- The radius of the Fermi sphere of free electrons in a monovalent metal with an fcc structure, in which the volume of the unit cell is a^3, is
- \left(\frac{12\pi^2}{a^3}\right)^{1/3}
- \left(\frac{3\pi^2}{a^3}\right)^{1/3}
- \left(\frac{\pi^2}{a^3}\right)^{1/3}
- \left(\frac{1}{a}\right)^{1/3}
- Deviation from Rutherford scattering formula for \alpha-particle scattering gives an estimate of:
- Size of an atom
- Thickness of target
- Size of a nucleus
- half life of \alpha-emitter
- If the critical magnetic field for aluminium is 7.9\times 10^3 A/m, the critical current which can flow through long thin superconducting wire of aluminium of diameter 1\times10^{-3}m is
- 5.1 A
- 1.6 A
- 4.0 A
- 2.48 A
- An elemental dielectric has a dielectric constant \epsilon=12 and it contains 5\times10^{18} atoms/m^3. Its electronic polarizability in uints of FM^2 is
- 4.17\times10^{-30}
- 8.34\times10^{-30}
- 6.32\times10^{-30}
- 12.64\times10^{-30}
- Mangetite (Fe_3O_4) has a cubic structure with a lattice constant of 8.4A^o. The saturation magnetization in this material in units of A/m is :
- 6.2\times10^5
- 6.2\times10^6
- 12.4\times10^6
- 12.4\times10^5
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Wednesday, 30 November 2016
Problem set 33
Monday, 28 November 2016
Problem set 32
- The Hamiltonian of a particle of unit mass moving in the xy-plane is given to be: H = xp_x- yp_y -\frac{1}{2}x^2 +\frac{1}{2} y^2 in suitable units. The initial values are given to be (x(0),y(0)) = (1,1) and (p_x(0),p_y(0) )=(\frac{1}{2}, -\frac{1}{2}). During the motion, the curves traced out by the particles in the xy-plane and the p_xp_y-plane are
- both straight lines
- a straight line and a hyperbola respectively
- a hyperbola and an ellipse, respectively
- both hyperbolas
- A static, spherically symmetric charge distribution is given by \rho(r) = \frac{A}{r} e^{-Kr} where A and K are positive constants. The electrostatic potential corresponding to this charge distribution varies with r as
- re^{-Kr}
- \frac{1}{r} e^{-Kr}
- \frac{1}{r^2} e^{-Kr}
- \frac{1}{r} (1-e^{-Kr})
- 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^{LP} and \omega_0^{HP} , respectively, the condition required to implement the band-pass and band-reject filters are, respectively,
- \omega_0^{HP}<\omega_0^{LP} and \omega_0^{HP}<\omega_0^{LP}
- \omega_0^{HP}<\omega_0^{LP} and \omega_0^{HP}>\omega_0^{LP}
- \omega_0^{HP}>\omega_0^{LP} and \omega_0^{HP}<\omega_0^{LP}
- \omega_0^{HP}>\omega_0^{LP} and \omega_0^{HP}>\omega_0^{LP}
- 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
- \hbar\omega/2
- \hbar\omega
- 3\hbar\omega/2
- 0
- Linearly independent solution of the differential equation \frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=0 are:
- e^{-x}, e^{-2x}
- e^{-x}, e^{2x}
- e^{-2x}, e^{x}
- e^{2x}, e^{x}
Hence, answer is (C)
\nabla^2V must be proportional to \frac{1}{r} e^{-Kr}. Taking \nabla^2=\frac{d^2}{dr^2}+\frac{2}{r}\frac{d}{dr}, for the option (B) we get \nabla^2\left(\frac{1}{r} e^{-Kr}\right)=\frac{K^2}{r} e^{-Kr}
Hence, option (B) is correct.
solutions are e^{-x}, e^{-2x}
Hence, answer is (A)
Friday, 25 November 2016
Problem set 31
- The two dimensional lattice of graphene is an arrangement of Carbon atoms forming a honeycomb lattice of lattice spacing a, as shown below. The Carbon atoms occupy the vertices.
