Problem set 90

1. A plane polarized EM wave of frequency $\omega$ is incident at an angle $\theta$ in a rectangular waveguide of resonant frequency $\omega_{mn}$. Then energy carried by the wave propagating inside the cavity will propagate with the group velocity of :
1. $\frac{c}{\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}}$
2. $c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)}$
3. $\frac{c}{\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)}}$
4. $c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}$

$$v_g=c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}$$

Hence, answer is (D)

2. The electric field of an electromagnetic wave propagating in the free space is given by : $\vec E(r,t)=E_0\hat z\cos{\left[200\sqrt{3}\pi x-200\pi y-\omega t\right]}$. Then the wave vector $\vec k$ is given by
1. $200\frac{\sqrt{3}}{2}\pi\hat x-200\pi\hat y$
2. $400\pi\left[\frac{\sqrt{3}}{2}\hat x-\frac{1}{2}\hat y\right]$
3. $200\sqrt{3}\pi\hat x$
4. $-200\pi\hat y$

General format of electromagnetic wave equation is $$\vec E(r,t)=E_0\hat z\cos{(\vec k\cdot\vec r-\omega t)}$$ $$\vec k\cdot\vec r= 200\sqrt{3}\pi x-200\pi y$$ $$\Rightarrow \vec k=400\pi\left[\frac{\sqrt{3}}{2}\hat x-\frac{1}{2}\hat y\right]$$

Hence, answer is (B)

3. The Ampere's law in the free space takes the form:
1. $\vec\nabla\times\vec B=\mu_0\vec J$
2. $\vec\nabla\times\vec B=\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$
3. $\vec\nabla\times\vec B=\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$
4. $\vec\nabla\times\vec B=\mu_0\vec J-\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$

$\vec\nabla\times\vec B=\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$

Hence, answer is (B)

4. An electric charge $+Q$ is placed at the center of a cube of sides 10 cm. The electric flux emanating from each of the face of the cube is :
1. $\frac{Q}{\epsilon_0}$
2. $\frac{Q}{10\epsilon_0}$
3. $\frac{Q}{6\epsilon_0}$
4. $\frac{10Q}{\epsilon_0}$

According to Gauss’ Theorem, the total electric flux through all the six faces of the cube is, $$\Phi = \frac{Q}{\epsilon_0}$$ Each face of the cube comprises $1/6$ of the total surface of the cube. Therefore, the flux passing through each face is, $$\Phi = \frac{Q}{6\epsilon_0}$$

Hence, answer is (C)

5. A field at certain point in the space is expressed as the potential function $V=3x^2z-xy^3+z$. Then the potential $V$ at point $(2,-1,1)$ is :
1. 15 V
2. 13 V
3. 0 V
4. 8 V

$$V=32^2\times1-2\times(-1)^3+1=15 V$$

Hence, answer is (A)

Problem set 89

1. A particle is at rest in rotating frame of reference. The pseudoforce(s) acting on the particle is(are)
1. None of these
2. Only the Coriolis force
3. only the centrifugal force
4. both Coriolis force and centrifugal force

Coriolis force is given by $F_{Co}=-2m(\vec\omega\times\vec v')$

Centrifugal force is given by $F_{c}=-2m\vec\omega\times(\vec\omega\times\vec r)$

Hence, in rotating frame, if particle is at rest only centrifugal force act on the particle. If particle is not at rest both forces act on the particle.

Hence, answer is (C)

2. A thin rigid rod of length $l$ is moving inside a sphere of radius $R(R>l)$ such that both of its ends are in contact with inner surface of the sphere. The degrees of freedom of the rod are:
1. Four
2. Three
3. Two
4. One

As both ends of the rigid rod are in contact with inner surface, the position of any one end can be determined with $\theta$ and $\phi$ coordinates with respect to coordinate system fixed at the center of the sphere. To determine position of other end we will require one more coordinate i.e. orientation of rod with respect to some fixed direction. Hence, degrees of freedom is three.

Hence, answer is (B)

3. For a system shown in figure given below, the Lagrangian function is given by : ($V=0$ at $y=0$)
   
1. $L=M\dot y^2+Mgy$
2. $L=\frac{1}{2}M\dot y^2+Mgy$
3. $L=\frac{1}{2}M\dot y^2-Mg(y-x)$
4. $L=M\dot y^2+Mg(y-x)$

$$L=T-V$$ $$T=\frac{1}{2}M\dot x^2+\frac{1}{2}M\dot y^2$$ But $\dot x=\dot y$ $$T=M\dot y^2$$ $$V=-Mgy$$ $$L=M\dot y^2+Mgy$$

Hence, answer is (A)

4. "Hamiltonian $H$ is not equal to the total energy $E$ (sum of kinetic and potential energies)", hold true for a system characterized with:
1. conservative forces and time-independent constraints
2. conservative forces and time-dependent constraints
3. dissipative forces and time-independent constraints
4. for every system irrespective of nature of forces and constraints

dissipative forces and time-independent constraints

Hence, answer is (C)

5. A particle moves under the action of force $\vec F=-\frac{1}{r^n}\hat r$. The particle moves in a closed orbit if:
1. n=-1 or n=2
2. n=1 or n=-2
3. n=-1 or n=-2
4. n=1 or n=2

According to Bertrand's theorem the central force of the form $F=\frac{-k}{r^{3-\beta^2}}$ can produce stable, closed, non-circular orbits if it satisfies the condition $\beta^2(\beta^2-4)(\beta^2-1)=0$. Hence, this condition is satisfied for $\beta=0,1,2$ i.e. $n=3-\beta^2=3,2,-1$

Hence, answer is (A)

Problem set 88

1. The moment of inertia of a thin disc of radius $R$ about an axis passing through its center and perpendicular to the plane of disc is:
1. $MR^2$
2. $\frac{2}{3}MR^2$
3. $\frac{3}{2}MR^2$
4. $\frac{1}{2}MR^2$

$\frac{1}{2}MR^2$

Hence, answer is (D)

2. A coin is tossed four times what is the probability of getting two heads and two tails?
1. $\frac{3}{8}$
2. $\frac{1}{2}$
3. $\frac{5}{8}$
4. $\frac{3}{4}$

When a coin is tossed once, there are two out comes possible. Hence, the coin is tossed four times, the total number of outcomes are $2^4=16$. Out of which we will two heads exactly $\frac{4!}{2!(4-2)!}=6$ times. Hence, probability of getting two heads and two tails is $\frac{6}{16}=\frac{3}{8}$

Hence, answer is (A)

3. Consider three vectors $\vec a=\hat i+\hat j+\hat k$, $\vec b=\hat i-\hat j+\hat k$ and $\vec c=\hat i-\hat j-\hat k$. Which of the following statement is true?
1. $\vec a$, $\vec b$, $\vec c$ are linearly independent
2. $\vec a$, $\vec b$ are linearly independent
3. $\vec b$ and $\vec c$ are right angle to each other
4. $\vec a$ and $\vec c$ are parallel

The vectors in a subset $S = \{ \vec v_1 ,\vec v_2 ,\cdots,\vec v_n \}$ of a vector space $V$ are said to be linearly dependent, if there exist a finite number of distinct vectors $ \vec {v}_{1},\vec {v}_{2},\dots ,\vec {v}_{k}$ in $S$ and scalars $a_{1},a_{2},\dots ,a_{k}$, not all zero, such that $$a_{1}\vec {v}_{1}+a_{2}\vec {v}_{2}+\cdots +a_{k}\vec {v}_{k}=\vec {0}$$ where zero denotes the zero vector. Let $$a_1\vec a+a_2\vec b+a_3\vec c=0$$ $$\begin{bmatrix}1&1&1\\1&-1&-1\\1&1&-1\end{bmatrix}\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ If $\vec a$, $\vec b$ and $\vec c$ are linearly independent, then the only solution to this system of equations is the trivial solution, $a_1=a_2=a_3=0$. For homogeneous systems this happens precisely when the determinant is non-zero. If determinant is zero then vectors are linearly dependent. $$\begin{vmatrix}1&1&1\\1&-1&-1\\1&1&-1\end{vmatrix}=4$$ Hence, $\vec a$, $\vec b$ and $\vec c$ are linearly independent.

