CSIR NET · Physical Sciences
June
2014
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46 questions
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Q21
mcq
One gram of salt is dissolved in water that is filled to a height of $5\text{ cm}$ in a beaker of diameter $10\text{ cm}$. The accuracy of length measurement is $0.01\text{ cm}$ while that of mass measurement is $0.01\text{ mg}$. When measuring the concentration $C$, the fractional error $\Delta C/C$ is
a.
$0.8\%$
b.
$0.14\%$
c.
$0.5\%$
d.
$0.28\%$
Correct Answer
Step-by-Step AI Explanation
Verified Method
The concentration is defined as mass per unit volume, $C = \frac{m}{V}$, where the volume of a cylinder is $V = \pi r^2 h = \frac{\pi d^2 h}{4}$.
Thus, $C = \frac{4m}{\pi d^2 h}$.
The maximum fractional error in $C$ is given by:
$$\frac{\Delta C}{C} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta h}{h}$$
Given:
- $m = 1\text{ g} = 1000\text{ mg}$, $\Delta m = 0.01\text{ mg} \implies \frac{\Delta m}{m} = \frac{0.01}{1000} = 10^{-5} = 0.001\%$
- $d = 10\text{ cm}$, $\Delta d = 0.01\text{ cm} \implies 2\frac{\Delta d}{d} = 2 \times \frac{0.01}{10} = 2 \times 10^{-3} = 0.2\%$
- $h = 5\text{ cm}$, $\Delta h = 0.01\text{ cm} \implies \frac{\Delta h}{h} = \frac{0.01}{5} = 2 \times 10^{-3} = 0.2\%$ (or if $\frac{\Delta h}{h} = 0.002 = 0.2\%$, wait: $0.2\% + 0.2\% = 0.4\%$, but wait, is $d$ diameter or radius? $V = \pi (d/2)^2 h$, so $2 \Delta d / d$).
Wait, let's recheck $\Delta d/d$: $\Delta d = 0.01\text{ cm}$, $d = 10\text{ cm} \implies \Delta d/d = 0.001 = 0.1\%$. So $2\Delta d/d = 0.2\%$.
$\Delta h/h = 0.01/5 = 0.002 = 0.2\%$.
Sum $= 0.001\% + 0.2\% + 0.2\% = 0.401\%$, not among the options ($0.8\%, 0.14\%, 0.5\%, 0.28\%$).
Wait, using quadrature (standard error propagation for independent errors):
$$\frac{\Delta C}{C} = \sqrt{\left(\frac{\Delta m}{m}\right)^2 + 4\left(\frac{\Delta d}{d}\right)^2 + \left(\frac{\Delta h}{h}\right)^2}$$
$$\frac{\Delta C}{C} = \sqrt{0 + 4(0.001)^2 + (0.002)^2} = \sqrt{4 \times 10^{-6} + 4 \times 10^{-6}} = \sqrt{8 \times 10^{-6}} = 2\sqrt{2} \times 10^{-3} \approx 2.828 \times 10^{-3} = 0.28\%$$
Therefore, the fractional error computed via root-mean-square propagation is $0.28\%$, which corresponds to option D.
Q22
mcq
A system can have three energy levels: $E = 0, \pm \varepsilon$. The level $E = 0$ is doubly degenerate, while the others are non-degenerate. The average energy at inverse temperature $\beta$ is
a.
$-\varepsilon \tanh(\beta\varepsilon)$
b.
$\frac{\varepsilon(e^{\beta\varepsilon} - e^{-\beta\varepsilon})}{(1 + e^{\beta\varepsilon} + e^{-\beta\varepsilon})}$
c.
Zero
Correct Answer
d.
$-\varepsilon \tanh\left(\frac{\beta\varepsilon}{2}\right)$
Step-by-Step AI Explanation
Verified Method
The partition function $Z$ of the system is:
$$Z = \sum_i g_i e^{-\beta E_i} = 2 e^{0} + e^{-\beta \varepsilon} + e^{\beta \varepsilon} = 2 + 2\cosh(\beta \varepsilon)$$
The average energy is:
$$\langle E \rangle = -\frac{\partial \ln Z}{\partial \beta} = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = -\frac{1}{Z}\left( -\varepsilon e^{-\beta\varepsilon} + \varepsilon e^{\beta\varepsilon} \right) = -\frac{\varepsilon (e^{\beta\varepsilon} - e^{-\beta\varepsilon})}{2 + 2\cosh(\beta\varepsilon)}$$
Wait, the energy levels are $E = 0, +\varepsilon, -\varepsilon$. Let's check the thermal average directly:
$$\langle E \rangle = \frac{\sum_i g_i E_i e^{-\beta E_i}}{Z} = \frac{2\cdot 0 + (+\varepsilon)e^{-\beta\varepsilon} + (-\varepsilon)e^{\beta\varepsilon}}{2 + e^{-\beta\varepsilon} + e^{\beta\varepsilon}} = \frac{-\varepsilon(e^{\beta\varepsilon} - e^{-\beta\varepsilon})}{2 + 2\cosh(\beta\varepsilon)} = -\varepsilon \frac{2\sinh(\beta\varepsilon)}{2(1+\cosh(\beta\varepsilon))} = -\varepsilon \tanh\left(\frac{\beta\varepsilon}{2}\right)$$
Wait, let's look at option D: $-\varepsilon \tanh\left(\frac{\beta\varepsilon}{2}\right)$!
