BE Civil Engineering (IOE, TU) Engineering Physics (IOE, SH 402) Question Paper 2078 Nepal

Differential equation. A particle of mass $m$ experiences a restoring force $-kx$ and a velocity-dependent damping force $-b\dfrac{dx}{dt}$. By Newton's second law:

$$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}$$ $$\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = 0,\qquad 2\beta=\frac{b}{m},\;\; \omega_0^2=\frac{k}{m}$$

Solution. Trying $x=e^{\alpha t}$ gives the auxiliary equation $\alpha^2 + 2\beta\alpha + \omega_0^2 = 0$, so

$$\alpha = -\beta \pm \sqrt{\beta^2-\omega_0^2}$$

For the under-damped case $\beta^2<\omega_0^2$, write $\omega = \sqrt{\omega_0^2-\beta^2}$. The general solution is

$$\boxed{x(t) = A_0\,e^{-\beta t}\cos(\omega t + \phi)}$$

This is an oscillation of angular frequency $\omega$ whose amplitude $A_0 e^{-\beta t}$ decays exponentially.

Relaxation time $\tau$ is the time for the amplitude to fall to $1/e$ of its initial value: $e^{-\beta\tau}=e^{-1}\Rightarrow \tau = 1/\beta = 2m/b$.

Quality factor $Q = \dfrac{\omega}{2\beta} = \omega\tau/2$, measuring how many radians the oscillator swings before its energy decays by $1/e$.

Numerical work.

$\omega_0^2 = k/m = 80/0.20 = 400\,\text{s}^{-2}$, so $\omega_0 = 20\,\text{rad/s}$.

$\beta = b/(2m) = 0.40/(2\times0.20) = 1.0\,\text{s}^{-1}$.

(i) $\omega = \sqrt{\omega_0^2-\beta^2} = \sqrt{400-1} = \sqrt{399} = 19.975\,\text{rad/s}$.

$\omega \approx 19.97\,\text{rad/s}$

(ii) $\tau = 1/\beta = 1/1.0 = 1.0\,\text{s}$.

Amplitude falls to $1/e$ in $\tau = 1.0\,\text{s}$.

Formation. A plano-convex lens of large radius of curvature $R$ rests on a flat glass plate, enclosing a thin air film of variable thickness $t$. Monochromatic light incident normally is partly reflected at the top of the air film and partly at the bottom; the two reflected beams interfere, producing concentric bright and dark rings (loci of constant $t$) centred on the point of contact.

Condition for dark rings (reflected light). The path difference is $2t + \lambda/2$ (the extra $\lambda/2$ from the phase reversal at the glass-to-air reflection at the lower surface). For a dark ring,

$$2t + \frac{\lambda}{2} = \left(m+\tfrac{1}{2}\right)\lambda \;\Rightarrow\; 2t = m\lambda,\quad m=0,1,2,\dots$$

Geometry of the film: for a ring of radius $r$, $r^2 = (2R-t)t \approx 2Rt$ (since $t\ll R$), so $t = r^2/2R$. Substituting,

$$2\cdot\frac{r^2}{2R} = m\lambda \;\Rightarrow\; r^2 = mR\lambda$$

With diameter $D=2r$:

$$\boxed{D_m^2 = 4mR\lambda}$$

Why the centre is dark. At the point of contact $t=0$, so the only path difference is the $\lambda/2$ phase change at the lower reflection: the two beams are exactly out of phase and interfere destructively. Hence the centre is dark.

Numerical work. Using two rings eliminates uncertainty in the contact point:

$$D_{20}^2 - D_{10}^2 = 4R\lambda(20-10) = 40R\lambda$$

$D_{20}^2 = (7.0\times10^{-3})^2 = 49.0\times10^{-6}\,\text{m}^2$

$D_{10}^2 = (5.0\times10^{-3})^2 = 25.0\times10^{-6}\,\text{m}^2$

$D_{20}^2 - D_{10}^2 = 24.0\times10^{-6}\,\text{m}^2$

$$R = \frac{D_{20}^2-D_{10}^2}{40\lambda} = \frac{24.0\times10^{-6}}{40\times589\times10^{-9}} = \frac{24.0\times10^{-6}}{2.356\times10^{-5}}$$ $$R = 1.0187\,\text{m}$$

Radius of curvature $R \approx 1.02\,\text{m}$.

Maxwell's equations (differential form, free space with sources).

