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A doughnut-shaped solenoid winding has dimensions $R=1$ meter, diameter of the loop $=10 \mathrm{~cm}$, and the number of windings $=1000$. If a current of $10$ amperes runs through the wire, what is the magnitude of the force on one loop? | The constitutive equation for electric fields in a dielectric medium moving with velocity $v$ in a magnetic field $B$ is
$$
\mathbf{D}=k \varepsilon_{0} \mathbf{E}+\varepsilon_{0}(k-1) \mathbf{v} \times \mathbf{B},
$$
where $k$ is its relative dielectric constant. For a point distance $r$ from the axis of rotation, $\mathbf{v}=\boldsymbol{\omega} \times \mathbf{r}$ and $\mathbf{v} \times \mathbf{B}=(\boldsymbol{\omega} \cdot \mathbf{B}) \mathbf{r}-(\mathbf{r} \cdot \mathbf{B}) \boldsymbol{\omega}=\omega B \mathbf{r}$ as $\mathbf{r}$ is perpendicular to $B$. As there are no free charges, Gauss' flux theorem $\oint \mathbf{D} \cdot d \mathbf{S}=0$ gives $\mathbf{D}=0$. Then from $\mathbf{D}=\varepsilon_{0} \mathbf{L}+\mathbf{P}$ we get
$$
\mathbf{P}=-\varepsilon_{0} \mathbf{E}=\varepsilon_{0}\left(1-\frac{1}{k}\right) \omega B \mathbf{r} .
$$
Hence the volume bound charge density is
$$
\rho^{\prime}=-\nabla \cdot \mathbf{P}=-\frac{1}{r} \frac{\partial}{\partial r}\left(r P_{r}\right)=-2 \varepsilon_{0}\left(1-\frac{1}{k}\right) \omega B
$$
and the surface bound charge density is
$$
\sigma^{\prime}=P_{r}=\varepsilon_{0}\left(1-\frac{1}{k}\right) \omega B a,
$$
as $r=a$ for the cylinder's surface.By symmetry and using Ampère's circuital law, we obtain the magnetic induction in a doughnut-shaped solenoid:
$$
B=\frac{\mu_{0} N I}{2 \pi r},
$$
where $r$ is the distance from the center of the doughnut. Consider a small section of length $d l$ of the solenoid. This section contains $\frac{N}{2 \pi R} d l$ turns of the winding, where $R$ is the radius of the doughnut. Take as current element a segment of this section which subtends an angle $d \theta$ at the axis of the solenoid:
$$
\Delta I=\frac{N I d l}{2 \pi R} \rho d \theta,
$$
where $\theta$ is the angle made by the radius from the axis to the segment and the line from the axis to center of the doughnut and $\rho$ is the radius of a loop of winding. The magnetic force on the current element is in the radial direction and has magnitude
$$
\begin{aligned}
d F &=\Delta I \cdot \frac{B}{2}=\frac{N I \rho}{4 \pi R} B d \theta d l \\
&=\frac{\mu_{0} N^{2} I^{2} \rho}{8 \pi^{2} R r} d \theta d l,
\end{aligned}
$$
where $B / 2$ is used, instead of $B$, because the magnetic field established by the current element itself has to be taken out from the total field. Note that $d F$ is perpendicular to the surface of the solenoid and only its component $d F \cdot \cos \theta$ along the line from the axis to the center of the doughnut is not canceled out with another element at $2 \pi-\theta$. As
$$
r=R+\rho \cos \theta,
$$
we have the total force on the doughnut
$$
\begin{aligned}
F &=\int \cos \theta d F \\
&=\frac{\mu_{0} N^{2} I^{2}}{8 \pi^{2} R} \int_{0}^{2 \pi R} d l \int_{0}^{2 \pi} \frac{\rho \cos \theta}{R+\rho \cos \theta} d \theta \\
&=\frac{\mu_{0} N^{2} I^{2}}{4 \pi} \int_{0}^{2 \pi}\left(1-\frac{R}{R+\rho \cos \theta}\right) d \theta \\
&=\frac{\mu_{0} N^{2} I^{2}}{4 \pi} \int_{0}^{2 \pi}\left[1-\left(1+\frac{\rho}{R} \cos \theta\right)^{-1}\right] d \theta \\
&=\frac{\mu_{0} N^{2} I^{2}}{2}\left\{1-\left[1-\left(\frac{\rho}{R}\right)^{2}\right]^{-\frac{1}{2}}\right\} \\
&=\frac{4 \pi \times 10^{-7} \times 1000^{2} \times 10^{2}}{2}\left[1-\frac{1}{\sqrt{1-0.05^{2}}}\right] \\
&=-0.079 \mathrm{~N}
\end{aligned}
$$
Hence, the force on one loop is
$$
\frac{F}{N}=-\frac{0.079}{1000}=-7.9 \times 10^{-5} \mathrm{~N}
$$
and points to the center of the doughnut. |
In a measurement of $e / m$ for electron using a Thomson type apparatus, i.e., crossed electric and magnetic fields in a cathode-ray tube, it is noted that if the accelerating potential difference is sufficiently large, the ratio $\mathrm{e} / \mathrm{m}$ becomes one-half as large as the accepted value. Take $e / m_{0}=1.8 \times 10^{11}$ $\mathrm{C} / \mathrm{kg}$.
Find the accelerating potential difference $V$ which causes $e / m$ to be one-half its accepted value. Take $c=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$. | A Thomson type apparatus is shown schematically in Fig. 2.79, where $V_{1}$ is the accelerating voltage and $V_{2}$ is the deflecting voltage.
Fig. $2.79$ With the addition of a magnetic field $B$ as shown, the electromagnetic field has the action of a velocity-filter. With given values of $V_{1}$ and $V_{2}$, we adjust the magnitude of $B$ so that the electrons strike the center $O$ of the screen. At this time the velocity of the electron is $v=E / B$ (since $e E=e v B$ ). Afterward the magnetic field $B$ is turned off and the displacement $y_{2}$ of the electrons on the screen is measured. The ratio $e / m$ is calculated as follows:
$$
\begin{aligned}
y_{1} &=\frac{1}{2} \cdot \frac{e E}{m}\left(\frac{L}{v}\right)^{2}, \\
y_{2}=\frac{D+\frac{L}{2}}{L / 2} y_{1} &=\frac{e E}{m v^{2}}\left(\frac{L^{2}}{2}+L D\right)=\frac{e}{m} \cdot \frac{d B^{2}}{V_{2}}\left(\frac{L^{2}}{2}+L D\right),
\end{aligned}
$$
giving
$$
e / m=\frac{V_{2} y_{2}}{d B^{2}\left(\frac{L^{2}}{2}+L D\right)} .
$$
When the accelerating voltage is very large, relativistic effects must be considered. From energy conversation
$$
e V_{1}+m_{0} c^{2}=m c^{2},
$$
we find
$$
V_{1}=\left(\frac{m}{e}-\frac{m_{0}}{e}\right) c^{2} .
$$
As $\frac{e}{m}=\frac{1}{2} \frac{e}{m_{0}}$, the accelerating voltage is
$$
V_{1}=\frac{m_{0} c^{2}}{e}=\frac{9 \times 10^{16}}{1.8 \times 10^{11}}=5 \times 10^{5} \mathrm{~V}
$$ |
In high energy proton-proton collisions, one or both protons may "diffractively dissociate" into a system of a proton and several charged pions. The reactions are
(1) $p+p \rightarrow p+(p+n \pi)$,
(2) $p+p \rightarrow(p+n \pi)+(p+m \pi)$.
Here $n$ and $m$ count the number of produced pions.
In the laboratory frame, an incident proton of total energy $E$ (the projectile) strikes a proton at rest (the target). Find the incident proton energy $E_{0}$ that is the minimum energy for reaction (2) to take place when both protons dissociate into a proton and 4 pions.
$$
m_{\pi}=0.140 \mathrm{GeV}, \quad m_{p}=0.938 \mathrm{GeV} .
$$ | The energy for the reaction
$$
p+p \rightarrow p+(p+4 \pi)
$$
is minimum when all the final particles are at rest in an inertial frame, particularly the center of mass frame $\Sigma^{\prime}$. Then in the laboratory frame $\Sigma$,
$$
E^{2}-p^{2}=\left(E_{0}+m_{p}\right)^{2}-\left(E_{0}^{2}-m_{p}^{2}\right)=2 m_{p} E_{0}+2 m_{p}^{2},
$$
and in $\Sigma^{\prime}$,
$$
E^{\prime 2}-p^{\prime 2}=\left(2 m_{p}+4 m_{\pi}\right)^{2}
$$
so that
$$
2 m_{p} E_{0}=2 m_{p}^{2}+16 m_{p} m_{\pi}+16 m_{\pi}^{2}
$$
giving
$$
E_{0}=\frac{m_{p}^{2}+8 m_{p} m_{\pi}+8 m_{\pi}^{2}}{m_{p}}=2.225 \mathrm{GeV}
$$
as the minimum energy the incident proton must have to cause the reaction.Since both the initial particles are protons and the final state particles are the same as before, the minimum energy remains the same, $2.225 \mathrm{GeV}$.For the reaction
$$
p+p \rightarrow(p+4 \pi)+(p+4 \pi)
$$
we have
$$
\left(E_{0}+m_{p}\right)^{2}-\left(E_{0}^{2}-m_{p}^{2}\right)=\left(2 m_{p}+8 m_{\pi}\right)^{2},
$$
giving the minimum incident energy as
$$
E_{0}=\frac{m_{p}^{2}+16 m_{p} m_{\pi}+32 m_{\pi}^{2}}{m_{p}}=3.847 \mathrm{GeV} .
$$ |
Consider a spinless particle represented by the wave function
$$
\psi=K(x+y+2 z) e^{-\alpha r}
$$
where $r=\sqrt{x^{2}+y^{2}+z^{2}}$, and $K$ and $\alpha$ are real constants.