Hence, answer is (A)

4. Which of the following defines a conservative force field?
1. $\vec\nabla\cdot\vec F=0$
2. $\vec\nabla\times\vec F=0$
3. $\oint\vec F\cdot\vec dr=0$
4. $\frac{d\vec F}{dt}=0$

A force field $\vec F$, is called a conservative force or conservative vector field if it satisfies following three equivalent conditions: $$\vec\nabla\times\vec F=0$$ $$\oint\vec F\cdot\vec dr=0$$ $$\vec F=-\vec\nabla\Phi$$ where $\Phi$ is scalar potential.

Hence, answer is (B and C)

5. $\nabla\left(\frac{1}{|\vec r|}\right)$ is given by
1. $\frac{1}{r}\hat r$
2. $\frac{1}{r^3}(\hat i+\hat j+\hat k)$
3. $-\frac{\vec r}{r^3}$
4. $r(\hat i-\hat j-\hat k)$

\begin{align*} \nabla\left(\frac{1}{|\vec r|}\right)&={\scriptstyle \left(\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z}\right)\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)}\\ &={\scriptstyle -\left(\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\right)\left(x\hat i+y\hat j+z\hat k\right)}\\ &=-\frac{\vec r}{r^3}\\ \end{align*}

Hence, answer is (C)

Problem set 87

1. Eigenvalues of the matrix $\begin{bmatrix}1&-1\\1&1\end{bmatrix}$
1. 1, -1
2. -1, $-i$
3. $i$, $-i$
4. $1+i$, $1-i$

Determinant of matrix is equal to product of eigenvalues. $$\begin{vmatrix}1&-1\\1&1\end{vmatrix}=2$$ Clearly $$(1+i)(1-i)=2$$

Hence, answer is (D)

2. Consider an $n$-MOSFET with the following parameters: current drive strength $K= 60 \:\mu A/V^2$, breakdown voltage $BV_{DS}=10\: V$, ratio of effective gate width to the channel length $\frac{W}{L}=5$ and threshold voltage $V_{th}=0.5V$. In the circuit given below, this $n$-MOSFET is operating in the
   
1. ohmic region
2. cut-off region
3. saturation region
4. breakdown region

The MOSFET is in

saturation when $V_{GS} > V_{th}$ and $V_{DS} > V_{GS} - V_{th}$.

cut-off region when $V_{GS} < V_{th}$

linear when $V_{GS} > V_{th}$ and $V_{DS} < V_{GS} - V_{th}$.

$V_{GS}=1.2 V$, $V_{th}=0.5V$, $V_{DS}=10 V$

Hence, answer is (C)

3. Particular integral of first order linear differential $\frac{dy}{dx}=x+y$ is given by:
   1. $y(x)=-x-1$
   2. $y(x)=x+1$
   3. $y(x)=x-1$
   4. $y(x)=-x+1$
   Linear differential equation is given by $$\frac{dy}{dx}+P(x)y=Q(x)$$ Multiply both sides by the Integrating Factor $e^{\int P(x)dx}$ $${\scriptstyle \frac{dy}{dx}e^{\int P(x)dx}+P(x)ye^{\int P(x)dx}=Q(x)e^{\int P(x)dx}}$$ $$\frac{d\left(ye^{\int P(x)dx}\right)}{dx}=Q(x)e^{\int P(x)dx}$$ On integrating we get $$ye^{\int P(x)dx}=\int Q(x)e^{\int P(x)dx}dx+c$$ $$\frac{dy}{dx}-y=x$$ $P(x)=-1$, $Q(x)=x$ $$ye^{-\int dx}=\int xe^{-\int dx}dx+c$$ $$ye^{-x}=\int xe^{-x}dx+c$$ Integrating by parts $$ye^{-x}=-xe^{-x}+\int e^{-x}dx+c$$ $$ye^{-x}=-xe^{-x}-e^{-x}+c$$ $$y=-x-1$$

   Hence, answer is (A)

4. The Fourier transform of a Gaussian function is of the form:
1. Exponential
2. Lorentzian
3. Gaussian
4. Screened coulomb

Hence, answer is (C)

5. The real part of $\log{(3+4i)}$ is :
1. $\log2$
2. $\log3$
3. $\log4$
4. $\log5$

Let $z=x+iy$, $r=\sqrt{x^2+y^2}$, $\theta=\tan^{-1}{\frac{y}{x}}$ and $z=re^{i\theta}$

For $z=3+4i$, $r=\sqrt{3^2+16^2}=5$, $\theta=\tan^{-1}{\frac{4}{3}}=0.92729$ and $z=5e^{i0.92729}$ \begin{align*} \log{(3+4i)}&=\log{5e^{i0.92729}}\\ &=\log5+i0.92729 \end{align*}

Hence, answer is (D)

Problem set 86

1. A function $f(x)$ satisfies the differential equation $\frac{d^2f}{dx^2}-\omega^2f=-\delta(x-a)$, where $\omega$ is positive. The Fourier transform $\tilde{f}(k)=\int_{-\infty}^{\infty}dx\:e^{ikx}f(x)$ of $f$, and the solution of the equation are, respectively,
1. $\frac{e^{ika}}{k^2+\omega^2}$ and $\frac{1}{2\omega}\left(e^{-\omega|x-a|}+e^{\omega|x-a|}\right)$
2. $\frac{e^{ika}}{k^2+\omega^2}$ and $\frac{1}{2\omega}e^{-\omega|x-a|}$
3. $\frac{e^{ika}}{k^2-\omega^2}$ and $\frac{1}{2\omega}\left(e^{-i\omega|x-a|}+e^{i\omega|x-a|}\right)$
4. $\frac{e^{ika}}{k^2-\omega^2}$ and $\frac{1}{2i\omega}\left(e^{-i\omega|x-a|}-e^{i\omega|x-a|}\right)$
$$\frac{d^2f(x)}{dx^2}-\omega^2f(x)=-\delta(x-a)$$ Taking Fourier transform $$F\!\!\left[\frac{d^2f(x)}{dx^2}\right]\!\!-\!\!F\!\left[\omega^2f(x)\right]\!\!=\!\!-F\!\left[\!\delta(\!x\!-\!a\!)\!\right]$$ $$F\left[\frac{d^2f(x)}{dx^2}\right]=-k^2f(k)$$ $$F\left[\delta(x-a)\right]=e^{ika}$$ $$-k^2f(k)-\omega^2f(k)=-e^{ika}$$ Hence, Fourier transform of $f(x)$ is $$f(k)=\frac{e^{ika}}{k^2+\omega^2}$$ Taking Inverse Fourier transform $$f(x)=F^{-1}\left[\frac{e^{ika}}{k^2+\omega^2}\right]$$ $$f(x)=\frac{1}{2\omega}F^{-1}\left[e^{ika}\frac{2\omega}{k^2+\omega^2}\right]$$ $$f(x)=\!\!\int F^{-1}\left[\frac{e^{ika}}{2\omega}\right]F^{-1}\left[\frac{2\omega}{k^2+\omega^2}\right]$$ $$F^{-1}\left[e^{ika}\right]=\delta(x-a)$$ $$F^{-1}\left[\frac{2\omega}{k^2+\omega^2}\right]=e^{-\omega |x|}$$ Using Convolution theorem, $$f(x)=\frac{1}{2\omega}\int \delta(k-a)e^{-\omega |x-k|}dk$$ $$f(x)=\frac{1}{2\omega}e^{-\omega |x-a|}$$

Hence, answer is (B)

2. Let $E_s$ denote the contribution of the surface energy per nucleon in the liquid drop model. The ratio $E_s\left(^{27}_{13}Al\right):E_s\left(^{64}_{30}Al\right)$ is
1. 2:3
2. 4:3
3. 5:3
4. 3:2

Surface energy $$E_s\propto A^{2/3}$$ Surface energy per nucleon is $$E_s\propto \frac{A^{2/3}}{A}=\frac{1}{A^{1/3}}$$ $$\frac{E_s\left(^{27}_{13}Al\right)}{E_s\left(^{64}_{30}Al\right)}=\frac{64^{1/3}}{27^{1/3}}=\frac{4}{3}$$

