BE Civil Engineering (IOE, TU) Engineering Physics (IOE, SH 402) Question Paper 2076 Nepal

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

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

Dividing by $m$ and writing $2\beta = b/m$ (damping constant) and $\omega_0^2 = k/m$ (natural frequency):

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

Solution (under-damped, $\beta < \omega_0$). Trying $x = e^{\lambda t}$ gives the auxiliary equation $\lambda^2 + 2\beta\lambda + \omega_0^2 = 0$, so $\lambda = -\beta \pm \sqrt{\beta^2 - \omega_0^2}$. For $\beta < \omega_0$ the root is complex, $\lambda = -\beta \pm i\omega_d$ with $\omega_d = \sqrt{\omega_0^2 - \beta^2}$. The real solution is

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

Amplitude decay. The envelope $A(t) = A_0 e^{-\beta t}$ decays exponentially; the oscillation frequency $\omega_d$ is slightly less than $\omega_0$.

 x
 |   .-.        envelope A0 e^{-bt}
 |  /   \   _
 | /     \ / \  _
 |/       V   \/ \___ ...
 +----------------------- t

Numerical part. Given $m = 0.20\ \text{kg}$, $k = 80\ \text{N/m}$, $b = 0.40\ \text{kg/s}$.

$\omega_0 = \sqrt{k/m} = \sqrt{80/0.20} = \sqrt{400} = 20\ \text{rad/s}$

$\beta = \dfrac{b}{2m} = \dfrac{0.40}{2(0.20)} = 1.0\ \text{s}^{-1}$

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

$\omega_d \approx 19.97\ \text{rad/s}$.

(ii) Amplitude $A(t) = A_0 e^{-\beta t}$ falls to $A_0/e$ when $\beta t = 1$, i.e. $t = 1/\beta = 1/1.0 = 1.0\ \text{s}$.

Time constant $\tau = 1.0\ \text{s}$.

Formation. A plano-convex lens of large radius of curvature $R$ rests on a flat glass plate, enclosing a thin air film whose thickness increases radially from the contact point. Monochromatic light incident from above is partly reflected at the top and bottom of the air film; the two reflected beams interfere. Because of the central symmetry, loci of equal thickness are circles, giving concentric bright and dark rings (Newton's rings). An extra phase change of $\pi$ occurs at the air-to-glass reflection at the lower surface.

Diameter of the $n$-th dark ring. For a film of thickness $t$, the path difference is $2t + \lambda/2$. Dark ring (destructive) requires $2t = n\lambda$. From the geometry of the lens, if $r$ is the ring radius, $r^2 = (2R - t)t \approx 2Rt$ for $t \ll R$, so $t = r^2/2R$. Hence

$$2\cdot\frac{r_n^2}{2R} = n\lambda \implies r_n^2 = nR\lambda$$

With diameter $D_n = 2r_n$:

$$D_n^2 = 4nR\lambda$$

Finding $\lambda$. Measuring two rings $n$ and $m$ eliminates uncertainty at the contact point:

$$D_m^2 - D_n^2 = 4(m-n)R\lambda \implies \lambda = \frac{D_m^2 - D_n^2}{4(m-n)R}$$

Numerical part. $D_{10} = 4.00\ \text{mm} = 4.00\times10^{-3}\ \text{m}$, so $D_{10}^2 = 16.00\times10^{-6}\ \text{m}^2$.

$D_{20} = 5.66\ \text{mm} = 5.66\times10^{-3}\ \text{m}$, so $D_{20}^2 = 32.0356\times10^{-6}\ \text{m}^2$.

$D_{20}^2 - D_{10}^2 = (32.0356 - 16.00)\times10^{-6} = 16.0356\times10^{-6}\ \text{m}^2$.

$m - n = 20 - 10 = 10$, $R = 1.00\ \text{m}$.

$$\lambda = \frac{16.0356\times10^{-6}}{4\times10\times1.00} = \frac{16.0356\times10^{-6}}{40} = 4.009\times10^{-7}\ \text{m}$$

$\lambda \approx 401\ \text{nm}$ (violet-blue light).