What is the total angular momentum of the particle?
You may find the following expressions for the first few spherical harmonics useful:
$$
\begin{aligned}
&Y_{0}^{0}=\sqrt{\frac{1}{4 \pi}}, \quad Y_{1}^{\pm 1}=\mp \sqrt{\frac{3}{8 \pi}} \sin \theta e^{\pm i \phi}, \\
&Y_{1}^{0}=\sqrt{\frac{3}{4 \pi}} \cos \theta, \quad Y_{2}^{\pm 1}=\mp \sqrt{\frac{15}{8 \pi}} \sin \theta \cos \theta e^{\pm i \phi} .
\end{aligned}
$$ | The wave function may be rewritten in spherical coordinates as
$$
\psi=K r(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta) e^{-\alpha r}
$$
its angular part being
$$
\psi(\theta, \phi)=K^{\prime}(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta),
$$
where $K^{\prime}$ is the normalization constant such that
$$
K^{\prime 2} \int_{\theta}^{\pi} d \theta \int_{0}^{2 \pi} \sin \theta(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta)^{2} d \phi=1 \text {. }
$$
Since
$$
\cos \phi=\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right), \quad \sin \phi=\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right)
$$
we have
$$
\begin{aligned}
\psi(\theta, \phi) &=K^{\prime}\left[\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right) \sin \theta+\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right) \sin \theta+\cos 2 \theta\right], \\
&=K^{\prime}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$
The normalization condition and the orthonormality of $Y_{l}^{m}$ then give
$$
K^{\prime 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}+\frac{1}{2} \cdot \frac{8 \pi}{3}+4 \cdot \frac{4 \pi}{3}\right]=1
$$
or
$$
K^{\prime}=\sqrt{\frac{1}{8 \pi}},
$$
and thus
$$
\begin{aligned}
\psi(\theta, \phi)=& \sqrt{\frac{1}{8 \pi}}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}\right.\\
&\left.+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$
The total angular momentum of the particle is
$$
\sqrt{\left\langle\mathbf{L}^{2}\right\rangle}=\sqrt{l(l+1)} \hbar=\sqrt{2} \hbar .
$$
as the wave function corresponds to $l=1$. |
In high energy proton-proton collisions, one or both protons may "diffractively dissociate" into a system of a proton and several charged pions. The reactions are
(1) $p+p \rightarrow p+(p+n \pi)$,
(2) $p+p \rightarrow(p+n \pi)+(p+m \pi)$.
Here $n$ and $m$ count the number of produced pions.
In the laboratory frame, an incident proton of total energy $E$ (the projectile) strikes a proton at rest (the target). Find the incident proton energy $E_{0}$ that is the minimum energy for reaction (1) to take place when the target dissociates into a proton and 4 pions.
$$
m_{\pi}=0.140 \mathrm{GeV}, \quad m_{p}=0.938 \mathrm{GeV} .
$$ | The quantity $E^{2}-p^{2}$ for a system, where we have taken $c=1$ for convenience, is invariant under Lorentz transformation. If the system undergoes a nuclear reaction that conserves energy and momentum, the quantity will also remain the same after the reaction. In particular for a particle of rest mass $m$,
$$
E^{2}-p^{2}=m^{2} .
$$
The energy for the reaction
$$
p+p \rightarrow p+(p+4 \pi)
$$
is minimum when all the final particles are at rest in an inertial frame, particularly the center of mass frame $\Sigma^{\prime}$. Then in the laboratory frame $\Sigma$,
$$
E^{2}-p^{2}=\left(E_{0}+m_{p}\right)^{2}-\left(E_{0}^{2}-m_{p}^{2}\right)=2 m_{p} E_{0}+2 m_{p}^{2},
$$
and in $\Sigma^{\prime}$,
$$
E^{\prime 2}-p^{\prime 2}=\left(2 m_{p}+4 m_{\pi}\right)^{2}
$$
so that
$$
2 m_{p} E_{0}=2 m_{p}^{2}+16 m_{p} m_{\pi}+16 m_{\pi}^{2}
$$
giving
$$
E_{0}=\frac{m_{p}^{2}+8 m_{p} m_{\pi}+8 m_{\pi}^{2}}{m_{p}}=2.225 \mathrm{GeV}
$$
as the minimum energy the incident proton must have to cause the reaction. |
A Fresnel sone plate is made by dividing a photographic image into 5 separate sones.
The first sone consists of an opaque circular disc of radius $r_{1}$. The second is a concentric transparent ring from $r_{1}$ to $r_{2}$ followed by an opaque ring from $r_{2}$ to $r_{3}$, a second transparent ring from $r_{3}$ to $r_{4}$ and a final sone from $r_{4}$ to infinity that is black. The radii $r_{1}$ to $r_{4}$ are in the ratio $r_{1}: r_{2}: r_{3}: r_{4}=1: \sqrt{2}: \sqrt{3}: \sqrt{4}$.
The sone plate is placed in the $x-y$ plane and illuminated by plane monochromatic light waves of wavelength $5,000 \AA$. The most intense spot of light behind the plate is seen on the axis of the sone plate 1 meter behind it.
What is the radius $r_{1}$? | The focal lengths of the sones are given by
$$
f=\frac{r_{j}^{2}}{j \lambda} .
$$
For $j=1, f=1 \mathrm{~m}, \lambda=5000 \AA$, we have $r_{1}=0.707 \mathrm{~mm}$. |
What is the drift velocity of electrons in a $1 \mathrm{~mm} \mathrm{Cu}$ wire carrying $10 \mathrm{~A}$? | It is $10^{-2} \mathrm{~cm} / \mathrm{sec}$. |
The ground state of the realistic helium atom is of course nondegenerate. However, consider a hypothetical helium atom in which the two electrons are replaced by two identical, spin-one particles of negative charge. Neglect spin-dependent forces. For this hypothetical atom, what is the degeneracy of the ground state? | The two new particles are Bosons; thus the wave function must be symmetrical. In the ground state, the two particles must stay in 1s orbit. Then the space wave function is symmetrical, and consequently the spin wave function is symmetrical too. As $s_{1}=1$ and $s_{2}=1$, the total $S$ has three possible values:
$S=2$, the spin wave function is symmetric and its degeneracy is $2 S+1=5$.
$S=1$, the spin wave function is antisymmetric and its degeneracy is $2 S+1=3$.
$S=0$, the spin wave function is symmetric and its degeneracy is $2 S+1=1$
If the spin-dependent forces are neglected, the degeneracy of the ground state is $5+3+1=9$. |
Consider the pion photoproduction reaction
$$
\gamma+p \rightarrow p+\pi^{\circ}
$$
where the rest energy is $938 \mathrm{MeV}$ for the proton and $135 \mathrm{MeV}$ for the neutral pion.
If the initial proton is at rest in the laboratory, find the laboratory threshold gamma-ray energy for this reaction to "go". | The quantity $E^{2}-P^{2} c^{2}$ is invariant under Lorentz transformation and for an isolated system is the same before and after a reaction. The threshold $\gamma$-ray energy is that for which the final state particles are all at rest in the center of mass frame. Thus
$$
\left(E_{\gamma}+m_{p} c^{2}\right)^{2}-\left(\frac{E_{\gamma}}{c}\right)^{2} c^{2}=\left(m_{p}+m_{\pi}\right)^{2} c^{4}
$$
where $E_{\gamma}$ is the energy of the photon and $\frac{E_{\gamma}}{c}$ its momentum, giving
$$
E_{\gamma}=\frac{\left(m_{\pi}^{2}+2 m_{\pi} m_{p}\right) c^{4}}{2 m_{p} c^{2}}=144.7 \mathrm{MeV}
$$
as the threshold $\gamma$-ray energy. |
A monochromatic transverse wave with frequency $\nu$ propagates in a direction which makes an angle of $60^{\circ}$ with the $x$-axis in the reference frame $K$ of its source. The source moves in the $x$-direction at speed $v=\frac{4}{5} c$ towards an observer at rest in the $K^{\prime}$ frame (where his $x^{\prime}$-axis is parallel to the $x$-axis). The observer measures the frequency of the wave.
