statmech

1. Calculate equivalent temperature of 1eV energy
$E=3/2kT$, $1eV=11604.52500617K$
2. For any process, the second law of thermodynamics requires that the change of entropy of the universe be
1. Positive only
2. Positive or zero
3. Zero only
4. Negative or zero
(B) because for reversible process entropy is zero and for irreversible process entropy is positive
3. For any process, the second law of thermodynamics requires that the change of entropy of the universe be
1. Positive only
2. Positive or zero
3. Zero only
4. Negative or zero
(B) because for reversible process entropy is zero and for irreversible process entropy is positive
4. A set of 15 distinguishable particles are placed in 3 energy states such that 2 particles in the first state, 12 in the second state and 1 in the third state. The number of distinct arrangements are:
1. 1365
2. 15
3. 455
4. $3^{15}$
Number of ways of arranging $N$ distinguishable particles into groups of $N_j$ each, such that $N_1$ in state 1, $N_2$ in state 2, etc. and $N_j$ in the $j^{th}$ state are given by $$W(N)=\frac{N!}{N_1!N_2!\dots N_j!}$$ $$W(15)=\frac{15!}{2!\times12!\times1!}=1365$$
5. The probability that two friends have the same birth month is:
1. $\frac{1}{6}$
2. $\frac{1}{12}$
3. $\frac{1}{36}$
4. $\frac{1}{144}$
A person can have birthday in any of the 12 months of the year. Another person can also have his/her birthday in any of the 12 months. Now,the total number of possible combinations is 12*12 , which equals 144. This set of all possible combinations contains 12 pairs of identical months {for example(Jan,Jan)}, which are our favourable outcomes.
Thus the probability =
(Number of favourable outcomes)/(Number of events in sample space)=12/(12*12)=1/12
6. The critical temperature for the Bose-Einstein condensation depends on the density of particles as :
   1. $n^{1/3}$
   2. $n^{2/3}$
   3. $n$
   4. $n^{1/2}$
   $N=\frac{V(2\pi mkT)^{3/2}}{h^3}2.162$. Hence (B)$n^{2/3}$
7. A partition function of two Bose particles each of which can occupy any of the two energy levels $0$ and $\epsilon$ is
1. $1+e^{-2\epsilon/kT}+2e^{-\epsilon/kT}$
2. $1+e^{-2\epsilon/kT}+e^{-\epsilon/kT}$
3. $2+e^{-2\epsilon/kT}+e^{-\epsilon/kT}$
4. $e^{-2\epsilon/kT}+e^{-\epsilon/kT}$
The partition function for Bose particles is given by $Z=\sum\limits_Re^{-(n_1\epsilon_1+n_2\epsilon_2+\dots)/kT}$, where $n_1$ is number of particles with energy $\epsilon_1$ and so on. Bose particles are indistinguishable particles and any number of particles can be in any state. Hence states of the system can be obtained as follow.
Single particle energy	0	$\epsilon$
	AA
		AA
	A	A
\begin{align*} Z&=e^{-(2\times0)/kT}+e^{-(2\times\epsilon)/kT}\\ &+e^{-(1\times0+1\times\epsilon)/kT}\\ &=1+e^{-2\epsilon/kT}+e^{-\epsilon/kT} \end{align*}
8. A one dimensional random walker takes steps to left or right with equal probability. The probability that the random walker starting from origin is back to origin after $N$ even number of steps is
   1. $\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}\left(\frac{1}{2}\right)^N$
   2. $\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}$
   3. $2N!\left(\frac{1}{2}\right)^{2N}$
   4. $N!\left(\frac{1}{2}\right)^N$
   The probability $p(m,N)$ that the particle will be at the position $x = ml$ after $N$ steps is given by $${\scriptstyle p(m,N)=\frac{N!}{\left(\frac{N+m}{2}\right)!\left(\frac{N-m}{2}\right)!}p^{\frac{1}{2}\left(N+m\right)}q^{\frac{1}{2}\left(N-m\right)}}$$ Here $p=q=\frac{1}{2}$, the probability of taking steps to right and left, $l$ is step length and $x=ml=0$. Hence $m=0$. Hence $$p(0,N)=\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}\left(\frac{1}{2}\right)^N$$
9. Five electrons (Fermions with spin $1/2\hbar$) are kept in a one-dimensional infinite potential well with width $a$. (Ground state energy of single electron well $=\frac{\hbar^2\pi^2}{2ma^2}$). The first absorption line corresponds to energy:
   1. $\frac{\hbar^2\pi^2}{2ma^2}$
   2. $\frac{5\hbar^2\pi^2}{2ma^2}$
   3. $\frac{7\hbar^2\pi^2}{2ma^2}$
   4. $\frac{11\hbar^2\pi^2}{2ma^2}$
   As electrons are fermions, there can be two electrons with same energy. Hence ground state configuration will be two electrons in $n=1$ state, two electrons in $n=2$ state and one electron in $n=3$ state. The first absorption line corresponds to transition of electron from $n=3$ state to $n=4$ state. Energy of $n=3$ state is $\frac{9\hbar^2\pi^2}{2ma^2}$ and energy of $n=4$ state is $\frac{16\hbar^2\pi^2}{2ma^2}$. Hence first absorption line corresponds to energy $\frac{16\hbar^2\pi^2}{2ma^2}-\frac{9\hbar^2\pi^2}{2ma^2}=\frac{7\hbar^2\pi^2}{2ma^2}$
10. An unbiased coin is tossed $n$ times. The probability that exactly $m$ heads will come up is
1. $\frac{n}{2^m}$
2. $\frac{1}{2^n}\frac{n!}{m!(n-m)!}$
3. $\frac{1}{2^m}\frac{n!}{m!(n-m)!}$
4. $\frac{m}{2^n}$
This problem is equivalent to 1D random walk problem. In a single toss probability getting head is $p=\frac{1}{2}$ and probability of getting tail is $q=\frac{1}{2}$. In a sequences of $n$ tosses probability of getting $m$ heads and $n-m$ tails is given by $$p^mq^{(n-m)}=\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}$$ However, total combinations of $m$ times head up and $(n-m)$ times tail up are given by $$\frac{n!}{m!(n-m)!}$$ Hence, probability of getting $m$ heads is \begin{align*} P&=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}\\ &=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^n \end{align*} Hence, answer is (B).
11. Consider the transition of liquid water to steam as water boils at a temperature of $100^oC$ under a pressure of 1 atmosphere. Which one of the following quantities does not change discontinuously at the transition?
1. The Gibbs free energy
2. The internal energy
3. The entropy
4. The specific volume
(A) The Gibbs free energy.
12. Consider two independently diffusing non-interacting particles in 3-dimensional space, both placed at the origin at time $t = 0$. These particles have different diffusion constants $D_1$ and $D_2$. The quantity $\left<\left[\vec R_1(t)- \vec R_2(t)\right]^2\right>$ where $\vec R_1(t)$ and $\vec R_2(t)$ are the positions of the particles at time $t$, behaves as:
1. $6t(D_1+D_2)$
2. $6t|D_1+D_2|$
3. $6t\sqrt{D_1^2+D_2^2}$
4. $6t\sqrt{D_1D_2}$