1. $\nabla\cdot\vec{E} = \dfrac{\rho}{\varepsilon_0}$ — Gauss's law: electric charge is the source of the electric field (field lines begin/end on charge).
2. $\nabla\cdot\vec{B} = 0$ — no magnetic monopoles; magnetic field lines are closed loops.
3. $\nabla\times\vec{E} = -\dfrac{\partial \vec{B}}{\partial t}$ — Faraday's law: a time-varying magnetic field induces a circulating electric field.
4. $\nabla\times\vec{B} = \mu_0\vec{J} + \mu_0\varepsilon_0\dfrac{\partial \vec{E}}{\partial t}$ — Ampere-Maxwell law: both conduction current and displacement current produce a magnetic field.

Wave equation in free space ($\rho=0,\;\vec{J}=0$). Take the curl of equation 3:

$$\nabla\times(\nabla\times\vec{E}) = -\frac{\partial}{\partial t}(\nabla\times\vec{B})$$

Use the identity $\nabla\times(\nabla\times\vec{E}) = \nabla(\nabla\cdot\vec{E}) - \nabla^2\vec{E}$. With $\nabla\cdot\vec{E}=0$ and substituting equation 4 (with $\vec{J}=0$):

$$-\nabla^2\vec{E} = -\frac{\partial}{\partial t}\left(\mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t}\right)$$ $$\boxed{\nabla^2\vec{E} = \mu_0\varepsilon_0\frac{\partial^2\vec{E}}{\partial t^2}}$$

This is the wave equation $\nabla^2\vec{E} = \dfrac{1}{v^2}\dfrac{\partial^2\vec{E}}{\partial t^2}$ with propagation speed

$$v = c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$$

An identical equation holds for $\vec{B}$.

Numerical value.

$\mu_0 = 4\pi\times10^{-7}\,\text{T·m/A} = 1.2566\times10^{-6}$, $\varepsilon_0 = 8.854\times10^{-12}\,\text{F/m}$.

$\mu_0\varepsilon_0 = 1.2566\times10^{-6}\times8.854\times10^{-12} = 1.1127\times10^{-17}\,\text{s}^2/\text{m}^2$.

$$c = \frac{1}{\sqrt{1.1127\times10^{-17}}} = \frac{1}{3.336\times10^{-9}} = 2.998\times10^{8}\,\text{m/s}$$

$c \approx 3.0\times10^{8}\,\text{m/s}$, equal to the measured speed of light, identifying light as an electromagnetic wave.

Setup. Inside the well ($0<x<L$) the potential $V=0$; outside, $V=\infty$ so $\psi=0$. The time-independent Schrödinger equation inside is

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \;\Rightarrow\; \frac{d^2\psi}{dx^2} + k^2\psi = 0,\quad k^2 = \frac{2mE}{\hbar^2}$$

General solution and boundary conditions. $\psi(x)=A\sin kx + B\cos kx$. The wavefunction must vanish at the walls:

• $\psi(0)=0 \Rightarrow B=0$.
• $\psi(L)=0 \Rightarrow A\sin kL=0 \Rightarrow kL = n\pi,\; n=1,2,3,\dots$

Thus $k_n = n\pi/L$ and $\psi_n(x)=A\sin\dfrac{n\pi x}{L}$.

Normalization. $\displaystyle\int_0^L A^2\sin^2\frac{n\pi x}{L}\,dx = A^2\frac{L}{2}=1 \Rightarrow A=\sqrt{2/L}$.

$$\boxed{\psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}}$$

Energy eigenvalues. From $k_n^2 = 2mE_n/\hbar^2$:

$$\boxed{E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}}$$

Numerical work ($m=9.11\times10^{-31}\,\text{kg}$, $h=6.626\times10^{-34}\,\text{J·s}$, $L=0.10\times10^{-9}\,\text{m}$):

$$E_1 = \frac{h^2}{8mL^2} = \frac{(6.626\times10^{-34})^2}{8\times9.11\times10^{-31}\times(1.0\times10^{-10})^2}$$

Numerator: $(6.626\times10^{-34})^2 = 4.390\times10^{-67}$.

Denominator: $8\times9.11\times10^{-31}\times1.0\times10^{-20} = 7.288\times10^{-50}$.

$$E_1 = \frac{4.390\times10^{-67}}{7.288\times10^{-50}} = 6.024\times10^{-18}\,\text{J}$$

Convert: $E_1 = \dfrac{6.024\times10^{-18}}{1.602\times10^{-19}} = 37.6\,\text{eV}$.