What is the angle of observation in the $K^{\prime}$ frame? | The frame $K$ of the light source moves with velocity $\beta$ c relative to $K^{\prime}$, the observer's frame. The (inverse) transformation of the components of the wave 4 -vector is given by
$$
k_{x}^{\prime} \mathrm{c}=\gamma\left(k_{x} \mathrm{c}+\beta \omega\right), \quad k_{y}^{\prime} \mathrm{c}=k_{y} \mathrm{c}, \quad k_{z}^{\prime} \mathrm{c}=k_{z} \mathrm{c}, \quad \omega^{\prime}=\gamma\left(\omega+\beta k_{x} \mathrm{c}\right)
$$
where $\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{2}}$. The angular frequency of the wave in $K$ is $\omega=2 \pi \nu$. If the angle between the light and the $x$-axis is $\theta$, then
$$
k_{x}=k \cos \theta, \quad k_{y}=k \sin \theta, \quad k_{z}=0, \quad \omega=k \mathrm{c} .
$$
Thus
$$
\omega^{\prime}=\gamma(\omega+\beta \omega \cos \theta)=\gamma(1+\beta \cos \theta) \omega,
$$
or
$$
\nu^{\prime}=\frac{(1+\beta \cos \theta) \nu}{\sqrt{1-\beta^{2}}} .
$$
The above can also be written as
$$
k^{\prime}=\gamma(1+\beta \cos \theta) k
$$
As
$$
k_{x}^{\prime}=\gamma\left(k_{x}+\frac{\beta \omega}{\mathrm{c}}\right)=\gamma k(\cos \theta+\beta),
$$
the angle $\mathbf{k}^{\prime}$ makes with the $x^{\prime}$-axis is given by
$$
\cos \theta^{\prime}=\frac{k_{x}^{\prime}}{k^{\prime}}=\frac{\cos \theta+\beta}{1+\beta \cos \theta} .
$$
With $\beta=0.8, \theta=60^{\circ}$, we have
$$
\nu^{\prime}=\left(\frac{1+0.8 \cos 60^{\circ}}{\sqrt{1-0.8^{2}}}\right) \nu=\frac{1.4}{0.6} \nu=\frac{7}{3} \nu,
$$
$$
\cos \theta^{\prime}=\frac{0.5+0.8}{1+0.8 \times 0.5}=\frac{13}{14},
$$
giving $\theta^{\prime}=21.8^{\circ}$ |
Suppose that the ionosphere is in thermal equilibrium at temperature $T$. It consists of electrons (mass $m$ ) and singly charged ions (mass $M$ ). If there were no large-scale electric field, the electrons and ions would have very different scale heights, $k T / m g$ and $k T / M g$ respectively. This would produce a large electric field, thereby modifying the density vs. height relation.
Taking into account the electrostatic field, find the numerical value of $E$ if the ions are protons.
(You may ignore the curvature of the Earth's surface. it may be helpful to know that the electric field turns out to be uniform.) | Use a coordinate system with origin at the Earth's surface and the $z$-axis vertically upward. The self-consistent electric field $\mathbf{E}$ is along the $z$ direction:
$$
E=-\frac{\partial V}{\partial z},
$$
or
$$
V=-E z
$$
assuming $E$ to be uniform. Then an electron and an ion at height $z$ will respectively have energies
$$
\begin{aligned}
&\varepsilon_{\mathrm{e}}=m g z+e E z \\
&\varepsilon_{\mathrm{i}}=M g z-e E z
\end{aligned}
$$
and number densities
$$
\begin{aligned}
&n_{\mathrm{e}}=n_{\mathrm{e} 0} e^{-\varepsilon_{\mathrm{e}} / k T}=n_{\mathrm{e} 0} e^{-(m g+e E) z / k T}, \\
&n_{\mathrm{i}}=n_{\mathrm{i} 0} e^{-\varepsilon_{\mathrm{i}} / k T}=n_{\mathrm{i} 0} e^{-(M g-e E) z / k T}
\end{aligned}
$$
as functions of height.Since
$$
\Delta \cdot \mathbf{E}=\frac{\rho}{\epsilon_{0}}=0
$$
for a uniform field, the net charge density $\rho=0$ everywhere. This means that
$$
-e n_{\mathrm{e}}+e n_{\mathrm{i}}=0,
$$
or
$$
n_{\mathrm{e} 0}=n_{\mathrm{i} 0}, \quad m g+e E=M g-e E .
$$
The last equation gives
$$
E=\frac{(M-m) g}{2 e} \approx \frac{M g}{2 e}
$$
as $M \gg m$.If the ions are protons, we have $M=1.672 \times 10^{-27} \mathrm{~kg}$, $e=$ $1.602 \times 10^{-19} \mathrm{C}$, and
$$
E=\frac{1.672 \times 10^{-27} \times 9.807}{2 \times 1.602 \times 10^{-19}}=5.12 \times 10^{-8} \mathrm{~V} \mathrm{~m}^{-2} .
$$ |
An electron (mass $m$, charge $e$ ) moves in a plane perpendicular to a uniform magnetic field. If energy loss by radiation is neglected the orbit is a circle of some radius $R$. Let $E$ be the total electron energy, allowing for relativistic kinematics so that $E \gg m c^{2}$.
Compute the needed field induction $B$ numerically, in Gauss, for the case where $R=30$ meters, $E=2.5 \times 10^{9}$ electron-volts. | In uniform magnetic field $B$ the motion of an electron is described in Gaussian units by
$$
\frac{d p}{d t}=\frac{e}{c} \mathbf{v} \times \mathbf{B},
$$
where $p$ is the momentum of the electron,
$$
\mathbf{p}=m \boldsymbol{v},
$$
with $\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{2}}, \beta=\frac{v}{c}$. Since $\frac{e}{c} \mathbf{v} \times \mathbf{B} \cdot \mathbf{v}=0$, the magnetic force does no work and the magnitude of the velocity does not change, i.e., $v$, and hence $\gamma$, are constant. For circular motion,
$$
\left|\frac{d v}{d t}\right|=\frac{v^{2}}{R} .
$$
Then
$$
m \gamma\left|\frac{d v}{d t}\right|=\frac{e}{c}|\mathbf{v} \times \mathbf{B}| .
$$
As $\mathbf{v}$ is normal to $B$, we have
$$
m \gamma \frac{v^{2}}{R}=\frac{e}{c} v B
$$
or
$$
B=\frac{p c}{e R} \text {. }
$$
With $E \gg m c^{2}, p c=\sqrt{E^{2}-m^{2} c^{4}} \approx E$ and
$$
B \approx \frac{E}{e R} \approx 0.28 \times 10^{4} \mathrm{Gs} .
$$ |
Two particles with the same mass $m$ are emitted in the same direction, with momenta $5 m c$ and $10 m c$ respectively. As seen from the slower one, what is the velocity of the faster particle? ($c=$ speed of light.) | In the laboratory frame $K_{0}$, the slower particle has momentum
$$
m \gamma_{1} v_{1}=m \gamma_{1} \beta_{1} c=5 m c
$$
giving
$$
\gamma_{1} \beta_{1}=\sqrt{\gamma_{1}^{2}-1}=5
$$
or
$$
\gamma_{1}^{2}=26
$$
Hence
$$
\beta_{1}^{2}=1-\frac{1}{26}=\frac{25}{26}, \quad \text { or } \quad v_{1}=\sqrt{\frac{25}{26}} c \text {. }
$$
Similarly for the faster particle, the velocity is
$$
v_{2}=\sqrt{\frac{100}{101}} c
$$
Let $K_{1}, K_{2}$ be the rest frames of the slower and faster particles respectively. The transformation for velocity between $K_{0}$ and $K$, which moves with velocity $v$ in the $x$ direction relative to $K_{0}$, is
$$
u_{x}^{\prime}=\frac{u_{x}-v}{1-\frac{u_{x} v}{c^{2}}} .
$$
Thus in $K_{1}$, the velocity of the faster particle is
$$
v_{2}^{\prime}=\frac{v_{2}-v_{1}}{1-\frac{u_{2} v_{1}}{c^{2}}}=\left(\frac{\sqrt{\frac{100}{101}}-\sqrt{\frac{25}{26}}}{1-\sqrt{\frac{100}{101} \cdot \frac{25}{26}}}\right) \mathrm{c}=0.595 \mathrm{c}
$$
In $K_{2}$, the velocity of the slower particle is
$$
v_{1}^{\prime}=\frac{v_{1}-v_{2}}{1-\frac{v_{1} v_{2}}{\mathrm{c}^{2}}}=-0.595 \mathrm{c} \text {. }
$$ |
When a $5000 \mathrm{lb}$ car driven at $60 \mathrm{mph}$ on a level road is suddenly put into neutral gear (i.e. allowed to coast), the velocity decreases in the following manner:
$$
V=\frac{60}{1+\left(\frac{t}{60}\right)} \mathrm{mph},
$$
where $t$ is the time in sec. Find the horsepower required to drive this car at $30 \mathrm{mph}$ on the same road.
Useful constants: $g=22 \mathrm{mph} / \mathrm{sec}, 1 \mathrm{H} . \mathrm{P} .=550 \mathrm{ft} . l \mathrm{~b} / \mathrm{sec}, 60 \mathrm{mph}=$ $88 \mathrm{ft} / \mathrm{sec}$. | Let $V_{0}=60 \mathrm{mph}$, then
$$
\frac{t}{60}=\frac{V_{0}}{V}-1 \text {. }
$$
Hence
$$
\frac{d V}{d t}=\frac{-V^{2}}{60 V_{0}},
$$
and the resistance acting on the car is $F=m V^{2} /\left(60 V_{0}\right)$, where $m$ is the mass of the car. The propulsive force must be equal to the resistance $F^{\prime}$ at the speed of $V^{\prime}=30 \mathrm{mph}$ in order to maintain this speed on the same road. It follows that the horsepower required is
$$
\begin{aligned}
P^{\prime} &=F^{\prime} V^{\prime}=\frac{m V^{\prime 3}}{60 V_{0}}=37500 \frac{\mathrm{mph}^{2} . l \mathrm{~b} .}{\mathrm{s}} \\
&=\frac{37500}{g} \frac{\mathrm{mph}^{2} . \mathrm{lb} \text { wt }}{\mathrm{s}}=\frac{37500}{22} \mathrm{mph} . \mathrm{lb} \text { wt } \\
&=\frac{37500}{22} \cdot \frac{88}{60} \frac{\mathrm{ft} . \mathrm{lb} \text { wt }}{\mathrm{s}} \\
&=2500 \frac{\mathrm{ft} . l \mathrm{~b} \text { wt }}{\mathrm{s}}=4.5 \text { H.P. }
\end{aligned}
$$
Note that pound weight (lb wt) is a unit of force and $1 \mathrm{lb} \mathrm{wt}=g \mathrm{ft} . \mathrm{lb} / \mathrm{s}^{2}$. The horsepower is defined as $550 \mathrm{ft} . l \mathrm{~b} \mathrm{wt} / \mathrm{s}$. |
One gram each of ice, water, and water vapor are in equilibrium together in a closed container. The pressure is $4.58 \mathrm{~mm}$ of $\mathrm{Hg}$, the temperature is $0.01^{\circ} \mathrm{C}$. Sixty calories of heat are added to the system. The total volume is kept constant. Calculate the masses of water vapor now present in the container.