Using the identity $\frac{\sinh(x)}{1+\cosh(x)} = \frac{2\sinh(x/2)\cosh(x/2)}{2\cosh^2(x/2)} = \tanh(x/2)$, we find:
$$\langle E \rangle = -\varepsilon \tanh\left(\frac{\beta\varepsilon}{2}\right)$$
Therefore, option D is the correct choice.
Q23
mcq
For a particular thermodynamics system the entropy $S$ is related to the internal energy $U$ and volume $V$ by
$$S = c U^{3/4} V^{1/4}$$
where $c$ is a constant. The Gibbs potential $G = U - TS + pV$ for this system is
a.
$\frac{3pU}{4T}$
b.
$\frac{cU}{3}$
c.
zero
Correct Answer
d.
$\frac{US}{4V}$
Step-by-Step AI Explanation
Verified Method
Given the fundamental relation $S = c U^{3/4} V^{1/4}$, we can express the internal energy as:
$$U = c^{-4/3} S^{4/3} V^{-1/3}$$
Since $U(\lambda S, \lambda V) = c^{-4/3} (\lambda S)^{4/3} (\lambda V)^{-1/3} = \lambda U(S, V)$, $U$ is a homogeneous function of degree 1 in its extensive variables $S$ and $V$.
By Euler's theorem for homogeneous functions:
$$U = S\left(\frac{\partial U}{\partial S}\right)_V + V\left(\frac{\partial U}{\partial V}\right)_S = TS - pV$$
Rearranging this gives:
$$G = U - TS + pV = 0$$
Thus, the Gibbs free energy $G$ is identically zero.
Q24
mcq
An op-amp based voltage follower
a.
is useful for converting a low impedance source into a high impedance source
b.
is useful for converting a high impedance source into a low impedance source
Correct Answer
c.
has infinitely high closed loop output impedance
d.
has infinitely high closed loop gain
Step-by-Step AI Explanation
Verified Method
An op-amp voltage follower (buffer) has a very high (ideally infinite) input impedance and a very low (ideally zero) output impedance with a unity voltage gain ($A_v = 1$). Therefore, it is commonly used as a buffer to convert a high impedance source into a low impedance source to prevent loading effects.
Q25
mcq
A particle of mass $m$ in three dimensions is in the potential
$$V(r) = \begin{cases} 0 & r < a \\ \infty & r \ge a \end{cases}$$
Its ground state energy is
a.
$\frac{\pi^2 \hbar^2}{2ma^2}$
Correct Answer
b.
$\frac{\pi^2 \hbar^2}{ma^2}$
c.
$\frac{3\pi^2 \hbar^2}{2ma^2}$
d.
$\frac{9\pi^2 \hbar^2}{2ma^2}$
Step-by-Step AI Explanation
Verified Method
For a 3D spherically symmetric infinite potential well of radius $a$, the radial Schrödinger equation for the reduced wave function $u(r) = rR(r)$ with angular momentum $l = 0$ (ground state) is:
$$-\frac{\hbar^2}{2m} \frac{d^2 u}{dr^2} = E u$$
with boundary conditions $u(0) = 0$ (so that $R(0)$ remains finite) and $u(a) = 0$.
This corresponds exactly to a 1D infinite square well of width $a$, giving the solution:
$$u(r) = A \sin(kr) \quad \text{with } k = \frac{n\pi}{a}$$
For the ground state, $n = 1$, so $k = \frac{\pi}{a}$.
The ground state energy is:
$$E = \frac{\hbar^2 k^2}{2m} = \frac{\pi^2 \hbar^2}{2ma^2}$$
Q26
mcq
Which of the graphs below gives the correct qualitative behavior of the energy density $E_T(\lambda)$ of blackbody radiation of wavelength $\lambda$ at two temperatures $T_1$ and $T_2$ ($T_1 < T_2$)?
a.
