CSIR NET · Physical Sciences
June
2021
·
55 questions
·
First 10 Solutions Free
Questions 1–10 have full step-by-step solutions visible.
Sign in to unlock all 55 →
Q1
mcq
A particle of mass $m$ moves in a potential that is $V = \frac{1}{2}m(\omega_1^2 x^2 + \omega_2^2 y^2 + \omega_3^2 z^2)$ in the coordinates of a non-inertial frame $F$. The frame $F$ is rotating with respect to an inertial frame with an angular velocity $\hat{\mathbf{k}}\Omega$, where $\hat{\mathbf{k}}$ is the unit vector along their common $z$-axis. The motion of the particle is unstable for all angular frequencies satisfying
a.
$(\Omega^2 - \omega_1^2)(\Omega^2 - \omega_2^2) > 0$
b.
$(\Omega^2 - \omega_1^2)(\Omega^2 - \omega_2^2) < 0$
Correct Answer
c.
$(\Omega^2 - (\omega_1 + \omega_2)^2)(\Omega^2 - |\omega_1 - \omega_2|^2) > 0$
d.
$(\Omega^2 - (\omega_1 + \omega_2)^2)(\Omega^2 - |\omega_1 - \omega_2|^2) < 0$
Step-by-Step AI Explanation
Verified Method
In the rotating frame with angular velocity $\vec{\Omega} = \Omega\hat{\mathbf{k}}$, the effective potential includes the centrifugal term:
$$V_{\text{eff}} = V - \frac{1}{2}m |\vec{\Omega} \times \vec{r}|^2 = \frac{1}{2}m [(\omega_1^2 - \Omega^2)x^2 + (\omega_2^2 - \Omega^2)y^2 + \omega_3^2 z^2]$$
The equations of motion including the Coriolis force $\vec{F}_{\text{cor}} = 2m(\vec{v} \times \vec{\Omega}) = 2m(\dot{y}\Omega \hat{\mathbf{i}} - \dot{x}\Omega \hat{\mathbf{j}})$ are:
$$\ddot{x} - 2\Omega \dot{y} + (\omega_1^2 - \Omega^2)x = 0$$
$$\ddot{y} + 2\Omega \dot{x} + (\omega_2^2 - \Omega^2)y = 0$$
$$\ddot{z} + \omega_3^2 z = 0$$
Assuming normal modes of the form $x, y \propto e^{i\omega t}$, the characteristic equation for the frequencies in the $xy$-plane is obtained from:
$$\begin{vmatrix} -\omega^2 + (\omega_1^2 - \Omega^2) & -2i\Omega\omega \\ 2i\Omega\omega & -\omega^2 + (\omega_2^2 - \Omega^2) \end{vmatrix} = 0$$
Expanding the determinant:
$$(\omega^2 - (\omega_1^2 - \Omega^2))(\omega^2 - (\omega_2^2 - \Omega^2)) - 4\Omega^2 \omega^2 = 0$$
$$\omega^4 - (\omega_1^2 + \omega_2^2 + 2\Omega^2)\omega^2 + (\omega_1^2 - \Omega^2)(\omega_2^2 - \Omega^2) = 0$$
Letting $\Lambda = \omega^2$, we have a quadratic equation:
$$\Lambda^2 - (\omega_1^2 + \omega_2^2 + 2\Omega^2)\Lambda + (\omega_1^2 - \Omega^2)(\omega_2^2 - \Omega^2) = 0$$
For stable oscillatory motion, all roots $\Lambda = \omega^2$ must be real and positive.
If the constant term of the quadratic equation is negative, i.e.,
$$(\omega_1^2 - \Omega^2)(\omega_2^2 - \Omega^2) = (\Omega^2 - \omega_1^2)(\Omega^2 - \omega_2^2) < 0$$
then the product of the two roots is negative. This guarantees that one root $\Lambda = \omega^2$ is strictly negative, which means $\omega = \pm i |\omega|$, leading to exponentially growing solutions $e^{|\omega| t}$, and thus the motion is definitively unstable. Hence, the motion is unstable for all angular frequencies satisfying $(\Omega^2 - \omega_1^2)(\Omega^2 - \omega_2^2) < 0$.
Q1
mcq
The Hamiltonian of a particle of mass $m$ in one-dimension is $H = \frac{1}{2m} p^2 + \lambda |x|^3$; where $\lambda > 0$ is a constant. If $E_1$ and $E_2$, respectively, denote the ground-state energies of the particle for $\lambda = 1$ and $\lambda = 2$ (in appropriate units) the ratio $E_2 / E_1$ is best approximated by
a.
1.260
b.
1.414
c.
1.516
d.
1.320
Correct Answer
Step-by-Step AI Explanation
Verified Method
To find how the energy scales with the coupling constant $\lambda$, we can use a scaling argument on the time-independent Schrödinger equation:
$$-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + \lambda |x|^k \psi = E \psi$$
Let $x = c y$. Then $\frac{d^2}{dx^2} = \frac{1}{c^2}\frac{d^2}{dy^2}$. The equation becomes:
$$-\frac{\hbar^2}{2m c^2} \frac{d^2\psi}{dy^2} + \lambda c^k |y|^k \psi = E \psi$$
To make the equation dimensionless or independent of $\lambda$, we set the scaling factor such that the coefficients of the kinetic and potential terms are proportional, meaning:
$$\frac{1}{c^2} \propto \lambda c^k \implies c^{k+2} \propto \lambda^{-1} \implies c \propto \lambda^{-1/(k+2)}$$
Then the energy eigenvalues scale as:
$$E \propto \frac{1}{c^2} \propto \lambda^{2/(k+2)}$$
Here, the potential has $k = 3$, so:
$$E \propto \lambda^{2/(3+2)} = \lambda^{2/5}$$
Therefore, the ratio of the ground-state energies for $\lambda_2 = 2$ and $\lambda_1 = 1$ is:
$$\frac{E_2}{E_1} = \left(\frac{\lambda_2}{\lambda_1}\right)^{2/5} = 2^{2/5} = 2^{0.4} = 4^{0.2} \approx 1.3195$$
Rounding to three decimal places gives $1.320$, which corresponds to Option D.