For 3-dimensional diffusion, according to Einstein-Smoluchowski relation or Einstein relation, diffusion coefficient is given by $$ D=\frac{L^2}{6t}\quad \text{Einstein-Smoluchowski}$$ where, $L$ is mean square displacement or diffusion length,and $t$ is diffusion time. $$D=\frac{< r^2 >}{g*t}\quad \text{Einstein}$$ where, $< r^2 >$ is the average of the square of the mean displacement, $g=1,2\text{ or }6$ depending if diffusion dimension is 1, 2, or 3.
Now, $\left[\vec R_1-\vec R_2\right]^2=R_1^2+R_2^2-2\vec R_1\cdot\vec R_2$ \begin{align*} &\left < \left[\vec R_1(t)- \vec R_2(t)\right]^2\right > \\ &=\left< \vec R_1(t)^2\right > + \left<\vec R_2(t)^2\right>\\ &-2\left < \vec R_1(t)\cdot\vec R_2(t)\right >\\ &=\left< \vec R_1(t)^2\right > + \left<\vec R_2(t)^2\right>\\ &=6t(D_1+D_2) \end{align*}
13. In a series of five Cricket matches, one of the captains calls "Heads" every time when the toss is taken. The probability that he will win 3 times and lose 2 times is
1. 1/8
2. 5/8
3. 3/16
4. 5/16
Here, the coin is tossed five times. In a single toss probability getting head is $p=\frac{1}{2}$ and probability of getting tail is $q=\frac{1}{2}$. In a sequences of $n=5$ tosses probability of getting $m=3$ heads and $n-m=2$ tails is given by $$p^mq^{(n-m)}=\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}$$ However, total combinations of $m$ times head up and $(n-m)$ times tail up are given by $$\frac{n!}{m!(n-m)!}$$ Hence, probability of getting $m$ heads is \begin{align*} P&=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}\\ &=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^n\\ &=\frac{5!}{3!(5-3)!}\left(\frac{1}{2}\right)^5\\ &=10\left(\frac{1}{2}\right)^5=\frac{5}{16} \end{align*} Hence, answer is (D).
14. 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 $\Gamma$
1. does not change during evolution to equilibrium
2. oscillates during evolution to equilibrium
3. is a maximum at equilibrium
4. is a minimum at equilibrium
Answer is (C) i.e. $\Gamma$ is a maximum at equilibrium. This is because, when system attains equilibrium its entropy and hence number of accessible states becomes maximum (equilibrium state corresponds to the most disordered state).
15. Let $\Delta W$ be the work done in a quasistatic reversible thermodynamic process. Which of the following statements about $\Delta W$ is correct?
1. $\Delta W$ is a perfect differential if the process is isothermal
2. $\Delta W$ is a perfect differential if the process is adiabatic
3. $\Delta W$ is always a perfect differential
4. $\Delta W$ cannot be a perfect differential
There are two types of functions viz. State functions and path functions. In case of state functions, the change in the state function depends only on the initial and final states of the system and not on the path from the initial to the final state (e.g. internal energy $U(S,V)$, enthalpy $H(S,P)$, Gibb's free energy $G(T,P)$ Helmholtz free energy $F(T,V)$ etc. are state functions). In case of path functions, the change in the path functions depends on the path followed from the initial to the final state (e.g..$dQ$, $dW$ etc. are path functions). State functions are exact differential while path functions are inexact differential.
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.
16. 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,
1. $J$ and 1
2. $-3J$ and 1
3. $-3J$ and 2
4. $-6J$ and 2
Clearly, for $S_1= S_2 = S_3=1$ or $S_1= S_2 = S_3=-1$, we get minimum energy $E=-3J$. Hence, there are 2 configurations. Hence, answer (C) is correct.
17. 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
1. $14\pi^2\hbar^2/mL^2$
2. $91\pi^2\hbar^2/mL^2$
3. $7\pi^2\hbar^2/mL^2$
4. $3\pi^2\hbar^2/mL^2$
For a particle in one dimensional infinite square well potential of width $L$, energy is given by $$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\quad n=1,2,\dots$$ In each level there can be two electrons with up and down spin. Hence, total energy of the system is \begin{align*} E_{total}&=2\frac{1^2\pi^2\hbar^2}{2mL^2}+2\frac{2^2\pi^2\hbar^2}{2mL^2}\\ &+2\frac{3^2\pi^2\hbar^2}{2mL^2}\\ &=\frac{14\pi^2\hbar^2}{mL^2} \end{align*} Hence, answer is (A)
18. 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
1. 20 J
2. 80 J
3. 220 J
4. 280 J
$$\delta Q=dU+\delta W$$ For path $A\rightarrow C$, $\delta W=100 J$ and $\delta Q=150 J$ $$150=du+100$$ $$dU=50 J$$ As, $dU$ is a state function, it does not depend on the path followed. Hence, for the path $A\rightarrow B\rightarrow C$, we have $$\delta Q=50+30=80 J$$ Hence, answer is (B)
19. 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
1. $-\alpha^2/\beta$
2. $-\alpha^2/2\beta$
3. $-3\alpha^2/\beta$
4. $-5\alpha^2/\beta$
For $\Delta F$ to be minimum $\frac{\partial(\Delta F)}{\partial\psi}=0$ i.e. \begin{align*} \frac{\partial(\Delta F)}{\partial\psi}&=2\alpha|\psi|+2\beta|\psi|^3\\ &=0\Rightarrow|\psi|^2\\ &=-\frac{\alpha}{\beta} \end{align*} \begin{align*} \Delta F&=-\alpha\times\frac{\alpha}{\beta}+\frac{\beta}{2}\frac{\alpha^2}{\beta^2}\\ &=-\frac{\alpha^2}{2\beta} \end{align*} Hence, answer is (B)
20. 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
1. $4\gamma T^4$
2. $2\gamma T^4$
3. $\gamma T^4$
4. $\gamma T^4/4$
Change in internal energy $dU$ is given by $dU=C_vdT$ $$U=\int C_vdT=\int \gamma T^3dT=\frac{\gamma T^4}{4}$$ Hence, answer is (D)
21. 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
1. $\frac{k_BT}{u_0}$
2. $\frac{(k_BT)^2}{u_0}$
3. $\sqrt{\frac{k_BT}{u_0}}$
4. $u_0k_BT$
$$ < x > = \frac{\int xe^{-\beta H}d\tau}{\int e^{-\beta H}d\tau}$$ where $\beta=\frac{1}{k_BT}$, and $H=\frac{p^2}{2m}+u(x)=\frac{p_x^2}{2m}+u_0x$ \begin{align*} &< x > =\frac{\int xe^{-\beta \left(\frac{p_x^2}{2m}+u_0x\right)}dxdp_x}{\int e^{-\beta \left(\frac{p_x^2}{2m}+u_0x\right)}dxdp_x}\\ &=\frac{\int_0^\infty xe^{-\beta u_0x}dx}{\int_0^\infty e^{-\beta u_0x}dx}\quad{\scriptstyle \int\limits_{0}^\infty x^{n}e^{-\alpha x}dx=\frac{n!}{\left(\alpha\right)^{n+1}}}\\ &=\frac{\frac{1}{(\beta u_0)^2}}{\frac{1}{\beta u_0}}={\frac{1}{\beta u_0}}={\frac{k_BT}{ u_0}} \end{align*} Hence, answer is (A).
22. 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
1. $\hbar\omega/2$
2. $\hbar\omega$
3. $3\hbar\omega/2$
4. $0$
For a three-dimensional isotropic simple harmonic potential, the lowest energy is $3\hbar\omega/2$ called as zero-point energy. Hence, option (C) is correct.
23. Consider a random walker on a square lattice. At each step the walker moves to a nearest neighbour site with equal probability for each of the four sites. The walker starts at the origin and takes 3 steps. The probability that during this walk no site is visited more than once is
1. 12/27
2. 27/64
3. 3/8
4. 9/16