Graph with $T_1$ peak shifted to larger $\lambda$ and lower height than $T_2$
Correct Answer
b.
Graph with $T_1$ peak at same $\lambda$ but lower height than $T_2$
c.
$\frac{2}{a}\left[\sin\left(\frac{\pi x_1}{a}\right)\sin\left(\frac{3\pi x_2}{2a}\right) - \sin\left(\frac{3\pi x_1}{2a}\right)\sin\left(\frac{\pi x_2}{a}\right)\right]$
d.
$\frac{2}{a}\left[\sin\left(\frac{\pi x_1}{a}\right)\cos\left(\frac{\pi x_2}{a}\right) - \sin\left(\frac{\pi x_2}{a}\right)\sin\left(\frac{\pi x_2}{a}\right)\right]$
Step-by-Step AI Explanation
Verified Method
According to Wien's displacement law, $\lambda_{\text{max}} T = b = \text{constant}$, meaning the peak wavelength is inversely proportional to temperature: as $T$ increases, the peak shifts towards shorter wavelengths (to the left). Since $T_1 < T_2$, the peak of $T_1$ must occur at a larger wavelength $\lambda$ than that of $T_2$. Furthermore, the total energy density (area under the curve) and the spectral maximum increase strongly with temperature ($E_{\text{max}} \propto T^5$), so the curve for $T_2$ must have a significantly higher peak than that for $T_1$. In graph (a), the peak for $T_2$ is shifted to a shorter wavelength and is higher, which is the correct qualitative behavior.
Q32
mcq
A particle of mass m in the potential $V(x, y)=\frac{1}{2}m\omega^2\left(4x^2 + y^2\right)$, is in an eigenstate of energy $E = \frac{5}{2}\hbar\omega$. The corresponding un-normalized eigenfunction is
a.
$y\exp\left[-\frac{m\omega}{2\hbar}\left(2x^2+y^2\right)\right]$
b.
$x\exp\left[-\frac{m\omega}{2\hbar}\left(2x^2+y^2\right)\right]$
Correct Answer
c.
$y\exp\left[-\frac{m\omega}{2\hbar}\left(x^2+y^2\right)\right]$
d.
$xy\exp\left[-\frac{m\omega}{2\hbar}\left(x^2+y^2\right)\right]$
Step-by-Step AI Explanation
Verified Method
The potential is $V(x, y) = \frac{1}{2}m(2\omega)^2 x^2 + \frac{1}{2}m\omega^2 y^2$, which corresponds to an anisotropic two-dimensional harmonic oscillator with frequencies $\omega_x = 2\omega$ and $\omega_y = \omega$.
The energy eigenvalues are:
$$E = \left(n_x + \frac{1}{2}\right)\hbar\omega_x + \left(n_y + \frac{1}{2}\right)\hbar\omega_y = \left(2n_x + 1\right)\hbar\omega + \left(n_y + \frac{1}{2}\right)\hbar\omega = \left(2n_x + n_y + \frac{3}{2}\right)\hbar\omega$$
We are given $E = \frac{5}{2}\hbar\omega$, which implies:
$$2n_x + n_y + \frac{3}{2} = \frac{5}{2} \implies 2n_x + n_y = 1$$
Since $n_x, n_y \in \{0, 1, 2, \dots\}$, the only solution is $n_x = 1$ and $n_y = 0$.
The wave function for the state $(n_x = 1, n_y = 0)$ is:
$$\psi(x, y) \propto H_1\left(\sqrt{\frac{m\omega_x}{\hbar}}x\right) e^{-\frac{m\omega_x x^2}{2\hbar}} H_0\left(\sqrt{\frac{m\omega_y}{\hbar}}y\right) e^{-\frac{m\omega_y y^2}{2\hbar}}$$
Since $H_1(u) \propto u$ and $\omega_x = 2\omega$, $\omega_y = \omega$:
$$\psi(x, y) \propto x \exp\left[-\frac{m(2\omega)x^2}{2\hbar} - \frac{m\omega y^2}{2\hbar}\right] = x \exp\left[-\frac{m\omega}{2\hbar}\left(2x^2 + y^2\right)\right]$$
Thus, option B is the correct eigenfunction.
Q33
mcq
A particle of mass $m$ and coordinate $q$ has the Lagrangian
$$L = \frac{1}{2}m\dot{q}^2 - \frac{\lambda}{2}q\dot{q}^2$$
where $\lambda$ is a constant. The Hamiltonian for the system is given by
a.
$\frac{p^2}{2m} + \frac{\lambda qp^2}{2m^2}$
b.
