} \ge 0$.

Statistical view. With $S = k_B\ln W$ ($W$ = number of microstates), a spontaneous change takes an isolated system to the macrostate with far more microstates, so $W$, and hence $S$, increases or stays the same.

Conclusion. The entropy of an isolated system, or the total entropy of system plus surroundings, can never decrease. The entropy of a single substance can fall (water freezing to ice), but only if its surroundings gain at least as much entropy.

Attempt any TWO questions. [2×3=6]

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a) Two examples of conservation of angular momentum

Angular momentum $L = I\omega$ remains constant when the net external torque on a system is zero.

1. A spinning ice skater (or ballet dancer): When the skater pulls her arms close to her body, the moment of inertia $I$ decreases. Since $L = I\omega$ is conserved, the angular velocity $\omega$ increases, so she spins faster. When she stretches her arms out, $I$ increases and $\omega$ decreases.

2. A planet revolving around the Sun: The gravitational force on a planet is directed towards the Sun (a central force), so the torque about the Sun is zero. Hence $L = mvr\sin\theta$ is constant. This means a planet moves faster when it is nearer the Sun (perihelion) and slower when it is farther away (aphelion) — which is Kepler's second law.

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b) Inverse square law of force

The inverse square law states that the magnitude of a force between two point objects is inversely proportional to the square of the distance between them:

$F \propto \frac{1}{r^{2}} \quad \text{or} \quad F = \frac{k}{r^{2}}$

where $k$ is a constant depending on the nature of the interaction.

Explanation: If the distance $r$ is doubled, the force becomes $\tfrac{1}{4}$ of its original value; if $r$ is tripled, the force falls to $\tfrac{1}{9}$. The force acts along the line joining the two objects.

Examples:

• Gravitational force: $F = \dfrac{G m_1 m_2}{r^{2}}$
• Electrostatic force (Coulomb's law): $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^{2}}$

Physical reason: The influence of a point source spreads uniformly over the surface of a sphere of radius $r$, whose area is $4\pi r^{2}$. Since the same "effect" is spread over an area growing as $r^{2}$, the intensity (force per unit area) falls as $1/r^{2}$.

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c) Law of equipartition of energy

Statement: In thermal equilibrium at absolute temperature $T$, the total energy of a system is shared equally among all its degrees of freedom, each degree of freedom receiving an average energy of $\tfrac{1}{2}k_BT$ (where $k_B$ is Boltzmann's constant).

Explanation: A molecule has translational, rotational and vibrational degrees of freedom. Each independent quadratic term in the expression for energy (such as $\tfrac{1}{2}mv_x^{2}$ or $\tfrac{1}{2}I\omega^{2}$) contributes $\tfrac{1}{2}k_BT$ to the average energy. A molecule with $f$ degrees of freedom therefore has average energy

$E = \frac{f}{2}k_BT$

Applications:

• Monatomic gas (He, Ar): $f = 3$ (translational only), so $E = \tfrac{3}{2}k_BT$ and molar specific heat at constant volume $C_V = \tfrac{3}{2}R$.
• Diatomic gas (H₂, O₂): $f = 5$ (3 translational + 2 rotational), so $C_V = \tfrac{5}{2}R$.
• Polyatomic gas: $f = 6$ or more, giving larger $C_V$.

This law successfully explains the molar specific heats of gases and the ratio $\gamma = C_p/C_V$.

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d) Significance of complex numbers for ac analysis

In ac circuits the voltage and current vary sinusoidally and are not always in phase; they differ by a phase angle $\phi$. Complex numbers allow both the magnitude and the phase of an ac quantity to be represented by a single symbol, called a phasor.

1. Single representation of magnitude and phase: A voltage $v = V_0\sin(\omega t + \phi)$ is written as the complex quantity $\tilde{V} = V_0 e^{j\phi} = V_0(\cos\phi + j\sin\phi)$, where $V_0$ gives the amplitude and $\phi$ the phase.

2. Impedance as a complex quantity: Resistance, inductive reactance and capacitive reactance combine as $Z = R + j\left(\omega L - \frac{1}{\omega C}\right)$ Its modulus gives the impedance $|Z|$ and its argument gives the phase difference between voltage and current.