During first step of random walk the probability that during this walk no site is visited more than once is 1

During second step of random walk the probability that during this walk no site is visited more than once is 3/4, because there are 3 sites which are unvisited and each site has probability 1/4

Similarly, during third step of random walk the probability is 3/4

Hence, the probability that during first, second and third step, no site is visited more than once is $$p=1\times\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}$$

Hence, answer is (D)

24. If the Planck's constant were to be zero, then the total energy contained in a box filled with radiation of all frequencies at temperature T would be (where $k$ is Boltzmann constant and $T\ne0$)
1. zero
2. infinite
3. $\frac{3}{2}kT$
4. $kT$
According to Stefan-Boltzmann law the the energy flux $\Phi$ emitted from a blackbody at temperature $T$, is $\Phi = \sigma T^4$, where $\sigma=\frac{2\pi^5k^4}{15h^3c^2}$. AS $\sigma\propto\frac{1}{h^3}$, energy will be infinite when $h=0$

Hence, answer is (B)

25. Two data sets A and B consist of 60 and 10 readings of a voltage measured using voltmeters of resolution of 1 mV and 0.5 mV respectively. The uncertainty in the mean voltage obtained from the data sets $A$ and $B$ are $U_A$ and $U_B$, respectively. If the uncertainty of the mean of the combined data sets is $U_{AB}$, then which of the following statements is correct?
1. $U_{AB} < U_A$ and $U_{AB} > U_B$
2. $U_{AB} < U_A$ and $U_{AB} < U_B$
3. $U_{AB} > U_A$ and $U_{AB} < U_B$
4. $U_{AB} > U_A$ and $U_{AB} > U_B$

For $N$ measurements $x_1,x_2,\cdots x_N$ with uncertainties $\sigma_1,\sigma_2\cdots\sigma_N$ weighted average is given by $$x_{avg}=\frac{\sum\limits_{i=1}^Nw_ix_i}{\sum\limits_{i=1}^Nw_i}\quad--(1)$$ where, $w_i=\frac{1}{\sigma_i}$. The uncertainty on $x_{avg}$ is $$\sigma_{avg}=\frac{1}{\sqrt{\sum\limits_{i=1}^Nw_i}}\quad--(2)$$

For an instrument having resolution $R$ has uncertainty $\sigma=\frac{R}{2}$

$\sigma_A=\frac{R_A}{2}=\frac{1}{2}=0.5 \:mV$

$\sigma_B=\frac{R_B}{2}=\frac{0.5}{2}=0.25 \:mV$

Using equation (2)

$U_A=\frac{\sigma_A}{\sqrt{N}}=\frac{0.5}{\sqrt{60}}=0.065$

$U_B=\frac{\sigma_B}{\sqrt{N}}=\frac{0.25}{\sqrt{10}}=0.08$

\begin{align*} U_{AB}&=\frac{1}{\sqrt{\sum\limits_{i=1}^Nw_i}}\\ &=\frac{1}{\sqrt{60\times\frac{1}{0.5}+10\times\frac{1}{0.25}}}\\ &=0.08 \end{align*}

Hence, answer is (B)

26. Using the Clausius-Clapeyron equation, the change in melting point of ice for 1 atmosphere rise in pressure is : (Given: The latent heat of fusion for water at $0^oC$ is $3.35\times10^5\:J/kg$, the volume of ice is $1.09070\:cc/g$ and the volume of water is $1.00013\: cc/gm$)
1. $-0.0075^oC$
2. $0.0075^oC$
3. $0.075^oC$
4. $-0.075^oC$
$$v_1=1.09070\: cc/gm$$ $$v_2=1.00013\: cc/gm$$ \begin{align*} L&=3.35\times10^5\:J/kg\\ &=80.069\: cal/gm\\ &=80.069\times4.2\times10^7 \:erg/gm \end{align*} $$T=0+273=273K$$ \begin{align*} dP&=1\: atm\\ &=76\times13.6\times980\:dyne/cm^2 \end{align*} Clausius-Clapeyron equation is $$\frac{dP}{dT}=\frac{L}{T(v_2-v_1)}$$ \begin{align*} dT&=\frac{T(v_2-v_1)dP}{L}\\ &={\textstyle\frac{273\times(1.00013-1.09070)\times76\times13.6\times980}{80.069\times4.2\times10^7}}\\ &=-0.0075\: ^oC \end{align*}