Hence, answer is (B)

3. According to the shell model, the nuclear magnetic moment of the $^{27}_{13}Al$ nucleus is (Given that for a proton $g_l=1$, $g_s=5.586$, and for a neutron $g_l=0$, $g_s=-3.826$)
1. $-1.913\mu_N$
2. $14.414\mu_N$
3. $4.793\mu_N$
4. 0

In $^{27}_{13}Al$ there are odd numbers of protons (13) and even numbers of neutrons (14). The unpaired proton is responsible for magnetic moment. $^{27}_{13}Al$ has an odd proton in the $j = 5/2 1d (l=2)$ state. Magnetic moment for single nucleon is given by $$\mu_j=\left(g_ll+\frac{1}{2}g_s\right)\mu_N$$ $$\mu_j=\left(1\times2+\frac{1}{2}5.586\right)\mu_N$$ $$\mu_j=4.793\mu_N$$

Hence, answer is (C)

4. The ground state electronic configuration of $^{22}Ti$ is $[Ar]3d^54s^2$. Which state, in the standard spectroscopic notations, is not possible in this configuration?
1. $^1F_{3}$
2. $^1S_{0}$
3. $^1D_{2}$
4. $^3P_{0}$

The rules governing the term symbol for the ground state according to L-S coupling scheme are given below:

1. If all shells and subshells are full then the term symbol is $^1S_0$.
2. The spin multiplicity is maximized i.e., the electrons occupy degenerate orbitals so as to retain parallel spins as long as possible (Hund’s rule).
3. The orbital angular momentum is also maximized i.e., the orbitals are filled with highest positive $m_l$ values first.
4. The overall $S$ is calculated by multiplying $\frac{1}{2}$ times the number of unpaired electrons. The overall $L$ is calculated by adding the $m_l$ values for each electron (so if there are two electrons in the same orbital, add twice that orbital's $m_l$).
5. If the sub-shell is less than half-filled, $J = L– S$ and if the sub-shell is more than half – filled, $J = L +S$.

The term symbol is given by $^{2S+1}L_J$. The left-hand superscript of the term is the spin multiplicity, given by $2S+1$ and the right- hand subscript is given by $J$.
The electron configuration of Ti is $1s^22s^22p^63s^23p^63d^24s^2$. For $d^2$ configuration we have

	$\uparrow$	$\uparrow$
$m_l=$	+2	+1	0	-1	-2

Hence, \begin{align*} L&=(+2)+(+1)=3\\ \end{align*} i.e. $F$ state \begin{align*} S&=(+1/2)+(+1/2)=1\\ \end{align*} $$2S+1=3$$ $$J=L-S=3-1=2$$ Term symbol for Ti is $^3F_{2}$

Let us determine which states are possible

If we have

	$\uparrow\downarrow$
$m_l=$	+2	+1	0	-1	-2

We get $^1S_0$ state

If we have

		$\uparrow\downarrow$
$m_l=$	+2	+1	0	-1	-2

We get $^1D_2$ state

If we have

	$\uparrow$			$\uparrow$
$m_l=$	+2	+1	0	-1	-2

We get $^3P_0$ state

For $^1F_3$ state we must have $S=0$, $L=3$ and $J=3$. However, it is not possible to arrange electrons to get $S=0$, $L=3$ and $J=3$.

Hence, answer is (A)

5. The band energy of an electron in a crystal for a particular $k$-direction has the form $\epsilon(k)=A-B\cos{2ka}$, where $A$ and $B$ are positive constants and $0 < ka <\pi$. The electron has a hole-like behaviour over the following range of $k$:
1. $\frac{\pi}{4} < ka < \frac{3\pi}{4}$
2. $\frac{\pi}{2} < ka < \pi$
3. $0 < ka < \frac{\pi}{4}$
4. $\frac{\pi}{2} < ka < \frac{3\pi}{4}$

The electron has a hole-like behaviour when it has negative effective mass. $$m*=\frac{\hbar^2}{\frac{\partial^2\epsilon}{\partial k^2}}=4a^2\cos{2ka}$$ Clearly, $m*$ will be negative in the range $\frac{\pi}{4} < ka < \frac{3\pi}{4}$

Hence, answer is (A)

Problem set 85

1. In a normal Zeeman effect experiment using a magnetic field of strength 0.3 T, the splitting between the components of a 660 nm spectral line is
1. 12 pm
2. 10 pm
3. 8 pm
4. 6 pm

$$\Delta\lambda=\frac{\mu_BB\lambda^2}{hc}$$ $$\Delta\lambda={\scriptstyle \frac{9.274\times10^{-24}\times0.3\times(660\times10^{-9})^2}{6.626\times10^{-34}\times3\times10^8}}$$ $$\Delta\lambda=6 pm$$

Hence, answer is (D)

2. What is the Fourier transform $\int dxe^{ikx}f(x)$ of $f(x)=\delta(x)+\sum\limits_{n=1}^\infty\frac{d^n}{dx^n}\delta(x)$, where $\delta(x)$ is the Dirac delta-function?
1. $\frac{1}{1-ik}$
2. $\frac{1}{1+ik}$
3. $\frac{1}{k+i}$
4. $\frac{1}{k-i}$

Using identity $\int dx f(x)\delta^{(n)}(x-y)=(-1)^nf^{(n)}(y)$, where $f^{(n)}(y)$ is $n^{th}$ derivative of $f$. \begin{align*} &\int\!\! dxe^{ikx}\!f(x)\!=\!{\scriptstyle \int dxe^{ikx}\left(\delta(x)+\sum\limits_{n=1}^\infty\frac{d^n}{dx^n}\delta(x)\right)}\\ &={\scriptstyle \int dxe^{ikx}\delta(x)+\int dxe^{ikx}\sum\limits_{n=1}^\infty\frac{d^n}{dx^n}\delta(x)}\\ &={\scriptstyle 1+\int dxe^{ikx}\left(\delta^{(1)}(x)+\delta^{(2)}(x)+\delta^{(3)}(x)+\dots\right)}\\ &={\scriptstyle 1+(-1)^1ik+(-1)^2(ik)^2+(-1)^3(ik)^3+\cdots}\\ &={\scriptstyle 1-ik-k^2+ik^3+k^4-ik^5\cdots}\\ &={\scriptstyle \left(1-k^2+k^4-k^6+\cdots\right)-ik+ik^3-ik^5\cdots}\\ &={\scriptstyle \left(1-k^2+k^4-k^6+\cdots\right)-ik\left(1-k^2+k^4-k^6+\cdots\right)}\\ &={\scriptstyle \left(1-k^2+k^4-k^6+\cdots\right)(1-ik)}\\ &={\scriptstyle \left(1+(ik)^2+(ik)^4+(ik)^6+\cdots\right)(1-ik)}\\ &=\frac{1}{1-(ik)^2}(1-ik)\\ &=\frac{1}{1+ik} \end{align*}

Hence, answer is (B)

3. A canonical transformation $(q,p)\rightarrow (Q,P)$ is made through the generating function $F(q,P)=q^2P$ on the Hamiltonian $H(q,p)=\frac{p^2}{2\alpha q^2}+\frac{\beta}{4}q^4$ where $\alpha$ and $\beta$ are constants. The equations of motion for $(Q,P)$ are
1. $\dot Q=P/\alpha$ and $\dot P=-\beta Q$
2. $\dot Q=4P/\alpha$ and $\dot P=-\beta Q/2$
3. $\dot Q=P/\alpha$ and $\dot P=-\frac{2P^2}{Q}-\beta Q$
4. $\dot Q=2P/\alpha$ and $\dot P=-\beta Q$

The generating function may be any of the following types: $F_1(q,Q)$, $F_2(q,P)$, $F_3(p,Q)$, $F_4(p,P)$. For these generating functions we have following relations between generating function and canonical variables