$E_2 = 4E_1 = 4\times37.6 = 150.4\,\text{eV}$.

$E_1 \approx 37.6\,\text{eV}$, $\;E_2 \approx 150.4\,\text{eV}$.

Capacitance is the charge stored per unit potential difference: $C = Q/V$. Its SI unit is the farad (F).

Derivation (partially filled). Let surface charge density on the plates be $\sigma = Q/A$. The free-space field between bare plates is $E_0 = \sigma/\varepsilon_0$; inside the dielectric the field is reduced to $E = E_0/K = \sigma/(K\varepsilon_0)$.

The potential difference is the sum of contributions across the air gap $(d-t)$ and the dielectric $(t)$:

$$V = E_0(d-t) + E\,t = \frac{\sigma}{\varepsilon_0}(d-t) + \frac{\sigma}{K\varepsilon_0}t = \frac{\sigma}{\varepsilon_0}\left[(d-t)+\frac{t}{K}\right]$$

Since $\sigma = Q/A$,

$$C = \frac{Q}{V} = \frac{Q}{\frac{Q}{A\varepsilon_0}\left[(d-t)+\frac{t}{K}\right]}$$ $$\boxed{C = \frac{\varepsilon_0 A}{(d-t)+\dfrac{t}{K}}}$$

Numerical work. $A = 200\,\text{cm}^2 = 200\times10^{-4} = 2.0\times10^{-2}\,\text{m}^2$; $d = 4.0\times10^{-3}\,\text{m}$; $t = 3.0\times10^{-3}\,\text{m}$; $K=5$; $\varepsilon_0 = 8.854\times10^{-12}\,\text{F/m}$.

Effective gap: $(d-t) + t/K = (4.0-3.0)\times10^{-3} + \dfrac{3.0\times10^{-3}}{5} = 1.0\times10^{-3} + 0.6\times10^{-3} = 1.6\times10^{-3}\,\text{m}$.

$$C = \frac{8.854\times10^{-12}\times2.0\times10^{-2}}{1.6\times10^{-3}} = \frac{1.7708\times10^{-13}}{1.6\times10^{-3}} = 1.107\times10^{-10}\,\text{F}$$

$C \approx 110.7\,\text{pF}$ (about $0.111\,\text{nF}$).

Reverberation time $T$ is the time taken for the steady-state sound intensity in a room to fall to $10^{-6}$ of its original value (a drop of 60 dB) after the source stops.

Sabine's formula:

$$T = \frac{0.161\,V}{A}$$

where $V$ is the room volume in $\text{m}^3$ and $A=\sum a_i S_i$ is the total absorption in metric sabine (m²).

Case 1.

$$T_1 = \frac{0.161\times1500}{120} = \frac{241.5}{120} = 2.0125\,\text{s}$$

$T_1 \approx 2.01\,\text{s}$.

Case 2. New total absorption $A_2 = 120 + 30 = 150\,\text{m}^2$.

$$T_2 = \frac{0.161\times1500}{150} = \frac{241.5}{150} = 1.61\,\text{s}$$

$T_2 = 1.61\,\text{s}$ — adding absorption shortens the reverberation time, improving speech clarity.

Principal maxima. A grating is a large number $N$ of equally spaced parallel slits of spacing $d$ (grating element). When a plane wave is incident normally, light from adjacent slits has a path difference $d\sin\theta$. Whenever this equals a whole number of wavelengths, the contributions from all slits add in phase, giving a sharp, intense principal maximum:

$$\boxed{d\sin\theta = n\lambda},\qquad n=0,1,2,\dots$$

Numerical work. Grating element:

$$d = \frac{1\,\text{cm}}{5000} = \frac{1\times10^{-2}}{5000} = 2.0\times10^{-6}\,\text{m}$$

For $n=2$, $\lambda = 546\times10^{-9}\,\text{m}$:

$$\sin\theta = \frac{n\lambda}{d} = \frac{2\times546\times10^{-9}}{2.0\times10^{-6}} = \frac{1.092\times10^{-6}}{2.0\times10^{-6}} = 0.546$$ $$\theta = \sin^{-1}(0.546) = 33.09^\circ$$

$\theta \approx 33.1^\circ$.