Advantages:

• Ohm's law in complex form: $\tilde{V} = \tilde{I}Z$, so ac circuit problems are solved exactly like dc circuits, using algebra of complex numbers instead of trigonometric equations.
• Series and parallel combinations of $R$, $L$ and $C$ can be added directly as complex numbers.
• Power factor $\cos\phi$ and resonance condition ($\omega L = 1/\omega C$) follow at once from the real and imaginary parts of $Z$.
• Graphical (phasor diagram) interpretation makes the phase relationship between different voltages and currents easy to visualise.

Thus complex numbers reduce complicated sinusoidal ac problems to simple algebraic operations.

(a) Magnetic susceptibility

When a material is placed in an external magnetic field $\vec{H}$, it becomes magnetized, acquiring a magnetization $\vec{M}$ (magnetic moment per unit volume). The magnetic susceptibility $\chi$ measures how readily a material responds to the applied field and is defined as

$\chi = \frac{M}{H}$

It is a dimensionless quantity. Physically, it tells how much magnetization is induced per unit applied magnetic field. Its sign and magnitude classify materials:

• Diamagnetic ($\chi$ small and negative, e.g. bismuth, copper): weakly repelled by the field.
• Paramagnetic ($\chi$ small and positive, e.g. aluminium, oxygen): weakly attracted.
• Ferromagnetic ($\chi$ large and positive, e.g. iron, cobalt): strongly attracted.

The relation between magnetic field quantities is $\vec{B} = \mu_0(\vec{H} + \vec{M}) = \mu_0(1+\chi)\vec{H} = \mu_0\mu_r\vec{H}$, so the relative permeability is $\mu_r = 1+\chi$.

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(b) Cascade cooling

Cascade cooling is a refrigeration technique in which two or more separate refrigeration cycles are coupled in series (a "cascade") to achieve very low temperatures that a single vapour-compression cycle cannot reach.

In a two-stage cascade system, the condenser of the lower-temperature cycle is thermally connected to the evaporator of the higher-temperature cycle through a cascade heat exchanger. The heat rejected by the low-temperature cycle is absorbed by the high-temperature cycle, which then rejects it to the surroundings.

Key points:

• Each cycle uses a refrigerant suited to its own temperature range (e.g. R-12 in the low stage, R-22 or ammonia in the high stage).
• The overlap of temperature ranges avoids extremely high pressure ratios in a single compressor.
• The intermediate temperature is chosen to minimize total work input, usually so that the work of the two cycles is equal.
• Applications: liquefaction of gases, ultra-low-temperature storage, and cryogenic processes.

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(c) Ideal and real gases

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(d) Non-conservative forces with examples

A force is non-conservative if the work done by it in moving a body between two points depends on the path followed, and the work done around a closed path is not zero. Equivalently, no potential energy function can be defined for such a force, so mechanical energy is not conserved (it is dissipated as heat, sound, etc.).

Examples:

1. Friction — The work done against friction depends on the length of the path. A block pushed along a curved path loses more energy to friction than along a straight path between the same two points. Work around a closed loop is non-zero.
2. Air resistance (drag) — A body moving through air loses energy continuously; the loss depends on the path and speed, and energy is converted into heat in the air.
3. Viscous force in fluids — A body moving through a viscous medium experiences a retarding force that dissipates mechanical energy as heat.
4. Tension in a string with friction at the pulley — Energy is lost at the contact, so the work is path-dependent.

In each case, the total mechanical energy of the system decreases, confirming the force is non-conservative.