Hence, answer is (A)

27. The free energy of a photon gas enclosed in a volume $V$ is given by $F=-\frac{1}{3}aVT^4$, where $a$ is constant and $T$ is the temperature of the gas. The chemical potential of the photon gas is
1. 0
2. $\frac{4}{3}aVT^4$
3. $\frac{1}{3}aT^4$
4. $aVT^4$

Chemical potential is given by $$\mu=\frac{\partial F}{\partial N}$$ $$\mu=\frac{\partial (-\frac{1}{3}aVT^4)}{\partial N}=0$$

Hence, answer is (A)

28. The wavefunctions of two identical particles in states $n$ and $s$ are given by $\phi_n(r_1)$ and $\phi_s(r_2)$, respectively. The particles obey Maxwell-Boltzmann statistics. The state of the combined two-particle system is expressed as
1. $\phi_n(r_1)+\phi_s(r_2)$
2. $\frac{1}{\sqrt{2}}\left[\phi_n(r_1)\phi_s(r_2)+\phi_n(r_2)\phi_s(r_1)\right]$
3. $\frac{1}{\sqrt{2}}\left[\phi_n(r_1)\phi_s(r_2)-\phi_n(r_2)\phi_s(r_1)\right]$
4. $\phi_n(r_1)\phi_s(r_2)$

In Maxwell-Boltzmann statistics particles are distinguishable, hence, $$\Phi=\phi_n(r_1)\phi_s(r_2)$$ or $$\Phi=\phi_n(r_2)\phi_s(r_1)$$

Hence, answer is (D)

29. A system of $N$ distinguishable particles, each of which can be in one of the two energy levels $0$ and $\epsilon$, has a total energy $n\epsilon$, where $n$ is an integer. The entropy of the system is proportional to
1. $N \ln{n}$
2. $n \ln{N}$
3. $\ln{\frac{N!}{n!}}$
4. $\ln{\left(\frac{N!}{n!(N-n)!}\right)}$

Let us consider three distinguishable particles A, B, C to be arranged in two energy levels $0$ and $\epsilon$, such that total energy is $n\epsilon$ ($n=0,1,3$). The total number of microstates are given as follow:

Total energy $n\epsilon$	0	$\epsilon$	Number of microstates
$0\epsilon$	ABC	-	1
$1\epsilon$	AB	C	3
	AC	B
	BC	A
$2\epsilon$	A	BC	3
	B	AC
	C	AB
$3\epsilon$	-	ABC	1

In general, the number microstates for $N$ particles for total energy $n\epsilon$ is given by $$\Omega=\frac{N!}{n!(N-n)!}$$ Hence, entropy of system is given by $$S=k_B\ln\Omega=k_B\ln{\left(\frac{N!}{n!(N-n)!}\right)}$$

Hence, answer is (D)

30. Three variables $a$, $b$, $c$ are each randomly chosen from uniform distribution in the interval $[0,1]$. The probability that $a+b>2c$ is
1. $\frac{3}{4}$
2. $\frac{2}{3}$
3. $\frac{1}{2}$
4. $\frac{1}{4}$

The condition $a+b>2c$ can be written as $\frac{a+b}{2}>c$. Here, $\frac{a+b}{2}$ is average value $a$ and $b$. Hence, $\frac{a+b}{2}$ lies in the interval $[0,1]$. Also, $c$ lies in the interval $[0,1]$. Hence, probability that $\frac{a+b}{2} > c$ is $\frac{1}{2}$.

Hence, answer is (C)

31. A system of $N$ non-interacting classical particles, each of mass $m$ is in a two-dimensional harmonic potential of the form $V(r)=\alpha(x^2+y^2)$ where $\alpha$ is a positive constant. The canonical partition function of the system at temperature $T$ $\left(\beta=\frac{1}{k_BT}\right)$:
1. $\left[\left(\frac{\alpha}{2m}\right)^2\frac{\pi}{\beta}\right]^N$
2. $\left(\frac{2m\pi}{\alpha\beta}\right)^{2N}$
3. $\left(\frac{\alpha\pi}{2m\beta}\right)^N$
4. $\left(\frac{2m\pi^2}{\alpha\beta^2}\right)^{N}$

The classical partition is given by $$Z=\frac{1}{h^s}\int\cdots\int e^{-\beta H(q,p)}\:dq\:dp$$ OR $$Z=\int\cdots\int e^{-\beta H(q,p)}\:dq\:dp$$ In second definition the factor $\frac{1}{h^s}$ is missing. Here, $s$ is dimensionality. This factor arises when treating the classical result as a limiting scenario of a quantum mechanical problem. In a purely classical context, however, this factor should not be included.

For 2D harmonic oscillator $$H=\frac{p_x^2}{2m}+\frac{p_x^2}{2m}+\alpha(x^2+y^2)$$ Hence, single particle canonical partition function is given by \begin{align*} Z_1&=\!\!\int\!\!\!\int\! e^{-\beta \left(\!\frac{p_x^2}{2m}\!+\!\frac{p_y^2}{2m}\!+\!\alpha(x^2\!+\!y^2)\!\right)}\!dxdydp_xdp_y\\ &=\int\limits_{-\infty}^{\infty}e^{-\beta \frac{p_x^2}{2m}}\:dp_x\int\limits_{-\infty}^{\infty}e^{-\beta \frac{p_y^2}{2m}}\:dp_y\\ &\int\limits_{-\infty}^{\infty} e^{-\beta \alpha x^2}\:dx\int\limits_{-\infty}^{\infty} e^{-\beta \alpha y^2}\:dy\\ &=\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{\pi }{\alpha\beta}}\sqrt{\frac{\pi }{\alpha\beta}}\\ &=\frac{2\pi^2 m}{\alpha\beta^2} \end{align*} Here, we have used formula $\int\limits_{-\infty}^{\infty} e^{-a x^2}\:dx=\sqrt{\frac{\pi }{a}}$

For $N$ particle system the partition function is given by $$Z=Z_1^N=\left(\frac{2\pi^2 m}{\alpha\beta^2}\right)^N$$

