A $2\times2$ matrix A has eigenvalues $e^{i\pi/5}$ and $e^{i\pi/6}$. The smallest value of ‘$n$’ such that $\text{A}^n = 1$ is:
a.
20
b.
30
c.
60
Correct Answer
d.
120
Step-by-Step AI Explanation
Verified Method
For $\text{A}^n = I$, both eigenvalues raised to the power $n$ must equal 1:
$$(e^{i\pi/5})^n = e^{i n\pi/5} = 1 \implies \frac{n}{5} \in 2\mathbb{Z} \implies n \text{ is a multiple of } 10$$
$$(e^{i\pi/6})^n = e^{i n\pi/6} = 1 \implies \frac{n}{6} \in 2\mathbb{Z} \implies n \text{ is a multiple of } 12$$
The smallest positive integer $n$ that is a multiple of both 10 and 12 is $\text{lcm}(10, 12) = 60$.
Q22mcq
The graph of the function $f(x)$ as shown below is best described by
a.
The bessel function $\text{J}_0(x)$
Correct Answer
b.
$\cos x$
c.
$e^{-x}\cos x$
d.
$\frac{1}{x}\cos x$
Step-by-Step AI Explanation
Verified Method
At $x = 0$, the function equals $1$. As $x$ increases, it exhibits damped oscillations. Looking at the roots, the first zero is at approximately $x \approx 2.4$, which is the well-known first zero of the zeroth-order Bessel function of the first kind, $J_0(x) \approx 2.4048$. In contrast, $\cos x$ has its first zero at $\pi/2 \approx 1.57$, and $x^{-1}\cos x$ diverges at $x = 0$. Hence, the graph best describes the Bessel function $J_0(x)$.
Q23mcq
In a series of five cricket matches, one of the captains calls “Heads” every time when the toss is taken. The probability that he will win 3 times and lose 2 times is
a.
1/8
b.
5/8
c.
3/16
d.
5/16
Correct Answer
Step-by-Step AI Explanation
Verified Method
Each toss has two equally likely outcomes (win or lose the toss) with $p = 1/2$. The number of wins follows a binomial distribution with $n = 5$ and $p = 1/2$. The probability of winning exactly $k = 3$ times is:
$$P(X = 3) = \binom{5}{3} p^3 (1-p)^{5-3} = 10 \left(\frac{1}{2}\right)^5 = \frac{10}{32} = \frac{5}{16}$$
Q24mcq
The unit normal vector at the point $\left(\frac{a}{\sqrt{3}}, \frac{b}{\sqrt{3}}, \frac{c}{\sqrt{3}}\right)$ on the surface of the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, is
Let the surface be defined by $F(x, y, z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1 = 0$. The normal vector is given by the gradient:
$$\nabla F = \left(\frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2}\right)$$
At the point $\left(\frac{a}{\sqrt{3}}, \frac{b}{\sqrt{3}}, \frac{c}{\sqrt{3}}\right)$, we have:
$$\nabla F = \frac{2}{\sqrt{3}} \left(\frac{1}{a}\hat{i} + \frac{1}{b}\hat{j} + \frac{1}{c}\hat{k}\right) = \frac{2}{\sqrt{3}abc} \left(bc\hat{i} + ca\hat{j} + ab\hat{k}\right)$$
The unit normal vector is:
$$\hat{n} = \frac{\nabla F}{|\nabla F|} = \frac{bc\hat{i} + ca\hat{j} + ab\hat{k}}{\sqrt{(bc)^2 + (ca)^2 + (ab)^2}}$$
In the question options, option (A) gives the numerator as $bc\hat{i} + ca\hat{j} + ab\hat{k}$, which corresponds to option A.
Q25mcq
A solid cylinder of height $H$, radius $R$ and density $\rho$, floats vertically on the surface of a liquid of density $\rho_0$. The cylinder will be set into oscillatory motion when a small instantaneous downward force is applied. The frequency of oscillation is
a.
$\frac{\rho g}{\rho_0 H}$
b.