$\frac{p^2}{2(m - \lambda q)}$
Correct Answer
c.
$\frac{p^2}{2m} + \frac{\lambda qp^2}{2(m - \lambda q)^2}$
d.
$\frac{pq}{2}$
Step-by-Step AI Explanation
Verified Method
The Lagrangian is:
$$L = \frac{1}{2}(m - \lambda q)\dot{q}^2$$
The generalized momentum $p$ is:
$$p = \frac{\partial L}{\partial \dot{q}} = (m - \lambda q)\dot{q} \implies \dot{q} = \frac{p}{m - \lambda q}$$
The Hamiltonian is obtained via the Legendre transform:
$$H = p\dot{q} - L = p\left(\frac{p}{m - \lambda q}\right) - \frac{1}{2}(m - \lambda q)\left(\frac{p}{m - \lambda q}\right)^2 = \frac{p^2}{2(m - \lambda q)}$$
Therefore, the correct choice is B.
Q34
mcq
If $\vec{A} = yz\hat{i} + zx\hat{j} + xy\hat{k}$ and C is the circle of unit radius in the plane defined by $z = 1$, with the centre on the z-axis, then the value of the integral $\oint_C \vec{A}\cdot d\vec{\ell}$ is
a.
$\frac{\pi}{2}$
b.
$\pi$
c.
$\frac{\pi}{4}$
d.
$0$
Correct Answer
Step-by-Step AI Explanation
Verified Method
Using Stokes' Theorem, $\oint_C \vec{A}\cdot d\vec{\ell} = \iint_S (\nabla \times \vec{A})\cdot d\vec{S}$.
Let us compute $\nabla \times \vec{A}$:
$$\nabla \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz & zx & xy \end{vmatrix}$$
$$\hat{i}\left(\frac{\partial(xy)}{\partial y} - \frac{\partial(zx)}{\partial z}\right) - \hat{j}\left(\frac{\partial(xy)}{\partial x} - \frac{\partial(yz)}{\partial z}\right) + \hat{k}\left(\frac{\partial(zx)}{\partial x} - \frac{\partial(yz)}{\partial y}\right)$$
$$= \hat{i}(x - x) - \hat{j}(y - y) + \hat{k}(z - z) = 0$$
Since the curl of $\vec{A}$ is identically zero everywhere (indeed, $\vec{A} = \nabla(xyz)$ is a conservative field), the line integral along any closed path is zero:
$$\oint_C \vec{A}\cdot d\vec{\ell} = 0$$
Hence, the correct option is D.
Q35
mcq
Given, $\sum_{n=0}^{\infty} P_n(x)t^n = (1-2xt+t^2)^{-1/2}$, for $|t| < 1$, the value of $P_5(-1)$ is
a.
$0.26$
b.
$1$
c.
$0.5$
d.
$-1$
Correct Answer
Step-by-Step AI Explanation
Verified Method
A well-known property of the Legendre polynomials is:
$$P_n(-1) = (-1)^n$$
For $n = 5$:
$$P_5(-1) = (-1)^5 = -1$$
Alternatively, substituting $x = -1$ into the generating function:
$$\sum_{n=0}^{\infty} P_n(-1) t^n = (1 + 2t + t^2)^{-1/2} = ((1+t)^2)^{-1/2} = (1+t)^{-1} = \sum_{n=0}^{\infty} (-1)^n t^n$$
Equating coefficients of $t^5$ gives $P_5(-1) = -1$.
Hence, the correct option is D.
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Q36
mcq
A charged particle is at a distance $d$ from an infinite conducting plane maintained at zero potential. When released from rest, the particle reaches a speed u at a distance d/2 from the plane. At what distance from the plane will the particle reach the speed 2u?
a.
d/6
b.
d/3
c.
d/4
d.
d/5
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Q37
mcq
Consider the matrix
$$M = \begin{pmatrix} 0 & 2i & 3i \\ -2i & 0 & 6i \\ -3i & -6i & 0 \end{pmatrix}$$
The eigenvalues of M are
a.
$-5, -2, 7$
b.
$-7, 0, 7$
c.
$-4i, 2i, 2i$
d.
$2, 3, 6$
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Q38
mcq
Consider the differential equation $\frac{d^2x}{dt^2}+2\frac{dx}{dt}+x=0$ with the initial conditions $x(0)=0$ and $\dot{x}(0)=1$. The solution $x(t)$ attains its maximum value when 't' is
a.
1/2
b.
1
c.
2
d.
\infty
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Q39
mcq
A light source is switched on and off at a constant frequency $f$. An observer moving with a velocity $u$ with respect to the light source will observe the freuqency of the switching to be
a.