Q2
mcq
Potassium chloride forms an FCC lattice, in which K and Cl occupy alternating sites. The density of KCl is $1.98\text{ g/cm}^3$ and the atomic weights of K and Cl are $39.1$ and $35.5$, respectively. The angles of incidence (in degrees) for which Bragg peaks will appear when X-ray of wavelength $0.4\text{ nm}$ is shone on a KCl crystal are
a.
$18.5$, $39.4$ and $72.2$
Correct Answer
b.
$19.5$ and $41.9$
c.
$12.5$, $25.7$, $40.5$ and $60.0$
d.
$13.5$, $27.8$, $44.5$ and $69.0$
Step-by-Step AI Explanation
Verified Method
Potassium chloride has the rock-salt (NaCl-type) structure, with 4 formula units per conventional cubic unit cell.
The mass of a unit cell is:
$$M_{\text{cell}} = \frac{4 \times (39.1 + 35.5)\text{ g/mol}}{6.022 \times 10^{23}\text{ mol}^{-1}} = \frac{4 \times 74.6}{6.022 \times 10^{23}}\text{ g} \approx 4.955 \times 10^{-22}\text{ g}$$
The volume of the unit cell is $a^3 = \frac{M_{\text{cell}}}{\rho} = \frac{4.955 \times 10^{-22}}{1.98}\text{ cm}^3 \approx 2.503 \times 10^{-22}\text{ cm}^3 = 2.503 \times 10^{-28}\text{ m}^3$.
Thus, the lattice parameter is:
$$a = (2.503 \times 10^{-28})^{1/3}\text{ m} \approx 0.630\text{ nm} = 6.30\text{ \AA}$$
However, since $\text{K}^+$ and $\text{Cl}^-$ have nearly identical numbers of electrons ($18$ electrons each), they scatter X-rays almost identically ($f_{\text{K}^+} \approx f_{\text{Cl}^-}$). Thus, the crystal behaves like a simple cubic lattice with lattice parameter $a' = a/2 \approx 0.315\text{ nm}$, or systematically reflections appear when all indices are even, etc.
Let us check Bragg's law $2d\sin\theta = \lambda = 0.4\text{ nm}$:
For (200), $d_{200} = a/2 = 0.315\text{ nm}$, $\sin\theta = \frac{\lambda}{2d} = \frac{0.4}{2 \times 0.315} = 0.6349 \implies \theta = 39.4^\circ$.
Wait, if (111) is considered: with $f_K \approx f_{Cl}$, (111) is very weak or absent. If we consider all FCC reflections:
For $d_{111} = a/\sqrt{3} = 0.630/\sqrt{3} = 0.364\text{ nm}$, $\sin\theta = 0.4/(2 \times 0.364) = 0.550 \implies \theta = 33.3^\circ$.
For (110) of simple cubic with $a' = 0.315\text{ nm}$, $d = 0.315/\sqrt{2} = 0.223\text{ nm}$, $\sin\theta = 0.4 / (2 \times 0.223) = 0.898 \implies \theta = 63.9^\circ$.
Wait, let's look at Option A: $18.5^\circ, 39.4^\circ, 72.2^\circ$.
$\sin(18.5^\circ) = 0.3173 \approx 0.4 / (2 \times 0.630) = 0.3175$, which corresponds to $d = 0.630\text{ nm}$, i.e., the $(100)$ plane with lattice parameter $a = 0.63\text{ nm}$!
$\sin(39.4^\circ) = 0.6347 \approx 2 \times 0.3175$, corresponding to $(200)$!
$\sin(72.2^\circ) = 0.952 \approx 3 \times 0.3175$, corresponding to $(300)$!
So these are successive orders of reflection from the (100) family with $d = a = 0.630\text{ nm}$ (or $n = 1, 2, 3$ in $2d\sin\theta = n\lambda$).
Thus, option A matches precisely.
Q2
mcq
A particle of mass $1\text{ GeV}/c^2$ and its antiparticle, both moving with the same speed $v$, produce a new particle $X$ of mass $10\text{ GeV}/c^2$ in a head-on collision. The minimum value of '$v$' required for this process is closest to
a.
$0.83\ c$
b.
$0.93\ c$
c.
$0.98\ c$
Correct Answer
d.
$0.88\ c$
Step-by-Step AI Explanation
Verified Method
Let the particle and antiparticle have rest mass $m = 1\text{ GeV}/c^2$. They move head-on with equal speed $v$, so the total momentum in the lab frame is zero. The new particle $X$ can be produced at rest with mass $M = 10\text{ GeV}/c^2$.
From conservation of energy:
$$2\gamma m c^2 = M c^2 \implies \gamma = \frac{M}{2m} = \frac{10}{2\times 1} = 5$$
Since $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = 5$:
$$1 - \frac{v^2}{c^2} = \frac{1}{25} = 0.04$$
$$\frac{v}{c} = \sqrt{1 - 0.04} = \sqrt{0.96} \approx 0.9798 \approx 0.98$$
Thus, the required speed is closest to $0.98\ c$.
Q3
mcq
A particle of mass $m$ is in a one-dimensional infinite potential well of length $L$, extending from $x = 0$ to $x = L$. When it is in the energy eigenstate labelled by $n, (n = 1, 2, 3, ....)$ the probability of finding it in the interval $0 \le x \le (L/8)$ is $1/8$. The minimum value of $n$ for which this is possible is
a.
4
Correct Answer
b.
2
c.
6
d.
8
Step-by-Step AI Explanation
Verified Method
The wavefunctions in an infinite square well of length $L$ are $\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$.