Generating function	Relations
$F_1(q,Q)$	$p_i=\frac{\partial F_1}{\partial q_i}$, $P_i=-\frac{\partial F_1}{\partial Q_i}$, $K=H+\frac{\partial F_1}{\partial t}$
$F_2(q,P)$	$p_i=\frac{\partial F_2}{\partial q_i}$, $Q_i=\frac{\partial F_2}{\partial P_i}$, $K=H+\frac{\partial F_2}{\partial t}$
$F_3(p,Q)$	$q_i=-\frac{\partial F_3}{\partial p_i}$, $P_i=-\frac{\partial F_3}{\partial Q_i}$, $K=H+\frac{\partial F_3}{\partial t}$
$F_4(p,P)$	$q_i=-\frac{\partial F_4}{\partial p_i}$, $Q_i=\frac{\partial F_4}{\partial P_i}$, $K=H+\frac{\partial F_4}{\partial t}$

In the present problem generating function $F(q,P)=q^2P$ is of second kind. Hence, $$p=\frac{\partial F}{\partial q}=2qP$$ $$Q=\frac{\partial F}{\partial P}=q^2$$ \begin{align*} K&=H+\frac{\partial F_2}{\partial t}=H\\ &=\frac{p^2}{2\alpha q^2}+\frac{\beta}{4}q^4\\ K&=\frac{2P^2}{\alpha}+\frac{\beta}{4}Q^2\\ \end{align*} Hence, equations of motions are $$\dot P=-\frac{\partial K}{\partial Q}=-\frac{\beta}{2}Q$$ $$\dot Q=\frac{\partial K}{\partial P}=\frac{4P}{\alpha}$$

Hence, answer is (B)

4. The internal energy $E(T)$ of a system at a fixed volume is found to depend on the temperature $T$ as $E(T)=aT^2+bT^4$. Then the entropy $S(T)$, as a function of temperature, is
1. $\frac{1}{2}aT^2+\frac{1}{4}bT^4$
2. $2aT^2+4bT^4$
3. $2aT+\frac{4}{3}bT^3$
4. $2aT+2bT^3$

$$dU=TdS-PdV$$ At constant volume $PdV=0$ $$dU=TdS$$ \begin{align*} S&=\int\frac{1}{T}dU\\ &=\int\frac{1}{T}dE\\ &=\int\frac{1}{T}\left(2aT\:dT+4bT^3dT\right)\\ &=\int\left(2a\:dT+4bT^2dT\right)\\ &=2aT+\frac{4}{3}bT^3 \end{align*}

Hence, answer is (C)

5. Consider a gas of Cs atoms at a number density of $10^{12}$ atoms/cc. When the typical inter-particle distance is equal to the thermal de Broglie wavelength of the particles, the temperature of the gas is nearest to (Take the mass of a Cs atom to be $22.7\times10^{-26}$ kg.)
1. $1\times10^{-9}$ K
2. $7\times10^{-5}$ K
3. $1\times10^{-3}$ K
4. $2\times10^{-8}$ K

Density of particles $n=\frac{N}{V}$. Hence, inter-particle separation is given by $$d=\left(\frac{V}{N}\right)^{\frac{1}{3}}=\left(\frac{1}{n}\right)^{\frac{1}{3}}$$ $$d=\left(\frac{1}{10^{12}}\right)^{\frac{1}{3}}=10^{-4}cm=10^{-6}m$$ $$\lambda_{th}=10^{-6}m$$ $$\lambda_{th}=\frac{h}{\sqrt{2\pi mk_BT}}$$ \begin{align*} T&=\frac{h^2}{2\pi mk_B\lambda_{th}^2}\\ &={\scriptstyle \frac{(6.626\times10^{-34})^{2}}{2\times3.14\times22.7\times10^{-26}\times 1.381 \times 10^{-23}\times10^{-12}}}\\ &=2.23009664\times10^{-8} K \end{align*}

Hence, answer is (D)

Problem set 84

1. In the schematic figure given below, assume that the propagation delay of each logic gate is $t_{gate}$.
   
   The propagation delay of the circuit will be maximum when the logic inputs A and B make the transition
1. $(0,1)\rightarrow(1,1)$
2. $(1,1)\rightarrow(0,1)$
3. $(0,0)\rightarrow(1,1)$
4. $(0,0)\rightarrow(0,1)$

When input of gate changes its state, the output of gate does not change instantaneously. Instead, the out put changes after a small delay. The delay in the circuit due to all gates is called propagation delay. The states of each gate for different inputs are as shown below.



During transition $(0,1)\rightarrow(1,1)$ from figures (b) and (c) we see that only first OR gate, AND gate, and second OR change their states. Hence, time delay is $3t_{gate}$

During transition $(1,1)\rightarrow(0,1)$ from figures (c) and (b) we see that only first OR gate, AND gate, and second OR change their states. Hence, time delay is $3t_{gate}$

During transition $(0,0)\rightarrow(1,1)$ from figures (a) and (c) we see that only NOT gate changes its state. Hence, time delay is $t_{gate}$

During transition $(0,0)\rightarrow(0,1)$ from figures (a) and (b) we see that all gates change their states. Hence, time delay is $4t_{gate}$

Hence, answer is (D)

2. Given the input voltage $V_i$, which of the following waveforms correctly represents the output voltage $V_0$ in the circuit shown below?
   
1. 
2. 
3. 
4. 

Output voltage is given by $$V_0=A(V_2-V_1)$$ where, $V_1$ and $V_2$ are voltages at inverting and non-inverting inputs and $A$ is gain given by $$A=\frac{R_f}{R_i}=\frac{10}{5}=2$$ $$V_0=2(V_2-V_1)$$ $$V_0=A(0.5-V_1)$$ If $V_1>V_2$, $V_0$ will be negative, but no option have negative voltage.

Hence, $V_1\leq V_2$

Since, $V_2=0.5\: V$, maximum value of $V_1=0.5$, hence, minimum output voltage is $V_{0_{min}}=0$

Also, this is inverting amplifier.

Hence, answer is (B)

3. In finding the roots of the polynomial $f(x)=3x^3-4x-5$ using the iterative Newton-Raphson method, the initial guess is taken to be $x=2$. In the next iteration its value is nearest to
1. 1.671
2. 1.656
3. 1.559
4. 1.551

Newton-Raphson formula is \begin{align*} x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\\ &=2-\frac{3\times2^3-4\times2-5}{9\times2^2-4}\\ &=1.65625 \end{align*}

Hence, answer is (B)

4. For a particle of energy $E$ and $P$ momentum (in a frame $F$), the rapidity $y$ is defined as $y=\frac{1}{2}\ln{\left(\frac{E+p_3c}{E-p_3c}\right)}$. In a frame $F'$ moving with velocity $v=(0,0,\beta c)$ with respect to $F$, the rapidity $y'$ will be
1. $y'=y+\frac{1}{2}\ln{\left(1-\beta^2\right)}$
2. $y'=y-\frac{1}{2}\ln{\left(\frac{1+\beta}{1-\beta}\right)}$
3. $y'=y+\ln{\left(\frac{1+\beta}{1-\beta}\right)}$
4. $y'=y+2\ln{\left(\frac{1+\beta}{1-\beta}\right)}$

$$y=\frac{1}{2}\ln{\left(\frac{E+p_3c}{E-p_3c}\right)}$$ $$y'=\frac{1}{2}\ln{\left(\frac{E'+p_3'c}{E'-p_3'c}\right)}$$ According to Lorentz transformation (boost) $$p_3'=\gamma\left(p_3-\frac{vE}{c^2}\right)$$ $$E'=\gamma\left(E-vp_3\right)$$ Using $v=\beta c$ $$p_3'=\gamma\left(p_3-\frac{\beta E}{c}\right)$$ $$E'=\gamma\left(E-\beta cp_3\right)$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\gamma\left(E-\beta cp_3\right)+\gamma\left(p_3-\frac{\beta E}{c}\right)c}{\gamma\left(E-\beta cp_3\right)-\gamma\left(p_3-\frac{\beta E}{c}\right)c}\right)}}$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\left(E-\beta cp_3\right)+\left(cp_3-\beta E\right)}{\left(E-\beta cp_3\right)-\left(cp_3-\beta E\right)}\right)}}$$ $$y'=\frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)-\beta\left(E+cp_3\right)}{\left(E- cp_3\right)+\beta\left(E- cp_3\right)}\right)}$$ $$y'=\frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)}{\left(E- cp_3\right)}\frac{1-\beta}{1+\beta}\right)}$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)}{\left(E- cp_3\right)}\right)} + \frac{1}{2}\ln{\left(\frac{1-\beta}{1+\beta}\right)}}$$ $$y'=y - \frac{1}{2}\ln{\left(\frac{1+\beta}{1-\beta}\right)}$$