- The Wigner-Seitz cell has an area of
- 2a^2
- \frac{\sqrt{3}}{2}a^2
- 6\sqrt{3}a^2
- \frac{3\sqrt{3}}{2}a^2
- The Bravais lattice for this array is a
- rectangular lattice with basis vectors \vec d_1 and \vec d_2
- rectangular lattice with basis vectors \vec c_1 and \vec c_2
- hexagonal lattice with basis vectors \vec a_1 and \vec a_2
- hexagonal lattice with basis vectors \vec b_1 and \vec b_2
- A narrow beam of X-rays with wavelength 1.5 A^\circ is reflected from an ionic crystal with an fcc lattice structure with a density of 3.32 gm cm^{-3}. The molecular weight is 108 AMU (1 AMU = 1.66 \times10^{-24} g).
- The lattice constant is
- 6.00A^\circ
- 4.56A^\circ
- 4.00A^\circ
- 2.56A^\circ
- The sine of the angle corresponding to (111)
- \sqrt{3}/4
- \sqrt{3}/8
- 1/4
- 1/8
- If an electron is in the ground state of the hydrogen atom, the probability that its distance from the proton is more than one Bohr radius is approximately
- 0.68
- 0.48
- 0.28
- 0.91
Hence, answer is (D)
The Bravais lattice for this array is a hexagonal lattice with basis vectors \vec b_1 and \vec b_2.
Hence, answer is (D)
As, N_0 molecules have molecular weight M, n molecules in the unit cell will have weight \frac{n\times M}{N_0}
{\scriptstyle\text{Density of unit cell}=\frac{\text{weight of unit cell}}{\text{volume of unit cell}}} \rho=\frac{n\times M}{N_0a^3} \begin{align*} a^3&=\frac{n\times M}{N_0\rho}\\ &=\frac{4\times 108}{6.02322\times 10^{23}\times 3320}\\ &=6A^o \end{align*} Hence, answer is (A)
According to Bragg's law 2d\sin\theta=\lambda\Rightarrow\sin\theta=\frac{\lambda}{2d} d=\frac{a}{\sqrt{h^2+k^2+l^2}} For (111) plane d=\frac{a}{\sqrt{3}} \sin\theta=\frac{\lambda\sqrt{3}}{2a}=\frac{1.5\times\sqrt{3}}{2\times6}=\frac{\sqrt{3}}{8} Hence, answer is (B)
Hence, answer is (A)
Wednesday, 23 November 2016
Problem set 30
- A particle is confined to the region x \ge 0 by a potential which increases linearly as u(x) = u_0x. The mean position of the particle at temperature T is
- \frac{k_BT}{u_0}
- \frac{(k_BT)^2}{u_0}
- \sqrt{\frac{k_BT}{u_0}}
- u_0k_BT
- A plane electromagnetic wave is propagating in a loss-less dielectric. The electric field is given by {\scriptstyle\vec E(x,y,z,t)=E_0(\hat x+A\hat z)\exp{\left[ik_0\left\{-ct+\left(x+\sqrt{3}z\right)\right\}\right]}} where c is the speed of light in vacuum, E_0 , A and k_0 are constants and \hat x and \hat z are unit vectors along the x- and z-, axes. The relative dielectric constant of the medium, \epsilon_r and the constant A are
- \epsilon_r=4 and A=-\frac{1}{\sqrt{3}}
- \epsilon_r=4 and A=+\frac{1}{\sqrt{3}}
- \epsilon_r=4 and A=\sqrt{3}
- \epsilon_r=4 and A=-\sqrt{3}
- In a system consisting of two spin-\frac{1}{2} particles labeled 1 and 2, let \vec S^{(1)} = \frac{\hbar}{2}\vec\sigma^{(1)} and \vec S^{(2)} = \frac{\hbar}{2}\vec\sigma^{(2)} denote the corresponding spin operators. Here \vec\sigma\equiv (\sigma_x ,\sigma_y ,\sigma_z) and \sigma_x ,\sigma_y ,\sigma_z are the three Pauli matrices.
- In the standard basis the matrices for the operators S^{(1)}_x S^{(2)}_y and S^{(1)}_y S^{(2)}_x are, respectively,
- {\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-1&0\\0&1\end{pmatrix}}
- {\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}i&0\\0&-i\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-i&0\\0&i\end{pmatrix}}
- {\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}},{\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}}
- {\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&-i\\0&0&i&0\end{pmatrix}},{\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&-i&0&0\\i&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}
- These two operators satisfy the relation
- {\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=S^{(1)}_z S^{(2)}_z}
- {\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=0}
- {\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=iS^{(1)}_z S^{(2)}_z}
- {\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=0}
- The radius of ^{64}_{29}Cu nucleus is measured to be 4.8\times10^{-13} cm.