The probability $P$ of finding the particle in $[0, L/8]$ is:
$$P = \int_0^{L/8} |\psi_n(x)|^2 dx = \frac{2}{L} \int_0^{L/8} \sin^2\left(\frac{n\pi x}{L}\right) dx$$
Using $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$:
$$P = \frac{2}{L} \left[ \frac{x}{2} - \frac{\sin(2n\pi x / L)}{4n\pi / L} \right]_0^{L/8} = \frac{2}{L} \left( \frac{L}{16} - \frac{\sin(n\pi / 4)}{4n\pi / L} \right) = \frac{1}{8} - \frac{\sin(n\pi/4)}{2n\pi}$$
For this probability to equal $1/8$, we must have:
$$\frac{\sin(n\pi/4)}{2n\pi} = 0 \implies \sin\left(\frac{n\pi}{4}\right) = 0$$
This requires $\frac{n\pi}{4} = k\pi$ for some integer $k \ge 1$, which gives $n = 4k$.
The minimum positive integer $n$ is for $k = 1$, giving $n = 4$.
Q3
mcq
The energies of a two-state quantum system are $E_0$ and $E_0 + \alpha\hbar$ (where $\alpha > 0$ is a constant) and the corresponding normalized state vectors are $|0\rangle$ and $|1\rangle$, respectively. At time $t = 0$, when the system is in the state $|0\rangle$, the potential is altered by a time independent term $V$ such that $\langle 1|V|0\rangle = (\hbar\alpha / 10)$. The transition probability to the state $|1\rangle$ at times $t \ll (1/\alpha)$, is
a.
$\frac{\alpha^2 t^2}{25}$
b.
$\frac{\alpha^2 t^2}{50}$
c.
$\frac{\alpha^2 t^2}{100}$
Correct Answer
d.
$\frac{\alpha^2 t^2}{200}$
Step-by-Step AI Explanation
Verified Method
From time-dependent perturbation theory, the transition probability from $|0\rangle$ to $|1\rangle$ under a perturbation $V$ is given by:
$$P_{0\to 1}(t) = \frac{1}{\hbar^2} \left| \int_0^t \langle 1|V|0\rangle e^{i\omega_{10} t'} dt' \right|^2$$
where $\omega_{10} = \frac{E_1 - E_0}{\hbar} = \frac{\alpha\hbar}{\hbar} = \alpha$.
Given that $\langle 1|V|0\rangle = \frac{\hbar\alpha}{10}$, we have:
$$P_{0\to 1}(t) = \frac{1}{\hbar^2} \left| \frac{\hbar\alpha}{10} \int_0^t e^{i\alpha t'} dt' \right|^2 = \frac{\alpha^2}{100} \left| \frac{e^{i\alpha t} - 1}{i\alpha} \right|^2 = \frac{\alpha^2}{100} \frac{4\sin^2(\alpha t / 2)}{\alpha^2} = \frac{1}{25}\sin^2\left(\frac{\alpha t}{2}\right)$$
For $t \ll 1/\alpha$, we have $\alpha t \ll 1$, so we can approximate $\sin\left(\frac{\alpha t}{2}\right) \approx \frac{\alpha t}{2}$:
$$P_{0\to 1}(t) \approx \frac{1}{25} \left(\frac{\alpha t}{2}\right)^2 = \frac{\alpha^2 t^2}{100}.$$
Therefore, the correct option is C.
Q4
mcq
A particle in one dimension executes oscillatory motion in a potential $V(x) = A|x|$, where $A > 0$ is a constant of appropriate dimension. If the time period $T$ of its oscillation depends on the total energy $E$ as $E^\alpha$, then the value of $\alpha$ is
a.
$1/3$
b.
$1/2$
Correct Answer
c.
$2/3$
d.
$3/4$
Step-by-Step AI Explanation
Verified Method
For a potential of the form $V(x) = A|x|^k$, the time period of oscillation scales with energy as $T \propto E^{\frac{1}{k} - \frac{1}{2}}$.
Here, the potential is $V(x) = A|x|^1$, so $k = 1$.
Therefore, the energy dependence is:
$$T \propto E^{\frac{1}{1} - \frac{1}{2}} = E^{1/2}$$
Hence, $\alpha = 1/2$, which corresponds to Option B.
Alternatively, calculating directly:
$$T = 4\int_0^{x_{\max}} \frac{dx}{\sqrt{\frac{2}{m}(E - Ax)}} = 4\sqrt{\frac{m}{2}} \left[ -\frac{2}{A} \sqrt{E - Ax} \right]_0^{E/A} = \frac{8\sqrt{m}}{A\sqrt{2}} E^{1/2} \propto E^{1/2}.$$
Thus, $\alpha = 1/2$.
Q4
mcq
The Newton-Raphson method is to be used to determine the reciprocal of the number $x = 4$. If we start with the initial guess $0.20$ then after the first iteration the reciprocal is
a.
0.23
b.
0.24
Correct Answer
c.
0.25
d.
0.26
Step-by-Step AI Explanation
Verified Method
To find the reciprocal of $N = 4$ using the Newton-Raphson method, we define the function whose zero is $1/N$:
$$f(y) = \frac{1}{y} - N = 0$$
The derivative is $f'(y) = -\frac{1}{y^2}$.
The Newton-Raphson iteration formula is:
$$y_{n+1} = y_n - \frac{f(y_n)}{f'(y_n)} = y_n - \frac{\frac{1}{y_n} - N}{-\frac{1}{y_n^2}} = y_n + y_n^2\left(\frac{1}{y_n} - N\right) = y_n(2 - Ny_n)$$
Given the initial guess $y_0 = 0.20$ and $N = 4$:
$$y_1 = 0.20 \times (2 - 4 \times 0.20) = 0.20 \times (2 - 0.80) = 0.20 \times 1.20 = 0.24$$
Thus, after the first iteration, the value is $0.24$.
Q5
mcq
Which of the following two physical quantities cannot be measured simultaneously with arbitrary accuracy for the motion of a quantum particle in three dimensions ?
a.
Square of the radial position and $z$-component of angular momentum ($r^2$ and $L_z$)
b.
$x$-components of linear and angular momenta ($p_x$ and $L_x$)
c.
$y$-components of position and $z$-component of angular momentum ($y$ and $L_z$)
Correct Answer
d.
Square of the magnitudes of the linear and angular momenta ($\mathbf{p}^2$ and $\mathbf{L}^2$)
Step-by-Step AI Explanation
Verified Method
Two observables cannot be measured simultaneously with arbitrary accuracy if their corresponding quantum operators do not commute, i.e., $[A, B] \neq 0$.