(Hint: For water at $0.01^{\circ} \mathrm{C}$, the latent heat of fusion is $80 \mathrm{cal} / \mathrm{g}$, the latent heat of vaporization is $596 \mathrm{cal} / \mathrm{g}$, and the latent heat of sublimation is $676 \mathrm{cal} / \mathrm{g}$. Also note that the volume of the vapor is much larger than the volume of the water or the volume of the ice.) | It is assumed that the original volume of water vapor is $V$, it volume is also $V$ after heating, and the masses of ice, water, and water vapor are respectively $x, y$ and $z$ at the new equilibrium. We have
$$
\begin{gathered}
x+y+z=3, \\
(1-x) L_{\mathrm{sub}}+(1-y) L_{\mathrm{vap}}=Q=60, \\
\frac{1-x}{\rho_{\mathrm{ice}}}+\frac{(1-y)}{\rho_{\text {water }}}+V_{0}=V . \\
V_{0}=\frac{R T}{\mu p} . \\
V=\frac{z}{\mu p} R T
\end{gathered}
$$
where $\mu=18 \mathrm{~g} / \mathrm{mole}, p=4.58 \mathrm{mmHg}, T=273.16 \mathrm{~K}, R=8.2 \times 10^{8}$ $\mathrm{m}^{3} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}, \rho_{\text {ice }}=\rho_{\text {water }}=1 \mathrm{~g} / \mathrm{cm}^{3}, L_{\text {sub }}=676 \mathrm{cal} / \mathrm{g}$, and $L_{\text {vap }}=$ $596 \mathrm{cal} / \mathrm{g}$. Solving the equations we find
$$
x=0.25 \mathrm{~g}, \quad y=1.75 \mathrm{~g}, \quad z=1.00 \mathrm{~g} .
$$
That is, the heat of $60 \mathrm{cal}$ is nearly all used to melt the ice. |
The entropy of water at atmospheric pressure and $100^{\circ} \mathrm{C}$ is $0.31$ $\mathrm{cal} / \mathrm{g} \cdot \mathrm{deg}$, and the entropy of steam at the same temperature and pressure is $1.76 \mathrm{cal} / \mathrm{g} \cdot \mathrm{deg}$.
What is the heat of vaporization at this temperature? | Heat of vaporization is
$$
L=T \Delta S=540 \mathrm{cal} / \mathrm{g} .
$$ |
For a particle of mass $m$ in a one-dimensional box of length $l$, the eigenfunctions and energies are
$$
\begin{aligned}
\psi_{n}(x) &=\sqrt{\frac{2}{l}} \sin \frac{n \pi x}{l}, \quad 0 \leq x \leq l \\
E_{n} &=\frac{1}{2 m}\left(\frac{n \pi \hbar}{l}\right)^{2}, \quad n=\pm 1, \pm 2, \ldots
\end{aligned}
$$
Suppose the particle is originally in a state $|n\rangle$ and the box length is increased to a length of $2 l(0 \leq x \leq 2 l)$ in a time $t \ll h / E_{n}$. Afterwards what is the probability that the particle will be found in an energy eigenstate with energy $E_{n}$ ? | First consider the process in which the box length is increased from $l$ to $2 l$. As $t \ll \frac{h}{E_{n}}$, it is reasonable to assume that the state of the particle in the box is unable to respond to the change during such a short time. Therefore the wave function of the particle after the change is completed is
$$
\psi(x)= \begin{cases}\sqrt{\frac{2}{l}} \sin \frac{n \pi x}{l}, & 0 \leq x \leq l \\ 0, & l \leq x \leq 2 l\end{cases}
$$
On the other hand, the eigenstates and eigenvalues of the same particle in the one-dimensional box of length $2 l$ would be, respectively,
$$
\begin{aligned}
\phi_{n^{\prime}}(x) &=\sqrt{\frac{1}{l}} \sin \frac{n^{\prime} \pi x}{2 l}, \quad(0 \leq x \leq 2 l), \\
E_{n^{\prime}} &=\frac{1}{2 m}\left(\frac{n^{\prime} \pi \hbar}{2 l}\right)^{2}, \quad\left(n^{\prime}=\pm 1, \pm 2, \ldots\right) .
\end{aligned}
$$
The energy $E_{n}$ of the particle corresponds to the energy level $E_{n^{\prime}}$ in the $2 l$ box, where $\frac{n}{l}=\frac{n^{\prime}}{2 l}$, i.e., $n^{\prime}=2 n$. The corresponding eigenstate is then $\phi_{2 n}$. Thus the probability amplitude is
$$
A=\int_{-\infty}^{\infty} \phi_{2 n}(x) \psi(x) d x=\frac{\sqrt{2}}{l} \int_{0}^{l} \sin ^{2} \frac{n \pi x}{l} d x, \frac{1}{\sqrt{2}},
$$
and the probability of finding the particle in an eigenstate with energy $E_{n}$ is
$$
P=|A|^{2}=\frac{1}{2} .
$$ |
Given a classical model of the tritium atom with a nucleus of charge $+1$ and a single electron in a circular orbit of radius $r_{0}$, suddenly the nucleus emits a negatron and changes to charge $+2$.
(The emitted negatron escapes rapidly and we can forget about it.) The electron in orbit suddenly has a new situation.
Find the ratio of the electron's energy after to before the emission of the negatron (taking the zero of energy, as usual, to be for zero kinetic energy at infinite distance). | As the negatron leaves the system rapidly, we can assume that its leaving has no effect on the position and kinetic energy of the orbiting electron.
From the force relation for the electron,
$$
\frac{m v_{0}^{2}}{r_{0}}=\frac{e^{2}}{4 \pi \varepsilon_{0} r_{0}^{2}},
$$
we find its kinetic energy
$$
\frac{m v_{0}^{2}}{2}=\frac{e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$
and its total mechanical energy
$$
E_{1}=\frac{m v_{0}^{2}}{2}-\frac{e^{2}}{4 \pi \varepsilon_{0} r_{0}}=\frac{-e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$
before the emission of the negatron. After the emission the kinetic energy of the electron is still $\frac{e^{2}}{8 \pi \varepsilon_{0} r_{0}}$, while its potential energy suddenly changes to
$$
\frac{-2 e^{2}}{4 \pi \varepsilon_{0} r_{0}}=-\frac{-e^{2}}{2 \pi \varepsilon_{0} r_{0}} .
$$
Thus after the emission the total mechanical energy of the orbiting electron is
$$
E_{2}=\frac{m v_{0}^{2}}{2}-\frac{2 e^{2}}{4 \pi \varepsilon_{0} r_{0}}=\frac{-3 e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$
giving
$$
\frac{E_{2}}{E_{1}}=3 .
$$
In other words, the total energy of the orbiting electron after the emission is three times as large as that before the emission. |
An air-spaced coaxial cable has an inner conductor $0.5 \mathrm{~cm}$ in diameter and an outer conductor $1.5 \mathrm{~cm}$ in diameter. When the inner conductor is at a potential of $+8000 \mathrm{~V}$ with respect to the grounded outer conductor, what is the charge per meter on the inner conductor? | Let the linear charge density for the inner conductor be $\lambda$. By symmetry we see that the field intensity at a point distance $r$ from the axis in the cable between the conductors is radial and its magnitude is given by Gauss' theorem as
$$
E=\frac{\lambda}{2 \pi \varepsilon_{0} r} .
$$
Then the potential difference between the inner and outer conductors is
$$
V=\int_{a}^{b} E d r=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln (b / a)
$$
with $a=1.5 \mathrm{~cm}, b=0.5 \mathrm{~cm}$, which gives
$$
\begin{aligned}
\lambda &=\frac{2 \pi \varepsilon_{0} V}{\ln (b / a)}=\frac{2 \pi \times 8.9 \times 10^{-12} \times 8000}{\ln (1.5 / 0.5)} \\
&=4.05 \times 10^{-7} \mathrm{C} / \mathrm{m} .
\end{aligned}
$$ |
A small celestial object, held together only by its self-gravitation, can be disrupted by the tidal forces produced by another massive body, if it comes near enough to that body. For an object of diameter $1 \mathrm{~km}$ and density $2 \mathrm{~g} / \mathrm{cm}^{3}$, find the critical distance from the earth (Roche limit). | Suppose the earth is fixed in space and the small celestial object orbits around it at a distance $l$ away as shown in Fig. 1.73. Let $M$ be the mass of the earth, $m$ the mass and $\rho$ the density of the small celestial object. Consider a unit mass of the object on the line $O C$ at distance $x$ from $C$. We have from the balance of forces on it
$$
(l-x) \omega^{2}=\frac{G M}{(l-x)^{2}}-\frac{G\left(\frac{4}{3}\right) \pi x^{3} \rho}{x^{2}} .
$$
We also have for the celestial body
$$
m l \omega^{2}=\frac{G M m}{l^{2}},
$$
which gives $\omega^{2}$ to be used in the above. Then as $\frac{x}{l} \ll 1$, retaining only the lowest order in $\frac{x}{l}$, we have
$$
l=\left(\frac{9 M}{4 \pi \rho}\right)^{\frac{1}{3}}
$$
With $M=6 \times 10^{27} \mathrm{~g}, \rho=2 \mathrm{~g} / \mathrm{cm}^{3}$, we find
$$
l=1.29 \times 10^{9} \mathrm{~cm}=1.29 \times 10^{4} \mathrm{~km} .
$$
If $l$ is less than this value, the earth's attraction becomes too large for the unit mass to be held by the celestial body and disruption of the latter occurs.