$\frac{\rho}{\rho_0}\sqrt{\frac{g}{H}}$
c.
$\sqrt{\frac{\rho g}{\rho_0 H}}$
d.
$\sqrt{\frac{\rho_0 g}{\rho H}}$
Correct Answer
Step-by-Step AI Explanation
Verified Method
Let $A$ be the cross-sectional area of the cylinder. The total mass of the cylinder is $M = \rho A H$.
When the cylinder is displaced downward by a small distance $z$, the extra restoring buoyant force is:
$$F = -\rho_0 A g z$$
The equation of motion is:
$$M \ddot{z} = -\rho_0 A g z \implies \rho A H \ddot{z} + \rho_0 A g z = 0$$
$$\ddot{z} + \left(\frac{\rho_0 g}{\rho H}\right) z = 0$$
Thus, the angular frequency of oscillation is $\omega = \sqrt{\frac{\rho_0 g}{\rho H}}$.
Q26mcq
Three particles of equal mass '$m$' are connected by two identical massless springs of stiffness constant '$k$' as shown in the figure.
If $x_1$, $x_2$ and $x_3$ denote the displacements of the masses from their respective equilibrium positions, the potential energy of the system is:
a.
$\frac{1}{2} k \left(x_1^2 + x_2^2 + x_3^2\right)$
The extension of the first spring is $(x_2 - x_1)$ and of the second spring is $(x_3 - x_2)$.
The total potential energy of the system is the sum of the potential energies of the two springs:
$$V = \frac{1}{2} k (x_2 - x_1)^2 + \frac{1}{2} k (x_3 - x_2)^2$$
Expanding the terms:
$$V = \frac{1}{2} k \left[(x_2^2 - 2x_1 x_2 + x_1^2) + (x_3^2 - 2x_2 x_3 + x_2^2)\right]$$
$$V = \frac{1}{2} k \left[x_1^2 + 2x_2^2 + x_3^2 - 2x_2(x_1 + x_3)\right]$$
This matches option D.
Q27mcq
Let $v$, $p$ and $E$ denotes the speed, the magnitude of the momentum, and the energy of a free particle of rest mass '$m$'. Then
In relativistic mechanics, the energy-momentum relation is $E = \sqrt{p^2 c^2 + m^2 c^4} = \gamma m c^2$ and momentum is $p = \gamma m v$.
From these relations, we have:
$$v = \frac{p c^2}{E} = \frac{p c^2}{\sqrt{p^2 c^2 + m^2 c^4}} = \frac{c p}{\sqrt{p^2 + m^2 c^2}}$$
Thus, option C is correct.
Q28mcq
A binary star system consists of two stars $\text{S}_1$ and $\text{S}_2$, with masses $m$ and $2m$ respectively separated by a distannce '$r$'. If both $\text{S}_1$ and $\text{S}_2$ individually follow circular orbits around the centre of the mass with intantaneous speeds $\text{v}_1$ and $\text{v}_2$ respectively, the ratio of speeds $\text{v}_1/\text{v}_2$ is:
a.
$\sqrt{2}$
b.
$1$
c.
$1/2$
d.
$2$
Correct Answer
Step-by-Step AI Explanation
Verified Method
For two stars orbiting their common centre of mass, both stars have the same angular frequency $\omega$.
The distances of the two stars from the centre of mass are:
$$r_1 = \frac{m_2}{m_1 + m_2} r = \frac{2m}{3m} r = \frac{2}{3} r$$
$$r_2 = \frac{m_1}{m_1 + m_2} r = \frac{m}{3m} r = \frac{1}{3} r$$
Since the speeds are given by $v_1 = \omega r_1$ and $v_2 = \omega r_2$, their ratio is:
$$\frac{v_1}{v_2} = \frac{r_1}{r_2} = \frac{2/3}{1/3} = 2$$
Q29mcq
Three charges are located on the circumference of a circle of radius '$R$' as shown in the figure below. The two charges $\text{Q}$ subtend an angle $90^\circ$ at the centre of the circle .The charge '$q$' is symmetrically placed with respect to the charges $\text{Q}$. If the electric field at the centre of the circle is zero, what is the magnitude of $\text{Q}$?
a.