Gauss's law. The net electric flux through any closed surface equals the charge enclosed divided by $\varepsilon_0$:

$$\oint \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$$

Field of a line charge. Consider an infinite wire with linear charge density $\lambda$. By symmetry $\vec E$ is radial. Take a coaxial Gaussian cylinder of radius $r$ and length $L$. Flux through the curved surface is $E(2\pi r L)$ (ends contribute nothing). Enclosed charge $= \lambda L$:

$$E\,(2\pi r L) = \frac{\lambda L}{\varepsilon_0} \implies E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

Coaxial capacitor. Inner cylinder radius $a$ carries $+\lambda$ per unit length, outer cylinder radius $b$ carries $-\lambda$. The potential difference is

$$V = \int_a^b E\,dr = \frac{\lambda}{2\pi\varepsilon_0}\int_a^b \frac{dr}{r} = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{b}{a}$$

Capacitance per unit length $C/L = \lambda/V$:

$$\boxed{\frac{C}{L} = \frac{2\pi\varepsilon_0}{\ln(b/a)}}$$

Numerical part. $a = 1.0\ \text{mm}$, $b = 4.0\ \text{mm}$, so $b/a = 4.0$ and $\ln 4.0 = 1.3863$.

$$\frac{C}{L} = \frac{2\pi(8.85\times10^{-12})}{1.3863} = \frac{5.561\times10^{-11}}{1.3863} = 4.011\times10^{-11}\ \text{F/m}$$

$C/L \approx 40.1\ \text{pF/m}$.

Maxwell's equations (integral form, free space).

1. Gauss's law (electric): $\displaystyle\oint \vec E\cdot d\vec A = \frac{Q}{\varepsilon_0}$ — electric flux originates from charge.
2. Gauss's law (magnetic): $\displaystyle\oint \vec B\cdot d\vec A = 0$ — no magnetic monopoles; field lines are closed.
3. Faraday's law: $\displaystyle\oint \vec E\cdot d\vec l = -\frac{d\Phi_B}{dt}$ — a changing magnetic flux induces an electric field.
4. Ampere-Maxwell law: $\displaystyle\oint \vec B\cdot d\vec l = \mu_0 I + \mu_0\varepsilon_0\frac{d\Phi_E}{dt}$ — currents and changing electric flux produce magnetic fields.

Displacement current. Consider Ampere's loop around the wire feeding a charging capacitor. A flat surface pierced by the wire gives conduction current $I$, but a bulging surface passing between the plates carries no conduction current — a contradiction. Maxwell resolved this by adding the displacement current

$$I_d = \varepsilon_0\frac{d\Phi_E}{dt}$$

Between the plates $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, so $\Phi_E = EA = Q/\varepsilon_0$ and $I_d = \varepsilon_0\,d\Phi_E/dt = dQ/dt = I$. Thus $I_d$ exactly equals the conduction current, making Ampere's law surface-independent.

Numerical part. $A = \pi r^2 = \pi(0.050)^2 = \pi(2.5\times10^{-3}) = 7.854\times10^{-3}\ \text{m}^2$.

Field: $E = V/d$, so $\dfrac{dE}{dt} = \dfrac{1}{d}\dfrac{dV}{dt} = \dfrac{2.0\times10^{6}}{2.0\times10^{-3}} = 1.0\times10^{9}\ \text{V/(m·s)}$.

$$I_d = \varepsilon_0 A\frac{dE}{dt} = (8.85\times10^{-12})(7.854\times10^{-3})(1.0\times10^{9})$$ $$I_d = 8.85\times10^{-12}\times7.854\times10^{-3}\times10^{9} = 6.95\times10^{-5}\ \text{A}$$

$I_d \approx 69.5\ \mu\text{A}$.

Schrodinger equation. Inside the well ($0 < x < L$) the potential $V = 0$:

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

General solution $\psi(x) = A\sin kx + B\cos kx$.

Boundary conditions. Walls are infinite, so $\psi(0) = \psi(L) = 0$. At $x = 0$: $B = 0$. At $x = L$: $A\sin kL = 0 \Rightarrow kL = n\pi$, $n = 1,2,3,\dots$ Hence $k_n = n\pi/L$.

Energy levels. From $k^2 = 2mE/\hbar^2$:

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

Normalization. $\int_0^L |A|^2\sin^2(n\pi x/L)\,dx = |A|^2 L/2 = 1 \Rightarrow A = \sqrt{2/L}$. Thus

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

Numerical part. $L = 0.10\ \text{nm} = 1.0\times10^{-10}\ \text{m}$.