Let us evaluate the commutators for each pair:
- **Option A:** $[r^2, L_z] = [x^2+y^2+z^2, x p_y - y p_x] = [x^2+y^2, x p_y - y p_x] = 0$. Since $r^2$ is a scalar (rotationally invariant), it commutes with all components of angular momentum.
- **Option B:** $[p_x, L_x] = [p_x, y p_z - z p_y] = 0$, because $p_x$ commutes with $y, z, p_y, p_z$.
- **Option C:** Using $L_z = x p_y - y p_x$, we find:
$$[y, L_z] = [y, x p_y - y p_x] = x [y, p_y] = i\hbar x \neq 0$$
Since the commutator is non-zero, $y$ and $L_z$ are incompatible observables and cannot be measured simultaneously with arbitrary precision.
- **Option D:** $\mathbf{p}^2$ is proportional to the kinetic energy operator for a free particle, which is a scalar under rotations, so $[\mathbf{p}^2, L_i] = 0$ for all components $i=x,y,z$. Consequently, $[\mathbf{p}^2, \mathbf{L}^2] = 0$.
Therefore, the correct pair is given in Option C.
Q5
mcq
A satellite of mass $m$ orbits around earth in an elliptic trajectory of semi-major axis $a$. At a radial distance $r = r_0$, measured from the centre of the earth, the kinetic energy is equal to half the magnitude of the total energy. If $M$ denotes the mass of the earth and the total energy is $-\frac{GMm}{2a}$, the value of $r_0/a$ is nearest to
a.
1.33
Correct Answer
b.
1.48
c.
1.25
d.
1.67
Step-by-Step AI Explanation
Verified Method
The total energy in an elliptical orbit is given by $E = -\frac{GMm}{2a}$.
The kinetic energy is $T$ and the potential energy is $V(r) = -\frac{GMm}{r}$.
We are given that at $r = r_0$, the kinetic energy is half the magnitude of the total energy:
$$T = \frac{1}{2} |E| = \frac{1}{2} \left(\frac{GMm}{2a}\right) = \frac{GMm}{4a}$$
Since $E = T + V(r_0)$, we have:
$$-\frac{GMm}{2a} = \frac{GMm}{4a} - \frac{GMm}{r_0}$$
Dividing by $GMm$:
$$-\frac{1}{2a} = \frac{1}{4a} - \frac{1}{r_0} \implies \frac{1}{r_0} = \frac{1}{4a} + \frac{1}{2a} = \frac{3}{4a}$$
Thus,
$$\frac{r_0}{a} = \frac{4}{3} \approx 1.33$$
Hence, the correct choice is A.
🎉 10 Free Sample Solutions Completed
Unlock All 55 Questions & Solutions
Sign up free in seconds to unlock step-by-step AI answers for the remaining 45 questions.
Q6
mcq
A two-state system evolves under the action of the Hamiltonian $H = E_0 |A\rangle\langle A| + (E_0 + \Delta)|B\rangle\langle B|$, where $|A\rangle$ and $|B\rangle$ are its two orthonormal states. These states transform to one another under parity i.e., $P|A\rangle = |B\rangle$ and $P|B\rangle = |A\rangle$. If at time $t = 0$ the system is in a state of definite parity $P = 1$, the earliest time $t$ at which the probability of finding the system in a state of parity $P = -1$ is one, is
a.
$\frac{\pi\hbar}{2\Delta}$
b.
$\frac{\pi\hbar}{\Delta}$
c.
$\frac{3\pi\hbar}{2\Delta}$
d.
$\frac{2\pi\hbar}{\Delta}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q6
mcq
A laser beam propagates from fiber 1 to fiber 2 in a cavity made up of two optical fibers (as shown in the figure). The loss factor of fiber 2 is $10\text{ dB/km}$. If $E_2(d)$ denotes the magnitude of the electric field in fiber 2 at a distance $d$ from the interface, the ratio $\frac{E_2(0)}{E_2(d)}$ for $d = 10\text{ km}$, is
a.
$10^2$
b.
$10^3$
c.
$10^5$
d.
$10^7$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q7
mcq
A generic $3 \times 3$ real matrix $A$ has eigenvalues $0$, $1$ and $6$, and $I$ is the $3 \times 3$ identity matrix. The quantity/quantities that cannot be determined from this information is/are the
a.
eigenvalues of $(I + A)^{-1}$
b.
eigenvalues of $(I + A^TA)$
c.
determinant of $A^TA$
d.
rank of $A$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q7
mcq
The $|3, 0, 0\rangle$ state (in the standard notation $|n, l, m\rangle$ of the H-atom in the non-relativistic theory decays to the state $|1, 0, 0\rangle$ via two dipole transitions. The transition route and the corresponding probability are
a.
$|3, 0, 0\rangle \rightarrow |2, 1, -1\rangle \rightarrow |1, 0, 0\rangle$ and $\frac{1}{4}$
b.
$|3, 0, 0\rangle \rightarrow |2, 1, 1\rangle \rightarrow |1, 0, 0\rangle$ and $\frac{1}{4}$
c.
$|3, 0, 0\rangle \rightarrow |2, 1, 0\rangle \rightarrow |1, 0, 0\rangle$ and $\frac{1}{3}$
d.
$|3, 0, 0\rangle \rightarrow |2, 1, 0\rangle \rightarrow |1, 0, 0\rangle$ and $\frac{2}{3}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q8
mcq
In the LCR circuit shown below, the resistance $R = 0.05\,\Omega$, the inductance $L = 1\text{ H}$ and the capacitance $C = 0.04\text{ F}$.
If the input $v_{\text{in}}$ is a square wave of angular frequency $1\text{ rad/sec}$, the output $v_{\text{out}}$ is best approximated by a
a.
square wave of angular frequency $1\text{ rad/sec}$.
b.
sine wave of angular frequency $1\text{ rad/sec}$.
c.
square wave of angular frequency $5\text{ rad/sec}$.
d.
sine wave of angular frequency $5\text{ rad/sec}$.