- The radius of ^{27}_{12}Mg nucleus can be estimated to be
- 2.86\times10^{-13} cm
- 5.2\times10^{-13} cm
- 3.6\times10^{-13} cm
- 8.6\times10^{-13} cm
- The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number A in its ground state varies as:
- A^{4/3}
- A^{1/3}
- A^{-1/3}
- A^{-2/3}
The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number A in its ground state varies as A^{-1/3}
Hence, answer is (C)
Monday, 21 November 2016
Problem set 29
- Let \psi_{nlm} denote the eigenstates of a hydrogen atom in the usual notation. The state \frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right] is an eigenstate of
- L^2 but not of the Hamiltonian or L_z
- the Hamiltonian, but not of L^2 or L_z
- the Hamiltonian, L^2 and L_z
- L^2 and L_z, but not of the Hamiltonian
- The Hamiltonian for a spin-1/2 particle at rest is given by H=E_0(\sigma_z+\alpha\sigma_x), where \sigma_x and \sigma_z are Pauli spin matrices and E_0 and \alpha are constants. The eigenvalues of this Hamiltonian are
- \pm E_0\sqrt{1+\alpha^2}
- \pm E_0\sqrt{1-\alpha^2}
- E_0 (doubly degenerate)
- E_0\left(1\pm\frac{1}{2}\alpha^2\right)
- For a system of independent non-interacting one-dimensional oscillators, the value of the free energy per oscillator, in the limit T\rightarrow0, is
- \frac{1}{2}\hbar\omega
- \hbar\omega
- \frac{3}{2}\hbar\omega
- 0
- If the reverse bias voltage of a silicon varactor is increased by a factor of 2, the corresponding transition capacitance
- increases by a factor of \sqrt{2}
- increases by a factor of 2
- decreases by a factor of \sqrt{2}
- decreases by a factor of 2
- A cavity contains black body radiation in equilibrium at temperature T. The specific heat per unit volume of the photon gas in the cavity is of the form C_v = \gamma T^3,where \gamma is a constant. The cavity is expanded to twice its original volume and then allowed to equilibrate at the same temperature T. The new internal energy per unit volume is
- 4\gamma T^4
- 2\gamma T^4
- \gamma T^4
- \gamma T^4/4
Hence, option (B) is correct.
Saturday, 19 November 2016
Problem set 28
- The solution of the differential equation \frac{dx}{dt}=2\sqrt{1-x^2}, with initial condition x=0 at t=0 is
- \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ \sinh{2t},\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
- \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*}
- \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ 1,\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
- x=1-\cos{2t},\quad t\geq 0
- Given a uniform magnetic field \vec B=B_0\hat k (where B_0 is a constant), a possible choice for the magnetic vector potential \vec A is
- B_0y\hat i
- -B_0y\hat i
- B_0(x\hat j+y\hat i)
- B_0(x\hat i-y\hat j)
- Consider a charge Q at the origin of 3-dimensional coordinate system. The flux of the electric field through the curved surface of a cone that has a height h and a circular base of radius R is
- \frac{Q}{\epsilon_0}
- \frac{Q}{2\epsilon_0}
- \frac{hQ}{R\epsilon_0}
- \frac{QR}{2h\epsilon_0}
- A Hermitian operator \hat O has two normalised eigenstates |1 > and |2 > with eigenvalues 1 and 2, respectively. The two states |u > =\cos\theta|1 > +\sin\theta|2 > and |v > =\cos\phi|1 > +\sin\phi|2 > are such that < v|\hat O|v > =7/4 and < u|v > =0. Which of the following are possible values of \theta and \phi?
