BE Civil Engineering (IOE, TU) Engineering Physics (IOE, SH 402) Question Paper 2077 Nepal

Differential equation. A mass $m$ on a spring (force constant $k$) experiencing a viscous damping force $-b\dot{x}$ obeys Newton's second law:

$$m\ddot{x} = -kx - b\dot{x}$$ $$\ddot{x} + \frac{b}{m}\dot{x} + \frac{k}{m}x = 0$$

Writing $2\beta = b/m$ (damping constant) and $\omega_0^2 = k/m$ (natural angular frequency):

$$\ddot{x} + 2\beta\dot{x} + \omega_0^2 x = 0$$

Solution. Try $x = e^{\alpha t}$, giving 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 $\beta^2 < \omega_0^2$. Put $\omega = \sqrt{\omega_0^2 - \beta^2}$. Then

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

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: $e^{-\beta\tau}=e^{-1}\Rightarrow \tau = 1/\beta = 2m/b$.

Logarithmic decrement $\lambda$ is the natural log of the ratio of successive amplitudes one period $T=2\pi/\omega$ apart: $\lambda = \ln\!\frac{A(t)}{A(t+T)} = \beta T$.

Numerical part.

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

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

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

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

(ii) Amplitude falls to $1/e$ when $t = \tau = 1/\beta = 1/1.0 = 1.0\,\text{s}$

$\tau = 1.0\,\text{s}$

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

Path difference. For an air film of thickness $t$, the path difference between the two reflected rays is $2t$. Reflection at the lower (denser glass) surface introduces an extra phase change of $\pi$, i.e. an additional path of $\lambda/2$:

$$\Delta = 2t + \frac{\lambda}{2}$$

Dark ring condition: destructive interference requires $\Delta = (2n+1)\lambda/2$, which gives

$$2t = n\lambda, \qquad n = 0,1,2,\dots$$

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

$$r_n^2 = nR\lambda \quad\Rightarrow\quad r_n = \sqrt{nR\lambda}$$

In terms of diameter $D_n = 2r_n$: $\;D_n^2 = 4nR\lambda$.

Centre dark: At the contact point $t=0$, so $\Delta = \lambda/2$ — a half-wave path difference — giving destructive interference, hence a dark central spot.

Numerical part. Using $D_n^2 = 4nR\lambda$ for two rings and subtracting eliminates contact errors:

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

$D_{20} = 6.36\,\text{mm}\Rightarrow D_{20}^2 = 40.4496\,\text{mm}^2$

$D_{10} = 4.50\,\text{mm}\Rightarrow D_{10}^2 = 20.25\,\text{mm}^2$

$D_{20}^2 - D_{10}^2 = 20.1996\,\text{mm}^2 = 20.1996\times10^{-6}\,\text{m}^2$

$$R = \frac{D_{20}^2 - D_{10}^2}{40\lambda} = \frac{20.1996\times10^{-6}}{40 \times 589\times10^{-9}}$$ $$R = \frac{20.1996\times10^{-6}}{2.356\times10^{-5}} = 0.8573\,\text{m}$$

$R \approx 0.857\,\text{m}$ (about $85.7\,\text{cm}$)

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

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

Field of an infinite line charge. Choose a coaxial cylindrical Gaussian surface of radius $r$ and length $L$. By symmetry $\vec{E}$ is radial and constant in magnitude over the curved surface; the flat ends contribute nothing. Enclosed charge $= \lambda L$:

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

Coaxial capacitor. Inner cylinder radius $a$ carries $+\lambda$ per unit length, outer cylinder radius $b$ carries $-\lambda$. In a dielectric, replace $\varepsilon_0$ by $\varepsilon = \varepsilon_r\varepsilon_0$. The potential difference is

$$V = -\int_b^a E\,dr = \int_a^b \frac{\lambda}{2\pi\varepsilon r}\,dr = \frac{\lambda}{2\pi\varepsilon}\ln\!\frac{b}{a}$$

Charge per length $Q/L = \lambda$, so capacitance per unit length is

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

Numerical part.

$a = 0.50\,\text{mm}$, $b = 3.5\,\text{mm}\Rightarrow b/a = 7.0$, $\ln 7 = 1.9459$.

$2\pi\varepsilon_0 = 2\pi(8.854\times10^{-12}) = 5.5632\times10^{-11}\,\text{F/m}$

$$\frac{C}{L} = \frac{2.3 \times 5.5632\times10^{-11}}{1.9459} = \frac{1.2795\times10^{-10}}{1.9459} = 6.575\times10^{-11}\,\text{F/m}$$

$C/L \approx 65.8\,\text{pF/m}$

Maxwell's equations (integral form).