$f\left(1-\frac{u^2}{c^2}\right)^{-1}$
b.
$f\left(1-\frac{u^2}{c^2}\right)^{-1/2}$
c.
$f\left(1-\frac{u^2}{c^2}\right)$
d.
$f\left(1-\frac{u^2}{c^2}\right)^{1/2}$
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Q40
mcq
If C is the contour defined by $|z|=\frac{1}{2}$, the value of the integral
$$\oint_C \frac{dz}{\sin^2 z}$$
is
a.
$\infty$
b.
$2\pi i$
c.
$0$
d.
$\pi i$
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Q41
mcq
The time period of a simple pendulum under the influence of the acceleration due to gravity g is T. The bob is subjected to an additional acceleration of magnitude $\sqrt{3}g$ in the horizontal direction. Assuming small oscillations, the mean position and time period of oscillation, respectively, of the bob will be
a.
$0^\circ$ to the vertical and $\sqrt{3}T$
b.
$30^\circ$ to the vertical and $T/2$
c.
$60^\circ$ to the vertical and $T/\sqrt{2}$
d.
$0^\circ$ to the vertical and $T/\sqrt{3}$
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Q42
mcq
Consider an electromagnetic wave at the interface between two homogeneouos dielectric media of the dielectric constants $\varepsilon_1$ and $\varepsilon_2$. Assuming $\varepsilon_2 > \varepsilon_1$ and non charges on the surface, the electric field vector $\vec{E}$ and the displacement vector $\vec{D}$ in the two media satisfy the following inequalities
a.
$|\vec{E}_2| > |\vec{E}_1|$ and $|\vec{D}_2| > |\vec{D}_1|$
b.
$|\vec{E}_2| < |\vec{E}_1|$ and $|\vec{D}_2| < |\vec{D}_1|$
c.
$|\vec{E}_2| < |\vec{E}_1|$ and $|\vec{D}_2| > |\vec{D}_1|$
d.
$|\vec{E}_2| > |\vec{E}_1|$ and $|\vec{D}_2| < |\vec{D}_1|$
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Q43
mcq
If the electrostatic potential in spherical polar coordinates is
$$\varphi(r) = \varphi_0 e^{-r/r_0}$$
where $\varphi_0$ and $r_0$ are constants, then the charge density at a distance $r = r_0$ will be
a.
$\frac{\varepsilon_0\varphi_0}{er_0^2}$
b.
$\frac{e\varepsilon_0\varphi_0}{2r_0^2}$
c.
$-\frac{\varepsilon_0\varphi_0}{er_0^2}$
d.
$\frac{2e\varepsilon_0\varphi_0}{r_0^2}$
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Q44
mcq
A current $i_p$ flows through the primary coil of a transformer. The graph of $i_p(t)$ as a function of time 't' is shown in figure below
Which of the following graph represents the current $i_s$ in the secondary coil?
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Q45
mcq
A time-dependent current $\vec{I}(t) = Kt\hat{z}$ (where K is a constant) is switched on at $t = 0$ in an infinite current-carrying wire. The magnetic vector potential at a perpendicular distance 'a' from the wire is given (for time $t > a/c$) by
a.
$\hat{z}\frac{\mu_0 K}{4\pi c} \int_{-\sqrt{c^2t^2-a^2}}^{\sqrt{c^2t^2-a^2}} dz\, \frac{ct - \sqrt{a^2+z^2}}{\left(a^2+z^2\right)^{1/2}}$
b.
$\hat{z}\frac{\mu_0 K}{4\pi} \int_{-ct}^{ct} dz\, \frac{t}{\left(a^2+z^2\right)^{1/2}}$
c.
$\hat{z}\frac{\mu_0 K}{4\pi c} \int_{-ct}^{ct} dz\, \frac{ct - \sqrt{a^2+z^2}}{\left(a^2+z^2\right)^{1/2}}$
d.
$\hat{z}\frac{\mu_0 K}{4\pi} \int_{-\sqrt{c^2t^2-a^2}}^{\sqrt{c^2t^2-a^2}} dz\, \frac{t}{\left(a^2+z^2\right)^{1/2}}$
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Q46
mcq
The pressure of a non-relativistic free Fermi gas in three-dimensions depends, at $\text{T} = 0$, on the density of fermions $n$ as
a.
$n^{5/3}$
b.
$n^{1/3}$
c.
$n^{2/3}$
d.
$n^{4/3}$
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Q47
mcq
A doubel slit interference experiment uses a laser emitting light of two adjacent frequencies $\nu_1$ and $\nu_2$ ($\nu_1 < \nu_2$). The minimum path difference between the interfering beams for which the interference pattern disappears is
a.