Hence, answer is (B)

5. The partition function of a single gas molecule is $Z_\alpha$. The partition function of $N$ such non-interacting gas molecules is given by
1. $\frac{(Z_\alpha)^N}{N!}$
2. $(Z_\alpha)^N$
3. $N(Z_\alpha)$
4. $\frac{(Z_\alpha)^N}{N}$

$Z_N=(Z_\alpha)^N$

Hence, answer is (B)

Problem set 83

1. Suppose that the Coulomb potential of the hydrogen atom is changed by adding an inverse-square term such that the total potential is $V(\vec r)=-\frac{Ze^2}{r}+\frac{g}{r^2}$, where $g$ is a constant. The energy eigenvalues $E_{nlm}$ in the modified potential
1. depend on $n$ and $l$, but not on $m$
2. depend on $n$ but not on $l$ and $m$
3. depend on $n$ and $m$, but not on $l$
4. depend explicitly on all three quantum numbers $n$, $l$ and $m$

Radial equation for hydrogen atom is $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{\hbar^2l(l+1)}{2\mu r^2}R+V(r)R=ER}$$ For $V(\vec r)=-\frac{Ze^2}{r}+\frac{g}{r^2}$ equation becomes $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{\hbar^2l(l+1)}{2\mu r^2}R+V(r)R+\frac{g}{r^2}R=ER}$$ $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left[\frac{\hbar^2\left(l(l+1)+\frac{2\mu}{\hbar^2}g\right)}{2\mu r^2}\right]R+V(r)R=ER}$$ Let $l_{eff}(l_{eff}+1)=l(l+1)+\frac{2\mu}{\hbar^2}g$ $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left[\frac{\hbar^2l_{eff}(l_{eff}+1)}{2\mu r^2}\right]R+V(r)R=ER}$$ $$E\propto\frac{1}{(N+l_{eff}+1)^2}$$

Hence, answer is (A)

2. When an ideal monatomic gas is expanded adiabatically from an initial volume $V_0$ to $3V_0$, its temperature changes from $T_0$ to $T$. Then the ratio $T/T_0$is
1. $\frac{1}{3}$
2. $\left(\frac{1}{3}\right)^{2/3}$
3. $\left(\frac{1}{3}\right)^{1/3}$
4. 3

For adiabatic expansion of ideal gas we have $$PV^{\gamma}=const$$ $$TV^{\gamma-1}=const$$ $$P^{1-\gamma}T^{\gamma}=const$$ $$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$ $$\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$ For monoatomic ideal gas $\gamma=\frac{5}{3}$ $$\frac{T_2}{T_1}=\left(\frac{1}{3}\right)^{\frac{5}{3}-1}$$ $$\frac{T_2}{T_1}=\left(\frac{1}{3}\right)^{\frac{2}{3}}$$

Hence, answer is (B)

3. A box of volume $V$ containing $N$ molecules of an ideal gas, is divided by a wall with a hole into two compartments. If the volume of the smaller compartment is $V/3$, the variance of the number of particles in it, is
1. $N/3$
2. $2N/9$
3. $\sqrt{N}$
4. $\sqrt{N}/3$

When a box of $V$, containing $N$ molecules, is divided into two compartments of volume $pV$ and $qV$ such that $pV+qV=V$ (or $p+q=1$) then variance in number of particles is given by $$\sigma=Npq$$ In our case compartments have volume $\frac{1}{3}V$ and $\frac{2}{3}V$. Hence, $p=\frac{1}{3}$ and $q=\frac{2}{3}$ $$\sigma=N\frac{1}{3}\frac{2}{3}=\frac{2N}{9}$$

Hence, answer is (B)

4. A gas of non-relativistic classical particles in one dimension is subjected to a potential $V(x)=\alpha|x|$ (where $\alpha$ is a constant). The partition function is ($\beta=\frac{1}{k_BT}$)
1. $\sqrt{\frac{4m\pi}{\beta^3\alpha^2h^2}}$
2. $\sqrt{\frac{2m\pi}{\beta^3\alpha^2h^2}}$
3. $\sqrt{\frac{8m\pi}{\beta^3\alpha^2h^2}}$
4. $\sqrt{\frac{3m\pi}{\beta^3\alpha^2h^2}}$

Classical partition function for single particle is $$Z_1=\frac{1}{h^3}\int d^3p\:d^3q\:e^{-\beta H(q,p)}$$ For one dimension $$Z_1=\frac{1}{h}\int dp_x\:dx\:e^{-\beta H(q,p)}$$ $$H=\frac{p_x^2}{2m}+\alpha|x|$$ \begin{align*} Z_1&=\frac{1}{h}\int dp_x\:dx\:e^{-\beta \left(\frac{p_x^2}{2m}+\alpha|x|\right)}\\ &=\frac{1}{h}\left\{\int dp_x\:e^{-\frac{\beta}{2m} p_x^2}\!\!\int dx\:e^{-\beta \alpha|x|}\!\right\}\\ \end{align*} Using $\int dx\:e^{-a x^2}=\sqrt{\frac{\pi}{a}}$ \begin{align*} I_1&=\int dp_x\:e^{-\frac{\beta}{2m} p_x^2}\\ &=\sqrt{\frac{2m\pi}{\beta}} \end{align*} \begin{align*} I_2&=\int dx\:e^{-\beta \alpha|x|}\\ &=2\int_0^\infty dx\:e^{-\beta \alpha x}\\ &=2\frac{1}{\beta\alpha} \end{align*} $$Z_1=\frac{1}{h}\sqrt{\frac{2m\pi}{\beta}}\frac{2}{\beta\alpha}$$ $$Z_1=\sqrt{\frac{8m\pi}{\beta^3\alpha^2h^2}}$$

Hence, answer is (C)

5. The dependence of current $I$ on the voltage $V$ of a certain device is given by $$I=I_0\left(1-\frac{V}{V_0}\right)^2$$ where $I_0$ and $V_0$ are constants. In an experiment the current $I$ is measured as the voltage $V$ applied across the device is increased. The parameters $V_0$ and $\sqrt{I_0}$ can be graphically determined as
1. the slope and the y-intercept of the $I-V^2$ graph
2. the negative of the ratio of the y-intercept and the slope, and the y-intercept of the $I-V^2$ graph
3. the slope and the y-intercept of the $\sqrt{I}-V$ graph
4. the negative of the ratio of the y-intercept and the slope, and the y-intercept of the $\sqrt{I}-V$ graph

The equation $$I=I_0\left(1-\frac{V}{V_0}\right)^2$$ can be written as $$\sqrt{I}=-\frac{\sqrt{I_0}}{V_0}V+\sqrt{I_0}$$ Hence, $\sqrt{I}-V$ graph is a straight line with slope $-\frac{\sqrt{I_0}}{V_0}$ and y-intercept $\sqrt{I_0}$

Hence, answer is (D)

Problem set 82

1. Two parallel plate capacitors, separated by distances $x$ and $1.1x$ respectively, have a dielectric material of dielectric constant 3.0 inserted between the plates, and are connected to a battery of voltage $V$. The difference in charge on the second capacitor compared to the first is
1. +66%
2. +20%
3. -3.3%
4. -10%

Capacitance of a parallel plate capacitor is inversely proportional to the separation between the plates i.e. $C\propto\frac{1}{d}$ $$C_1\propto\frac{1}{x}$$ $$C_2\propto\frac{1}{1.1x}$$ Let us take area of plates such that $$C_1=\frac{1}{x}$$ $$C_2=\frac{1}{1.1x}$$ $$C_2-C_1=\frac{1}{1.1x}-\frac{1}{x}=-\frac{0.1}{1.1x}$$ $$\% difference=\frac{100\times\left(-\frac{0.1}{1.1x}\right)}{\frac{1}{1.1x}}$$ $$\% difference=-10\%$$