Acceptance angle $\theta_a$ is the maximum half-angle of the cone of light (measured from the fibre axis) that can enter the core and still undergo total internal reflection, so that it is guided along the fibre.

Numerical aperture (NA) is the sine of the acceptance angle; it measures the light-gathering capacity of the fibre:

$$\text{NA} = \sin\theta_a = \sqrt{n_1^2 - n_2^2}$$

(for a fibre in air, where the outer medium $n_0=1$).

Numerical work.

$n_1^2 = 1.50^2 = 2.2500$

$n_2^2 = 1.46^2 = 2.1316$

$n_1^2 - n_2^2 = 2.2500 - 2.1316 = 0.1184$

$$\text{NA} = \sqrt{0.1184} = 0.3441$$

NA $\approx 0.344$.

Acceptance angle:

$$\theta_a = \sin^{-1}(0.3441) = 20.13^\circ$$

$\theta_a \approx 20.1^\circ$.

Biot–Savart law. A current element $I\,d\vec{l}$ produces, at a point with position vector $\vec{r}$ from the element, a magnetic field

$$d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$$

directed perpendicular to both $d\vec{l}$ and $\vec{r}$.

Field at the centre of a circular loop. For every element, $d\vec{l}\perp\vec{r}$ (so $|d\vec{l}\times\hat{r}| = dl$) and $r=R$. All elements give $d\vec{B}$ in the same direction (along the axis through the centre), so magnitudes add:

$$B = \int dB = \frac{\mu_0 I}{4\pi R^2}\int_0^{2\pi R} dl = \frac{\mu_0 I}{4\pi R^2}(2\pi R)$$ $$\boxed{B = \frac{\mu_0 I}{2R}}$$

Numerical work. $\mu_0 = 4\pi\times10^{-7}\,\text{T·m/A}$, $I = 3.0\,\text{A}$, $R = 5.0\times10^{-2}\,\text{m}$.

$$B = \frac{4\pi\times10^{-7}\times3.0}{2\times5.0\times10^{-2}} = \frac{1.2566\times10^{-6}\times3.0}{0.10} = \frac{3.770\times10^{-6}}{0.10}$$ $$B = 3.770\times10^{-5}\,\text{T}$$

$B \approx 3.77\times10^{-5}\,\text{T} = 37.7\,\mu\text{T}$.

Spontaneous emission. An atom in an excited state $E_2$ decays to a lower state $E_1$ on its own, after a random time, emitting a photon of energy $h\nu = E_2 - E_1$. The emitted photons are random in direction, phase and polarization, so the light is incoherent (ordinary light sources).

Stimulated emission. An incident photon of energy exactly $h\nu = E_2 - E_1$ induces an excited atom to drop to $E_1$, emitting a second photon. This emitted photon is identical to the incident one — same frequency, phase, direction and polarization. The two coherent photons can then stimulate further emissions, giving the multiplication (light amplification) at the heart of a laser.

Population inversion. In thermal equilibrium more atoms occupy the lower state ($N_1 > N_2$, Boltzmann distribution), so absorption dominates over stimulated emission. Population inversion is the non-equilibrium condition $N_2 > N_1$, achieved by pumping. Only then does an incident beam gain more photons by stimulated emission than it loses by absorption, so net light amplification — laser action — becomes possible. Hence population inversion is the essential prerequisite for lasing.

Role of the metastable state. A metastable level has an unusually long lifetime (microseconds to milliseconds) compared with ordinary excited states (nanoseconds). Atoms pumped to a higher level decay quickly into the metastable state and accumulate there, allowing $N_2$ to build up above $N_1$. Without a metastable state, atoms would de-excite too fast for population inversion to be sustained.

Meissner effect. When a material is cooled below its critical temperature $T_c$ in a magnetic field, it expels the magnetic flux from its interior, so that $\vec{B}=0$ inside the bulk. A superconductor is therefore a perfect diamagnet ($\chi = -1$), not merely a perfect conductor; the flux is pushed out even if the field was applied before cooling.

Type I vs Type II superconductors.

Intrinsic semiconductor conductivity vs temperature. The electrical conductivity increases as temperature rises. Reason: thermal energy excites more electrons across the band gap from the valence band to the conduction band, generating additional electron–hole pairs. The carrier concentration grows roughly as $n \propto e^{-E_g/2k_BT}$, and this rapid exponential rise in carrier number outweighs the small fall in mobility, so $\sigma$ increases — opposite to the behaviour of metals.