- \theta=-\frac{\pi}{6} and \phi=\frac{\pi}{3}
- \theta=\frac{\pi}{6} and \phi=\frac{\pi}{3}
- \theta=-\frac{\pi}{4} and \phi=\frac{\pi}{4}
- \theta=\frac{\pi}{3} and \phi=-\frac{\pi}{6}
- The ground state energy of a particle of mass m in the potential V(x)=V_0\cosh{\left(\frac{x}{L}\right)}, where L and V_0 are constants (and V_0>>\frac{\hbar^2}{2mL^2}) is approximately
- V_0+\frac{\hbar}{L}\sqrt{\frac{2V_0}{m}}
- V_0+\frac{\hbar}{L}\sqrt{\frac{V_0}{m}}
- V_0+\frac{\hbar}{4L}\sqrt{\frac{V_0}{m}}
- V_0+\frac{\hbar}{2L}\sqrt{\frac{V_0}{m}}
(A)\vec B=\vec\nabla \times B_0y\hat i=-B_0\hat k
(B)\vec B=\vec\nabla \times (-B_0y\hat i)=B_0\hat k
(C)\vec B=\vec\nabla \times (B_0(x\hat j+y\hat i))=0
(D)\vec B=\vec\nabla \times (B_0(x\hat i-y\hat j))=2B_0\hat k
Hence, answer is (B)
Hence, answer is (B)
Friday, 18 November 2016
Problem set 27
- The Fourier transform of f(x) is \tilde{f}(k)=\int_{-\infty}^{\infty}dx\:e^{ikx}f(x). If f(x)=\alpha\delta(x)+\beta\delta'(x)+\gamma\delta''(x), where \delta(x) is the Dirac delta-function (and prime denotes derivative), what is \tilde{f}(k)?
- \alpha+i\beta k+i\gamma k^2
- \alpha+\beta k-\gamma k^2
- \alpha-i\beta k-\gamma k^2
- i\alpha+\beta k-i\gamma k^2
- A particle moves in three-dimensional space in a central potential V(r)=kr^4, where k is a constant. The angular frequency \omega for a circular orbit depends on its radius R as
- \omega\propto R
- \omega\propto R^{-1}
- \omega\propto R^{1/4}
- \omega\propto R^{-2/3}
- The Lagrangian of a system is given by {\scriptstyle L=\frac{1}{2}m\dot q_1^2+2m\dot q_2^2-k\left(\frac{5}{4}q_1^2+2q_2^2-2q_1q_2\right)} where m and k are positive constants. The frequencies of its normal modes are
- \sqrt{\frac{k}{2m}}\sqrt{\frac{3k}{m}}
- \sqrt{\frac{k}{2m}}(13\pm\sqrt{73})
- \sqrt{\frac{5k}{2m}}\sqrt{\frac{k}{m}}
- \sqrt{\frac{k}{2m}}\sqrt{\frac{6k}{m}}
- Consider a particle of mass m moving with a speed v. If T_R denotes the relativistic kinetic energy and T_N its non-relativistic approximation, then the value of (T_R-T_N)/T_R for v=0.01\:c, is
- 1.25\times10^{-5}
- 5.0\times10^{-5}
- 7.5\times10^{-5}
- 1.0\times10^{-4}
- Two masses, m each, are placed at the points (x,y)=(a,a) and (-a,-a). Two masses, 2m each, are placed at the points (a,-a) and (-a,a). The principal moments of inertia of the system are
- 2ma^2, 4ma^2
- 4ma^2, 8ma^2
- 4ma^2, 4ma^2
- 8ma^2, 8ma^2
For a two-dimensional case, we have
{\scriptstyle{I}=\begin{pmatrix}
\sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\
-\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right)
\end{pmatrix}
}
\begin{align*}
I_{11}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)\\
&=ma^2+ma^2+2ma^2+2ma^2\\
&=6ma^2
\end{align*}
\begin{align*}
I_{12}&= -\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\
&=-(ma^2+ma^2-2ma^2-2ma^2)\\
&=2ma^2
\end{align*}
\begin{align*}
I_{21}&= -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}\\
&=-(ma^2+ma^2-2ma^2-2ma^2)\\
&=2ma^2
\end{align*}
\begin{align*}
I_{22}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right)\\
&=ma^2+ma^2+2ma^2+2ma^2\\
&=6ma^2
\end{align*}
{I}=\begin{pmatrix}
6ma^2&2ma^2\\
2ma^2&6ma^2
\end{pmatrix}
The principal moments of inertia are obtained by solving
{I}=\begin{vmatrix}
6ma^2-\lambda&2ma^2\\