Hence, answer is (D)

2. The state of a particle of mass $m$ in a one-dimensional rigid box in the interval 0 to $L$ is given by the normalised wavefunction $\psi(x)\!=\!\!\sqrt{\frac{2}{L}}\!\!\left(\frac{3}{5}\sin{\left(\frac{2\pi x}{L}\right)}+\frac{4}{5}\sin{\left(\frac{4\pi x}{L}\right)}\!\right)$. If its energy is measured, the possible outcomes and the average value of energy are, respectively
1. $\frac{h^2}{2mL^2}$, $\frac{2h^2}{mL^2}$ and $\frac{73}{50}\frac{h^2}{mL^2}$
2. $\frac{h^2}{8mL^2}$, $\frac{h^2}{2mL^2}$ and $\frac{19}{40}\frac{h^2}{mL^2}$
3. $\frac{h^2}{2mL^2}$, $\frac{2h^2}{mL^2}$ and $\frac{19}{10}\frac{h^2}{mL^2}$
4. $\frac{h^2}{8mL^2}$, $\frac{2h^2}{mL^2}$ and $\frac{73}{200}\frac{h^2}{mL^2}$

Eigen function and energy of a particle in a 1D box are given by $$\psi_n(x)=\sqrt{\frac{2}{L}}\sin{\left(\frac{n\pi x}{L}\right)}$$ $$E_n=\frac{n^2h^2}{8mL^2}$$ Hence, $$\psi(x)=c_2\psi_2(x)+c_4\psi_4(x)$$ where, $c_2=\frac{3}{5}$ and $c_4=\frac{4}{5}$ are the expansion coefficients.

Particle may be in its one of the eigenstate, hence possible values of energies are $$E_2=\frac{2^2h^2}{8mL^2}=\frac{h^2}{2mL^2}$$ $$E_4=\frac{4^2h^2}{8mL^2}=\frac{2h^2}{mL^2}$$ Average value of energy is \begin{align*} < E > &= < \psi |H|\psi > \\ &=|c_2|^2E_2+|c_4|^2E_4\\ &=\frac{9}{25}\frac{h^2}{2mL^2}+\frac{16}{25}\frac{2h^2}{mL^2}\\ &=\frac{73}{50}\frac{h^2}{mL^2} \end{align*}

Hence, answer is (A)

3. If $\hat L_x$, $\hat L_y$ and $\hat L_z$ are the components of the angular momentum operator in three dimensions, the commutator $\left[\hat L_x, \hat L_x\hat L_y\hat L_z\right]$ may be simplified to
1. $i\hbar\hat L_x\left(\hat L_z^2-\hat L_y^2\right)$
2. $i\hbar\hat L_z\hat L_y\hat L_x$
3. $i\hbar\hat L_x\left(2\hat L_z^2-\hat L_y^2\right)$
4. 0

\begin{align*} \left[\hat L_x, \hat L_x\hat L_y\hat L_z\right]&= {\scriptstyle \hat L_x \left[\hat L_x, \hat L_y\hat L_z\right]+\left[\hat L_x, \hat L_x\right]\hat L_y\hat L_z}\\ &=\hat L_x \left[\hat L_x, \hat L_y\hat L_z\right]\\ &={\scriptstyle \hat L_x \left\{\hat L_y\left[\hat L_x, \hat L_z\right]+\left[\hat L_x, \hat L_y\right]\hat L_z\right\}}\\ &={\scriptstyle \hat L_x \left\{\hat L_y\left(-i\hbar \hat L_y\right)+\left(i\hbar\hat L_z\right)\hat L_z\right\}}\\ &=i\hbar\hat L_x\left(\hat L_z^2-\hat L_y^2\right) \end{align*}

Hence, answer is (A)

4. The eigenstates corresponding to eigen-values $E_1$ and $E_2$ of a time-independent Hamiltonian are $|1 > $ and $|2 > $ respectively. If at $t=0$, the system is in a state $ |\psi(t=0) > =\sin\theta |1 > +\cos\theta |2 > $ the value of $< \psi(t)|\psi(t) > $ at time $t$ will be
1. 1
2. $(E_1\!\sin^2\theta\!+\!E_2\!\cos^2\theta)/\!\sqrt{E_1^2\!+\!E_2^2}$
3. $e^{iE_1t/\hbar}\sin\theta+e^{iE_2t/\hbar}\cos\theta$
4. $e^{-iE_1t/\hbar}\sin^2\theta+e^{-iE_2t/\hbar}\cos^2\theta$

Time evolution of quantum systems is given by Unitary Transformations, \begin{align*} \left|\psi(t)\right > &=\hat U\left|\psi(t=0)\right > \\ &=e^{-iHt/\hbar}\left\{\sin\theta \left|1\right > +\cos\theta \left|2\right > \right\}\\ &=\sin\theta e^{-\frac{iHt}{\hbar}}\left|1\right > +\cos\theta e^{-\frac{iHt}{\hbar}}\left|2\right > \end{align*} Using property $f(\hat A)\psi_n=f(a_n)\psi_n$ we get \begin{align*} \left|\psi(t)\right > &=\sin\theta e^{-\frac{iE_1t}{\hbar}}\left|1\right > +\cos\theta e^{-\frac{iE_2t}{\hbar}}\left|2\right > \\ \end{align*} Hence, $$\left < \psi(t)\right|\left.\psi(t)\right > =\sin^2\theta+\cos^2\theta=1$$

Hence, answer is (A)

5. The specific heat per molecule of a gas of diatomic molecules at high temperatures is
1. $8k_B$
2. $3.5k_B$
3. $4.5k_B$
4. $3k_B$

At high temperatures, the specific heat at constant volume $C_v$ has three degrees of freedom from rotation, two from translation, and two from vibration. Hence, $f=7$. $$E=\frac{7}{2}RT$$ $$C_v=\frac{dE}{dT}=\frac{7}{2}R$$

Hence, answer is (B)

Problem set 81

1. The Gauss hypergeometric function $F(a,b,c;z)$, defined by the Taylor series expansion around $z=0$ as $${\scriptstyle F(a,b,c;z)=\sum\limits_{n=0}^\infty\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)n!}z^n}$$ satisfies the recursion relation
1. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{c}{ab}F(a-1,b-1,c-1;z)}$
2. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{c}{ab}F(a+1,b+1,c+1;z)}$
3. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}F(a-1,b-1,c-1;z)}$
4. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}F(a+1,b+1,c+1;z)}$

$${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=1}^\infty n\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)n!}z^{n-1}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=1}^\infty\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)(n-1)!}z^{n-1}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=0}^\infty\frac{a(a+1)\cdots(a+n)b(b+1)\cdots(b+n)}{c(c+1)\cdots(c+n)(n)!}z^{n}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}\sum\limits_{n=0}^\infty\frac{(a+1)\cdots(a+n)(b+1)\cdots(b+n)}{(c+1)\cdots(c+n)(n)!}z^{n}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}\frac{d}{dz}F(a+1,b+1,c+1;z)}$$

Hence, answer is (D)

2. Let $(x,y)$ and $(x',y')$ be the coordinate systems used by the observers $O$ and $O'$, respectively. Observer moves with a velocity $v= \beta c$ along their common positive $x$-axis. If $x_+=x+ct$ and $x_-=x-ct$ are the linear combinations of the coordinates, the Lorentz transformation relating $O$ and $O'$ takes the form
1. $x_+'=\frac{x_--\beta x_+}{\sqrt{1-\beta^2}}$ and $x_-'=\frac{x_+-\beta x_-}{\sqrt{1-\beta^2}}$
2. $x_+'=\sqrt{\frac{1+\beta}{1-\beta}}x_+$ and $x_-'=\sqrt{\frac{1-\beta}{1+\beta}}x_-$
3. $x_+'=\frac{x_+-\beta x_-}{\sqrt{1-\beta^2}}$ and $x_-'=\frac{x_--\beta x_+}{\sqrt{1-\beta^2}}$
4. $x_+'=\sqrt{\frac{1-\beta}{1+\beta}}x_+$ and $x_-'=\sqrt{\frac{1+\beta}{1-\beta}}x_-$