If the unit mass is located to the right of $C$ on the extended line of $O C$, $x$ is negative but the above conclusion still holds true. We may also consider a unit mass located off the line $O C$ such as the point $P$ in Fig. 1.74. We now have
$$
\sqrt{(l-x)^{2}+y^{2}} \omega^{2} \cos \theta=\frac{G M}{(l-x)^{2}+y^{2}} \cos \theta-\frac{4}{3} \pi \rho G \sqrt{x^{2}+y^{2}} \cos \varphi,
$$
with
$$
\cos \theta=\frac{l-x}{\sqrt{(l-x)^{2}+y^{2}}}, \quad \cos \varphi=\frac{x}{\sqrt{x^{2}+y^{2}}} \text {. }
$$
As $x / l \ll 1, y / l \ll 1$, and retaining only the first-order terms we would obtain the same result. |
A self-luminous object of height $h$ is $40 \mathrm{~cm}$ to the left of a converging lens with a focal length of $10 \mathrm{~cm}$. A second converging lens with a focal length of $20 \mathrm{~cm}$ is $30 \mathrm{~cm}$ to the right of the first lens.
Calculate the distance between the final image and the second lens. | From $\frac{1}{f_{1}}=\frac{1}{u_{1}}+\frac{1}{v_{1}}$, where $f_{1}=10 \mathrm{~cm}, u_{1}=40 \mathrm{~cm}$, we obtain $v_{1}=13 \frac{1}{3} \mathrm{~cm}$.
From $\frac{1}{f_{2}}=\frac{1}{u_{2}}+\frac{1}{v_{2}}$, where $f_{2}=20 \mathrm{~cm}, u_{2}=\left(30-\frac{10}{3}\right) \mathrm{cm}$, we obtain $v_{2}=-100 \mathrm{~cm}$, i.e., the final image is $100 \mathrm{~cm}$ to the left of the second lens. |
At room temperature, $k_{\mathrm{B}} T / e=26 \mathrm{mV}$. A sample of cadmium sulfide displays a mobile carrier density of $10^{16} \mathrm{~cm}^{-3}$ and a mobility coefficient $\mu=10^{2} \mathrm{~cm}^{2} /$ volt sec.
Calculate the electrical conductivity of this sample. | The electrical conductivity is given by $\sigma=n e \mu$. With $n=10^{22} \mathrm{~m}^{-3}$, $e=1.6 \times 10^{-19} \mathrm{C}, \mu=10^{-2} \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$, we have for the material $\sigma=$ $16 \Omega^{-1} \mathrm{~m}^{-1}$. |
Consider the situation which arises when a negative muon is captured by an aluminum atom (atomic number $Z=13$ ). After the muon gets inside the "electron cloud" it forms a hydrogen-like muonic atom with the aluminum nucleus. The mass of the muon is $105.7 \mathrm{MeV}$.
(Slide rule accuracy; neglect nuclear motion)
Compute the mean life of the above muonic atom in the $3 d$ state, taking into account the fact that the mean life of a hydrogen atom in the $3 d$ state is $1.6 \times 10^{-8}$ sec.
| For spontaneous transitions from the $3 d$ state, the largest probability is for $3 d \rightarrow 2 p$. In nonrelativistic approximation, the photon energy is given by
$$
\begin{aligned}
h \nu &=\frac{m_{\mu} Z^{2} e^{4}}{2 \hbar^{2}}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \\
&=\frac{m_{\mu} c^{2} Z^{2}}{2}\left(\frac{e^{2}}{\hbar c}\right)^{2}\left(\frac{5}{36}\right) \\
&=\frac{105.7 \times 13^{2}}{2}\left(\frac{1}{137}\right)^{2}\left(\frac{5}{36}\right)=6.61 \times 10^{-2} \mathrm{MeV} .
\end{aligned}
$$
The corresponding wavelength is
$$
\lambda=\frac{c}{\nu}=\frac{h c}{h \nu}=\frac{4.135 \times 10^{-15} \times 3 \times 10^{10}}{6.61 \times 10^{4}}=1.88 \times 10^{-9} \mathrm{~cm} .
$$The transition probability per unit time is
$$
A \propto \omega^{3}\left|\mathbf{r}_{k k^{\prime}}\right|^{2} .
$$
For hydrogen-like atoms, as
$$
\left|\mathbf{r}_{k k^{\prime}}\right| \propto \frac{1}{Z}, \quad \omega \propto m Z^{2}, \quad \text { and so } \quad A \propto m^{3} Z^{4},
$$
the mean life of the muonic atom in the $3 d$ state is
$$
\begin{aligned}
T_{\mu} &=\left(\frac{A_{0}}{A}\right) T_{0}=\left(\frac{m_{0}}{m_{\mu}}\right)^{3} \frac{T_{0}}{Z^{4}} \\
&=\left(\frac{0.51}{105.7}\right)^{3} \times \frac{1}{13^{4}} \times 1.6 \times 10^{-8}=6.3 \times 10^{-20} \mathrm{~s} .
\end{aligned}
$$ |
The $z$-component of the spin of an electron in free space (no electromagnetic fields) is measured and found to be $+\hbar / 2$. What is the probability of this occuring?
| In the $\sigma_{z}$ representation, the spin wave function is $\left(\begin{array}{l}1 \\ 0\end{array}\right)$, the eigenfunctions of $\sigma_{x}$ are $\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ -1\end{array}\right)$, corresponding to eigenvalues $+1,-1$ respectively. Expanding $\left(\begin{array}{l}1 \\ 0\end{array}\right)$ in these two states, we see that the possible results of a measurement of $s_{x}$ are $\pm \hbar / 2$ since $\hat{s}_{x}=\frac{1}{2} \sigma_{x}$, the mean value being zero:
$$
\left\langle s_{x}\right\rangle=(1,0) \hat{s}_{x}\left(\begin{array}{l}
1 \\
0
\end{array}\right)=0 .
$$
The probabilities of finding the result to be $+\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ are $P_{+}$and $P_{-}$respectively:
$$
\begin{aligned}
&P_{+}=\left|\frac{1}{\sqrt{2}}(1,1)\left(\begin{array}{l}
1 \\
0
\end{array}\right)\right|^{2}=\frac{1}{2}, \\
&P_{-}=\left|\frac{1}{\sqrt{2}}(1,-1)\left(\begin{array}{l}
1 \\
0
\end{array}\right)\right|^{2}=\frac{1}{2} .
\end{aligned}
$$ |
While sitting in front of a color TV with a $25 \mathrm{kV}$ picture tube potential, you have an excellent chance of being irradiated with $\mathrm{X}$-rays.
For the resulting continuous distribution, calculate the shortest wavelength (maximum energy) X-ray. | When a high voltage is applied to the picture tube, electrons emitted from the negative electrode will be accelerated by the electric field to strike the screen target. If the energy of the electrons exceeds a certain value, they can knock off inner-shell electrons in the target atoms and make holes in the inner shells. Then as the outer-shell electrons fall in to fill up these holes, X-rays are emitted.
The maximum energy of the X-ray photons produced, $h \nu_{\max }$, is equal to the energy $\mathrm{eV}$ of the incident electrons. Hence the minimum wavelength of the X-rays is
$$
\lambda_{\min }=\frac{h c}{e V}=\frac{12000}{V}=\frac{12000}{25 \times 10^{3}}=0.48 \AA .
$$ |
Twenty grams of ice at $0^{\circ} \mathrm{C}$ are dropped into a beaker containing 120 grams of water initially at $70^{\circ} \mathrm{C}$. Find the final temperature of the mixture neglecting the heat capacity of the beaker. Heat of fusion of ice is $80$ $\mathrm{cal} / \mathrm{g}$. | We assume the temperature of equilibrium to be $T$ after mixing. Thus
$$
M_{1} L_{\text {fusion }}+M_{1} C_{p, \text { water }} T=M_{2} C_{p, \text { water }}\left(T_{0}-T\right) .
$$
We substitute $M_{1}=20 \mathrm{~g}, M_{2}=120 \mathrm{~g}, T_{0}=70^{\circ} \mathrm{C}, L_{\text {fusion }}=80 \mathrm{cal} / \mathrm{g}$ and $C_{p, \text { water }}=1 \mathrm{cal} / \mathrm{g}$, and obtain the final temperature $T=48.57^{\circ} \mathrm{C}$. |
Two conductors are embedded in a material of conductivity $10^{-4} \Omega / \mathrm{m}$ and dielectric constant $\varepsilon=80 \varepsilon_{0}$. The resistance between the two conductors is measured to be $10^{5} \Omega$. Calculate the capacitance between the two conductors. | Suppose that the two conductors carry free charges $Q$ and $-Q$. Consider a closed surface enclosing the conductor with the charge $Q$ (but not the other conductor). We have, using Ohm's and Gauss' laws,
$$
I=\oint \mathbf{j} \cdot d \mathbf{S}=\oint \sigma \mathbf{E} \cdot d \mathbf{S}=\sigma \oint \mathbf{E} \cdot d \mathbf{S}=\sigma \frac{Q}{\varepsilon} .