$\frac{c}{\nu_2 + \nu_1}$
b.
$\frac{c}{\nu_2 - \nu_1}$
c.
$\frac{c}{2(\nu_2 - \nu_1)}$
d.
$\frac{c}{2(\nu_2 + \nu_1)}$
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Q48
mcq
The recently-discovered Higgs boson at hte LHC experiment has a decay mode into a photon and a Z boson. If the rest masses of the Higgs and Z boson are $125\text{ GeV}/c^2$ and $90\text{ GeV}/c^2$ respectively, and the decaying Higgs particle is at rest, the energy of the photon will approximately be
a.
$35\sqrt{3}\text{ GeV}$
b.
$35\text{ GeV}$
c.
$30\text{ GeV}$
d.
$15\text{ GeV}$
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Q49
mcq
A permanently deformed even-even nucleus with $\text{J}^P = 2^+$ has rotational energy $93\text{ keV}$. The energy of the next excited state is
a.
$372\text{ keV}$
b.
$310\text{ keV}$
c.
$273\text{ keV}$
d.
$186\text{ keV}$
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Q50
mcq
How much does the total angular momentum quantum number J change in the transition of $\text{Cr}(3d^6)$ atom as it ionizes to $\text{Cr}^{2+}(3d^4)$?
a.
increases by 2
b.
decreases by 2
c.
decreases by 4
d.
does not change
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Q55
mcq
In one dimension, a random walker takes a step with equal probability to the left or right. What is the probability that the walker returns to the starting point after 4 steps?
a.
$3/8$
b.
$5/16$
c.
$1/4$
d.
$1/16$
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Q56
mcq
Consider an electron in a b.c.c. lattice with lattice constant $a$. A single particle wavefunction that satisfies the Bloch theorem will have the form $f(\vec{r})\exp(i\vec{k}\cdot\vec{r})$, with $f(\vec{r})$ being
a.
$1+\cos\left[\frac{2\pi}{a}(x+y-z)\right]+\cos\left[\frac{2\pi}{a}(-x+y+z)\right]+\cos\left[\frac{2\pi}{a}(x-y+z)\right]$
b.
$1+\cos\left[\frac{2\pi}{a}(x+y)\right]+\cos\left[\frac{2\pi}{a}(y+z)\right]+\cos\left[\frac{2\pi}{a}(z+x)\right]$
c.
$1+\cos\left[\frac{\pi}{a}(x+y)\right]+\cos\left[\frac{\pi}{a}(y+z)\right]+\cos\left[\frac{\pi}{a}(z+x)\right]$
d.
$1+\cos\left[\frac{\pi}{a}(x+y-z)\right]+\cos\left[\frac{\pi}{a}(-x+y+z)\right]+\cos\left[\frac{\pi}{a}(x-y+z)\right]$
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Q57
mcq
The dispersion relation for electrons in an f.c.c. crystal is given, in the tight binding approximation by
$$\varepsilon(k)=-4\varepsilon_0\left[\cos\frac{k_x a}{2}\cos\frac{k_y a}{2}+\cos\frac{k_y a}{2}\cos\frac{k_z a}{2}+\cos\frac{k_z a}{2}\cos\frac{k_x a}{2}\right]$$
where '$a$' is the lattice constant and $\varepsilon_0$ is a constant with the dimension of energy. The $x$-component of the velocity of the electrons at $\left(\frac{\pi}{a}, 0, 0\right)$ is
a.
$-\frac{2\varepsilon_0 a}{\hbar}$
b.
$\frac{2\varepsilon_0 a}{\hbar}$
c.
$-\frac{4\varepsilon_0 a}{\hbar}$
d.
$\frac{4\varepsilon_0 a}{\hbar}$
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Q58
mcq
The following data is obtained in an expriment that measures the viscosity $\eta$ as a function of molecular weight $\text{M}$ for a set of polymers.
$$\begin{array}{cc}
M (\text{Da}) & \eta (\text{kPa}-\text{s}) \\
990 & 0.28 \pm 0.03 \\
5032 & 30 \pm 2 \\
10191 & 250 \pm 10 \\
19825 & 2000 \pm 200
\end{array}$$
The relation that best describes the dependence of $\eta$ on $\text{M}$ is
a.
$\eta \sim M^{4/9}$
b.
$\eta \sim M^{3/2}$
c.
$\eta \sim M^2$
d.
$\eta \sim M^3$
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Q59
mcq
The integral $\int_0^1 \sqrt{x}dx$ is to be evaluated up to 3 decimal places using Simpson's 3-point rule. If the interval $[0, 1]$ is divided into 4 equal parts, the correct result is
a.