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q8
mcq
The dispersion relation of a gas of non-interacting bosons in $d$ dimensions is $E(k) = ak^s$, where $a$ and $s$ are positive constants. Bose-Einstein condensation will occur for all values of
a.
$d > s$
b.
$d + 2 > s > d - 2$
c.
$s > 2$ independent of $d$
d.
$d > 2$ independent of $s$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q9
mcq
In an experiment, the velocity of a non-relativistic neutron is determined by measuring the time ($\sim 50\text{ ns}$) it takes to travel from the source to the detector kept at a distance $L$. Assume that the error in the measurement of $L$ is negligibly small. If we want to estimate the kinetic energy $T$ of the neutron to within $5\%$ accuracy, i.e., $|\delta T/T| \le 0.05$, the maximum permissible error $|\delta t|$ in measuring the time of flight is nearest to
a.
$1.75\text{ ns}$
b.
$0.75\text{ ns}$
c.
$2.25\text{ ns}$
d.
$1.25\text{ ns}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q9
mcq
A linear diatomic molecule consists of two identical small electric dipoles with an equilibrium separation $R$, which is assumed to be a constant. Each dipole has charges $+q$ of mass $m$ separated by $r$ when the molecule is at equilibrium. Each dipole can execute simple harmonic motion of angular frequency $\omega$. Recall that the interaction potential between two dipoles of moments $\boldsymbol{p}_1$ and $\boldsymbol{p}_2$, separated by $\mathbf{R}_{12} = R_{12}\hat{\mathbf{n}}$ is $(\boldsymbol{p}_1 \cdot \boldsymbol{p}_2 - 3(\boldsymbol{p}_1 \cdot \hat{\mathbf{n}})(\boldsymbol{p}_2 \cdot \hat{\mathbf{n}}))/(4\pi \varepsilon_0 R_{12}^3)$.
Assume that $R \gg r$ and let $\Omega^2 = \frac{q^2}{4\pi \varepsilon_0 m R^3}$. The angular frequencies of small oscillations of the diatomic molecule are
a.
$\sqrt{\omega^2 + \Omega^2}$ and $\sqrt{\omega^2 - \Omega^2}$
b.
$\sqrt{\omega^2 + 3\Omega^2}$ and $\sqrt{\omega^2 - 3\Omega^2}$
c.
$\sqrt{\omega^2 + 4\Omega^2}$ and $\sqrt{\omega^2 - 4\Omega^2}$
d.
$\sqrt{\omega^2 + 2\Omega^2}$ and $\sqrt{\omega^2 - 2\Omega^2}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q10
mcq
The door of an X-ray machine room is fitted with a sensor D (0 is open and 1 is closed). It is also equipped with three fire sensors F1, F2 and F3 (each is 0 when disabled and 1 when enabled). The X-ray machine can operate only if the door is closed and at least 2 fire sensors are enabled. The logic circuit to ensure that the machine can be operated is
a.
Circuit (a)
b.
Circuit (b)
c.
Circuit (c)
d.
Circuit (d)
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q10
mcq
In the reaction $p + n \to p + K^+ + X$, mediated by strong interaction, the baryon number $B$, strangeness $S$ and the third component of isospin $I_3$ of the particle $X$ are, respectively.
a.
$-1, -1$ and $-1$
b.
$+1, -1$ and $-1$
c.
$+1, -2$ and $-1/2$
d.
$-1, -1$ and $0$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q11
mcq
A monochromatic source emitting radiation with a certain frequency moves with a velocity $v$ away from a stationary observer $A$. It is moving towards another observer $B$ (also at rest) along a line joining the two. The frequencies of the radiation recorded by $A$ and $B$ are $\nu_A$ and $\nu_B$, respectively. If the ratio $\frac{\nu_B}{\nu_A} = 7$, then the value of $v/c$ is
a.
$\frac{1}{2}$
b.
$\frac{1}{4}$
c.
$\frac{3}{4}$
d.
$\frac{\sqrt{3}}{2}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q11
mcq
The pressure of a gas in a vessel needs be maintained between $1.5\text{ bar}$ to $2.5\text{ bar}$ in an experiment. The vessel is fitted with a pressure transducer that generates $4\text{ mA}$ to $20\text{ mA}$ current for pressure in the range $1\text{ bar}$ to $5\text{ bar}$. The current output of the transductor has a linear dependence on the pressure.
The reference voltage $\text{V}_1$ and $\text{V}_2$ in the comparators in the circuit (shown in figure above) suitable for the desired operating conditions are, respectively
a.
$2\text{ V}$ and $10\text{ V}$
b.
$2\text{ V}$ and $5\text{ V}$
c.
$3\text{ V}$ and $10\text{ V}$
d.
$3\text{ V}$ and $5\text{ V}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q12
mcq
The ratio $C_p/C_v$ of the specific heats at constant pressure and volume of a monoatomic ideal gas in two dimension is
a.
$3/2$
b.
$2$
c.
$5/3$
d.
$5/2$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q12
mcq
A ${}^{60}\text{Co}$ nucleus $\beta$-decays from its ground state with $J^P = 5^+$ to a state of ${}^{60}\text{Ni}$ with $J^P = 4^+$. From the angular momentum selection rules, the allowed values of the orbital angular momentum $L$ and the total spin $S$ of the electron-antineutrino pair are
a.
$L = 0$ and $S = 1$
b.
$L = 1$ and $S = 0$
c.
$L = 0$ and $S = 0$
d.
$L = 1$ and $S = 1$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q13
mcq
A discrete random variable $X$ takes a value from the set $\{-1, 0, 1, 2\}$ with the corresponding probabilities $p(X) = \frac{3}{10}, \frac{2}{10}, \frac{2}{10}\text{ and } \frac{3}{10}$, respectively. The probability distribution $q(Y) = (q(0), q(1), q(4))$ of the random variable $Y = X^2$ is
a.
$\left(\frac{1}{5}, \frac{3}{5}, \frac{1}{5}\right)$
b.
$\left(\frac{1}{5}, \frac{1}{2}, \frac{3}{10}\right)$
c.