Ground state ($n=1$):

$$E_1 = \frac{h^2}{8mL^2} = \frac{(6.63\times10^{-34})^2}{8(9.11\times10^{-31})(1.0\times10^{-10})^2}$$

Numerator: $(6.63\times10^{-34})^2 = 4.396\times10^{-67}\ \text{J}^2\text{s}^2$.

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

$$E_1 = \frac{4.396\times10^{-67}}{7.288\times10^{-50}} = 6.032\times10^{-18}\ \text{J}$$

In eV: $E_1 = \dfrac{6.032\times10^{-18}}{1.60\times10^{-19}} = 37.7\ \text{eV}$.

First excited state ($n=2$): $E_2 = 4E_1 = 4\times37.7 = 150.8\ \text{eV}$.

$E_1 \approx 37.7\ \text{eV}$, $\;E_2 \approx 150.8\ \text{eV}$.

Reverberation time is the time taken for the sound intensity in an enclosure to fall to $10^{-6}$ of its steady value, i.e. for the sound level to drop by $60\ \text{dB}$ after the source stops.

Sabine's formula:

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

where $V$ is the room volume in $\text{m}^3$ and $A = \sum \alpha_i S_i$ is the total absorption in sabine ($\text{m}^2$).

Initial reverberation time. $V = 1500\ \text{m}^3$, $A = 120\ \text{m}^2$:

$$T = \frac{0.161\times1500}{120} = \frac{241.5}{120} = 2.01\ \text{s}$$

$T \approx 2.01\ \text{s}$.

Absorption needed for $T' = 1.5\ \text{s}$. Rearranging:

$$A' = \frac{0.161\,V}{T'} = \frac{241.5}{1.5} = 161\ \text{m}^2$$

Additional absorption $= A' - A = 161 - 120 = 41\ \text{m}^2$ (sabine).

Extra absorption required $\approx 41\ \text{m}^2$.

Action of a grating. A plane transmission grating is a large number of equally spaced parallel slits. When a plane wave is incident normally, each slit diffracts light; the diffracted wavelets from successive slits interfere. Constructive interference (principal maxima) occurs when the path difference between adjacent slits equals an integral number of wavelengths.

Grating equation:

$$d\sin\theta = n\lambda$$

where $d$ is the grating spacing, $\theta$ the diffraction angle and $n$ the order.

Grating spacing. $5000$ lines/cm means

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

First order ($n = 1$). $\lambda = 589\ \text{nm} = 5.89\times10^{-7}\ \text{m}$.

$$\sin\theta_1 = \frac{n\lambda}{d} = \frac{5.89\times10^{-7}}{2.0\times10^{-6}} = 0.2945$$ $$\theta_1 = \sin^{-1}(0.2945) = 17.1^\circ$$

$\theta_1 \approx 17.1^\circ$.

Highest order. Maximum when $\sin\theta = 1$:

$$n_{max} = \frac{d}{\lambda} = \frac{2.0\times10^{-6}}{5.89\times10^{-7}} = 3.40$$

Since $n$ must be an integer, the highest observable order is $n_{max} = 3$.

Acceptance angle. The maximum half-angle of the cone of light, measured from the fibre axis, within which incident rays will undergo total internal reflection at the core-cladding boundary and so be guided along the fibre. Rays entering at larger angles escape into the cladding.

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

Derivation. Let core index $n_1$, cladding index $n_2$, surrounding medium $n_0$. A ray entering at angle $\theta_a$ refracts into the core at angle $\theta_r$ and strikes the core-cladding interface at angle $(90^\circ - \theta_r)$. For total internal reflection this must equal the critical angle $\theta_c$ where $\sin\theta_c = n_2/n_1$.

At the entry face (Snell): $n_0\sin\theta_a = n_1\sin\theta_r = n_1\cos\theta_c = n_1\sqrt{1 - (n_2/n_1)^2}$.

$$n_0\sin\theta_a = \sqrt{n_1^2 - n_2^2}$$

For air $n_0 = 1$:

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

Numerical part. $n_1 = 1.50$, $n_2 = 1.46$.

$n_1^2 = 2.2500$, $n_2^2 = 2.1316$, difference $= 0.1184$.

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

NA $\approx 0.344$.

Acceptance angle: $\theta_a = \sin^{-1}(0.344) = 20.1^\circ$.

$\theta_a \approx 20.1^\circ$ (full acceptance cone $\approx 40.2^\circ$).