According to Lorentz transformation $$x'=\gamma(x-vt)$$ $$y'=y$$ $$z'=z$$ $$t'=\gamma(t-xv/c^2)$$ where, $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$. Hence, for $v= \beta c$, $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ \begin{align*} x_+'&=x'+ct'\\ &=\gamma(x-\beta ct)+c\gamma(t-x\beta/c)\\ &=\gamma\left\{x-\beta ct+ct-x\beta\right\}\\ &=\gamma\left\{x_+-\beta (x+ct)\right\}\\ &=\gamma\left\{1-\beta \right\}x_+\\ &=\sqrt{\frac{1-\beta}{1+\beta}}x_+ \end{align*} \begin{align*} x_-'&=x'-ct'\\ &=\gamma(x-\beta ct)-c\gamma(t-x\beta/c)\\ &=\gamma\left\{x-\beta ct-ct+x\beta\right\}\\ &=\gamma\left\{x_-+\beta (x-ct)\right\}\\ &=\gamma\left\{1+\beta \right\}x_-\\ &=\sqrt{\frac{1+\beta}{1-\beta}}x_- \end{align*}

Hence, answer is (D)

3. A particle of mass $m$ is constrained to move in a vertical plane along a trajectory given by $x=A\cos\theta$, $y=A\sin\theta$, where $A$ is constant.
1. The Lagrangian of the particle is
1. $\frac{1}{2}mA^2\dot\theta^2-mgA\cos\theta$
2. $\frac{1}{2}mA^2\dot\theta^2-mgA\sin\theta$
3. $\frac{1}{2}mA^2\dot\theta^2$
4. $\frac{1}{2}mA^2\dot\theta^2+mgA\cos\theta$
2. The equation of motion of particle is
1. $\ddot\theta-\frac{g}{A}\cos\theta=0$
2. $\ddot\theta+\frac{g}{A}\sin\theta=0$
3. $\ddot\theta=0$
4. $\ddot\theta-\frac{g}{A}\sin\theta=0$

1. $$L=\frac{1}{2}mA^2\dot\theta^2-mgA\cos\theta$$

   Hence, answer is (A)

2. $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot\theta}\right)-\frac{\partial L}{\partial\theta}=0$$ gives $$\ddot\theta-\frac{g}{A}\sin\theta=0$$

   Hence, answer is (D)

4. The $x$- and $z$-components of a static magnetic field in a region are $B_x=B_0(x^2-y^2)$ and $B_z=0$, respectively. Which of the following solutions for its $y$-component is consistent with the Maxwell equations?
1. $B_y=B_0xy$
2. $B_y=-2B_0xy$
3. $B_y=B_0(x^2-y^2)$
4. $B_y=B_0(\frac{1}{3}x^3-xy^2)$

$\vec\nabla\cdot\vec B=0$ gives $$2B_0x+\frac{\partial B_y}{\partial y}=0$$ $$\frac{\partial B_y}{\partial y}=-2B_0x$$ $$B_y=-2B_0xy$$

Hence, answer is (B)

5. A magnetic field $\vec B$ is $B\hat z$ in the region $x > 0$ and zero elsewhere. A rectangular loop, in the $xy$-plane, of sides $l$ (along the $x$-direction) and $h$ (along the $y$-direction) is inserted into the $x > 0$ region from the $x < 0$ region at a constant velocity $\vec v =v \hat x$. Which of the following values of $l$ and $h$ will generate the largest EMF?
1. $l=8$, $h=3$
2. $l=4$, $h=6$
3. $l=6$, $h=4$
4. $l=12$, $h=2$

Let us consider an area element $\vec A=hx\hat z$ passing through magnetic field $\vec B=B\hat z$. Magnetic flux passing through area element is $$\Phi=\vec B\cdot\vec A=Bhx$$ Rate of change of magnetic flux $$\frac{d\Phi}{dt}=Bh\frac{dx}{dt}=Bhv$$ $$\epsilon=-\frac{d\Phi}{dt}=-Bhv$$ Hence, EMF $\epsilon$ will be largest for largest value of $h$

Hence, answer is (B)

Problem set 80

1. A dielectric sphere of radius $R$ carries polarization $\vec P = kr^2\hat r$, where $r$ is the distance from the centre and $k$ is a constant. In the spherical polar coordinate system, $\hat r$, $\hat \theta$ and $\hat \phi$ are the unit vectors.
1. The bound volume charge density inside the sphere at a distance $r$ from the centre is
1. $-4kR$
2. $-4kr$
3. $-4kr^2$
4. $-4kr^3$
2. The electric field inside the sphere at a distanced $d$ from the centre is
1. $\frac{-kd^2}{\epsilon_0}\hat r$
2. $\frac{-kR^2}{\epsilon_0}\hat r$
3. $\frac{-kd^2}{\epsilon_0}\hat\theta$
4. $\frac{-kR^2}{\epsilon_0}\hat\theta$

1. The unit vector $\hat n$ on the surface of the sphere is equal to the radial unit vector. The bound surface charge is equal to $\sigma_b=\vec P\cdot\hat n|_{r=R}$

   The bound volume charge is equal to \begin{align*} \rho_b&=-\vec\nabla\cdot\vec P\\ &=-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\:kr^2\right)\\ &=-4kr \end{align*}

   Hence, answer is (B)

2. The electric field inside the sphere at a distanced $d$ from the centre is \begin{align*} \vec E_{volume}&=\frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi d^3\rho_b}{d^2}\hat r\\ &=\frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi d^3(-4kd)}{d^2}\hat r\\ &=-\frac{4kd^2}{3\epsilon_0}\hat r \end{align*}

   Hence, answer is ()

2. Let $X$ and $Y$ be two independent random variables, each of which follow a normal distribution with the same standard deviation $\sigma$, but with means $+\mu$ and $-\mu$, respectively. Then the sum follows a
1. distribution with two peaks at $\pm\mu$ and mean $0$ and standard deviation $\sigma\sqrt{2}$
2. normal distribution with mean 0 and standard deviation $2\sigma$
3. distribution with two peaks at $\pm\mu$ and mean 0 and standard deviation $2\sigma$
4. normal distribution with mean 0 and standard deviation $\sigma\sqrt{2}$

The convolution of two normal densities with means $\mu_1$ and $\mu_2$ and variances $\sigma_1$ and $\sigma_2$ is again a normal density, with mean $\mu_1+\mu_2$ and variance $\sigma_1^2+\sigma_2^2$. $$\mu_t=-\mu+\mu=0$$ $$\sigma^2_t=\sigma^2+\sigma^2=2\sigma^2$$ $$\sigma_t=\sqrt{2}\sigma$$

Hence, answer is (D)

3. Using dimensional analysis, Planck defined a characteristic temperature $T_p$ from powers of the gravitational constant $G$, Planck’s constant $h$, Boltzmann constant $k_B$ and the speed of light $c$ in vacuum. The expression for $T_p$ is proportional to
1. $\sqrt{\frac{hc^5}{k_B^2G}}$
2. $\sqrt{\frac{hc^3}{k_B^2G}}$
3. $\sqrt{\frac{G}{hc^4k_B^2}}$
4. $\sqrt{\frac{hk_B^2}{Gc^3}}$

The Planck temperature is defined as: $$T_p=\frac{m_pc^2}{k_B}$$ where, $m_p$ is Plank mass.

