WBSST Physics PG Examination Quiz (2016)
Solution: The cross product of two vectors, $\vec{a} \times \vec{b}$, by definition, results in a new vector that is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$. This property is fundamental to the vector cross product. Option A and B are incorrect applications of vector calculus. Option D is true but not the most fundamental property of the cross product of two vectors, which is what the question is asking.
Solution: For the sum of three vectors to be zero, they must form a closed triangle when placed head-to-tail. A triangle is a two-dimensional shape, so all three vectors must lie in the same plane. Hence, they are co-planar.
Solution: The trajectory of a comet is a conic section under the influence of the sun's gravitational field. The shape depends on its total energy. If the energy is negative, the orbit is elliptical. If it is zero, the orbit is parabolic. If it is positive, the orbit is hyperbolic. Since all three are possible, "All of the above" is the most comprehensive answer.
Solution: Centripetal force is the force that acts on an object moving in a circular path and is directed towards the center of the circle. The formula for this force is $F = \frac{mv^2}{r}$.
Solution: In an elastic collision, both momentum and kinetic energy are conserved. For a head-on elastic collision of two particles of equal mass, with one initially at rest, the moving particle transfers all its momentum and kinetic energy to the stationary particle and comes to rest. This is a classic characteristic of an elastic collision.
Solution: The gravitational force between the Earth and the satellite provides the necessary centripetal force to keep the satellite in its orbit. Without this force, the satellite would fly off in a straight line according to Newton's First Law.
Solution: The Second Law of Thermodynamics states that for a reversible process, the total entropy change of the universe (system + surroundings) is zero. For an irreversible process, the total entropy of the universe increases.
Solution: This is an application of the conservation of angular momentum ($L = I\omega$). Since the angular momentum is conserved, $I_{initial}\omega_{initial} = I_{final}\omega_{final}$. The moment of inertia for a solid sphere is $I = \frac{2}{5}MR^2$, and angular velocity is $\omega = \frac{2\pi}{T}$. If the radius is halved, $R_{final} = R/2$, so the new moment of inertia is $I_{final} = \frac{2}{5}M(\frac{R}{2})^2 = \frac{1}{4}I_{initial}$. Equating the initial and final angular momenta: $I_{initial}\frac{2\pi}{T_{initial}} = I_{final}\frac{2\pi}{T_{final}} \implies I_{initial}\frac{1}{T_{initial}} = \frac{1}{4}I_{initial}\frac{1}{T_{final}} \implies T_{final} = \frac{1}{4}T_{initial}$.
Solution: For a diagonal matrix, the eigenvalues are simply the elements on the main diagonal. In this case, the eigenvalues are 1 and 2.
Solution: The relationship between kinetic energy ($K.E$) and momentum ($p$) is $K.E = \frac{p^2}{2m}$. If the momentum is doubled, $p' = 2p$. The new kinetic energy is $K.E' = \frac{(2p)^2}{2m} = \frac{4p^2}{2m} = 4 K.E$. The kinetic energy becomes four times its original value.
Solution: In Lagrangian mechanics, the generalized momentum $p_i$ conjugate to the generalized coordinate $q_i$ is defined as the partial derivative of the Lagrangian ($L$) with respect to the generalized velocity ($\dot{q}_i$).
Solution: The work done is calculated using the line integral $W = \int \vec{F} \cdot d\vec{r}$. The infinitesimal displacement vector is $d\vec{r} = dx\hat{i} + dy\hat{j}$. So, $\vec{F} \cdot d\vec{r} = (3x^2\hat{i} + 4y\hat{j}) \cdot (dx\hat{i} + dy\hat{j}) = 3x^2dx + 4ydy$. $W = \int_{(0,0)}^{(1,1)} (3x^2dx + 4ydy) = \int_0^1 3x^2dx + \int_0^1 4ydy = [x^3]_0^1 + [2y^2]_0^1 = (1^3 - 0^3) + (2(1^2) - 2(0^2)) = 1 + 2 = 3$ J.