$$
If the potential difference between the two conductors is $V$, we have $V=$ $I R=\frac{\sigma Q}{\varepsilon} R$, whence
$$
C=\frac{Q}{V}=\frac{\varepsilon}{\sigma R} .
$$
Numerically the capacitance between the conductors is
$$
C=\frac{80 \times 8.85 \times 10^{-12}}{10^{-4} \times 10^{5}}=7.08 \times 10^{-11} \mathrm{~F} .
$$ |
Consider a large number of $N$ localized particles in an external magnetic field $\mathbf{H}$. Each particle has spin $1 / 2$. Find the number of states accessible to the system as a function of $M_{8}$, the $z$-component of the total spin of the system. Determine the value of $M_{\mathrm{s}}$ for which the number of states is maximum. | The spin of a particle has two possible orientations $1 / 2$ and $-1 / 2$. Let the number of particles with spin $1 / 2$ whose direction is along $\mathbf{H}$ be $N_{\uparrow}$ and the number of particles with spin $-1 / 2$ whose direction is opposite to $\mathbf{H}$ be $N_{\downarrow}$; then the component of the total spin in the direction of $\mathbf{H}$ is $M_{\mathrm{s}}=\frac{1}{2}\left(N_{\uparrow}-N_{\downarrow}\right)$. By $N_{\uparrow}+N_{\downarrow}=N$, we can obtain $N_{\uparrow}=\frac{N}{2}+M_{\mathrm{s}}$ and $N_{\downarrow}=\frac{N}{2}-M_{\mathrm{s}}$. The number of states of the system is
$$
Q=\frac{N !}{N_{\uparrow} ! N_{\downarrow} !} \frac{N !}{\left[\frac{N}{2}+M_{\mathrm{s}}\right] !\left[\frac{N}{2}-M_{\mathrm{s}}\right] !}
$$
Using Stirling's formula, one obtains
$$
\begin{aligned}
\ln Q &=\ln \frac{N !}{N_{\uparrow} ! N_{\downarrow} !} \\
&=N \ln N-N_{\uparrow} \ln N_{\uparrow}-N_{\downarrow} \ln N_{\downarrow} \\
&=N \ln N-N_{\uparrow} \ln N_{\uparrow}-\left(N-N_{\uparrow}\right) \ln \left(N-N_{\uparrow}\right) .
\end{aligned}
$$
By
$$
\frac{\partial \ln Q}{\partial N_{\uparrow}}=-\ln N_{\uparrow}+\ln \left(N-N_{\uparrow}\right)=0,
$$
we get $N_{\uparrow}=\frac{N}{2}$, i.e., $M_{\mathrm{s}}=0$ when the number of states of the system is maximum. |
Assume a visible photon of $3 \mathrm{eV}$ energy is absorbed in one of the cones (light sensors) in your eye and stimulates an action potential that produces a $0.07$ volt potential on an optic nerve of $10^{-9} \mathrm{~F}$ capacitance.
Calculate the energy of the action potential. | $Q=V C=0.07 \times 10^{-9}=7 \times 10^{-11}$ Coulomb.$E=\frac{Q V}{2}=\frac{1}{2} \times 7 \times 10^{-11} \times 0.07=2.5 \times 10^{-12}$ joule. |
A spherical dust particle falls through a water mist cloud of uniform density such that the rate of accretion onto the droplet is proportional to the volume of the mist cloud swept out by the droplet per unit time. If the droplet starts from rest in the cloud, find the value of the acceleration of the drop for large times. | Suppose the spherical dust particle initially has mass $M_{0}$ and radius $R_{0}$. Take the initial position of the dust particle as the origin and the $x$-axis along the downward vertical. Let $M(t)$ and $R(t)$ be the mass and radius of the droplet at time $t$ respectively. Then
$$
M(t)=M_{0}+\frac{4}{3} \pi\left(R^{3}-R_{0}^{3}\right) \rho,
$$
where $\rho$ is the density of the water mist, giving
$$
\frac{d M}{d t}=\rho 4 \pi R^{2} \frac{d R}{d t} .
$$
The droplet has a cross section $\pi R^{2}$ and sweeps out a cylinder of volume $\pi R^{2} \dot{x}$ in unit time, where $\dot{x}$ is its velocity. As the rate of accretion is proportional to this volume, we have
$$
\frac{d M}{d t}=\alpha \pi R^{2} \dot{x},
$$
$\alpha$ being a positive constant. Hence
$$
\dot{x}=\frac{4 \rho}{\alpha} \dot{R} .
$$
The momentum theorem gives
$$
M(t+d t) \dot{x}(t+d t)-M(t) \dot{x}(t)=M g d t .
$$
Using Taylor's theorem to expand $M(t+d t)$ and $\dot{x}(t+d t)$ and retaining only the lowest-order terms, we obtain
$$
\dot{x} \frac{d M}{d t}+M \ddot{x}=M g .
$$
For large $t, M(t) \approx \frac{4}{3} \pi R^{3} \rho, d M / d t \approx 3 M \dot{R} / R$, and the above becomes
$$
\ddot{R}+\frac{3 \dot{R}^{2}}{R}=\frac{\alpha g}{4 \rho} .
$$
For a particular solution valid for large $t$, setting
$$
R(t)=a t^{2},
$$
where $a$ is a constant, in the above we obtain
$$
a=\frac{\alpha g}{56 \rho} .
$$
Thus for large $t$,
$$
\dot{x}=\frac{4 \rho}{\alpha} \cdot 2 a t=\frac{g t}{7} .
$$
Hence the acceleration for large times is $g / 7$. |
A solenoid is designed to generate a magnetic field over a large volume. Its dimensions are as follows: length $=2$ meters, radius $=0.1$ meter, number of turns $=1000$.
(Edge effects should be neglected.)
Calculate the self-inductance of the solenoid in Henrys. | inside it is
Suppose the solenoid carries a current $I$. The magnetic induction
$$
B=\mu_{0} n I=\mu_{0} N I / l,
$$
and the magnetic flux linkage is
$$
\psi=N B S=N \frac{\mu_{0} N I}{l} \cdot \pi r^{2}=\frac{I \mu_{0} N^{2} \pi r^{2}}{l} .
$$
Hence the self-inductance is
$$
\begin{aligned}
L=\frac{\psi}{I} &=\frac{\mu_{0} N^{2} \pi r^{2}}{l}=\frac{4 \pi \times 10^{-7} \times 1000^{2} \times \pi \times 0.1^{2}}{2} \\
&=1.97 \times 10^{-2} \mathrm{H} .
\end{aligned}
$$ |
Suppose two compact stars, each of one solar mass, are in a circular orbit around each other with a radius of one solar radius. What is the time scale for decay of this orbit? Take
$$
\begin{aligned}
\text { solar mass } &=2 \times 10^{33} \mathrm{gm}, \\
\text { solar radius } &=7 \times 10^{10} \mathrm{~cm}, \\
G &=6.7 \times 10^{-8} \mathrm{~cm}^{3} \mathrm{gm}^{-1} \mathrm{sec}^{-2}, \\
c &=3 \times 10^{10} \mathrm{~cm} \mathrm{sec}^{-1} .
\end{aligned}
$$ | The total mass-energy of an isolated system is conserved, so it is not possible to radiate monopole gravitational radiation. Also as the total momentum of an isolated system is conserved, which means that the second time derivative of the mass dipole moment $\sum m r$ is zero, it is not possible to radiate dipole gravitational radiation. The lowest multipole gravitational radiation is quadrupole.
For a system of two stars in a circular orbit around their common center of mass, the rate of energy loss by radiation of gravitational waves is
$$
-\frac{d E}{d t}=\frac{32 G^{4}}{5 c^{5} r^{5}}\left(m_{1} m_{2}\right)^{2}\left(m_{1}+m_{2}\right)
$$
where $r$ is their mutual distance, which is constant for motion in a circular orbit. With $m_{1}=m_{2}=m$ and the data given we have
$$
-\frac{d E}{d t}=\frac{64 G^{4}}{5 c^{5} r^{5}} m^{5}=1.57 \times 10^{31} \mathrm{erg} \mathrm{s}^{-1} .
$$
With
$$
\begin{gathered}
E=-\frac{G m^{2}}{2 r}, \\
\frac{d r}{d t}=\frac{2 r^{2}}{G m^{2}} \frac{d E}{d t}=-\frac{128 G m^{3}}{5 c^{5} r^{3}} .
\end{gathered}
$$
As $\frac{d r}{d t}$ is the rate at which the two stars approach each other, the time taken for the complete collapse of the orbit is
$$
\begin{aligned}
\tau=\int_{r}^{0} \frac{d r}{\left(\frac{d r}{d t}\right)} &=-\frac{5 c^{5}}{128 G^{3} m^{3}} \int_{r}^{0} r^{3} d r \\
&=\frac{5 c^{5}}{512 G^{3} m^{3}} r^{4}=2.4 \times 10^{15} \mathrm{~s} .
\end{aligned}
$$ |
A beam of neutral atoms passes through a Stern-Gerlach apparatus. Five equally-spaced lines are observed. What is the total angular momentum of the atom? | When unpolarized neutral atoms of total angular momentum $J$ pass through the Stern-Gerlach apparatus, the incident beam will split into $2 J+1$ lines. Thus $2 J+1=5$, giving $J=2$. |
Calculate the ratio of the mean densities of the earth and the sun from the following approximate data:
$\theta=$ angular diameter of the sun seen from the earth $=\frac{1}{2}^{\circ}$.
$l=$ length of $1^{\circ}$ of latitude on the earth's surface $=100 \mathrm{~km}$.
$t=$ one year $=3 \times 10^{7} \mathrm{~s}$.