Hence, answer is (D)

32. In a two-state system, the transition rate of a particle from state 1 to state 2 is $t_{12}$, and the transition rate of a particle from state 2 to state 1 is $t_{21}$. In the steady state, the probability of finding the particle in state 1 is
1. $\frac{t_{21}}{t_{12}+t_{21}}$
2. $\frac{t_{12}}{t_{12}+t_{21}}$
3. $\frac{t_{12}t_{21}}{t_{12}+t_{21}}$
4. $\frac{t_{12}-t_{21}}{t_{12}+t_{21}}$

Let $P_1$ be the probability of transition from state 1 to state 2 and $P_2$ be the probability of transition from state 2 to state 1. Then the rate equations can be written as $$\frac{d}{dt}\begin{pmatrix}P_1\\P_2\end{pmatrix}=\begin{pmatrix}-t_{21}&t_{12}\\t_{21}&-t_{12}\end{pmatrix}\begin{pmatrix}P_1\\P_2\end{pmatrix}$$ In steady state, we have,$$\frac{d}{dt}\begin{pmatrix}P_1\\P_2\end{pmatrix}=0$$ Hence, $$\begin{pmatrix}-t_{21}&t_{12}\\t_{21}&-t_{12}\end{pmatrix}\begin{pmatrix}P_1\\P_2\end{pmatrix}=0$$ This gives, $$t_{21}P_1=t_{12}P_2$$ Hence, $$P_1=\frac{t_{12}}{t_{21}}P_2$$ $$P_1=\frac{t_{12}}{t_{21}}(1-P_1)$$ $$P_1\left(1+\frac{t_{12}}{t_{21}}\right)=\frac{t_{12}}{t_{21}}$$ $$P_1=\frac{t_{21}}{t_{12}+t_{21}}$$

Hence, answer is (A)

33. The viscosity $\eta$ of a liquid is given by Poiseuille's formula $\eta=\frac{\pi Pa^4}{8lV}$. Assume that $l$ and $V$ can be measured very accurately, but the pressure $P$ has an rms error of 1% and the radius $a$ has an independent rms error of 3%. The rms error of viscosity is closest to
1. 2%
2. 4%
3. 12%
4. 13%

If $\Delta x$ and $\Delta y$ are rms errors in $x$ and $y$ then error in $z=x^my^n$ is given by $$\frac{\Delta z}{z}=\sqrt{\left(\frac{m\Delta x}{x}\right)^2+\left(\frac{n\Delta y}{y}\right)^2}$$ Let us take $\frac{\pi }{8lV}=1$. Hence, $\eta=Pa^4$. $$\Delta \eta=\eta\sqrt{\left(\frac{\Delta P}{P}\right)^2+\left(\frac{4\Delta a}{a}\right)^2}$$ Let $P=1$ and $a=1$. Hence, $\eta=1$ $$\Delta \eta=\sqrt{\left(\Delta P\right)^2+\left(4\Delta a\right)^2}$$ $$\Delta \eta=\sqrt{\left(\frac{1}{100}\right)^2+\left(4\times\frac{3}{100}\right)^2}$$ $$\Delta \eta=\frac{1}{100}\sqrt{145}$$ $$\Delta \eta=\frac{12.04}{100}$$ $$\Delta \eta=12.04\%$$

Hence, answer is (C)

34. The first order diffraction peak of a crystalline solid occurs at a scattering angle of $30^0$ when the diffraction pattern is recorded using an x-ray beam of wavelength 0.15 nm. If the error in measurements of the wavelength and the angle are 0.01 nm and $1^0$ respectively, then the error in calculating the inter- planar spacing will approximately be
1. $1.1\times 10^{-2}\:nm$
2. $1.3\times 10^{-4}\:nm$
3. $2.5\times 10^{-2}\:nm$
4. $2.0\times 10^{-3}\:nm$

$$2d\sin\theta=\lambda$$ $$d=\frac{\lambda}{2\sin\theta}$$ Let $z=\sin\theta$. $$d=\frac{\lambda}{2z}$$ \begin{align*} \Delta z&=\cos\theta\Delta\theta\\ &=\sin{30}\times 1^0\\ &=\frac{\sqrt{3}}{2}\frac{\pi}{180} \end{align*} \begin{align*} \Delta d&=\left|\frac{\lambda}{2z}\right|\sqrt{\left(\frac{\Delta\lambda}{\lambda}\right)^2+\left(\frac{\Delta z}{z}\right)^2}\\ &={\textstyle\left|\frac{0.15}{2\sin{30}}\right|\sqrt{\left(\frac{0.01}{0.15}\right)^2+\left(\frac{\frac{\sqrt{3}}{2}\frac{\pi}{180}}{\sin{30}}\right)^2}}\\ &=\left|0.15\right|\sqrt{\frac{1}{225}+\frac{3\pi^2}{180^2}}\\ &=0.011=1.1\times 10^{-2}\:nm \end{align*}

Hence, answer is (A)

35. The partition function of a system of $N$ Ising spins is $Z=\lambda_1^N+\lambda_2^N$, where $\lambda_1$ and $\lambda_2$ are functions of temperature, but are independent of $N$. If $\lambda_1 > \lambda_2$, the free energy per spin in the limit $N\rightarrow\infty$ is
1. $-k_BT\ln{\left(\frac{\lambda_1}{\lambda_2}\right)}$
2. $-k_BT\ln{\lambda_2}$
3. $-k_BT\ln{\left(\lambda_1\lambda_2\right)}$
4. $-k_BT\ln{\lambda_1}$

\begin{align*} F&=-kT\ln{Z}\\ &=-kT\ln{\left(\lambda_1^N+\lambda_2^N\right)}\\ F&=-kT\ln{\left(\lambda_1^N\left[1+\left(\frac{\lambda_2}{\lambda_1}\right)^N\right]\right)}\\ \end{align*} For $\lambda_1 > \lambda_2$, $\frac{\lambda_2}{\lambda_1} < 1$, in the limit $N\rightarrow\infty$, $\frac{\lambda_2}{\lambda_1} = 0$ $$F=-kT\ln{\left(\lambda_1^N\right)}=-NkT\ln{\lambda_1^N}$$ $$\frac{F}{N}=-kT\ln{\lambda_1}$$

Hence, answer is (D)