$q/\sqrt{2}$
Correct Answer
b.
$\sqrt{2}q$
c.
$2q$
d.
$4q$
Step-by-Step AI Explanation
Verified Method
Let the centre of the circle be the origin. The two identical charges $Q$ subtend an angle of $90^\circ$ at the centre, so they can be placed at angles $+45^\circ$ and $-45^\circ$ (or $\pm \pi/4$) relative to the $x$-axis. The charge $q$ is symmetrically placed, which means it is located along the opposite direction on the circumference, at an angle of $180^\circ$. The electric field at the centre due to the two charges $Q$ is directed away from them along the $-x$-axis (assuming $Q > 0$) with magnitude $E_Q = 2 \times \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2} \cos(45^\circ) = \frac{\sqrt{2}Q}{4\pi\varepsilon_0 R^2}$. The electric field due to $q$ at the centre is along the $+x$-axis with magnitude $E_q = \frac{q}{4\pi\varepsilon_0 R^2}$. For the net field to be zero, these magnitudes must be equal: $\sqrt{2}Q = q \implies Q = q/\sqrt{2}$.
Q30mcq
Consider a hollow charged shell of inner radius '$a$' and outer radius '$b$'. The volume charge density is $\rho(r) = \frac{k}{r^2}$ (where $k$ is a constant) in the region $a < r < b$. The magnitude of the electric field produced at distance $r > a$ is:
a.
$\frac{k(b-a)}{\varepsilon_0 r^2} \text{ for } r > a$
b.
$\frac{k(b-a)}{\varepsilon_0 r^2} \text{ for } a < r < b \text{ and } \frac{kb}{\varepsilon_0 r^2} \text{ for } r > b$
c.
$\frac{k(r-a)}{\varepsilon_0 r^2} \text{ for } a < r < b \text{ and } \frac{k(b-a)}{\varepsilon_0 r^2} \text{ for } r > b$
Correct Answer
d.
$\frac{k(r-a)}{\varepsilon_0 a^2} \text{ for } a < r < b \text{ and } \frac{k(b-a)}{\varepsilon_0 a^2} \text{ for } r > b$
Step-by-Step AI Explanation
Verified Method
By Gauss's law, the electric field at a radius $r$ is given by $\oint \vec{E} \cdot d\vec{A} = E(r) 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0}$. For $a < r < b$, the enclosed charge is $Q_{\text{enc}} = \int_a^r \rho(r') 4\pi r'^2 dr' = \int_a^r \frac{k}{r'^2} 4\pi r'^2 dr' = 4\pi k (r - a)$. Thus, $E(r) = \frac{k(r - a)}{\varepsilon_0 r^2}$. For $r > b$, the total enclosed charge is $Q_{\text{enc}} = 4\pi k (b - a)$, so $E(r) = \frac{k(b - a)}{\varepsilon_0 r^2}$.
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Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by $\vec{E}_1 = \hat{i}E_0 \cos \omega t$ and $\vec{E}_2 = \hat{j}E_0 \cos (\omega t + \varphi)$ where $\varphi$ is the phase difference. The intensity of the resulting wave is given by $\frac{\varepsilon_0}{2} \left\langle E^2 \right\rangle$, where $\left\langle E^2 \right\rangle$ is the time average of $E^2$. The total intensity is
a.
$0$
b.
$\varepsilon_0 E_0^2$
c.
$\varepsilon_0 E_0^2 \sin^2 \varphi$
d.
$\varepsilon_0 E_0^2 \cos^2 \varphi$
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Four charges (two $+q$ and two $-q$) are kept fixed at the four vertices of a square of side '$a$' as shown
At the point $P$ which is at a distance $R$ from the centre ($R \gg a$), the potential is proportional to
a.
$\frac{1}{R}$
b.
$\frac{1}{R^2}$
c.
$\frac{1}{R^3}$
d.