$g=10 \mathrm{~ms}^{-2}$. | Let $r$ be the distance between the sun and the earth, $M_{e}$ and $M_{s}$ be the masses and $R_{e}$ and $R_{s}$ be the radii of the earth and the sun respectively, and $G$ be the gravitational constant. We then have
$$
\begin{aligned}
&\frac{G M_{e} M_{s}}{r^{2}}=M_{e} r \omega^{2}, \\
&\frac{2 R_{s}}{r}=\frac{1}{2} \frac{2 \pi}{360}=\frac{\pi}{360} \mathrm{rad},
\end{aligned}
$$
i.e.
$$
r=\frac{720 R_{s}}{\pi} .
$$
The above gives
$$
\frac{G M_{s}}{\left(720 R_{s} / \pi\right)^{3}}=\omega^{2}
$$
or
$$
\frac{G M_{s}}{R_{s}^{3}}=\left(\frac{720}{\pi}\right)^{3}\left(\frac{2 \pi}{3 \times 10^{7}}\right)^{2} .
$$
For a mass $m$ on the earth's surface,
$$
\frac{G m M_{e}}{R_{e}^{2}}=m g,
$$
giving
$$
\frac{G M_{e}}{R_{e}^{3}}=\frac{g}{R_{e}}=\frac{g}{\left(\frac{360 \times 100}{2 \pi}\right)}=\frac{g \pi}{18 \times 10^{3}}
$$
Hence
$$
\frac{\rho_{e}}{\rho_{s}}=\frac{g \pi}{18 \times 10^{3}}\left(\frac{720}{\pi}\right)^{-3}\left(\frac{2 \pi}{3 \times 10^{7}}\right)^{-2}=3.31
$$ |
The latent heat of vaporization of water is about $2.44 \times 10^{6} \mathrm{~J} / \mathrm{kg}$ and the vapor density is $0.598 \mathrm{~kg} / \mathrm{m}^{3}$ at $100^{\circ} \mathrm{C}$. Find the rate of change of the boiling temperature with altitude near sea level in ${ }^{\circ} \mathrm{C}$ per $\mathrm{km}$. Assume the temperature of the air is $300 \mathrm{~K}$.
(Density of air at $0^{\circ} \mathrm{C}$ and 1 atm is $1.29 \mathrm{~kg} / \mathrm{m}^{3}$ ). | The Boltzmann distribution gives the pressure change with height:
$$
p(z)=p(0) \exp -\frac{m g z}{k T},
$$
where $p(0)$ is the pressure at sea level $z=0, m$ is the molecular weight of air, and $T_{0}=300 \mathrm{~K}$ is the temperature of the atmosphere. The ClausiusClapeyron equation can be written as
$$
\frac{d p}{d T}=\frac{L}{T\left(V_{2}-V_{1}\right)}=\frac{L}{T M\left(\frac{1}{\rho_{2}}-\frac{1}{\rho_{1}}\right)}=\frac{\alpha}{T} .
$$
with $\rho_{1}=1000 \mathrm{~kg} / \mathrm{m}^{3}, \rho_{2}=0.598 \mathrm{~kg} / \mathrm{m}^{3}$ and $L / M=2.44 \times 10^{6} \mathrm{~J} / \mathrm{kg}$, we have
$$
\alpha=\frac{L \rho_{1} \rho_{2}}{M\left(\rho_{1}-\rho_{2}\right)}=1.40 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} .
$$
So the rate of change of the boiling point with height is
$$
\frac{d T}{d z}=\frac{d T}{d p} \cdot \frac{d p}{d z}=\frac{T}{\alpha} \cdot\left(\frac{-m g}{k T_{0}}\right) p(z) .
$$
Using the equation of state for ideal gas $p=\rho k T_{0} / m$, we have near the sea level
$$
\frac{d T}{d z}=-\rho g T(0) / \alpha,
$$
where $\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}$ is the density of air, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ and $T(0)=100^{\circ} \mathrm{C}$.
Thus $\frac{d T}{d z}=-0.87^{\circ} \mathrm{C} / \mathrm{km}$. |
A 55 year old man can focus objects clearly from $100 \mathrm{~cm}$ to $300 \mathrm{~cm}$. Representing the eye as a simple lens $2 \mathrm{~cm}$ from the retina, what is the focal length of the lens at the far point (focussed at $300 \mathrm{~cm}$)? | $\frac{1}{f_{\text {far }}}=\frac{1}{u_{\text {far }}}+\frac{1}{v}, \quad u_{\text {far }}=300 \mathrm{~cm}, v=2 \mathrm{~cm}$.
Solving the equation yields $f_{\text {far }}=1.987 \mathrm{~cm}$. |
A defective satellite of mass $950 \mathrm{~kg}$ is being towed by a spaceship in empty space. The two vessels are connected by a uniform $50 \mathrm{~m}$ rope whose mass per unit length is $1 \mathrm{~kg} / \mathrm{m}$. The spaceship is accelerating in a straight line with acceleration $5 \mathrm{~m} / \mathrm{sec}^{2}$.
What is the force exerted by the spaceship on the rope? | $$
\begin{aligned}
F &=\left(m_{\text {rope }}+m_{\text {gatellite }}\right) \cdot a \\
&=(950+50) \times 5=5 \times 10^{3} \mathrm{~N} .
\end{aligned}
$$ |
The index of refraction of air at $300 \mathrm{~K}$ and 1 atmosphere pressure is $1.0003$ in the middle of the visible spectrum. Assuming an isothermal atmosphere at $300 \mathrm{~K}$, calculate by what factor the earth's atmosphere would have to be more dense to cause light to bend around the earth with the earth's curvature at sea level. (In cloudless skies we could then watch sunset all night, in principle, but with an image of the sun drastically compressed vertically.) You may assume that the index of refraction $n$ has the property that $n-1$ is proportional to the density. (Hint: Think of Fermat's Principle.) The 1/e height of this isothermal atmosphere is 8700 metres. | We are given that
$$
n(r)-1=\rho e^{-\frac{r-f}{s} r b},
$$
where $R=6400 \times 10^{3} \mathrm{~m}$ is the earth's radius and $\rho$ is the density coefficient of air. Then
$$
\begin{gathered}
n(r)=1+\rho e^{-\frac{5 \pi}{500}}, \\
\frac{d n(r)}{d r}=n^{\prime}(r)=-\frac{1}{8700} \rho e^{-\frac{5-8}{500}} .
\end{gathered}
$$
It is also given that air is so dense that it makes light bend around the earth with the earth's curvature at sea level, as shown in Fig. 1.11.
Fig. $1.11$
The optical path length from $A$ to $B$ is
$$
l=n(r) r \theta .
$$
According to Fermat's Principle, the optical path length from A to B should be an extremum. Therefore,
$$
\frac{d l}{d r}=\left[n^{\prime}(r) r+n(r)\right] \theta=0,
$$
i.e.,
$$
n^{\prime}(r)=\frac{-n(r)}{r} .
$$
Substituting (3) into (2) yields
$$
\frac{1}{8700} \rho e^{-\frac{r-\pi}{800}}=\frac{n(r)}{r} .
$$
At sea level, $r=R=6400 \times 10^{3} \mathrm{~m}$. This with (1) and (4) yields
$$
\frac{\rho \times 6400 \times 10^{3}}{8700}=1+\rho
$$
giving
$$
\rho=0.00136 .
$$
At sea level, i.e., at $300 \mathrm{~K}$ and 1 atmosphere pressure, $n_{0}-1=\rho_{0}=0.0003$. Therefore
$$
\frac{\rho}{\rho_{0}}=4.53 .
$$
Thus only if the air were $4.53$ times as dense as the real air would light bend around the earth with the earth's curvature at sea level. |
A 100-ohm resistor is held at a constant temperature of $300 \mathrm{~K}$. A current of 10 amperes is passed through the resistor for $300 \mathrm{sec}$.
What is the change in the internal energy of the universe? | As the temperature of the resistor is constant, its state does not change. The entropy is a function of state. Hence the change in the entropy of the resistor is zero: $\Delta S_{1}=0$.The heat that flows from the resistor to the external world (a heat source of constant temperature) is
$$
I^{2} R t=3 \times 10^{6} \mathrm{~J} .
$$
The increase of entropy of the heat source is $\Delta S_{2}=3 \times 10^{6} / 300=10^{4} \mathrm{~J} / \mathrm{K}$. Thus the total change of entropy is $\Delta S=\Delta S_{1}+\Delta S_{2}=10^{4} \mathrm{~J} / \mathrm{K}$.The increase of the internal energy of the universe is
$$
\Delta U=3 \times 10^{6} \mathrm{~J} \text {. }
$$ |
An electron is confined in the ground state of a one-dimensional harmonic oscillator such that $\sqrt{\left\langle(x-\langle x\rangle)^{2}\right\rangle}=10^{-10} \mathrm{~m}$. Find the energy (in $\mathrm{eV})$ required to excite it to its first excited state.