$\left(\frac{2}{5}, \frac{2}{5}, \frac{1}{5}\right)$
d.
$\left(\frac{3}{10}, \frac{3}{10}, \frac{2}{5}\right)$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q13
mcq
Diffuse hydrogen gas within a galaxy may be assumed to follow a Maxwell distribution at temperature $10^6\text{ K}$, while the temperature appropriate for the H gas in the inter-galactic space, following the same distribution, may be taken to be $10^4\text{ K}$. The ratio of thermal broadening $\Delta v_G/\Delta v_{IG}$ of the Lyman-$\alpha$ line from the H-atoms within the galaxy to that from the inter-galactic space is closest to
a.
$100$
b.
$1/100$
c.
$10$
d.
$1/10$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q14
mcq
A polymer made up of $N$ monomers, is in thermal equilibrium at temperature $T$. Each monomer could be of length $a$ or $2a$. The first contributes zero energy, while the second one contributes $\varepsilon$. The average length (in units of $Na$) of the polymer at temperature $T = (\varepsilon/k_B)$ is
a.
$\frac{5+e}{4+e}$
b.
$\frac{4+e}{3+e}$
c.
$\frac{3+e}{2+e}$
d.
$\frac{2+e}{1+e}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q14
mcq
The total number of phonon modes in a solid of volume $V$ is $\int_0^{\omega_D} g(\omega) d\omega = 3N$, where $N$ is the number of primitive cells, $\omega_D$ is the Debye frequency and density of photon modes is $g(\omega) = AV\omega^2$ (with $A > 0$ a constant). If the density of the solid doubles in a phase transition, the Debye temperature $\theta_D$ will
a.
increase by a factor of $2^{2/3}$
b.
increase by a factor of $2^{1/3}$
c.
decrease by a factor of $2^{2/3}$
d.
decrease by a factor of $2^{1/3}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q15
mcq
The volume of the region common to the interiors of two infinitely long cylinders defined by $x^2 + y^2 = 25$ and $x^2 + 4z^2 = 25$ is best approximated by
a.
$225$
b.
$333$
c.
$423$
d.
$625$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q15
mcq
A perfectly conducting fluid, of permittivity $\varepsilon$ and permeability $\mu$, flows with a uniform velocity $v$ in the presence of time dependent electric and magnetic fields $\boldsymbol{E}$ and $\boldsymbol{B}$, respectively. If there is a finite current density in the fluid, then
a.
$\nabla \times (\boldsymbol{v} \times \boldsymbol{B}) = \frac{\partial \boldsymbol{B}}{\partial t}$
b.
$\nabla \times (\boldsymbol{v} \times \boldsymbol{B}) = -\frac{\partial \boldsymbol{B}}{\partial t}$
c.
$\nabla \times (\boldsymbol{v} \times \boldsymbol{B}) = \sqrt{\varepsilon\mu} \frac{\partial \boldsymbol{E}}{\partial t}$
d.
$\nabla \times (\boldsymbol{v} \times \boldsymbol{B}) = -\sqrt{\varepsilon\mu} \frac{\partial \boldsymbol{E}}{\partial t}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q16
mcq
In an experiment to measure the charge to mass ratio $e/m$ of the electron by Thomson's method, the value of the deflecting electric field and the accelerating potential are $6 \times 10^6\text{ N/C}$ (Newton per coulomb) and $150\text{ V}$, respectively. The magnitude of the magnetic field that leads to zero deflection of the electron beam is closest to
a.
$0.6\text{ T}$
b.
$1.2\text{ T}$
c.
$0.4\text{ T}$
d.
$0.8\text{ T}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q16
mcq
The energy levels of a non-degenerate quantum system are $\varepsilon_n = nE_0$, where $E_0$ is a constant and $n = 1, 2, 3,...$ At a temperature $T$, the free energy $F$ can be expressed in terms of the average energy $E$ by
a.
$E_0 + k_B T \ln \frac{E}{E_0}$
b.
$E_0 + 2k_B T \ln \frac{E}{E_0}$
c.
$E_0 - k_B T \ln \frac{E}{E_0}$
d.
$E_0 - 2k_B T \ln \frac{E}{E_0}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q17
mcq
The volume and temperature of a spherical cavity filled with black body radiation are $V$ and $300\text{ K}$, respectively. If it expands adiabatically to a volume $2V$, its temperature will be closest to
a.
$150\text{ K}$
b.
$300\text{ K}$
c.
$250\text{ K}$
d.
$240\text{ K}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q17
mcq
The unnormalized wavefunction of a particle in one dimension in an infinite square well with walls at $x = 0$ and $x = a$, is $\psi(x) = x(a - x)$. If $\psi(x)$ is expanded as a linear combination of the energy eigenfunctions, $\int_0^a |\psi(x)|^2 dx$ is proportional to the infinite series
[You may use: $\int_0^a t \sin t \, dt = -a \cos a + \sin a$ and $\int_0^a t^2 \sin t \, dt = -2 - (a^2 - 2)\cos a + 2a \sin a$ ]
a.
$\sum_{n=1}^\infty (2n-1)^{-6}$
b.
$\sum_{n=1}^\infty (2n-1)^{-4}$
c.
$\sum_{n=1}^\infty (2n-1)^{-2}$
d.
$\sum_{n=1}^\infty (2n-1)^{-8}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q18
mcq
A particle, thrown with a speed $v$ from the earth's surface, attains a maximum height $h$ (measured from the surface of the earth). If $v$ is half the escape velocity and $R$ denotes the radius of earth, then $h/R$ is
a.
$2/3$
b.
$1/3$
c.
$1/4$
d.
$1/2$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q18
mcq
The Legendre polynomials $P_n(x), n = 0, 1, 2,...$, satisfying the orthogonality condition $\int_{-1}^1 P_n(x) P_m(x) dx = \frac{2}{2n+1} \delta_{nm}$ on the interval $[-1, +1]$, may be defined by the Rodrigues formula
$$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n$$
The value of the definite integral $\int_{-1}^1 (4 + 2x - 3x^2 + 4x^3) P_3(x) dx$ is
a.