2ma^2&6ma^2-\lambda
\end{vmatrix}=0
Solving this determinant we get
\lambda^2-12ma^2\lambda+32m^2a^4=0
Solving this quadratic equation in \lambda we get,
\lambda_1=8ma^2 and \lambda_2=4ma^2
Hence, answer is (B)
Wednesday, 16 November 2016
Problem set 26
- 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 is
- -\alpha^2/\beta
- -\alpha^2/2\beta
- -3\alpha^2/\beta
- -5\alpha^2/\beta
- Consider a hydrogen atom undergoing a 2P\rightarrow 1S transition. The lifetime t_{sp} of the 2P state for spontaneous emission is 1.6 ns and the energy difference between the levels is 10.2eV. Assuming that the refractive index of the medium n_0 = 1, the ratio of Einstein coefficients for stimulated and spontaneous emission B_{21}(\omega)/A_{21}(\omega) is given by
- 0.683\times10^{12} m^3J^{-1}s^{-1}
- 0.146\times10^{-12} Jsm^{-3}
- 6.83\times10^{12} m^3J^{-1}s^{-1}
- 1.46\times10^{-12} Jsm^{-3}
- In the scattering of some elementary particles, the scattering cross-section is found to depend on the total energy and the fundamental constants h (Planck’s constant) and c (the speed of light in vacuum). Using dimensional analysis, the dependence of \sigma on these quantities is given by
- \sqrt{\frac{hc}{E}}
- \frac{hc}{E^{3/2}}
- \left(\frac{hc}{E}\right)^2
- \frac{hc}{E}
- If y=\frac{1}{\tanh x}, then x is
- \ln{\left(\frac{y+1}{y-1}\right)}
- \ln{\left(\frac{y-1}{y+1}\right)}
- \ln{\sqrt{\frac{y-1}{y+1}}}
- \ln{\sqrt{\frac{y+1}{y-1}}}
- The function \frac{z}{\sin{\pi z^2}} of a complex variable z has
- a simple pole at 0 and poles of order 2 at \pm\sqrt{n} for n=1,2,3,\dots
- a simple pole at 0 and poles of order 2 at \pm\sqrt{n} and \pm i\sqrt{n} for n=1,2,3,\dots
- poles of order 2 at \pm\sqrt{n} for n=0,1,2,3,\dots
- poles of order 2 at \pm n for n=0,1,2,3,\dots
Monday, 14 November 2016
Problem set 25
- 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
- \psi_0(x)=\sqrt{sech~ x}
- \psi_0(x)=sech~ x
- \psi_0(x)=sech^2~ x
- \psi_0(x)=sech^3~ x
- The perturbation \begin{align*} H'=\begin{cases} b(a-x),&-a < x < a\\ 0&\text{otherwise} \end{cases} \end{align*} acts on a particle of mass m confined in an infinite square well potential \begin{align*} V(x)=\begin{cases} 0,&-a < x < a\\ \infty&\text{otherwise} \end{cases} \end{align*} The first order correction to the ground-state energy of the particle is
- \frac{ba}{2}
- \frac{ba}{\sqrt{2}}
- 2ba
- ba
- Let |0> and |1> 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}}\left(|0>+|1>\right) is
- \Delta x=\sqrt{\hbar/2m\omega}
- \Delta x=\sqrt{\hbar/m\omega}
- \Delta x=\sqrt{2\hbar/m\omega}
- \Delta x=\sqrt{\hbar/4m\omega}
- What would be the ground state energy of the Hamiltonian H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}-\alpha \delta(x) if variational principle is used to estimate it with the trial wavefunction \psi(x)= Ae^{-bx^2} with b as the variational parameter? [Hint: {\scriptstyle\int_{-\infty}^{\infty} x^{2n} e^{-2bx^2}\:dx = (2b)^{-n-\frac{1}{2}}\Gamma\left(n+\frac{1}{2}\right)}]
- -m\alpha^2/2\hbar^2
- -2m\alpha^2/\pi\hbar^2
- -m\alpha^2/\pi\hbar^2
- m\alpha^2/\pi\hbar^2
- A given quantity of gas is taken from the state A \rightarrow C reversibly, by two paths, A\rightarrow C
directly and A\rightarrow B\rightarrow C as shown in the figure below.