$\frac{1}{R^4}$
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A point charge '$q$' of mass '$m$' is kept at a distance '$d$' below a grounded infinite conducting sheet which lies in the $xy$-plane. What is the value of '$d$' for which the charge remains stationary?
a.
$q/4\sqrt{mg\pi\varepsilon_0}$
b.
$q/\sqrt{mg\pi\varepsilon_0}$
c.
There is no finite value of '$d$'
d.
$\sqrt{mg\pi\varepsilon_0}/q$
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The wave function of a state of the hydrogen atom is given by
$$\psi = \psi_{200} + 2\psi_{211} + 3\psi_{210} + \sqrt{2}\psi_{21-1}$$
where $\psi_{n\ell m}$ denotes the normalized eigen function of the state with quantum numbers $n$, $l$ and $m$ in the usual notation. The expectation value of $L_z$ in the state $\psi$ is:
a.
$\frac{15\hbar}{16}$
b.
$\frac{11\hbar}{16}$
c.
$\frac{3\hbar}{8}$
d.
$\frac{\hbar}{8}$
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If a particle is represented by the normalized wave function
$$\psi(x) = \begin{cases} \frac{\sqrt{15}(a^2 - x^2)}{4a^{5/2}} & \text{for } -a < x < a \\ 0 & \text{otherwise} \end{cases}$$
the uncertainty $\Delta p$ in its momentum is
a.
$\frac{2\hbar}{5a}$
b.
$\frac{5\hbar}{2a}$
c.
$\frac{\sqrt{10}\,\hbar}{a}$
d.
$\frac{\sqrt{5}\,\hbar}{\sqrt{2}\,a}$
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Given the usual canonical commutation relations, the commutator $[\text{A}, \text{B}]$ of $\text{A} = \text{i}(x p_y - y p_x)$ and $\text{B} = (y p_z + z p_y)$ is :
a.
$\hbar(x p_z - p_x z)$
b.
$-\hbar(x p_z - p_x z)$
c.
$\hbar(x p_z + p_x z)$
d.
$-\hbar(x p_z + p_x z)$
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The entropy of a system, $\text{S}$, is related to the accessible phase space volume $\Gamma$ by $\text{S} = \text{k}_\text{B} \ln \Gamma(\text{E}, \text{N}, \text{V})$ where $\text{E}$, $\text{N}$ and $\text{V}$ are the energy, number of particles and volume respectively. From this one can conclude that $\Gamma$
a.
does not change during evolution to equilibrium
b.
Oscillates during evolution to equilibrium
c.
Is a maximum in equilibrium
d.
Is a minimum in equilibrium
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Let $\Delta \text{W}$ be the work done in a quasistatic reversible thermodynamics process. Which of the following statements about $\Delta \text{W}$ is correct?
a.
$\Delta \text{W}$ is a perfect differential if the process is isothermal
b.
$\Delta \text{W}$ is a perfect differential if the process is adiabatic
c.
$\Delta \text{W}$ is always a perfect differential
d.
$\Delta \text{W}$ cannot be a perfect differential
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Consider a system of three spins $\text{S}_1$, $\text{S}_2$ and $\text{S}_3$ each of which can take values $+1$ and $-1$. The energy of the system is given by $\text{E} = -\text{J}[\text{S}_1\text{S}_2 + \text{S}_2\text{S}_3 + \text{S}_3\text{S}_1]$, where $\text{J}$ is a positive constant. The minimum energy and the corresponding number of spin configurations are, respectively,
a.
$\text{J}$ and $1$
b.
$-3\text{J}$ and $1$
c.
$-3\text{J}$ and $2$
d.
$-6\text{J}$ and $2$
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The minimum energy of a collection of 6 non-interacting electrons of spin $-\frac{1}{2}$ placed in a one dimensional infinite square well potential of width $\text{L}$ is
a.
$14\pi^2\hbar^2 / \text{mL}^2$
b.
$91\pi^2\hbar^2 / \text{mL}^2$
c.
$7\pi^2\hbar^2 / \text{mL}^2$
d.