1. Gauss's law (electricity): $\displaystyle\oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ — electric flux originates from charge.
2. Gauss's law (magnetism): $\displaystyle\oint \vec{B}\cdot d\vec{A} = 0$ — no magnetic monopoles; magnetic 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 (emf).
4. Ampere–Maxwell law: $\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0\left(I + \varepsilon_0\frac{d\Phi_E}{dt}\right)$ — magnetic fields arise from conduction current and from changing electric flux.

Displacement current. In a charging capacitor no conduction current flows through the gap, yet a magnetic field exists there. Maxwell resolved the inconsistency in Ampere's law by adding the displacement current

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

which arises from the time-varying electric flux. With it, Ampere's law becomes $\oint\vec B\cdot d\vec l = \mu_0(I + I_d)$, making the current continuous across the gap.

Numerical part. Between the plates $E = V/d$, so $\Phi_E = EA = \dfrac{V}{d}A$ and

$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \frac{\varepsilon_0 A}{d}\frac{dV}{dt} = C\frac{dV}{dt}$$

Area $A = \pi r^2 = \pi(0.040)^2 = 5.0265\times10^{-3}\,\text{m}^2$.

$$C = \frac{\varepsilon_0 A}{d} = \frac{8.854\times10^{-12}\times5.0265\times10^{-3}}{2.0\times10^{-3}} = 2.225\times10^{-11}\,\text{F}$$ $$I_d = C\frac{dV}{dt} = 2.225\times10^{-11}\times1.5\times10^{6} = 3.34\times10^{-5}\,\text{A}$$

$I_d \approx 33.4\,\mu\text{A}$

Time-independent Schrodinger equation. Start from the time-dependent equation for $\Psi(x,t)$:

$$i\hbar\frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi$$

For a stationary state separate variables: $\Psi(x,t)=\psi(x)\,e^{-iEt/\hbar}$. Substituting and dividing through by $e^{-iEt/\hbar}$:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi$$

or $\displaystyle\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E-V)\psi = 0$. This is the time-independent Schrodinger equation.

Infinite well. Inside the well $V=0$ for $0<x<L$, and $\psi=0$ outside (infinite walls). The equation becomes $\psi'' + k^2\psi = 0$ with $k^2 = 2mE/\hbar^2$, giving $\psi = A\sin kx + B\cos kx$.

Boundary conditions: $\psi(0)=0\Rightarrow B=0$; $\psi(L)=0\Rightarrow \sin kL = 0\Rightarrow kL = n\pi$, $n=1,2,3,\dots$

Hence $k_n = n\pi/L$ and

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

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

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

Numerical part. With $E_n = \dfrac{n^2 h^2}{8mL^2}$:

Compute $E_1$. $h = 6.626\times10^{-34}\,\text{J s}$, $m = 9.11\times10^{-31}\,\text{kg}$, $L = 0.20\times10^{-9}\,\text{m}$.

$h^2 = 4.390\times10^{-67}$; $L^2 = 4.0\times10^{-20}\,\text{m}^2$.

$8mL^2 = 8\times9.11\times10^{-31}\times4.0\times10^{-20} = 2.9152\times10^{-49}$

$$E_1 = \frac{4.390\times10^{-67}}{2.9152\times10^{-49}} = 1.506\times10^{-18}\,\text{J} = \frac{1.506\times10^{-18}}{1.602\times10^{-19}} = 9.40\,\text{eV}$$

Ground state $E_1 \approx 9.40\,\text{eV}$

First excited state $E_2 = 4E_1 = 37.6\,\text{eV}$. $E_2 \approx 37.6\,\text{eV}$

Photon energy for $2\to1$: $\Delta E = E_2 - E_1 = 3E_1 = 28.2\,\text{eV} = 28.2\times1.602\times10^{-19} = 4.518\times10^{-18}\,\text{J}$

$$\lambda = \frac{hc}{\Delta E} = \frac{6.626\times10^{-34}\times3.0\times10^{8}}{4.518\times10^{-18}} = \frac{1.988\times10^{-25}}{4.518\times10^{-18}} = 4.40\times10^{-8}\,\text{m}$$

$\lambda \approx 44.0\,\text{nm}$ (ultraviolet)

Reverberation time $T$ is the time taken for the sound intensity (energy density) in a room to fall to $10^{-6}$ of its initial value after the source stops, i.e. a drop of $60\,\text{dB}$.