During the process A\rightarrow C the work done by the gas is 100 J and the heat absorbed is 150 J. If during the process A\rightarrow B\rightarrow C the workdone by the gas is 30 J, the heat absorbed is - 20 J
- 80 J
- 220 J
- 280 J
Saturday, 12 November 2016
Problem set 24
- The Hamiltonian of a simple pendulum consisting of a mass m attached to a massless string of length l is H = \frac{p_\theta^2}{2ml^2}+ mgl(1- \cos\theta). If L denotes the Lagrangian, the value \frac{dL}{dt} is:
- -\frac{2g}{l}p_\theta\sin\theta
- -\frac{g}{l}p_\theta\sin{2\theta}
- \frac{g}{l}p_\theta\cos{\theta}
- lp_\theta^2\cos{\theta}
- Two bodies of equal mass 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
- ml^2\omega/4
- ml^2\omega/2
- ml^2\omega
- 2ml^2\omega
- 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
- is constant inside and varies as R outside the solenoid
- varies as R inside and is constant outside the solenoid
- varies as \frac{1}{R} inside and as R outside the solenoid
- varies as R inside and as \frac{1}{R} outside the solenoid
- Consider an infinite conducting sheet in the xy-plane with a time dependent cunent density Kt\hat i, where K is a constant. The vector potential at (x, y,z) is given by \vec A=\frac{\mu_0K}{4c}(ct-z)^2\hat i. The magnetic field \vec B is
- \frac{\mu_0Kt}{2}\hat j
- -\frac{\mu_0Kz}{2c}\hat j
- -\frac{\mu_0K}{2c}(ct-z)\hat i
- -\frac{\mu_0K}{2c}(ct-z)\hat j
- 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?
- n = 1 and m=1
- n = 2 and m=2
- n = 1 and m=2
- n = 2 and m=4
Magnetic filed inside the solenoid is B=\mu_0 n I, where n is number of turns per unit length, and, I is current flowing through the wire. Magnetic filed outside the solenoid is zero.
Vector potential inside the solenoid:
Now, according to Stoke's law, we have \oint \vec A\cdot d\vec l=\int_S\left(\nabla\times\vec A\right)\cdot d\vec S But \left(\nabla\times\vec A\right)=\vec B \oint \vec A\cdot d\vec l=\int_S\vec B\cdot d\vec S=\Phi_B=flux Let us calculate flux \Phi_B. For this let us consider a circular path of radius R centered along its axis. The flux through the circular area is \Phi_B=\pi R^2B=\pi R^2\mu_0 n I Hence, \oint \vec A\cdot d\vec l=\pi R^2\mu_0 n I But, \oint \vec A\cdot d\vec l=2\pi R A_\Phi, this is because \vec A is along the direction of current which is circumferential. 2\pi R A_\Phi=\pi R^2\mu_0 n I A_\Phi=\mu_0 n I\frac{R}{2}
Vector potential outside the solenoid:
Let us consider the point R>r. The flux through circular area is \Phi_B=\pi r^2B=\pi r^2\mu_0 n I \oint \vec A\cdot d\vec l=\pi r^2\mu_0 n I But, \oint \vec A\cdot d\vec l=2\pi R A_\Phi 2\pi R A_\Phi=\pi r^2\mu_0 n I A=\mu_0 n I\frac{r^2}{2R}
Hence, \vec A varies as R inside and as \frac{1}{R} outside the solenoid
Hence, answer is (D)
Hence, answer is (C)
Thursday, 10 November 2016
Problem set 23
- A diode D as shown in the circuit has an i-V relation that can be approximated by \begin{align*} i_{_D}=\begin{cases} v^2_{_D}+2v_{_D},&\text{for }v_{_D}>0\\ 0,&\text{for }v_{_D}\leq 0 \end{cases} \end{align*} The value of v_{_D} in the circuit is
- (-1+\sqrt{11})
- 8V
- 5V
- 2V
- The Taylor expansion of the function \ln{(\cosh x)}, where x is real, about the point x= 0 starts with the following terms:
- -\frac{1}{2}x^2+\frac{1}{12}x^4+\cdots