Solution: Young's modulus is the ratio of stress to strain, $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{Strain}}$. We can rearrange this to find the area: $A = \frac{F}{Y \cdot \text{Strain}}$. Given: $F = 100$ N, $Y = 10^{11} \text{N/m}^2$, Strain = $0.01 = 10^{-2}$. $A = \frac{100}{10^{11} \cdot 10^{-2}} = \frac{10^2}{10^9} = 10^{-7} \text{m}^2$.
Solution: Terminal velocity is reached when the downward gravitational force equals the upward viscous drag force. The gravitational force is $F_g \propto r^3$, and the viscous force according to Stokes' Law is $F_v \propto rv_t$. Equating these forces, we get $r^3 \propto rv_t$, which implies $v_t \propto r^2$.
Solution: The angular frequency of a simple harmonic motion for a mass-spring system is $\omega = \sqrt{\frac{k}{m}}$. The linear frequency is related to the angular frequency by $f = \frac{\omega}{2\pi}$. Therefore, $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$.
Solution: A simple harmonic motion is a sinusoidal oscillation with a single angular frequency. A superposition of two simple harmonic motions with the same angular frequency, like $x = a \cos(\omega t) + b \sin(\omega t)$, can be expressed in the form $x = A \sin(\omega t + \phi)$, which is the general form of SHM. Options A and B involve different frequencies and therefore do not represent a single SHM. Option D also represents SHM, but option C is a more general and common representation.
Solution: A Carnot cycle is a completely reversible cycle. Since entropy is a state function, its net change over any closed reversible cycle is zero.
Solution: We use the magnification formula $m = \frac{v}{u}$ and the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Given $u = -10$ cm and $m = 2$. From magnification: $2 = \frac{v}{-10} \implies v = -20$ cm. From lens formula: $\frac{1}{-20} - \frac{1}{-10} = \frac{1}{f} \implies \frac{1}{20} = \frac{1}{f} \implies f = 20$ cm.
Solution: Brewster's angle (or the polarization angle) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. The reflected component is completely linearly polarized.
Solution: The standard differential equation for SHM is $\frac{d^2x}{dt^2} + \omega^2x = 0$, where $\omega$ is the angular frequency. By comparing the given equation with the standard form, we find $\omega^2 = 4$, so $\omega = 2$ rad/s. The time period $T$ is given by $T = \frac{2\pi}{\omega}$. Therefore, $T = \frac{2\pi}{2} = \pi$ seconds.
Solution: This phenomenon is known as resonance. When the frequency of the external driving force matches the natural frequency of a system, the amplitude of the oscillations becomes maximum.
Solution: The Joule-Thomson effect is an isenthalpic process, meaning it occurs at a constant enthalpy. This is because the process involves no heat exchange with the surroundings and no external work is done on the gas.
Solution: The power gain in decibels (dB) is defined as $10 \log_{10}(\frac{P_{out}}{P_{in}})$. The factor of 20 is used for voltage or current gain, not power gain.
Solution: The equation for a transverse wave is $y(x,t) = A \sin(kx - \omega t)$. The particle velocity is the derivative with respect to time, $v_p = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t)$, so the maximum particle velocity is $A\omega$. The wave velocity is $v_w = \frac{\omega}{k}$. The ratio is $\frac{v_{p,max}}{v_w} = \frac{A\omega}{\omega/k} = Ak$.
Solution: The relationship between intensity and amplitude is $I \propto A^2$. Given $I_1 = I$ and $I_2 = 4I$, we have $A_1 \propto \sqrt{I}$ and $A_2 \propto \sqrt{4I} = 2\sqrt{I}$. The maximum intensity is due to constructive interference, $I_{max} \propto (A_1+A_2)^2 \propto (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$. The minimum intensity is due to destructive interference, $I_{min} \propto (A_2-A_1)^2 \propto (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$. The ratio is $\frac{I_{max}}{I_{min}} = \frac{9I}{I} = 9:1$.
Solution: The Zeeman effect is the phenomenon of splitting of spectral lines into several components in the presence of a static magnetic field. The Stark effect is a similar phenomenon but occurs in an electric field.