$3\pi^2\hbar^2 / \text{mL}^2$
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A live music broadcast consists of a radio-wave of frequency $7\text{ MHz}$, amplitude-modulated by a microphone output consisting of signals with a maximum frequency of $10\text{ KHz}$. The spectrum of modulated output will be zero outside the frequency band
a.
$7.00\text{ MHz to } 7.01\text{ MHz}$
b.
$6.99\text{ MHz to } 7.01\text{ MHz}$
c.
$6.99\text{ MHz to } 7.00\text{ MHz}$
d.
$6.995\text{ MHz to } 7.005\text{ MHz}$
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In the op-amp circuit shown in the figure, $\text{V}_i$ is a sinusoidal input signal of frequency $10\text{ Hz}$ and $\text{V}_0$ is the output signal. The magnitude of the gain and the phase shift, respectively, are close to the values
a.
$5\sqrt{2}\text{ and }\frac{\pi}{2}$
b.
$5\sqrt{2}\text{ and }\frac{-\pi}{2}$
c.
$10\text{ and zero}$
d.
$10\text{ and }\pi$
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A diode $\text{D}$ as shown in the circuit as an $\text{i}-\text{v}$ relation which can be proximated by $$i_\text{D} = \begin{cases} v_\text{D}^2 + 2v_\text{D}, & \text{for } v_\text{D} > 0 \\ 0, & \text{for } v_\text{D} \le 0 \end{cases}$$ The value of $v_\text{D}$ in the circuit is:
a.
$(-1+\sqrt{11})\text{V}$
b.
$8\text{V}$
c.
$5\text{V}$
d.
$2\text{V}$
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Given a $2 \times 2$ unitary matrix $\text{U}$ satisfying $\text{U}'\text{U} = \text{UU}' = 1$ with $\det\text{U} = \text{e}^{i\varphi}$, one can construct a unitary matrix $\text{V} (\text{V}'\text{V} = \text{VV}' = 1)$ with $\det\text{V} = 1$ from it by
a.
Multiplying $\text{U}$ by $\text{e}^{-i\varphi/2}$
b.
Multiplying any single element of $\text{U}$ by $\text{e}^{-i\varphi}$
c.
Multiplying any row or column of $\text{U}$ by $\text{e}^{-i\varphi/2}$
d.
Multiplying $\text{U}$ by $\text{e}^{-i\varphi}$.
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The value of the integral $\int_\text{C} \frac{z^3\text{d}z}{z^2 - 5z + 6}$, where $\text{C}$ is a closed contour defined by the equation $2|z| - 5 = 0$, traversed in the anti-clockwise direction, is:
a.
$-16\pi i$
b.
$16\pi i$
c.
$8\pi i$
d.
$2\pi i$
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A function $\text{f}(x)$ obeys the differential equation $\frac{\text{d}^2\text{f}}{\text{d}x^2} - (3 - 2i)\text{f} = 0$ and satisfies the conditions $\text{f}(0) = 1$ and $\text{f}(x) \rightarrow 0$ as $x \rightarrow \infty$. The value of $\text{f}(\pi)$ is :
a.
$\text{e}^{2\pi}$
b.
$\text{e}^{-2\pi}$
c.
$-\text{e}^{-2\pi}$
d.
$-\text{e}^{2\pi i}$
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A planet of mass '$\text{m}$' moves in the gravitational field of the Sun (mass $\text{M}$). If the semi-major and semi-minor axes of the orbit are '$\text{a}$' and '$\text{b}$' respectively, the angular momentum of the planet is:
The Hamiltonian of a simple pendulum consisting of a mass '$\text{m}$' attached to a massless string of length $l$ is $\text{H} = \frac{\text{p}_\theta^2}{2\text{m}\ell^2} + \text{mg}\ell(1 - \cos\theta)$. If $\text{L}$ denotes the Lagrangian, the value of $\frac{\text{dL}}{\text{dt}}$ is :
Which of the following set of phase-space trajectories which one is not possible for a particle obeying Hamilton's equations of motion (for a time-independent Hamiltonian)?