36. The Hamiltonian of a system of $N$ non-interacting spin-1/2 particles is $H=-\mu_0B\sum_iS_i^z$, where $S_i^z=\pm1$ are the components of $i^{th}$ spin along an external magnetic field $B$. At a temperature $T$ such that $e^{\mu_0B/k_BT}=2$, the specific heat per particle is
1. $\frac{16}{25}k_B$
2. $\frac{8}{25}k_B\ln2$
3. $k_B\left(\ln2\right)^2$
4. $\frac{16}{25}k_B\left(\ln2\right)^2$

Because each spin is independent of the others and distinguishable, we can find the partition function for one spin, $Z_1$, and use the relation $Z_N=Z_1^N$ to obtain, the partition function for $N$ spins. \begin{align*} Z_1&=\sum\limits_{s=\pm1}e^{-\beta\mu_0 Bs}\\ &=e^{-\beta\mu_0 B}+e^{\beta\mu_0 B}\\ &=2\cosh{\beta\mu_0 B} \end{align*} $$Z_N=\left(2\cosh{\beta\mu_0 B}\right)^N$$ Mean energy \begin{align*} E&=-\frac{\partial\ln{Z_N}}{\partial\beta}\\ &=-N\mu B\tanh{\beta\mu_0 B} \end{align*} \begin{align*} C_V&=\frac{\partial E}{\partial T}\\ &=\left(\frac{\mu_0 B}{k T}\right)^2Nk\:sech^2{\beta\mu_0 B} \end{align*} $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{4}{\left(e^{\frac{\mu_0 B}{kT}}+e^{-\frac{\mu_0 B}{kT}}\right)^2}$$ Since, $e^{\mu_0B/k_BT}=2$ $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{4}{\left(2+\frac{1}{2}\right)^2}$$ $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{16}{25}$$ $$\frac{C_v}{N}=\left(\ln2\right)^2k\frac{16}{25}$$

Hence, answer is (D)

37. The ground state energy of a particle in an infinite square well is 1eV. If four particles obeying Bose-Einstein statistics are kept in this well, then the ground state energy will be :
1. 30 eV
2. 10 eV
3. 4 eV
4. $\frac{1}{4}$ eV

As particles obey Bose-Einstein statistics, all particles will be the ground state. Hence, energy will be 4 eV.

Hence, answer is (C)

38. Consider a system in contact with a heat and particle reservoir. It may be unoccupied or occupied by one particle with energy 0 and $\epsilon$. The grand partition function will be ($\beta=1/kT$)
1. $Z(\mu,T)=e^{-\epsilon\beta}$
2. $Z(\mu,T)=\left(1+e^{-\epsilon\beta}\right)^{-1}$
3. $Z(\mu,T)=1+e^{-\epsilon\beta}$
4. $Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}$

$$Z(\mu,T)=\sum\limits_{i} e^{-\beta n_i(\epsilon_i-\mu)}$$ where $i$ runs over every microstate, $n_i$ number particles occupying the microstate $i$, $\epsilon_i$ is energy of a particles in the $i^{th}$ microstate

Let us write all microstates

Microstate no. $i$	State with enrgy 0	State with enrgy $\epsilon$	energy of microstate $\epsilon_i$	$n_i$
1	0	0	0	0
2	1	0	0	1
3	0	1	$\epsilon$	1

\begin{align*} Z(\mu,T)&= e^{-\beta \times0\times(0-\mu)}+e^{-\beta \times1\times(0-\mu)}\\ &+e^{-\beta\times 1\times(\epsilon-\mu)} \end{align*} $$Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}$$

Hence, answer is (D)

39. Consider a system of $N$ linear polyatomic molecules. Each molecule consists of $n$ atoms. At high temperature the vibrational contribution to the specific heat is
1. $(3n-5)kN$
2. $(3n-5)\frac{kN}{2}$
3. $(3n-6)kN$
4. $(3n-6)\frac{kN}{2}$

For a polyatomic molecule containing $n$ atoms, the total number of coordinates is $3n$. Out of these, 3 coordinates are taken up for the translational motion of the molecule as a whole and 3 for rotational motion. Hence, vibrational degrees of freedom is $(3n - 6)$. However, for a linear molecule, the rotational motion along the molecular axis is not meaningful as the rotated configuration is indistinguishable from the original configuration. Therefore, for a linear molecule, there are two rotational degrees of freedom and $(3n- 5)$ vibrational degrees of freedom. According to equipartition theorem, each degree of freedom has energy $\frac{1}{2}kT$. Hence, vibrational energy consider of a linear molecule is given by $(3n-5)\frac{kT}{2}$. Hence energy of $N$ linear molecules is $$E=(3n-5)\frac{NkT}{2}$$ $$C_v=\frac{\partial E}{\partial T}=(3n-5)\frac{Nk}{2}$$

Hence, answer is (B)

40. The partition function $z(T)$ of a linear quantum mechanical harmonic oscillator in thermal equilibrium with a heat reservoir at temperature $T$ is given by:
1. $\frac{e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}$
2. $\frac{e^{-\beta\hbar\omega}}{1+e^{-\beta\hbar\omega}}$
3. $\frac{e^{-\beta\hbar\omega/2}}{1+e^{-\beta\hbar\omega}}$
4. $\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}$
where, $\hbar\omega > kT$

$$E_n=(n+\frac{1}{2})\hbar\omega$$ \begin{align*} z&=\sum\limits_{n=0}^{\infty}e^{-\beta E_n}\\ &=\sum\limits_{n=0}^{\infty}e^{-\beta (n+\frac{1}{2})\hbar\omega}\\ &=e^{ \frac{-\beta\hbar\omega}{2}}\sum\limits_{n=0}^{\infty}e^{-\beta n\hbar\omega}\\ z&=e^{ \frac{-\beta\hbar\omega}{2}}\left(1+e^{-\beta \hbar\omega}+e^{-2\beta \hbar\omega}+\cdots\right)\\ \end{align*} This is a geometric series where each term is obtained from previous term by multiplying $e^{-\beta \hbar\omega}$. Hence, $$\left(1+e^{-\beta \hbar\omega}+e^{-2\beta \hbar\omega}+\cdots\right)=\frac{1}{1-e^{-\beta\hbar\omega}}$$ $$z=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}$$

Hence, answer is (D)

41. Consider a configuration of a system of 10 distinguishable particles in which there are 3 particles in state 1, 3 particles in state 2 and 4 particles in state 3. The total number of microstates is :
1. 4200
2. 864
3. 102060
4. 360