- \frac{1}{2}x^2-\frac{1}{12}x^4+\cdots
- -\frac{1}{2}x^2+\frac{1}{6}x^4+\cdots
- \frac{1}{2}x^2+\frac{1}{6}x^4+\cdots
- Given 2\times2 unitary matrix U satisfying U^\dagger U=UU^\dagger = 1 with \det U = e^{i\phi}, one can construct a unitary matrix V \:( V^\dagger V = VV^\dagger= 1) with \det V=1 from it by
- multiplying U by e^{i\phi/2}
- multiplying any single element of U by e^{i\phi}
- multiplying any row or column of U by e^{i\phi/2}
- multiplying U by e^{i\phi}
- The value of the integral \int_C \frac{z^3dz}{z^2-5z+6}, where C is a closed contour defined by the equation 2|z|- 5 = 0, traversed in the anti-clockwise direction, is
- -16\pi i
- 16\pi i
- 8\pi i
- 2\pi i
- The function f(x) obeys the differential equation \frac{d^2f}{dx^2}-(3 - 2i)f = 0 and satisfies the conditions f(0) = 1 and f(x)\rightarrow \infty as x\rightarrow 0. The value of f(\pi) is
- e^{2\pi}
- e^{-2\pi}
- -e^{-2\pi}
- -e^{2\pi i}
Hence, answer is (D)
Let us take option (A) i.e. multiplying U by e^{-i\phi/2} \begin{equation*} V=e^{-i\phi}U=\begin{pmatrix} e^{-i\phi/2}a&e^{-i\phi/2}b\\ -e^{i\phi/2}b^*&e^{i\phi/2}a^* \end{pmatrix} \end{equation*} Clearly, we get, \det{V}=1.
Hence, option (A) is correct
Wednesday, 9 November 2016
Problem set 22
- Given the usual canonical commutation relations, the commutator [A,B] of A=i(xp_y-yp_z) and B=i(yp_z+zp_y) is
- \hbar(xp_z-p_xz)
- -\hbar(xp_z-p_xz)
- \hbar(xp_z+p_xz)
- -\hbar(xp_z+p_xz)
- The entropy of a system, S, is related to the accessible phase space volume \Gamma by S = k\ln \Gamma(E, N,V) where E, N and V are the energy, number of particles and volume respectively. From this one can conclude that \Gamma
- does not change during evolution to equilibrium
- oscillates during evolution to equilibrium
- is a maximum at equilibrium
- is a minimum at equilibrium
- Let \Delta W be the work done in a quasistatic reversible thermodynamic process. Which of the following statements about \Delta W is correct?
- \Delta W is a perfect differential if the process is isothermal
- \Delta W is a perfect differential if the process is adiabatic
- \Delta W is always a perfect differential
- \Delta W cannot be a perfect differential
- Consider a system of three spins S_1, S_2 and S_3 each of which can take values +1 and -1. The energy of the system is given by E = -J\left[ S_1 S_2 + S_2 S_3 + S_3 S_1\right], where J is a positive constant. The minimum energy and the corresponding number of spin configurations are, respectively,
- J and 1
- -3J and 1
- -3J and 2
- -6J and 2
- The minimum energy of a collection of 6 non-interacting electrons of spin-\frac{1}{2} and mass m placed in a one dimensional infinite square well potential of width L is
- 14\pi^2\hbar^2/mL^2
- 91\pi^2\hbar^2/mL^2
- 7\pi^2\hbar^2/mL^2
- 3\pi^2\hbar^2/mL^2
According to first law of thermodynamics dU=dQ+dW Here dU is exact, but, dQ and dW are inexact. However, for adiabatic process dQ=0. Hence, we have dU=dW. As dU is always exact, for adiabatic process, dW is exact.
Hence answer (B) is correct.
Monday, 7 November 2016
Problem set 21
- 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 > > a), the potential is proportional to
- 1/R
- 1/R^2
- 1/R^3
- 1/R^4
- 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. For what value of d will the charge remains stationary?