Two bodies of equal mass '$\text{m}$' are connected by a massless rigid rod of length '$l$' lying in the xy-plane with the centre of the rod at the origin. If this system is rotating about the z-axis with a frequency $\omega$, its angular momentum is
a.
$\text{m}l^2\omega/4$
b.
$\text{m}l^2\omega/2$
c.
$\text{m}l^2\omega$
d.
$2\text{m}l^2\omega$
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An infinite solenoid with its axis of symmetry along the z-direction carries a steady current I.
The vector potential $\vec{A}$ at a distance R from the axis.
a.
Is constant inside and varies as R outside the solenoid.
b.
Varies as R inside and is constant outside the solenoid.
c.
Varies as 1/R inside and as R outside the solenoid.
d.
Varies as R inside and as 1/R outside the solenoid.
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Consider an infinite conducting sheet in the xy-plane with a time dependent current density $Kt\hat{i}$, where K is a constant. The vector potential at $(x, y, z)$ is given by
$$\vec{A} = \frac{\mu_0 K}{4c}(ct-z)^2\hat{i}$$
The magnetic field $\vec{B}$ is:
a.
$\frac{\mu_0 K t}{2}\hat{j}$
b.
$-\frac{\mu_0 K z}{2c}\hat{j}$
c.
$-\frac{\mu_0 K}{2c}(ct-z)\hat{i}$
d.
$-\frac{\mu_0 K}{2c}(ct-z)\hat{j}$
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When a charged particle emits electromagnetic radiation, the electric field $\vec{E}$ and the Poynting vector $\vec{S} = \frac{1}{\mu_0}\vec{E}\times\vec{B}$ at a large distance 'r' from the emitter vary as $\frac{1}{r^n}$ and $\frac{1}{r^m}$ respectively. Which of the following choices for 'n' and 'm' are correct?
a.
$n = 1$ and $m = 1$
b.
$n = 2$ and $m = 2$
c.
$n = 1$ and $m = 2$
d.
$n = 2$ and $m = 4$
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The energies in the ground state and first excited state of a particle of mass $m = \frac{1}{2}$ in a potential $V(x)$ are $-4$ and $-1$, respectively, (in units in which $\hbar = 1$). If the corresponding wavefunctions are related by $\psi_1(x) = \psi_0(x)\sinh x$, then the ground state eigenfunction is
a.
$\psi_0(x) = \sqrt{\operatorname{sech} x}$
b.
$\psi_0(x) = \operatorname{sech} x$
c.
$\psi_0(x) = \operatorname{sech}^2 x$
d.
$\psi_0(x) = \operatorname{sech}^3 x$
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The perturbation
$$H' = \begin{cases} b(a-x) & -a < x < a \\ 0 & \text{otherwise} \end{cases}$$
acts on a particle of mass 'm' confined in an infinite square well potential
$$V(x) = \begin{cases} 0 & -a < x < a \\ \infty & \text{otherwise} \end{cases}$$ The first order correction to the ground state energy of the particle is
a.
$\frac{ba}{2}$
b.
$\frac{ba}{\sqrt{2}}$
c.
$2ba$
d.
$ba$
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Let $|0\rangle$ and $|1\rangle$ denote the normalized eigenstates corresponding to the ground and the first excited states of a one-dimensional harmonic oscillator. The uncertainty $\Delta x$ in the state $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ is:
a.
$\Delta x = \sqrt{\hbar / 2m\omega}$
b.
$\Delta x = \sqrt{\hbar / m\omega}$
c.
$\Delta x = \sqrt{2\hbar / m\omega}$
d.
$\Delta x = \sqrt{\hbar / 4m\omega}$
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What would be the ground state energy of the Hamiltonian
$$H = -\frac{\hbar^2}{2m} \frac{\mathrm{d}^2}{\mathrm{d}x^2} - \alpha\delta(x)$$
if vibrational principle is used to estimate it with the trial wavefunction $\psi(x) = A \mathrm{e}^{-bx^2}$ with b as the variational parameter?
$$\left[\text{Hint: } \int_{-\infty}^\infty x^{2n} \mathrm{e}^{-2bx^2} \mathrm{d}x = (2b)^{-n-\frac{1}{2}} \Gamma\left(n+\frac{1}{2}\right)\right]$$
a.