Number of ways of arranging $N$ distinguishable particles into groups of $N_j$ each, such that $N_1$ in state 1, $N_2$ in state 2, etc. and $N_j$ in the $j^{th}$ state are given by $$W(N)=\frac{N!}{N_1!N_2!\dots N_j!}$$ $$W(10)=\frac{10!}{3!\times 3!\times 4!}=4200$$

Hence, answer is (A)

42. If the temperature of a black body enclosure is tripled, the number of photons will increase by a factor of :
1. 2
2. 9
3. 8
4. 27

Number of photons enclosed in black body are given by $$N=\left(\frac{16\pi k^3\zeta(3)}{c^3h^3}\right)VT^3$$ where, $V$ is volume and $T$ is temperature of black body and $\zeta(3)$ is Riemann zeta function. Hence, $$N\propto VT^3$$

Hence, answer is (D)

43. In a counting experiment to determine the statistics obeyed by the $\beta$ particles emitted by a radioactive substance, the number of $\beta$ particles counted in 50 seconds time interval was repeatedly measured 100 times. The statistical ensemble in this case consists of the following number of members:
1. 50
2. 100
3. 5000
4. 2

As the reading is repeated 100 times, there will be 100 members in the ensembles.

Hence, answer is (B)

44. Each of the two isolated vessels, $A$ and $B$ of fixed volumes, contains $N$ molecules of a perfect monatomic gas at a pressure $P$. The temperatures of $A$ and $B$ are $T_1$ and $T_2$, respectively. The two vessels are brought into thermal contact. At equilibrium, the change in entropy is
1. $\frac{3}{2}Nk_B\ln{\left[\frac{T_1^2+T_2^2}{4T_1T_2}\right]}$
2. $Nk_B\ln{\left(\frac{T_2}{T_1}\right)}$
3. $\frac{3}{2}Nk_B\ln{\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]}$
4. $2Nk_B$

Energy of $N$ monatomic gas molecules is given by $$E=\frac{3}{2}Nk_BT$$ $$E_A=\frac{3}{2}Nk_BT_1$$ $$E_B=\frac{3}{2}Nk_BT_2$$ $$C_A=C_B=C=\frac{3}{2}Nk_B$$ When two bodies are in thermal contact, final temperature is given by $$T_f=\frac{C_AT_1+C_BT_2}{C_A+C_B}$$ $$T_f=\frac{T_1+T_2}{2}$$ Change in entropy of body $A$ is \begin{align*} \Delta S_A&=\int_{T_1}^{T_f}\frac{dQ}{T}\\ &=\int_{T_1}^{T_f}\frac{CdT}{T}\\ &=\frac{3}{2}Nk_B\ln{\frac{T_f}{T_1}} \end{align*} $$\Delta S_B=\frac{3}{2}Nk_B\ln{\frac{T_f}{T_2}}$$ Total change in entropy is $$\Delta S=\Delta S_A+\Delta S_B$$ \begin{align*} \Delta S&=\frac{3}{2}Nk_B\left(\ln{\frac{T_f}{T_1}}+\ln{\frac{T_f}{T_2}} \right)\\ &=\frac{3}{2}Nk_B\ln{\left[\frac{T_f^2}{T_1T_2}\right]} \\ &=\frac{3}{2}Nk_B\ln{\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]} \\ \end{align*}

Hence, answer is (C)

45. The mean internal energy of a one-dimensional classical harmonic oscillator in equilibrium with a heat bath of temperature $T$ is
1. $\frac{1}{2}k_BT$
2. $k_BT$
3. $\frac{3}{2}k_BT$
4. $3k_BT$

Energy of harmonic oscillator is given by $$E=\frac{p_x^2}{2m}+\frac{1}{2}kx^2$$ According to equipartition theorem each quadratic degrees of freedom has energy $\frac{1}{2}k_BT$. Energy is quadratic in $p_x$ and also in $x$. Hence, $$E=\frac{1}{2}k_BT+\frac{1}{2}k_BT=k_BT$$

Hence, answer is (B)

46. The internal energy of $n$ moles of a gas is given by $E = \frac{3}{2}nRT- \frac{a}{V}$, where $V$ is the volume of the gas at temperature $T$ and $a$ is a positive constant. One mole of the gas in state $(T_1, V_1)$ is allowed to expand adiabatically into vacuum to a final state $(T_2, V_2)$. The temperature $T_2$ is
1. $T_1+Ra\left(\frac{1}{V_2}+\frac{1}{V_1}\right)$
2. $T_1-\frac{2}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$
3. $T_1+\frac{2}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$
4. $T_1-\frac{1}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$

According to first of thermodynamics $$dU=dQ+dW$$ For adiabatic process there is no heat transfer. Hence, $dQ=$ $$dU=dW$$ For a free expansion of gas in vacuum, work done is zero, $dW=0$ $$dU=E_2-E_1=0$$ $$E_2=E_1$$ $$\frac{3}{2}nRT_2- \frac{a}{V_2}=\frac{3}{2}nRT_1- \frac{a}{V_1}$$ $$\frac{3}{2}nRT_2 =\frac{3}{2}nRT_1- \frac{a}{V_1}+\frac{a}{V_2}$$ $$T_2 =T_1+\frac{2a}{3nR}\left( \frac{1}{V_2}-\frac{1}{V_1}\right)$$ for one mole $n=1$ $$T_2 =T_1+\frac{2a}{3R}\left( \frac{1}{V_2}-\frac{1}{V_1}\right)$$

Hence, answer is ()

47. A monatomic crystalline solid comprises of $N$ atoms, out of which $n$ atoms are in interstitial positions. If the available interstitial sites are $N'$, the number of possible microstates is
1. $\frac{(N'+n)!}{n!N!}$
2. $\frac{N!}{n!(N+n)!}\frac{N'!}{n!(N'+n)!}$
3. $\frac{N!}{n!(N'-n)!}$
4. $\frac{N!}{n!(N-n)!}\frac{N'!}{n!(N'-n)!}$

$\frac{N!}{n!(N-n)!}\frac{N'!}{n!(N'-n)!}$

Hence, answer is (D)