- q/4\sqrt{mg\pi\epsilon_0}
- q/\sqrt{mg\pi\epsilon_0}
- There is no finite value of d
- \sqrt{mg\pi\epsilon_0}/q
- The wave function of a state of the hydrogen atom is given by {\scriptstyle\Psi=\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}}, where \psi_{nlm} is the normalized eigenfunction 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
- 15\hbar/16
- 11\hbar/16
- 3\hbar/8
- \hbar/8
- The energy eigenvalues of a particle in the potential V(x) =\frac{1}{2}m\omega^2x^2-ax are
- E_n=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{2m\omega^2}
- E_n=\left(n+\frac{1}{2}\right)\hbar\omega+\frac{a^2}{2m\omega^2}
- E_n=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{m\omega^2}
- E_n=\left(n+\frac{1}{2}\right)\hbar\omega
- If a particle is represented by the normalized wave function \begin{align*} \psi(x)=\begin{cases}\frac{\sqrt{15}\left(a^2-x^2\right)}{4a^{5/2}}&{\scriptstyle\text{for} -a < x < a}\\ 0 &\text{otherwise} \end{cases} \end{align*} the uncertainty \Delta p in its momentum is
- 2\hbar/5a
- 5\hbar/2a
- \sqrt{10}\hbar/a
- \sqrt{5}\hbar/\sqrt{2}a
For monopole E\propto 1/r^2 and V\propto 1/r
For dipole E\propto 1/r^3 and V\propto 1/r^2
For quadrupole E\propto 1/r^4 and V\propto 1/r^3
For octapole E\propto 1/r^5 and V\propto 1/r^4
Hence, answer is (C).
Sunday, 6 November 2016
Problem set 20
- Let v, p and E denote the speed, the magnitude of the momentum, and the energy of a free particle of rest mass m. Then
- dE/dp=constant
- p=mv
- v=cp/\sqrt{p^2+m^2c^2}
- E=mc^2
- A binary star system consists of two stars S_1 and S_2, with masses m and 2m, respectively, separated by a distance r. If both S_1 and S_2 individually follow circular orbits around the centre of mass with instantaneous speeds V_1 and V_2 respectively, the speeds ratio V_1/V_2 is
- \sqrt{2}
- 1
- 1/2
- 2
- Three charges are located on the circumference of a circle of radius R as shown in the figure below. The two charges Q subtend an angle 90^o at the centre of the circle. The charge q is symmetrically placed with respect to the charges Q. If the electric field at the centre of the circle is zero, what is the magnitude of Q?
- q/\sqrt{2}
- \sqrt{2}q
- 2q
- 2q
- Consider a hollow charged shell of inner radius a and outer radius b. The volume charge density is \rho(r) =\frac{k}{r^2} (k is a constant) in the region a < r < b. The magnitude of the electric field produced at distance r > a is
- \frac{k(b-a)}{\epsilon_0r^2} for all r > a
- \frac{k(b-a)}{\epsilon_0r^2} for a < r < b and \frac{kb}{\epsilon_0r^2} for r > b
- \frac{k(r-a)}{\epsilon_0r^2} for a < r < b and \frac{k(b-a)}{\epsilon_0r^2} for r > b
- \frac{k(r-a)}{\epsilon_0a^2} for a < r < b and \frac{k(b-a)}{\epsilon_0a^2} for r > b
- 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+\phi)} and \vec E_2 = \hat j E_0 \cos{(\omega t+\phi)} where \phi is the phase difference. The intensity of the resulting wave is given by \frac{\epsilon_0}{2}\left < E^2\right >, where \left < E^2\right > is the time average of E^2. The total intensity is
- 0
- \epsilon_0E_0^2
- \epsilon_0E_0^2\sin^2{\phi}
- \epsilon_0E_0^2\cos^2{\phi}
The rest mass energy and kinetic energy of particle are mc^2 and K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2, respectively.
Here, option (D) is wrong, because, total energy E, is sum of kinetic energy plus rest mass energy i.e. E=\frac{mc^2}{\sqrt{1-v^2/c^2}}.
Option (A) is wrong, because, velocity is involved in the expression for E.
For region a\leq r\leq b, using Gauss law, we have, \int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0} \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^r\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^r\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(r-a) \end{align*} E=\frac{k(r-a)}{\epsilon_0 r^2} For region r > b, using Gauss law, we have, \int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0} \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^b\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^b\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(b-a) \end{align*} E=\frac{k(b-a)}{\epsilon_0 r^2} Hence, answer is (C)