$3/5$
b.
$11/15$
c.
$23/32$
d.
$16/35$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q19
mcq
The vector potential for an almost point like magnetic dipole located at the origin is $A = \frac{\mu \sin\theta}{4\pi r^2}\hat{\phi}$, where $(r, \theta, \phi)$ denote the spherical polar coordinates and $\hat{\phi}$ is the unit vector along $\phi$. A particle of mass $m$ and charge $q$, moving in the equatorial plane of the dipole, starts at time $t = 0$ with an initial speed $v_0$ and an impact parameter $b$. Its instantaneous speed at the point of closest approach is
a.
$v_0$
b.
$0/0$
c.
$v_0 + \frac{\mu q}{4\pi m b^2}$
d.
$\sqrt{v_0^2 + \left(\frac{\mu q}{4\pi m b^2}\right)^2}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q19
mcq
The figure below shows an ideal capacitor consisting of two parallel circular plates of radius $R$. Points $P_1$ and $P_2$ are at a transverse distance $r_1 > R$ from the line joining the centres of the plates, while points $P_3$ and $P_4$ are at a transverse distance $r_2 < R$.
If $B(x)$ denotes the magnitude of the magnetic fields at these points, which of the following holds while the capacitor is charging ?
a.
$B(P_1) < B(P_2)$ and $B(P_3) < B(P_4)$
b.
$B(P_1) > B(P_2)$ and $B(P_3) > B(P_4)$
c.
$B(P_1) = B(P_2)$ and $B(P_3) < B(P_4)$
d.
$B(P_1) = B(P_2)$ and $B(P_3) > B(P_4)$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q20
mcq
A conducting wire in the shape of a circle lies on the $(x, y)$-plane, with its centre at the origin. A bar magnet moves with a constant velocity towards the wire along the $z$-axis (as shown in the figure below).
We take $t = 0$ to be the instant at which the midpoint of the magnet is at the centre of the wire loop and the induced current to be positive when it is counter-clockwise as viewed by the observer facing the loop and the incoming magnet. In these conventions, the best schematic representation of the induced current $I(t)$ as a function of $t$, is
a.
(a)
b.
(b)
c.
(c)
d.
(d)
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q20
mcq
Balls of ten different colours labeled by $a = 1, 2, \dots, 10$ are to be distributed among different coloured boxes. A ball can only go in a box of the same colour, and each box can contain at most one ball. Let $n_a$ and $N_a$ denote, respectively, the numbers of balls and boxes of colour $a$. Assuming that $N_a \gg n_a \gg 1$, the total entropy (in units of the Boltzmann constant) can be best approximated by
a.
$\sum_a (N_a \ln N_a + n_a \ln n_a - (N_a - n_a) \ln (N_a - n_a))$
b.
$\sum_a (N_a \ln N_a - n_a \ln n_a - (N_a - n_a) \ln (N_a - n_a))$
c.
$\sum_a (N_a \ln N_a - n_a \ln n_a + (N_a - n_a) \ln (N_a - n_a))$
d.
$\sum_a (N_a \ln N_a + n_a \ln n_a + (N_a - n_a) \ln (N_a - n_a))$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q21
mcq
To measure the height $h$ of a column of liquid helium in a container, a constant current $I$ is sent through an NbTi wire of length $l$, as shown in the figure. The normal state resistance of the NbTi wire is $R$.
If the superconducting transition temperature of NbTi is $\approx 10\text{ K}$, then the measured voltage $V(h)$ is best described by the expression.
a.
$IR\left(\frac{1}{2} - \frac{2h}{l}\right)$
b.
$IR\left(1 - \frac{h}{l}\right)$
c.
$IR\left(\frac{1}{2} - \frac{h}{l}\right)$
d.
$IR\left(1 - \frac{2h}{l}\right)$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q21
mcq
The components of the electric field, in a region of space devoid of any charge or current sources, are given to be $E_i = a_i + \sum_{j=1,2,3} b_{ij} x_j$, where $a_i$ and $b_{ij}$ are constants independent of the coordinates.
The number of independent components of the matrix $b_{ij}$ is
a.
5
b.
6
c.
3
d.
4
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q22
mcq
The equation of motion of a one-dimensional forced harmonic oscillator in the presence of a dissipative force is described by $\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 16x = 6te^{-8t} + 4t^2 e^{-2t}$. The general form of the particular solution, in terms of constants $A, B$ etc., is
a.
$t(At^2 + Bt + C)e^{-2t} + (Dt + E)e^{-8t}$
b.
$(At^2 + Bt + C)e^{-2t} + (Dt + E)e^{-8t}$
c.
$t(At^2 + Bt + C)e^{-2t} + t(Dt + E)e^{-8t}$
d.
$(At^2 + Bt + C)e^{-2t} + t(Dt + E)e^{-8t}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q22
mcq
A particle of mass $m$ in one dimensions is in the ground state of a simple harmonic oscillator described by a Hamiltonian $H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2$ in the standard notation. An impulsive force at time $t = 0$ suddenly imparts a momentum $p_0 = \sqrt{\hbar m\omega}$ to it. The probability that the particle remains in the original ground state is
a.
$e^{-2}$
b.
$e^{-3/2}$
c.
$e^{-1}$
d.
$e^{-1/2}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q23
mcq
The position of a particle in one dimension changes in discrete steps. With each step it moves to the right, however, the length of the step is drawn from a uniform distribution from the interval $\left[\lambda - \frac{1}{2}w, \lambda + \frac{1}{2}w\right]$, where $\lambda$ and $w$ are positive constants. If $X$ denotes the distance from the starting point after $N$ steps, the standard deviation $\sqrt{\langle X^2 \rangle - \langle X \rangle^2}$ for large values of $N$ is
a.
$\frac{\lambda}{2} \times \sqrt{N}$
b.
$\frac{\lambda}{2} \times \sqrt{\frac{N}{3}}$
c.
$\frac{w}{2} \times \sqrt{N}$
d.