Sabine's formula:

$$T = \frac{0.161\,V}{\sum a_i S_i} = \frac{0.161\,V}{\bar{a}\,S}$$

where $V$ is the room volume (m$^3$), $S$ the total surface area (m$^2$), and $\bar a$ the average absorption coefficient; $0.161$ has units s/m.

Numerical. Volume:

$$V = 20\times15\times8 = 2400\,\text{m}^3$$

Total surface area (six faces):

$$S = 2(20\times15) + 2(20\times8) + 2(15\times8) = 600 + 320 + 240 = 1160\,\text{m}^2$$

Total absorption $A = \bar a S = 0.25\times1160 = 290\,\text{m}^2\text{(sabin)}$.

$$T = \frac{0.161\times2400}{290} = \frac{386.4}{290} = 1.333\,\text{s}$$

$T \approx 1.33\,\text{s}$

Action of a grating. A plane transmission grating consists of a large number of equally spaced, identical parallel slits ruled on a transparent plate. When a plane wavefront falls normally on it, each slit diffracts light; the diffracted beams from all slits interfere. Constructive interference (a sharp principal maximum) occurs in directions where the path difference between adjacent slits, $d\sin\theta$, equals a whole number of wavelengths. The large number of slits makes the maxima very sharp, allowing the grating to act as a spectroscope.

Grating equation:

$$d\sin\theta = m\lambda, \qquad m = 0,1,2,\dots$$

where $d$ is the grating spacing (distance between adjacent slits) and $m$ the order. If $(e+d)$ is the grating element with opaque width $d$ and transparent width $e$, then absent spectra occur when an order of the grating coincides with a minimum of the single-slit diffraction pattern, i.e. when $\frac{e+d}{e}=\frac{m}{p}$ for integers $m,p$.

Numerical. Grating spacing:

$$d = \frac{1\,\text{cm}}{5000} = \frac{10^{-2}\,\text{m}}{5000} = 2.0\times10^{-6}\,\text{m}$$

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

$$\sin\theta = \frac{m\lambda}{d} = \frac{2\times546\times10^{-9}}{2.0\times10^{-6}} = \frac{1.092\times10^{-6}}{2.0\times10^{-6}} = 0.546$$ $$\theta = \sin^{-1}(0.546) = 33.10^\circ$$

$\theta \approx 33.1^\circ$

Spontaneous vs stimulated emission.

In stimulated emission an excited atom is induced by an incident photon to drop to a lower state, emitting a second photon identical to the first — the basis of laser action.

Population inversion is the non-equilibrium condition in which more atoms occupy a higher energy level than a lower one ($N_2 > N_1$). It is essential for laser action because only then does stimulated emission dominate over absorption, giving net light amplification; it is achieved by pumping and the use of a metastable level.

Numerical aperture (NA) measures the light-gathering ability of a fibre; it is the sine of the maximum acceptance angle. For a step-index fibre with core index $n_1$ and cladding index $n_2$ (in a medium of index $n_0$):

$$\text{NA} = \sqrt{n_1^2 - n_2^2}, \qquad n_0\sin\theta_a = \text{NA}$$

Acceptance angle $\theta_a$ is the maximum half-angle of the entrance cone within which incident light is guided by total internal reflection.

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

$$\text{NA} = \sqrt{1.50^2 - 1.46^2} = \sqrt{2.25 - 2.1316} = \sqrt{0.1184} = 0.3441$$

NA $\approx 0.344$

In air $n_0 = 1$, so $\sin\theta_a = 0.3441$:

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

$\theta_a \approx 20.1^\circ$

Biot–Savart law. The magnetic field $d\vec B$ produced at a point by a current element $I\,d\vec l$ at displacement $\vec r$ from it is

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

with magnitude $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\phi}{r^2}$, where $\phi$ is the angle between $d\vec l$ and $\hat r$.

Field at centre of a circular loop. For every element of a loop of radius $R$, the point (centre) is at distance $r=R$ and $d\vec l \perp \hat r$ so $\sin\phi = 1$. All $d\vec B$ point in the same direction (along the axis), so they add:

$$B = \frac{\mu_0 I}{4\pi R^2}\oint dl = \frac{\mu_0 I}{4\pi R^2}(2\pi R) = \frac{\mu_0 I}{2R}$$

For $N$ turns: $B = \dfrac{\mu_0 N I}{2R}$.