$-m\alpha^2 / 2\hbar^2$
b.
$-2m\alpha^2 / \pi\hbar^2$
c.
$-m\alpha^2 / \pi\hbar^2$
d.
$m\alpha^2 / \pi\hbar^2$
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The free energy difference between the superconducting and the normal states of a material is given by $\Delta F = F_S - F_N = \alpha |\psi|^2 + \frac{\beta}{2}|\psi|^4$, where $\psi$ is an order parameter and $\alpha$ and $\beta$ are constants such that $\alpha > 0$ in the normal and $\alpha < 0$ in the superconducting state, while $\beta > 0$ always. The minimum value of $\Delta F$ in the superconducting state is
a.
$-\alpha^2 / \beta$
b.
$-\alpha^2 / 2\beta$
c.
$-3\alpha^2 / 2\beta$
d.
$-5\alpha^2 / 2\beta$
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A given quantity of gas is taken from the state $\text{A} \rightarrow \text{C}$ reversibly, by two paths, $\text{A} \rightarrow \text{C}$ directly and $\text{A} \rightarrow \text{B} \rightarrow \text{C}$ as shown in the figure below.
During the $\text{A} \rightarrow \text{C}$ the work done by the gas is $100\text{ J}$ and the heat absorbed is $150\text{ J}$. If during the process $\text{A} \rightarrow \text{B} \rightarrow \text{C}$ the work done by the gas is $30\text{ J}$, the heat absorbed is:
a.
$20\text{J}$
b.
$80\text{J}$
c.
$220\text{ J}$
d.
$280\text{ J}$
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Consider a one-dimensional Ising model with N spins, at very low temperatures when almost all the spins are aligned parallel to each other. There will be a few spin flips with each flip costing an energy 2J. In a configuration with r spin flips, the energy of the system is $\text{E} = -\text{NJ} + 2r\text{J}$ and the number of configuration is $^N\text{C}_r$; r varies from 0 to N. The partition function is
A magnetic field sensor based on the Hall effect is to be fabricated by implanting. As into a Si film of thickness $1\text{ }\mu\text{m}$. The specifications require a magnetic field sensitivity of $500\text{ mV/Tesla}$ at an excitation current of $1\text{ mA}$. The implanation dose is to be adjusted such that the average carrier density, after activation, is
a.
$1.25\times10^{26}\text{ m}^{-3}$
b.
$1.25\times10^{22}\text{ m}^{-3}$
c.
$4.1\times10^{21}\text{ m}^{-3}$
d.
$4.1\times10^{20}\text{ m}^{-3}$
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Band-pass and band-reject filters can be implemented by combining a low pass and a high pass filter in series and in parallel, respectively. If the cut-off frequencies of the low pass and high pass filters are $\omega_0^{\text{LP}}$ and $\omega_0^{\text{HP}}$, respectively, the condition required to implement the band-pass and band-reject filters are respectively.
a.
$\omega_0^{\text{HP}} < \omega_0^{\text{LP}}$ and $\omega_0^{\text{HP}} < \omega_0^{\text{LP}}$
b.
$\omega_0^{\text{HP}} < \omega_0^{\text{LP}}$ and $\omega_0^{\text{HP}} > \omega_0^{\text{LP}}$
c.
$\omega_0^{\text{HP}} > \omega_0^{\text{LP}}$ and $\omega_0^{\text{HP}} < \omega_0^{\text{LP}}$
d.
$\omega_0^{\text{HP}} > \omega_0^{\text{LP}}$ and $\omega_0^{\text{HP}} > \omega_0^{\text{LP}}$
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The output characteristics of a solar panel at a certain level of irradiance is shown in the figure below.
If the solar cell is to power a load of $5\text{ }\Omega$, the power drawn by the load is:
a.
$97\text{W}$
b.
$73\text{ W}$
c.
$50\text{ W}$
d.