$\frac{w}{2} \times \sqrt{\frac{N}{3}}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q23
mcq
A particle in two dimensions is found to trace an orbit $r(\theta) = r_0 \,\theta^2$. If it is moving under the influence of a central potential $V(r) = c_1 r^{-a} + c_2 r^{-b}$, where $r_0$, $c_1$ and $c_2$ are constants of appropriate dimensions, the values of ‘$a$’ and ‘$b$’, respectively, are
a.
2 and 4
b.
2 and 3
c.
3 and 4
d.
1 and 3
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q24
mcq
The figures below depict three different wavefunctions of a particle confined to a one-dimensional box $-1 \le x \le 1$.
The wavefunctions that correspond to the maximum expectation values $|\langle x \rangle|$ (absolute value of the mean position) and $\langle x^2 \rangle$, respectively, are
a.
B and C
b.
B and A
c.
C and B
d.
A and B
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q24
mcq
In the following circuit the input voltage $V_{\text{in}}$ is such that $|V_{\text{in}}| < |V_{\text{sat}}|$, where $V_{\text{sat}}$ is the saturation voltage of the op-amp. (Assume that the diode is an ideal one and $R_LC$ is much larger than the duration of the measurement) For the input voltage as shown in the figure above, the output voltage $\text{V}_{\text{out}}$ is best represented by
a.
(a)
b.
(b)
c.
(c)
d.
(d)
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q25
mcq
The volume integral $I = \iiint_V \mathbf{A} \cdot (\nabla \times \mathbf{A})\, d^3x$ is over a region $V$ bounded by a surface $\Sigma$ (an infinitesimal area element being $\hat{\mathbf{n}} dS$, where $\hat{\mathbf{n}}$ is the outward unit normal). If it changes to $I + \Delta I$, when the vector $\mathbf{A}$ is changed to $\mathbf{A} + \nabla \Lambda$, then $\Delta I$ can be expressed as:
a.
$\iiint_V \nabla \cdot (\nabla \Lambda \times \mathbf{A})\, d^3x$
b.
$-\iiint_V \nabla^2 \Lambda\, d^3x$
c.
$-\oiint_\Sigma (\nabla \Lambda \times \mathbf{A}) \cdot \hat{\mathbf{n}} dS$
d.
$\oiint_\Sigma \nabla \Lambda \cdot \hat{\mathbf{n}} dS$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q25
mcq
Lead is superconducting below $7\text{ K}$ and has a critical magnetic field $800 \times 10^{-4}\text{ Tesla}$ close to $0\text{ K}$. At $2\text{ K}$ the critical current that flows through a long lead wire of radius $5\text{ mm}$ is closest to
a.
$1760\text{ A}$
b.
$1670\text{ A}$
c.
$1950\text{ A}$
d.
$1840\text{ A}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q26
mcq
If we use the Fourier transform $\phi(x, y) = \int e^{ikx} \phi_k(y) dk$ to solve the partial differential equation:
$$-\frac{\partial^2\phi(x, y)}{\partial y^2} - \frac{1}{y^2} \frac{\partial^2\phi(x, y)}{\partial x^2} + \frac{m^2}{y^2}\phi(x, y) = 0$$
in the half-plane $\{(x, y): -\infty < x < \infty, 0 < y < \infty\}$ the Fourier modes $\phi_k(y)$ depend on $y$ as $y^\alpha$ and $y^\beta$. The values of $\alpha$ and $\beta$ are
a.
$\frac{1}{2} + \sqrt{1 + 4(k^2 + m^2)}$ and $\frac{1}{2} - \sqrt{1 + 4(k^2 + m^2)}$
b.
$1 + \sqrt{1 + 4(k^2 + m^2)}$ and $1 - \sqrt{1 + 4(k^2 + m^2)}$
c.
$\frac{1}{2} + \frac{1}{2}\sqrt{1 + 4(k^2 + m^2)}$ and $\frac{1}{2} - \frac{1}{2}\sqrt{1 + 4(k^2 + m^2)}$
d.
$1 + \frac{1}{2}\sqrt{1 + 4(k^2 + m^2)}$ and $1 - \frac{1}{2}\sqrt{1 + 4(k^2 + m^2)}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q27
mcq
The fulcrum of a simple pendulum (consisting of a particle of mass $m$ attached to the support by a massless string of length $l$ ) oscillates vertically as $\sin z(t) = a\sin\omega t$, where $\omega$ is a constant. The pendulum moves in a vertical plane and $\theta(t)$ denotes its angular position with respect to the $z$-axis.
If $l\frac{d^2\theta}{dt^2} + \sin\theta(g - f(t)) = 0$, (where $g$ is the acceleration due to gravity) describes the equation of motion of the mass, then $f(t)$ is
a.
$a\omega^2\cos\omega t$
b.
$a\omega^2\sin\omega t$
c.
$-a\omega^2\cos\omega t$
d.
$-a\omega^2\sin\omega t$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q28
mcq
The $Q$-value of the $\alpha$-decay of $^{232}\text{Th}$ to the ground state of $^{228}\text{Ra}$ is $4082\text{ keV}$. The maximum possible kinetic energy of the $\alpha$-particle is closest to
a.
$4082\text{ keV}$
b.
$4050\text{ keV}$
c.
$4035\text{ keV}$
d.
$4012\text{ keV}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q29
mcq
The nuclei of $^{137}\text{Cs}$ decay by the emission of $\beta$-particles with a half-life of $30.08\text{ years}$. The activity (in units of disintegrations per second or $\text{Bq}$) of a $1\text{ mg}$ source of $^{137}\text{Cs}$, prepared on January 1, 1980, as measured on January 1, 2021 is closest to
a.
$1.79 \times 10^{16}$
b.
$1.79 \times 10^9$
c.
$1.24 \times 10^{16}$
d.
$1.24 \times 10^9$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q30
mcq
In an elastic scattering process at an energy $E$, the phase shifts satisfy $\delta_0 \approx 30^\circ, \delta_1 \approx 10^\circ$, while the other phase shifts are zero. The polar angle at which the differential cross-section peaks is closest to
a.
$20^\circ$
b.
$10^\circ$
c.
$0^\circ$
d.
$30^\circ$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.