Hall effect. When a current-carrying conductor is placed in a transverse magnetic field, the moving charge carriers experience a Lorentz force $\vec F = q\vec v\times\vec B$ that deflects them to one face. Charge accumulates there, setting up a transverse electric field (Hall field) until the electric force balances the magnetic force. The resulting transverse voltage is the Hall voltage $V_H$.

Derivation. At equilibrium $eE_H = e v_d B$, so $E_H = v_d B$. The drift velocity relates to current by $I = n e v_d A = n e v_d (w t)$ where $w$ is width, $t$ thickness, so $v_d = I/(n e w t)$. The Hall voltage across width $w$ is $V_H = E_H w = v_d B w = \dfrac{I B}{n e t}$.

The Hall coefficient $R_H = \dfrac{1}{ne}$ (for electrons, sign negative), giving

$$V_H = \frac{R_H I B}{t},\qquad R_H = \frac{V_H\,t}{I B}$$

Numerical part. $I = 5.0\ \text{A}$, $B = 0.40\ \text{T}$, $t = 1.0\times10^{-3}\ \text{m}$, $V_H = 1.25\times10^{-3}\ \text{V}$.

From $V_H = \dfrac{IB}{net}$:

$$n = \frac{IB}{e\,t\,V_H} = \frac{(5.0)(0.40)}{(1.60\times10^{-19})(1.0\times10^{-3})(1.25\times10^{-3})}$$

Numerator: $5.0\times0.40 = 2.0$.

Denominator: $1.60\times10^{-19}\times1.0\times10^{-3}\times1.25\times10^{-3} = 2.0\times10^{-25}$.

$$n = \frac{2.0}{2.0\times10^{-25}} = 1.0\times10^{25}\ \text{m}^{-3}$$

Carrier concentration $n \approx 1.0\times10^{25}\ \text{m}^{-3}$.

Spontaneous vs stimulated emission.

In stimulated emission an excited atom is struck by a photon of energy $h\nu = E_2 - E_1$ and drops to the lower level, emitting a second identical photon, giving light amplification.

Population inversion. Normally, by the Boltzmann distribution, lower levels are more populated than upper ones, so absorption dominates. Lasing requires the opposite: more atoms in the upper level $E_2$ than in $E_1$ ($N_2 > N_1$). This non-equilibrium state is called population inversion and is achieved by pumping (optical or electrical).

Metastable state. A metastable level has a long lifetime ($\sim10^{-3}\ \text{s}$ vs $\sim10^{-8}\ \text{s}$ for ordinary excited states). Atoms accumulate there long enough to build up the inversion before stimulated emission de-excites them.

Two properties of laser light: (1) high monochromaticity (very narrow spectral width); (2) high coherence and directionality (the beam is nearly parallel and in phase). (Also high intensity/brightness.)

Three-level vs four-level. In a three-level laser the lower lasing level is the ground state, which is heavily populated; more than half the atoms must be pumped above ground state to achieve inversion, requiring a high pumping threshold. In a four-level laser the lower lasing level lies above the ground state and is normally almost empty (it depopulates rapidly to the ground state), so inversion is achieved with much less pumping. Hence the four-level system is more efficient and can operate continuously.

Superconductivity is the phenomenon in which certain materials, cooled below a characteristic temperature, exhibit exactly zero electrical resistance and expel magnetic flux from their interior.

Critical temperature ($T_c$) is the temperature below which a material becomes superconducting; above $T_c$ it behaves as a normal conductor.

Meissner effect. When a superconductor is cooled below $T_c$ in a magnetic field, it expels the magnetic flux from its interior, becoming a perfect diamagnet ($B = 0$ inside). This is a distinct property beyond mere zero resistance — it shows superconductivity is a true thermodynamic state, not just perfect conductivity.

Type I vs Type II.

Numerical part. $H_c(0) = 0.0640\ \text{T}$, $T_c = 7.2\ \text{K}$, $T = 4.2\ \text{K}$.

$$\frac{T}{T_c} = \frac{4.2}{7.2} = 0.5833,\qquad \left(\frac{T}{T_c}\right)^2 = 0.3403$$ $$H_c(T) = 0.0640\,[1 - 0.3403] = 0.0640\times0.6597 = 0.04222\ \text{T}$$

$H_c(4.2\ \text{K}) \approx 0.0422\ \text{T}$ (about $42.2\ \text{mT}$).