Solution: The electronic configuration of a sodium atom is $1s^2 2s^2 2p^6 3s^1$. For the valence electron in the $3s$ orbital, the orbital quantum number is $l=0$, so the total orbital angular momentum is $L=0$. The spin quantum number is $s=\frac{1}{2}$, so the total spin angular momentum is $S=\frac{1}{2}$. The total angular momentum is $J = L+S = 0+\frac{1}{2} = \frac{1}{2}$. The spin multiplicity is $2S+1 = 2(\frac{1}{2})+1 = 2$. The spectroscopic term is ${}^{2S+1}L_J = {}^2S_{1/2}$.
Solution: Bose-Einstein statistics apply to bosons, which are particles with integer spin. Photons have a spin of 1 and are bosons. Electrons, protons, and neutrons are fermions and have a half-integer spin, so they obey Fermi-Dirac statistics.
Solution: The energy density of an electric field in a medium is given by $u = \frac{1}{2}\epsilon E^2$. The permittivity of the medium is $\epsilon = \epsilon_r\epsilon_0$, where $\epsilon_r$ is the relative permittivity or dielectric constant and $\epsilon_0$ is the permittivity of free space. Substituting this, we get $u = \frac{1}{2}\epsilon_0\epsilon_r E^2$.
Solution: The Curie temperature is the temperature at which a ferromagnetic material loses its permanent magnetic properties and becomes paramagnetic. Above this temperature, the thermal energy is sufficient to disrupt the alignment of magnetic domains.
Solution: A transformer works on the principle of mutual induction, which requires a constantly changing magnetic flux. A DC source provides a constant current and a constant magnetic field. Since there is no change in magnetic flux, no e.m.f. is induced in the secondary coil, and thus no current is supplied.
Solution: The electric field is the negative gradient of the electric potential. This means that the electric field points in the direction of the steepest decrease in potential.
Solution: According to classical electromagnetism, a charged particle radiates electromagnetic energy only when it is accelerating. Constant linear velocity and being at rest do not involve acceleration. Uniform circular motion (option D) is a form of acceleration (centripetal acceleration), so it also causes radiation. However, option (C) "Moving with constant acceleration" is a more general statement covering all cases of radiation.
Solution: The ratio of specific heats is given by $\gamma = 1 + \frac{2}{f}$, where $f$ is the number of degrees of freedom. For a linear triatomic gas molecule (e.g., $CO_2$) at normal temperatures, there are 3 translational and 2 rotational degrees of freedom, giving a total of $f = 5$. Thus, $\gamma = 1 + \frac{2}{5} = 1.40$.
Solution: The rate of cooling is given by $\frac{dT}{dt} \propto \frac{\text{Area}}{\text{Volume}}$. For a sphere, the surface area is $A=4\pi r^2$ and the volume is $V=\frac{4}{3}\pi r^3$. The rate of cooling is proportional to the ratio of area to mass, which is $\frac{A}{m} = \frac{4\pi r^2}{\rho \cdot \frac{4}{3}\pi r^3} \propto \frac{1}{r}$. Therefore, the ratio of the rates of cooling for spheres of radius $r$ and $2r$ is $\frac{(dT/dt)_1}{(dT/dt)_2} = \frac{1/r}{1/(2r)} = \frac{2}{1}$.
Solution: This is described by Wien's displacement law, which states that the peak wavelength of blackbody radiation is inversely proportional to the absolute temperature, $\lambda_{max} \propto T^{-1}$.
Solution: A state function is a property whose value depends only on the current state of the system, not on the path taken to reach that state. Work is a path function, meaning the amount of work done depends on the specific process (path) between the initial and final states. Temperature, internal energy, and entropy are state functions.
Solution: One of the key postulates of Bohr's atomic model is that the angular momentum of an electron in a stable orbit is quantized, and is an integral multiple of $\frac{h}{2\pi}$. The formula is $L = n\frac{h}{2\pi}$, where $n$ is the principal quantum number.
Solution: We use the relativistic velocity addition formula: $u' = \frac{u+v}{1 + \frac{uv}{c^2}}$. Here, $u = 0.8c$ and $v = 0.8c$. $u' = \frac{0.8c + 0.8c}{1 + \frac{(0.8c)(0.8c)}{c^2}} = \frac{1.6c}{1 + 0.64} = \frac{1.6c}{1.64} \approx 0.9756c$. This is less than the speed of light, as required by special relativity.