Given:

• Height reached above Earth’s surface, (h = 4R_e)
• Radius of Earth, (R_e = 6.4 \times 10^6\ \text{m})
• Acceleration due to gravity at surface, (g = 9.8\ \text{m/s}^2)

At maximum height, final velocity (v_f = 0).
Distance from Earth’s centre at that point:

[ r = R_e + h = R_e + 4R_e = 5R_e ]

By conservation of mechanical energy,

[ \frac12 mv^2 - \frac{GM_em}{R_e}

-\frac{GM_em}{5R_e} ]

[ \frac12 v^2 = \frac{GM_e}{R_e} - \frac{GM_e}{5R_e} ]

[ v^2 = \frac{8GM_e}{5R_e} ]

Using (GM_e = gR_e^2),

[ v^2 = \frac{8gR_e}{5} ]

[ v = \sqrt{\frac{8gR_e}{5}} ]

Substituting values:

[ v = \sqrt{\frac{8 \times 9.8 \times 6.4 \times 10^6}{5}} ]

[ v = \sqrt{1.00352 \times 10^8} ]

[ v \approx 1.002 \times 10^4\ \text{m/s} ]

[ \boxed{v \approx 10.02\ \text{km/s}} ]

So, the body must be projected upward with a velocity of about (10.02\ \text{km/s}) to reach a height of (4R_e) from the Earth’s surface.

Given:

• Molecular diameter, (d = 3.5,\text{Å} = 3.5 \times 10^{-10},\text{m})
• Temperature, (T = 27^\circ\text{C} = 300,\text{K})
• Pressure, (P = 1,\text{atm} = 1.013 \times 10^5,\text{N/m}^2)
• Boltzmann constant, (k = 1.38 \times 10^{-23},\text{J/K})

Formula: [ \lambda = \frac{kT}{\sqrt{2},\pi d^2 P} ]

Substitution: [ \lambda = \frac{(1.38 \times 10^{-23})(300)} {\sqrt{2} \times 3.14 \times (3.5 \times 10^{-10})^2 \times (1.013 \times 10^5)} ]

[ \lambda = \frac{4.14 \times 10^{-21}} {1.414 \times 3.14 \times 1.225 \times 10^{-19} \times 1.013 \times 10^5} ]

[ \lambda = \frac{4.14 \times 10^{-21}}{5.51 \times 10^{-14}} ]

[ \lambda = 7.51 \times 10^{-8},\text{m} ]

Result: [ \boxed{\lambda = 7.51 \times 10^{-8},\text{m} \approx 75.1,\text{nm} \approx 751,\text{Å}} ]

So, the mean free path of nitrogen at (27^\circ\text{C}) and one atmosphere pressure is approximately (7.51 \times 10^{-8},\text{m}) (or (751,\text{Å})).

Assuming the two cells are distinct (labelled) and the four particles are distinguishable, each particle has 2 independent choices.

Number of microstates is

[ \Omega = 2^4 = 16 ]

So, the number of microstates is 16.

(If the particles were treated as indistinguishable, the number of distributions would be (\binom{4+2-1}{2-1}=5), but in the usual classical statistical-mechanics treatment the answer is 16.)

Given:

• Length of solenoid, $l = 0.5\ \text{m}$
• Cross-sectional area, $A = 20\ \text{cm}^2 = 20 \times 10^{-4}\ \text{m}^2 = 2.0 \times 10^{-3}\ \text{m}^2$
• Number of turns, $N = 500$
• Permeability of free space, $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$

Formula: For an air-cored solenoid, $L = \frac{\mu_0 N^2 A}{l}$

Substitution: $L = \frac{(4\pi \times 10^{-7})(500)^2 (2.0 \times 10^{-3})}{0.5}$

$L = \frac{4\pi \times 10^{-7} \times 250000 \times 2.0 \times 10^{-3}}{0.5}$

$L = 4\pi \times 10^{-4}\ \text{H}$

$L = 1.2566 \times 10^{-3}\ \text{H}$

Result: $\boxed{L \approx 1.26 \times 10^{-3}\ \text{H} = 1.26\ \text{mH}}$

So, the inductance of the solenoid is approximately 1.26 mH.