Comparison with axial field. On the axis at distance $x$ from the centre, $B_x = \dfrac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}$. Setting $x=0$ recovers the centre value $\mu_0 N I/2R$, which is the maximum; the field decreases monotonically as $x$ increases.

Numerical. $\mu_0 = 4\pi\times10^{-7}\,\text{T m/A}$, $N=200$, $I=2.5\,\text{A}$, $R = 0.060\,\text{m}$.

$$B = \frac{4\pi\times10^{-7}\times200\times2.5}{2\times0.060}$$

Numerator: $4\pi\times10^{-7}\times500 = 6.2832\times10^{-4}$. Denominator: $0.120$.

$$B = \frac{6.2832\times10^{-4}}{0.120} = 5.236\times10^{-3}\,\text{T}$$

$B \approx 5.24\,\text{mT}$

Hall effect. When a current-carrying conductor (current along $x$) is placed in a magnetic field $B$ along $z$, the moving charge carriers experience a transverse Lorentz force $q\vec v\times\vec B$ that pushes them to one side. Charge accumulates on that face, setting up a transverse electric field $E_H$ (along $y$) until the electric force balances the magnetic force. The resulting transverse potential difference is the Hall voltage $V_H$.

Derivation. At equilibrium $qE_H = qvB \Rightarrow E_H = vB$. The drift speed relates to current density $J = nqv$, so $v = J/(nq) = I/(nqA)$ where $A = w t$ (width $w$, thickness $t$). With $E_H = V_H/w$:

$$\frac{V_H}{w} = \frac{I B}{nq\,w t}\quad\Rightarrow\quad V_H = \frac{IB}{nqt}$$

The Hall coefficient is defined as $R_H = \dfrac{1}{nq}$, giving

$$R_H = \frac{V_H\, t}{I B}, \qquad n = \frac{1}{R_H q} = \frac{IB}{V_H\,t\,q}$$

Numerical. $t = 0.50\times10^{-3}\,\text{m}$, $I = 3.0\,\text{A}$, $B = 0.40\,\text{T}$, $V_H = 6.0\times10^{-3}\,\text{V}$.

$$R_H = \frac{V_H t}{IB} = \frac{6.0\times10^{-3}\times0.50\times10^{-3}}{3.0\times0.40} = \frac{3.0\times10^{-6}}{1.20} = 2.5\times10^{-6}\,\text{m}^3/\text{C}$$

$R_H = 2.5\times10^{-6}\,\text{m}^3/\text{C}$

Carrier concentration ($q = 1.602\times10^{-19}\,\text{C}$):

$$n = \frac{1}{R_H q} = \frac{1}{2.5\times10^{-6}\times1.602\times10^{-19}} = \frac{1}{4.005\times10^{-25}} = 2.497\times10^{24}\,\text{m}^{-3}$$

$n \approx 2.50\times10^{24}\,\text{m}^{-3}$

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

Meissner effect. When a superconductor is cooled below $T_c$ in a magnetic field, it actively expels the magnetic field from its interior ($B = 0$ inside), behaving as a perfect diamagnet. This is distinct from (and stronger than) merely having zero resistance, and it shows superconductivity is a true thermodynamic state.

Type-I vs Type-II. Type-I superconductors show a single critical field $H_c$ above which superconductivity is destroyed abruptly (complete Meissner effect up to $H_c$); they are usually pure metals with low $H_c$. Type-II superconductors have two critical fields $H_{c1}$ and $H_{c2}$: below $H_{c1}$ they fully expel flux, between $H_{c1}$ and $H_{c2}$ they enter a mixed (vortex) state allowing partial flux penetration while remaining superconducting, and above $H_{c2}$ they go normal; they are usually alloys/compounds with high $H_{c2}$, useful for high-field magnets.

Critical field vs temperature. A superconductor returns to the normal state if the applied field exceeds the temperature-dependent critical field, which follows the parabolic law

$$H_c(T) = H_c(0)\left[1 - \left(\frac{T}{T_c}\right)^2\right]$$

Numerical. $H_c(0) = 6.5\times10^{4}\,\text{A/m}$, $T = 4.0\,\text{K}$, $T_c = 7.2\,\text{K}$.

$$\left(\frac{T}{T_c}\right)^2 = \left(\frac{4.0}{7.2}\right)^2 = (0.5556)^2 = 0.3086$$ $$H_c(4) = 6.5\times10^{4}\,(1 - 0.3086) = 6.5\times10^{4}\times0.6914 = 4.494\times10^{4}\,\text{A/m}$$

$H_c(4\,\text{K}) \approx 4.49\times10^{4}\,\text{A/m}$