Solution: The wave function is normalized, meaning $\int |\psi(x)|^2 dx = 1$. The integral on the left must be dimensionless. The dimension of the integral is (dimension of $|\psi|^2$) $\times$ (dimension of $x$). Let the dimension of $\psi(x)$ be $L^a$. Then the dimension of the integral is $(L^a)^2 \cdot L = L^{2a+1}$. For this to be dimensionless, we must have $2a+1=0$, which gives $a = -1/2$. So the dimension of $\psi(x)$ is $L^{-1/2}$.
Solution: The hexadecimal number is likely a typo and meant to be 'A' (which represents the decimal number 10). The binary representation of the decimal number 10 is $1010$.
Solution: The energy of an electron in the nth orbit of a hydrogen atom is given by the Bohr model as $E_n = -\frac{13.6 \text{ eV}}{n^2}$. Therefore, the energy is inversely proportional to the square of the principal quantum number, $E_n \propto \frac{1}{n^2}$.
Solution: According to Planck's relation, the energy of a photon is directly proportional to its frequency, with the constant of proportionality being Planck's constant, $h$. So, $E = h\nu$.
Solution: The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$. For the same velocity, the wavelength is inversely proportional to the mass. Since a proton is much more massive than an electron ($m_p > m_e$), its de Broglie wavelength will be shorter.
Solution: The minimum wavelength ($\lambda_{min}$) of the continuous X-ray spectrum is determined by the maximum energy of the electrons striking the target. This energy is given by $E_{max} = eV$, where $e$ is the electron charge and $V$ is the accelerating voltage. The energy of the photon is $E = \frac{hc}{\lambda}$, so $\lambda_{min} = \frac{hc}{eV}$. It depends only on the accelerating voltage.
Solution: Alpha ($\alpha$) is the current gain for a common base configuration, and beta ($\beta$) is the current gain for a common emitter configuration. They are related by the formula $\beta = \frac{\alpha}{1-\alpha}$.
Solution: In a synchrotron, the magnetic field is used to bend the trajectory of charged particles and keep them moving in a circular path. The electric field is used to accelerate the particles, increasing their kinetic energy.
Solution: Leptons are a class of fundamental particles that do not undergo strong interactions. Examples of leptons are electrons, muons, and neutrinos. Protons and neutrons are baryons (hadrons), and pions are mesons (hadrons).
Solution: Time dilation is the phenomenon where a moving clock is observed to tick more slowly than a stationary clock. It is observed in a moving frame of reference relative to a stationary observer.
Solution: A NAND gate is the logical inverse of an AND gate. The Boolean expression for AND is $A \cdot B$. The negation is represented by an overline. Thus, the expression for NAND is $\overline{A \cdot B}$.
Solution: The work function ($\phi$) of a metal is the minimum energy needed to remove an electron from the surface of the metal. This energy must be supplied by the incident photon. Any excess energy is converted into the kinetic energy of the ejected electron.
Solution: According to Gauss's law, the electric flux through any closed surface is proportional to the net charge enclosed within that surface. For a hollow spherical shell with charge distributed on its surface, any closed surface inside the shell encloses no charge. Therefore, the electric field inside is zero.
Solution: The electric field of a dipole at a point on its axis is given by the formula $E = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$ for a point at a large distance $r$ from the dipole's center. Thus, the electric field is proportional to $r^{-3}$.
Solution: According to Ampere's law and the right-hand rule, two parallel conductors carrying current in the same direction will generate magnetic fields that result in an attractive force between them. The force per unit length is given by $F/L = \frac{\mu_0 I_1 I_2}{2\pi d}$.
Solution: The magnetic force (Lorentz force) on a charged particle is given by the cross product formula $\vec{F} = q(\vec{v} \times \vec{B})$. The result of a cross product is a vector that is perpendicular to the plane containing both of the original vectors. Therefore, the force is perpendicular to both the velocity of the electron and the magnetic field.
No comments:
Post a Comment