Problem set 57

1. 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)

2. Which one of the following sets of Maxwell's equations for time independent charge density $\rho$ and current density $\vec J$ is correct?
1. $$\begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=-\frac{\partial\vec B}{\partial t}\\ \vec\nabla\times\vec B&=\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*}$$
2. $$\begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}$$
3. $$\begin{align*} \vec\nabla\cdot\vec E&=0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}$$
4. $$\begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=\mu_0|\vec J |\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*}$$

Maxwell equations are given by $$\begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=-\frac{\partial\vec B}{\partial t}\\ \vec\nabla\times\vec B&=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*}$$ For time independent case $\frac{\partial\vec B}{\partial t}=0$ and $\frac{\partial\vec E}{\partial t}=0$. Hence, Maxwell equations become $$\begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}$$

Hence, answer is (B)

3. 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)

4. 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)

5. 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)