Given:

• Charge of electron, $q = 1.6 \times 10^{-19}\ \text{C}$
• Speed, $v = 10^6\ \text{m/s}$
• Magnetic field, $B = 2\ \text{T}$
• Angle, $\theta = 30^\circ$

Formula: The magnetic force on a charged particle is $F = qvB\sin\theta$

Substitution: $F = (1.6 \times 10^{-19})(10^6)(2)\sin 30^\circ$ $F = (1.6 \times 10^{-19})(10^6)(2)(0.5)$ $F = 1.6 \times 10^{-13}\ \text{N}$

Result: The force on the electron is $\boxed{F = 1.6 \times 10^{-13}\ \text{N}}$

Direction: The force acts perpendicular to the plane containing $\vec v$ and $\vec B$. Since the electron is negatively charged, the direction is opposite to that given by $\vec v \times \vec B$.

N-coupled oscillator and normal modes of two coupled particles

A system of $N$ particles of masses connected to one another by springs (or other elastic couplings), so that the motion of each particle depends on the motion of its neighbours, is called an N-coupled oscillator. It has $N$ degrees of freedom and therefore $N$ normal modes. In a normal mode all particles oscillate with the same frequency and fixed phase relations. Any general motion is a superposition of the normal modes.

Two particles connected by a spring

Two particles each of mass $m$ move along a line with displacements $x_1$ and $x_2$ from their equilibrium positions. They are connected by a spring of constant $k$ and are free at the other ends. The extension of the spring is $x_2 - x_1$, so

$m\ddot{x}_1 = k(x_2 - x_1), \qquad m\ddot{x}_2 = -k(x_2 - x_1)$

Adding: $\ddot{x}_1 + \ddot{x}_2 = 0$, so $x_1 + x_2 = 2(A + Bt)$. This is uniform motion of the centre of mass, a mode of zero frequency ($\omega_1 = 0$).

Subtracting: $\ddot{(x_1 - x_2)} = -\dfrac{2k}{m}(x_1 - x_2)$, which is simple harmonic motion with

$\omega_2 = \sqrt{\frac{2k}{m}}$

The possible modes

1. In-phase (symmetric) mode: $x_1 = x_2$. The spring is neither stretched nor compressed and the particles move together with frequency $\omega_1 = 0$, a pure translation.
2. Out-of-phase (antisymmetric) mode: $x_1 = -x_2$. The particles move in opposite directions, the centre of mass stays at rest, and the frequency is $\omega_2 = \sqrt{2k/m}$.

General solution (any motion is a combination of these modes):

$x_1 = A + Bt + C\cos(\omega_2 t + \phi), \qquad x_2 = A + Bt - C\cos(\omega_2 t + \phi)$

If the centre of mass is at rest ($B = 0$), this reduces to $x_1 = A + C\cos(\omega_2 t + \phi)$ and $x_2 = A - C\cos(\omega_2 t + \phi)$.

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OR

Plane progressive wave and energy density

A plane progressive wave is a wave that travels in a fixed direction through a medium with constant speed, without change of form, so that every particle executes the same simple harmonic motion with a phase lag that depends on its distance from the source. The wavefronts are planes. For a wave along $+x$:

$y = a\sin(\omega t - kx), \qquad k = \frac{2\pi}{\lambda}, \quad v = \frac{\omega}{k}$

Energy density

Consider a medium of density $\rho$ and elasticity $E$, with $v = \sqrt{E/\rho}$.

Kinetic energy per unit volume:

$u_K = \tfrac12 \rho \left(\frac{\partial y}{\partial t}\right)^2 = \tfrac12 \rho a^2\omega^2\cos^2(\omega t - kx)$

Potential energy per unit volume:

$u_P = \tfrac12 E \left(\frac{\partial y}{\partial x}\right)^2 = \tfrac12 \rho v^2 a^2 k^2\cos^2(\omega t - kx) = \tfrac12 \rho a^2\omega^2\cos^2(\omega t - kx)$

since $vk = \omega$. Thus $u_K = u_P$ at every point and instant.

Total energy density:

$u = u_K + u_P = \rho a^2\omega^2\cos^2(\omega t - kx)$

The average of $\cos^2$ over a cycle is $\tfrac12$, so the average energy density is

$\bar{u} = \tfrac12 \rho a^2 \omega^2 = 2\pi^2\rho a^2 n^2$

where $n$ is the frequency. The energy is shared equally between kinetic and potential forms.