48. A system of $N$ localized, non-interacting spin $1/2$ ions of magnetic moment $\mu$ each is kept in an external magnetic field $H$. If the system is in equilibrium at temperature $T$, the Helmholtz free energy of the system is
1. $Nk_BT\ln{\left(\cosh{\frac{\mu H}{k_BT}}\right)}$
2. $-Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)}$
3. $Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)}$
4. $-Nk_BT\ln{\left(2\sinh{\frac{\mu H}{k_BT}}\right)}$

A spin $\frac{1}{2}$ has two possible orientations. (These are the two possible values of the projection of the spin on the z axis: $m_s=\pm1/2$.) Associated with each spin is a magnetic moment which has the two possible values $\pm\mu$. Hence, in a magnetic field the two states are of different energy ($E=-\vec\mu\cdot\vec H$). Hence, single particle partition function is given by \begin{align*} z&=\sum e^{-E/k_BT}\\ &=e^{-\mu H/k_BT}+e^{\mu H/k_BT}\\ &=2\cosh{\frac{\mu H}{k_BT}} \end{align*} Hence, N particle partition function is $$Z=z^N=\left(2\cosh{\frac{\mu H}{k_BT}}\right)^N$$ Helmholtz free energy is given by \begin{align*} F&=-k_BT\ln{Z}\\ &=-Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)} \end{align*}

Hence, answer is (B)

49. The phase diagram of a free particle of mass $m$ and kinetic energy $E$, moving in a one-dimensional box with perfectly elastic walls at $x = 0$ and $x = L$, is given by


For free particle $E=\frac{p_x^2}{2m}$ $$p_x=\pm\sqrt{2mE}$$ Particle bounce forth and back between $x=0$ and $x=L$

Hence, answer is (A)

50. 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)

51. 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 $$M^1L^0T^0=\left[M^0L^{n_1}T^{-n_1}\right]\left[M^{-n_2}L^{3n_2}T^{-2n_2}\right]\left[M^{n_3}L^{2n_3}T^{-n_3}\right]$$ $$n_{3}-n_{2}=1$$ $$ n_{1}+3n_{2}+2n_{3}=0$$ $$ -n_{1}-2n_{2}-n_{3}=0$$ $$\Rightarrow n_{1}=1/2,n_{2}=-1/2,n_{3}=1/2$$ $$m_p=\sqrt{\frac{c\hbar}{G}}$$ $$T_p=\sqrt{\frac{\hbar c^5}{k_B^2G}}$$

Hence, answer is (A)

52. 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)

53. 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)

54. 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)

55. 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)

56. 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)

57. 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)

58. If energy of a two-dimensional simple harmonic oscillator $E=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{1}{2}m\omega^2(x^2+y^2)$ is fixed to be $3\hbar\omega$, the entropy is given by ($k_B$ is Boltzmann constant):
1. $k_B\ln3$
2. $2k_B\ln3$
3. $k_B\ln2$
4. zero

The energy of a two dimensional oscillator is $$E=\left(n_x+n_y+1\right)\hbar\omega$$ The energy is fixed to be $3\hbar\omega$. Clearly, following combinations of $n_x$, $n_y$ will give energy $3\hbar\omega$
$$n_x=2, n_y=0$$ $$n_x=1, n_y=1$$ $$n_x=0, n_y=2$$ Hence, number of miscrstates accessible is $\Omega=3$ $$S=k_B\ln\Omega=k_B\ln3$$

Hence, answer is (A)

59. The equation of state for photon gas is:
1. $pV=\frac{5}{3}E$
2. $pV=\frac{2}{3}E$
3. $pV=\frac{1}{3}E$
4. $pV=\hbar\omega$ for some fixed frequency $\omega$

$pV=\frac{1}{3}E$

Hence, answer is (C)

60. The energy density for photons in a cavity is proportional to
1. $T^3$
2. $T$
3. $T^4$
4. $T^{4/3}$

According to Stefan Boltzmann law energy density is proportional to fourth power of temperature.

Hence, answer is (C)

61. Let $\rho$ be the density matrix for a system. Then
1. $Tr(\rho)=0$
2. $Tr(\rho) < 0$
3. $0\leq Tr(\rho) < 1$
4. $Tr(\rho)=1$

The Density matrix for the pure state $|\psi>$ is given by $$\rho=|\psi><\psi|$$ This density matrix has the following properties:

1. $\hat\rho^2=\hat\rho$ projector
2. $\hat\rho^\dagger=\hat\rho$ hermiticity
3. $T_r\hat\rho=1$ normalization
4. $\hat\rho\geq 0$ positivity

Hence, answer is (D)

62. A system has only two energy levels $E_1$ and $E_2$. In equilibrium at temperature $T$, the number of particles occupying level $E_1$ is double of those occupying level $E_2$. The value of $E_2-E_1$ must be ($k$ is Boltzmann constant):
1. $kT\ln2$
2. $kT\ln3$
3. $3kT$
4. $2kT$

$$n_1\propto e^{-E_1/kT}$$ $$n_2\propto e^{-E_2/kT}$$ $$\frac{n_1}{n_2}=e{(E_2-E_1)/kT}=2$$ $$E_2-E_1=kT\ln2$$

Hence, answer is (A)

63. The quantities (i) isothermal compressibility (ii) volume coefficient of expansion are :
1. Extensive and intensive respectively
2. Intensive and extensive respectively
3. Both extensive
4. Both intensive

The properties that increase proportionally with the size of the system, such as $n$ and $V$, called extensive properties, and those defined for each small region in the system, such as $P$ and $T$, called intensive properties. Terms that are added together or are on opposite sides of an equal sign must contain the same number of extensive variables. The quotient of two extensive variables is an intensive variable.

Isothermal compressibility is given by $K=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$ and volume coefficient of expansion is given by $\alpha=-\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$.

Both of these are intensive properties.

Hence, answer is (D)

64. The chemical potential in classical limit is:
1. Zero
2. Negative
3. Positive
4. Complex quantity

Negative

Hence, answer is (B)

65. Van der Waals equation for one mole is $\left(p+\frac{a}{V^2}\right)(V-b)=RT$. The equation for $n$ moles would be:
1. $\left(p+\frac{an^2}{V^2}\right)(V-nb)=RT$
2. $\left(p+\frac{a^2}{V^2}\right)(V-b)=nRT$
3. $\left(p+\frac{an^2}{V^2}\right)(V-nb)=nRT$
4. $\left(p+\frac{a}{n2V^2}\right)(nV-b)=nRT$

$\left(p+\frac{an^2}{V^2}\right)(V-nb)=nRT$

Hence, answer is (C)