$45\text{ W}$
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Consider the energy level diagram shown below, which corresponds to the molecular nitrogen laser.
If the pump rate $\text{R}$ is $10^{20}\text{ atoms cm}^{-3}\text{ s}^{-1}$ and the decay routes are as shown with $\tau_{21} = 20\text{ ns}$ and $\tau_1 = 1\text{ }\mu\text{s}$, the equilibrium populations of states 2 and 1 are, respectively,
a.
$10^{14}\text{ cm}^{-3}$ and $2\times10^{12}\text{ cm}^{-3}$
b.
$2\times10^{12}\text{ cm}^{-3}$ and $10^{14}\text{ cm}^{-3}$
c.
$2\times10^{12}\text{ cm}^{-3}$ and $2\times10^{6}\text{ cm}^{-3}$
d.
zero and $10^{20}\text{ cm}^{-3}$
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Consider a hydrogen atom undergoing a $2\text{P}\rightarrow 1\text{S}$ transition. The lifetime $\text{t}_{\text{sp}}$ of the $2\text{P}$ state for spontaneous emission is $1.6\text{ ns}$ and the energy difference between the levels is $10.2\text{ eV}$. Assuming that the refractive index of the medium $\text{n}_0 = 1$, the ratio of the Einstein coefficients for stimulated emission $\text{B}_{21}(\omega)/\text{A}_{21}(\omega)$ is given by
Consider a He-Ne laser cavity consisting of two mirrors of reflectivities $R_1 = 1$ and $R_2 = 0.98$. The mirrors are separated by a distance $d = 20\text{ cm}$ and the medium in between has a refractive index $n_0 = 1$ and absorption coefficient $\alpha = 0$. The values of the separation between the modes $\delta\nu$ and the width $\Delta\nu_p$ of each mode of the laser cavity are:
Non-interacting bosons undergo Bose-Einstein Condensation (BEC) when trapped in a three-dimensional isotropic simple harmonic potential. For BEC to occur, the chemical potential must be equal to
a.
$\hbar\omega/2$
b.
$\hbar\omega$
c.
$3\hbar\omega/2$
d.
$0$
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In a band structure calculation, the dispersion relation for electrons is found to be
$$\varepsilon_k = \beta(\cos k_x a + \cos k_y a + \cos k_z a)$$
where $\beta$ is a constant and $a$ is the lattice constant. The effective mass at the boundary of the first Brilliouin zone is
a.
$\frac{2\hbar^2}{5\beta a^2}$
b.
$\frac{4\hbar^2}{5\beta a^2}$
c.
$\frac{\hbar^2}{2\beta a^2}$
d.
$\frac{\hbar^2}{3\beta a^2}$
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The muon has mass $105\text{ MeV/c}^2$ and mean lifetime $2.2\text{ }\mu\text{s}$ in its rest frame. The mean distance traversed by muon of energy $315\text{ MeV/c}^2$ before decaying is approximately
a.
$3\times 10^5\text{ km}$
b.
$2.2\text{ cm}$
c.
$6.6\text{ }\mu\text{m}$
d.
$1.98\text{ km}$
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Consider the following particles: the proton $\text{p}$, the neutron $\text{n}$, the neutral pion $\pi^0$ and the delta resonance $\Delta^+$. When ordered in terms of decreasing lifetime, the correct arrangement is as follows:
a.
$\pi^0, \text{n}, \text{p}, \Delta^+$
b.
$\text{p}, \text{n}, \Delta^+, \pi^0$
c.
$\text{p}, \text{n}, \pi^0, \Delta^+$
d.
$\Delta^+, \text{n}, \pi^0, \text{p}$
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The single particle energy difference between the p-orbitals (i.e. $\text{p}_{3/2}$ and $\text{p}_{1/2}$) of the nucleus $^{114}_{\ 50}\text{Sn}$ is $3\text{ MeV}$. The energy difference between the states in its $1\text{f}$ orbital is
a.
$-7\text{ MeV}$
b.
$7\text{ MeV}$
c.
$5\text{ MeV}$
d.
$-5\text{ MeV}$
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