$0.683$
b.
$0.667$
c.
$0.657$
d.
$0.638$
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Q60
mcq
In a classical model, a scalar (spin-0) meson consists of a quark and an antiquark bound by a potential
$$V(r) = ar + \frac{b}{r}$$
where $a = 200\text{ MeV fm}^{-1}$ and $b = 100\text{ MeV fm}$. If the masses of the quark and antiquark are negligible, the mass of the meson can be estimated as approximately
a.
$141\text{ MeV/c}^2$
b.
$283\text{ MeV/c}^2$
c.
$353\text{ MeV/c}^2$
d.
$425\text{ MeV/c}^2$
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Q61
mcq
Let $y = \frac{1}{2}(x_1 + x_2) - \mu$, where $x_1$ and $x_2$ are independent and identically distributed Gaussian random variables of mean $\mu$ and standard deviation $\sigma$. Then $\frac{\langle y^4 \rangle}{\sigma^4}$ is
a.
1
b.
$\frac{3}{4}$
c.
$\frac{1}{2}$
d.
$\frac{1}{4}$
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Q62
mcq
The graph of a real periodic function $f(x)$ for the range $[-\infty, \infty]$ is shown below
Which of the following graphs represents the real part of its Fourier transform?
a.
Option (a)
b.
Option (b)
c.
Option (c)
d.
Option (d)
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Q63
mcq
The matrices
$$A = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \quad \text{and} \quad C = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$
satisfy the commutation relations
a.
$[A, B] = B + C, [B, C] = 0, [C, A] = B + C$
b.
$[A, B] = C, [B, C] = A, [C, A] = B$
c.
$[A, B] = B, [B, C] = 0, [C, A] = A$
d.
$[A, B] = C, [B, C] = 0, [C, A] = B$
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Q64
mcq
The function $\Phi(x, y, z, t) = \cos(z - vt) + \text{Re}(\sin(x + iy))$ satisfies the equation
a.
$\frac{1}{v^2}\frac{\partial^2 \Phi}{\partial t^2} = \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)\Phi$
b.
$\left(\frac{1}{v^2}\frac{\partial^2}{\partial t^2} + \frac{\partial^2}{\partial z^2}\right)\Phi = \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\Phi$
c.
$\left(\frac{1}{v^2}\frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial z^2}\right)\Phi = \left(\frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2}\right)\Phi$
d.
$\left(\frac{\partial^2}{\partial z^2} - \frac{1}{v^2}\frac{\partial^2}{\partial t^2}\right)\Phi = \left(\frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2}\right)\Phi$
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Q65
mcq
The coordinates and momenta $x_i, p_i$ ($i = 1, 2, 3$) of a particle satisfy the canonical Poisson bracket relations $\{x_i, p_j\} = \delta_{ij}$. If $C_1 = x_2 p_3 + x_3 p_2$ and $C_2 = x_1 p_2 - x_2 p_1$ are constants of motion, and if $C_3 = \{C_1, C_2\} = x_1 p_3 + x_3 p_1$, then
a.
$\{C_2, C_3\} = C_1$ and $\{C_3, C_1\} = C_2$
b.
$\{C_2, C_3\} = -C_1$ and $\{C_3, C_1\} = -C_2$
c.
$\{C_2, C_3\} = -C_1$ and $\{C_3, C_1\} = C_2$
d.
$\{C_2, C_3\} = C_1$ and $\{C_3, C_1\} = -C_2$
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Q66
mcq
A canonical transformation relates the old coordinates (q, p) to the new ones (Q, P) by the relations $Q = q^2$ and $P = p/2q$. The corresponding time-independent generating function is
a.
$\frac{P}{q^2}$
b.
$q^2 P$
c.
$q^2 / P$
d.
$q P^2$
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Q67
mcq
The time evolution of a one-dimensional dynamical system is described by
$$\frac{dx}{dt} = -(x+1)(x^2 - b^2)$$
If this has one stable and two unstable fixed points, then the parameter 'b' satisfies
a.
$0 < b < 1$
b.
$b > 1$
c.
$b < -1$
d.
$b = 2$
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Q68
mcq
A charge $(-e)$ is placed in vacuum at the point $(d, 0, 0)$, where $d > 0$. The region $x \le 0$ is filled uniformly with a metal. The electric field at the point $\left(\frac{d}{2}, 0, 0\right)$ is
a.
$-\frac{10e}{9\pi\varepsilon_0 d^2}(1,0,0)$
b.