Now, let $$m_p=c^{n_1}G^{n_2}\hbar^{n_3}$$ Using dimensional analysis we have $${\scriptstyle M^1L^0T^0=\left[M^0L^{n_1}T^{-n_1}\right]\left[M^{-n_2}L^{3n_2}T^{-2n_2}\right]\left[M^{n_3}L^{2n_3}T^{-n_3}\right]}$$ $$n_{3}-n_{2}=1$$ $$ n_{1}+3n_{2}+2n_{3}=0$$ $$ -n_{1}-2n_{2}-n_{3}=0$$ $$\Rightarrow n_{1}=1/2,n_{2}=-1/2,n_{3}=1/2$$ $$m_p=\sqrt{\frac{c\hbar}{G}}$$ $$T_p=\sqrt{\frac{\hbar c^5}{k_B^2G}}$$

Hence, answer is (A)

4. A ball of mass $m$, initially at rest, is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately (take $g = 9.8\: m/s^2$)
1. 9.80 m/s
2. 9.10 m/s
3. 8.91 m/s
4. 7.02 m/s

For an object bouncing off a stationary object, such as a floor, coefficient of restitution, $e=\sqrt{\frac{h'}{h}}$ $$h'=e^2h=4.05\:m$$ $$v=\sqrt{2gh}$$ $$v=\sqrt{2\times 9.8\times4.05}=8.91\:m/s$$

Hence, answer is (C)

Problem set 79

1. An unperturbed two-level system has energy eigenvalues $E_1$ and $E_2$, and eigenfunctions $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ When perturbed, its Hamiltonian is represented by $\begin{pmatrix}E_1&A\\A^*&E_2\end{pmatrix}$
1. The first-order correction to $E_1$ is
1. $4A$
2. $2A$
3. $A$
4. 0
2. The second-order correction to $E_1$ is
1. 0
2. $A$
3. $\frac{A^2}{E_2-E_1}$
4. $\frac{A^2}{E_1-E_2}$
3. The first-order correction to the eigenfunetion $\begin{pmatrix}1\\0\end{pmatrix}$ is
1. $\begin{pmatrix}0\\\frac{A^*}{E_1-E_2}\end{pmatrix}$
2. $\begin{pmatrix}0\\1\end{pmatrix}$
3. $\begin{pmatrix}\frac{A^*}{E_1-E_2}\\0\end{pmatrix}$
4. $\begin{pmatrix}1\\1\end{pmatrix}$

1. The perturbed Hamiltonian is given by $H=H_0+H'$ $$H=\begin{pmatrix}E_1&A\\A^*&E_2\end{pmatrix}$$ As $E_1$ and $E_2$ are eigenvalues $$H_0=\begin{pmatrix}E_1&0\\0&E_2\end{pmatrix}$$ $$H'=\begin{pmatrix}0&A\\A^*&0\end{pmatrix}$$ First order correction to energy $E_1$ is \begin{align*} E_1^{(1)}&=\left < \psi_1|H'|\psi_1 \right > \\ &=\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}0&A\\A^*&0\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}\\ &=0 \end{align*}

   Hence, answer is (D)

2. The second order correction to the energy is given by \begin{align*} E_n^{(2)}&=\sum\limits_{m\neq n}\frac{\left|\left < m|H'|n \right > \right|^2}{E_n^{(0)}-E_m^{(0)}}\\ E_1^{(2)}&=\sum\limits_{m\neq 1}\frac{\left|\left < m|H'|1 \right > \right|^2}{E_1^{(0)}-E_m^{(0)}}\\ &=\frac{\left|\left < 2|H'|1 \right > \right|^2}{E_1^{(0)}-E_2^{(0)}}\\ \left < 2|H'|1 \right >&=\!\begin{pmatrix}\!0\!&\!1\!\end{pmatrix}\!\begin{pmatrix}\!0\!&\!A\!\\\!A^*\!&\!0\!\end{pmatrix}\! \begin{pmatrix}\!1\!\\\!0\!\end{pmatrix}\\ &=A^*\\ E_1^{(2)}&=\frac{A^2}{E_1-E_2} \end{align*}

   Hence, answer is (D)

3. Wave corrected upto first order is given by $$\psi_n^{(1)}=\!\psi_n^{(0)}\!+\!\sum\limits_{m\neq n}\!\frac{\left|\left < m|H'|n \right > \right|^2}{E_n^{(0)}-E_m^{(0)}}\psi_m^{(0)}$$ $$\psi_1^{(1)}=\psi_1^{(0)}+\frac{\left|\left < 2|H'|1 \right > \right|^2}{E_1^{(0)}-E_2^{(0)}}\psi_2^{(0)}$$ $$\psi_1^{(1)}=\begin{pmatrix}1\\0\end{pmatrix}+\frac{A^*}{E_1-E_2}\begin{pmatrix}0\\1\end{pmatrix}$$ Hence first order correction to the eigenfunction $\begin{pmatrix}1\\0\end{pmatrix}$ is $\begin{pmatrix}0\\\frac{A^*}{E_1-E_2}\end{pmatrix}$

   Hence, answer is (A)

2. One of the eigen values of the matrix $\begin{pmatrix}2&3&0\\3&2&0\\0&0&1\end{pmatrix}$ is 5
1. The other two eigenvalues are
1. 0 and 0
2. 1 and 1
3. 1 and -1
4. -1 and -1
2. The normalized eigenvector corresponding to the eigenvalue 5 is
1. $\frac{1}{\sqrt{2}} \begin{pmatrix}0\\-1\\1\end{pmatrix}$
2. $\frac{1}{\sqrt{2}} \begin{pmatrix}-1\\1\\0\end{pmatrix}$
3. $\frac{1}{\sqrt{2}} \begin{pmatrix}1\\0\\-1\end{pmatrix}$
4. $\frac{1}{\sqrt{2}} \begin{pmatrix}1\\1\\0\end{pmatrix}$

1. Determinant of matrix = product of eigenvalues.

   Let $\lambda_1$ and $\lambda_2$ be other eigenvalues $$Det=5\times\lambda_1\times\lambda_2=-5$$ $$\lambda_1\times\lambda_2=-1\quad---(1)$$ Trace of matrix=sum of eigenvalues $$5+\lambda_1+\lambda_2=5$$ $$\lambda_1+\lambda_2=0\quad---(2)$$ Solving equation (1) and (2) we get $$\lambda_1=1\quad\lambda_2=-1$$

   Hence, answer is (C)

2. Eigenvalue equation for $\lambda=5$ is $A-\lambda=0$ $$\begin{pmatrix}-3&3&0\\3&-3&0\\0&0&-4\end{pmatrix}\!\begin{pmatrix}x\\y\\z\end{pmatrix}\!=\!\begin{pmatrix}0\\0\\0\end{pmatrix}$$ $$-3x+3y=0$$ $$3x-3y=0$$ $$-4z=0$$ $$\Rightarrow z=0, x=y=c$$ Applying normalization condition $$|c|^2+|c|^2+0=1$$ $$\Rightarrow c=\frac{1}{\sqrt{2}}$$ $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\\0\end{pmatrix}$$

   Hence, answer is (d)

3. The powder diffraction pattern of a body centred cubic crystal is recorded by using $Cu K_\alpha$ X-rays of wavelength $1.54 \:A^o$.
1. If the (002) planes diffract at $60^o$, the lattice parameter is
1. $2.67 A^o$
2. $3.08 A^o$
3. $3.56 A^o$
4. $5.34 A^o$
2. Assuming the atomic mass of the constituent atoms to be 50.94 amu, the density of the crystal in units of kg m$^{-3}$ is
1. $3.75 \times 10^3$
2. $4.45 \times 10^3$
3. $5.79 \times 10^3$
4. $8.89 \times 10^3$

Bragg's law is $$2d\sin\theta=n\lambda$$ For first order pattern $n=1$ \begin{align*} d&=\frac{\lambda}{2\sin\theta}\\ &=\frac{1.54}{2\times\sin30}\\ &=1.54 \end{align*} For cubic system $$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$ $$a=d\sqrt{h^2+k^2+l^2}=3.08$$

Hence, answer is (B)

For body centered cubic crystal there two atoms per unit cell. $$\rho=\frac{mass~of~two~atoms}{volume~of~cell}$$ 1 atomic mass unit = $1.66053886 \times 10^{-27}$ kilograms $$\rho={\textstyle \frac{2\times50.94\times 1.66053886 \times 10^{-27}}{(3.08\times 10^{-10})^3}}$$ $$\rho=5.79 \times 10^{3}$$

Hence, answer is (C)