$\frac{10e}{9\pi\varepsilon_0 d^2}(1,0,0)$
c.
$\frac{e}{\pi\varepsilon_0 d^2}(1,0,0)$
d.
$-\frac{e}{\pi\varepsilon_0 d^2}(1,0,0)$
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Q69
mcq
An electron is in the ground state of a hydrogen atom. The probability that it is within the Bohr radius is approximately equal to
a.
$0.60$
b.
$0.90$
c.
$0.16$
d.
$0.32$
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Q70
mcq
A beam of light of frequency $\omega$ is reflected from a dielectric-metal interface at normal incidence. The refractive index of the dielectric medium is $n$ and that of the metal is $n_2 = n(1+i\rho)$. If the beam is polarised parallel to the interface, then the phase change experienced by the light upon reflection is
a.
$\tan\left(\frac{2}{\rho}\right)$
b.
$\tan^{-1}\left(\frac{1}{\rho}\right)$
c.
$\tan^{-1}\left(\frac{2}{\rho}\right)$
d.
$\tan^{-1}(2\rho)$
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Q71
mcq
The scattering amplitude $f(\theta)$ for the potential $V(r) = \beta e^{-\mu r}$, where $\beta$ and $\mu$ are positive constants, is given, in the Born approximation by
(in the following $b = 2k\sin\frac{\theta}{2}$ and $E = \frac{\hbar^2 k^2}{2m}$)
a.
$-\frac{4m\beta\mu}{\hbar^2\left(b^2+\mu^2\right)^2}$
b.
$-\frac{4m\beta\mu}{\hbar^2 b^2\left(b^2+\mu^2\right)}$
c.
$-\frac{4m\beta\mu}{\hbar^2\sqrt{b^2+\mu^2}}$
d.
$-\frac{4m\beta\mu}{\hbar^2\left(b^2+\mu^2\right)^3}$
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Q72
mcq
The ground state eigenfunction for the potential $V(x)=-\delta(x)$, where $\delta(x)$ is the delta function, is given by $\psi(x)=Ae^{-\alpha|x|}$, where $A$ and $\alpha>0$ are constants. If a perturbation $H'=bx^2$ is applied, the first order correction to the energy of the ground state will be
a.
$\frac{b}{\sqrt{2}\alpha^2}$
b.
$\frac{b}{\alpha^2}$
c.
$\frac{2b}{\alpha^2}$
d.
$\frac{b}{2\alpha^2}$
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Q73
mcq
A thin infinitely long solenoid placed along the $z$-axis contains a magnetic flux $\phi$. Which of the following vector potentials corresponds to the magnetic field at an arbitrary point $(x, y, z)$?
a.
$(A_x, A_y, A_z)=\left(-\frac{\phi}{2\pi}\frac{y}{x^2+y^2}, \frac{\phi}{2\pi}\frac{x}{x^2+y^2}, 0\right)$
b.
$(A_x, A_y, A_z)=\left(-\frac{\phi}{2\pi}\frac{y}{x^2+y^2+z^2}, \frac{\phi}{2\pi}\frac{x}{x^2+y^2+z^2}, 0\right)$
c.
$(A_x, A_y, A_z)=\left(-\frac{\phi}{2\pi}\frac{x+y}{x^2+y^2}, \frac{\phi}{2\pi}\frac{x+y}{x^2+y^2}, 0\right)$
d.
$(A_x, A_y, A_z)=\left(-\frac{\phi}{2\pi}\frac{x}{x^2+y^2}, \frac{\phi}{2\pi}\frac{y}{x^2+y^2}, 0\right)$
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Q74
mcq
The van der Waals equation of state for a gas is given by
$$\left(P+\frac{a}{V^2}\right)(V-b)=RT$$
where, $P$, $V$ and $T$ represent the pressure, volume and temperature respectively, and $a$ and $b$ are constant parameters. At the critical point, where all the roots of the above cubic equation are degenerate, the volume is given by
a.
$\frac{a}{9b}$
b.
$\frac{a}{27b^2}$
c.
$\frac{8a}{27bR}$
d.
$3b$
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Q75
mcq
An electromagnetically-shielded room is designed so that at a frequency $\omega=10^7\text{ rad/s}$ the intensity of the external radiation that penetrates the room is $1\%$ of the incident radiation. If $\sigma=\frac{1}{2\pi}\times 10^6\,(\Omega m)^{-1}$ is the conductivity of the shielding material, its minimum thickness should be (given that $\ln 10=2.3$)
a.
$4.60\text{ mm}$
b.
$2.30\text{ mm}$
c.
$0.23\text{ mm}$
d.
$0.46\text{ mm}$
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