[Hint: The virial theorem can help.] | The virial theorem for a one-dimensional harmonic oscillator states that $\langle T\rangle=\langle V\rangle$. Thus $E_{0}=\langle H\rangle=\langle T\rangle+\langle V\rangle=2\langle V\rangle=m_{e} \omega^{2}\left\langle x^{2}\right\rangle$, or, for the ground state,
$$
\frac{\hbar \omega}{2}=m_{e} \omega^{2}\left\langle x^{2}\right\rangle,
$$
giving
$$
\omega=\frac{\hbar}{2 m_{e}\left\langle x^{2}\right\rangle} .
$$
As $\langle x\rangle=0$ for a harmonic oscillator, we have
$$
\sqrt{\left\langle(x-\langle x\rangle)^{2}\right\rangle}=\sqrt{\left\langle x^{2}\right\rangle-\langle x\rangle^{2}}=\sqrt{\left\langle x^{2}\right\rangle}=10^{-10} \mathrm{~m} .
$$
The energy required to excite the electron to its first excited state is therefore
$$
\begin{aligned}
\Delta E=\hbar \omega &=\frac{\hbar^{2}}{2 m_{e}\left\langle x^{2}\right\rangle}=\frac{\hbar^{2} c^{2}}{2 m_{e} c^{2}\left\langle x^{2}\right\rangle} \\
&=\frac{\left(6.58 \times 10^{-16}\right)^{2} \times\left(3 \times 10^{8}\right)^{2}}{2 \times 0.51 \times 10^{-20}}=3.8 \mathrm{eV} .
\end{aligned}
$$ |
$6.0 \times 10^{22}$ atoms of helium gas occupy $2.0$ litres at atmospheric pressure. What is the temperature of the gas?
| Using the equation of state for an ideal gas, we get
$$
T=p V / n k=241 \mathrm{~K} .
$$ |
What is the cohesive energy with respect to separated ions for crystalline $\mathrm{NaCl}$ ? Give the approximate value and a derivation using a very simple model. Lattice constant $a=5.6 \AA$. | For a crystal consisting of $N$ ions, each of charge $\pm e$, the cohesive energy is
$$
U(r)=-\frac{N}{2} \sum_{j}^{\prime}\left[\pm \frac{e^{2}}{r_{l j}}-\frac{b}{r_{l j}^{n}}\right]
$$
where $b$ is the Madelung constant and the prime indicates that $j=l$ is excluded from the summation. With $r_{l j}=\alpha_{j} R$ the above becomes
$$
U(R)=-\frac{N}{2}\left(\frac{\alpha e^{2}}{R}-\frac{\beta}{R^{n}}\right),
$$
where
$$
\alpha=\sum_{j}^{\prime}(\pm) \alpha_{j}^{-1}, \quad B=b \sum_{j}^{\prime} \alpha_{j}^{-n}
$$
At equilibrium $U(R)$ is a minimum and $R=R_{0}$ is given by
$$
\left(\frac{\partial U}{\partial R}\right)_{R=R_{0}}=0
$$
which yields
$$
R_{0}=\left(\frac{\alpha e^{2}}{n B}\right)^{\frac{1}{n-1}} .
$$
and thus
$$
U\left(R_{0}\right)=-\frac{N \alpha e^{2}}{2 R_{0}}\left(1-\frac{1}{n}\right) .
$$
The two terms in $U(R)$ are a Coulomb potential and a repulsive potential. By comparing the calculated Coulomb potential and the observed total binding energy, $n$ can be estimated to be about 10 . So to a $10 \%$ accuracy we have
$$
U\left(R_{0}\right) \approx-\frac{N \alpha e^{2}}{2 R_{0}} .
$$
Applying Ewald's method to the known $\mathrm{NaCl}$ structure, we can calculate $\alpha$ and obtain $\alpha=1.7476$. Then with $R_{0}=\frac{a}{2}=2.8 \AA$, we find
$$
U\left(R_{0}\right)=178 \mathrm{kcal} / \mathrm{mol},
$$
as for crystalline $\mathrm{NaCl}, n=8$. |
Sunlight is normally incident on the surface of water with index of refraction $n=1.33$.
If the incident flux is $1 \mathrm{~kW} / \mathrm{m}^{2}$, what is the pressure that sunlight exerts on the surface of the water? | Fresnel's formulae give for normally incidence
$$
\begin{aligned}
&R=\left(\frac{n_{2}-n_{1}}{n_{2}+n_{1}}\right)^{2}=\left(\frac{n-1}{n+1}\right)^{2}=0.02 \\
&T=\frac{4 n_{1} n_{2}}{\left(n_{2}+n_{1}\right)^{2}}=\frac{4 n}{(n+1)^{2}}=0.98 \\
&R+T=1
\end{aligned}
$$
From the point of view of the photon theory, light pressure is the result of a transfer of the momentum of the incoming light to the target. If $W$ is the incident flux density and $R$ the reflection coefficient, the change of momentum per unit area is
$$
\frac{W R}{c}-\left(-\frac{W}{c}\right)=\frac{(1+R) W}{c} .
$$
By definition this is the light pressure $p$
$$
p=\frac{(1+0.02) \times 1000}{3 \times 10^{8}}=3.34 \times 10^{-6} \mathrm{~N} / \mathrm{m}^{2} .
$$ |
A given type of fuel cell produces electrical energy $W$ by the interaction of $\mathrm{O}_{2}$ fed into one electrode and $\mathrm{H}_{2}$ to the other. These gases are fed in at 1 atmosphere pressure and $298 \mathrm{~K}$, and react isothermally and isobarically to form water. Assuming that the reaction occurs reversibly and that the internal resistance of the cell is negligible, calculate the e.m.f. of the cell. Given: one Faraday $=96,500$ coulombs $/ \mathrm{g}$ mole.
Enthalpies in joules/g mole at 1 atmospheric and $298 \mathrm{~K}$ for oxygen, hydrogen, and water are respectively $17,200,8,100$ and $-269,300$.
Entropies in joules/mole. $\mathrm{K}$ at 1 atmosphere and $298 \mathrm{~K}$ for oxygen, hydrogen, and water are respectively 201,128 and $66.7$. | The chemical equation is
$$
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}=\mathrm{H}_{2} \mathrm{O} \text {. }
$$
In the reversible process at constant temperature and pressure, the decrease of Gibbs function of the system is equal to the difference between the total external work and the work the system does because of the change of volume. Thus
$$
-\Delta g=\varepsilon \Delta q,
$$
or
$$
-\sum_{\mathrm{i}}\left(\Delta h_{\mathrm{i}}-T \Delta S_{\mathrm{i}}\right)=\varepsilon \Delta q \text {. }
$$
If 1 mole of water forms, there must have been electric charges of $2 \mathrm{~F}$ flowing in the circuit, i.e., $\Delta g=2 \mathrm{~F}$. Thus the e.m.f. is
$$
\varepsilon=\frac{1}{2 \mathrm{~F}}\left[T S_{\mathrm{W}}-h_{\mathrm{W}}-T\left(S_{\mathrm{H}}+\frac{1}{2} S_{0}\right)+\left(h_{\mathrm{H}}+\frac{1}{2} h_{0}\right)\right]
$$
As given $, S_{0}=201 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, S_{\mathrm{H}}=128 \mathrm{~J} / \mathrm{mol} \mathrm{K}$, $S_{\mathrm{W}}=66.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, \dot{h}_{0}=-17200 \mathrm{~J} / \mathrm{mol}$ $h_{\mathrm{H}}=8100 \mathrm{~J} / \mathrm{mol}, h_{\mathrm{W}}=-269300 \mathrm{~J} / \mathrm{mol}$, and $T=298 \mathrm{~K}$, We have $\varepsilon=1.23 \mathrm{~V}$ |
Calculate the ratio of the intensities of the reflected wave and the incident wave for a light wave incident normally on the surface of a deep body of water with index of refraction $n=1.33$. | Fresnel's equations give for light incident normally
$$
R=\frac{I_{\mathrm{r}}}{I_{\mathrm{i}}}=\left(\frac{n_{2}-n_{1}}{n_{2}+n_{1}}\right)^{2}=0.02 .
$$ |
Čerenkov radiation is emitted by a high energy charged particle which moves through a medium with a velocity greater than the velocity of electromagnetic wave propagation in the medium.
A Čerenkov radiation particle detector is made by fitting a long pipe of one atmosphere, $20^{\circ} \mathrm{C}$ hydrogen gas with an optical system capable of detecting the emitted light and of measuring the angle of emission $\theta$ to an accuracy of $\delta \theta=10^{-3}$ radian. A beam of charged particles with momentum of $100 \mathrm{GeV} / \mathrm{c}$ are passed through the counter. Since the momentum is known, the measurement of the Čerenkov angle is, in effect, a measurement of the particle rest mass $m_{0}$. For particles with $m_{0}$ near $1 \mathrm{GeV} / \mathrm{c}^{2}$, and to first order in small quantities, what is the fractional error (i.e., $\delta m_{0} / m_{0}$ ) in the determination of $m_{0}$ with the Čerenkov counter? | As shown in Fig. 5.13, the radiation emitted by the charge at $Q^{\prime}$ at time $t^{\prime}$ arrives at $P$ at time $t$ when the charge is at Q. As the radiation propagates at the speed $c / n$ and the particle has speed $v$ where $v>c / n$, we have
$$
\mathrm{Q}^{\prime} \mathrm{P}=\frac{c}{n}\left(t-t^{\prime}\right), \quad \mathrm{Q}^{\prime} \mathrm{Q}=v\left(t-t^{\prime}\right),
$$
or
$$
\frac{\mathrm{Q}^{\prime} \mathrm{P}}{\mathrm{Q}^{\prime} \mathrm{Q}}=\cos \theta=\frac{c}{v n}=\frac{1}{\beta n},
$$
where $\beta=\frac{v}{c}$. At all the points intermediate between $Q^{\prime}$ and $Q$ the radiation emitted will arrive at the line QP at time $t$. Hence QP forms the wavefront of all radiation emitted prior to $t$.
Fig. $5.13$
As $|\cos \theta| \leq 1$, we require $\beta \geq \frac{1}{n}$ for emission of Čerenkov radiation. Hence we require
$$
\gamma=\frac{1}{\sqrt{1-\beta^{2}}} \geq\left(1-\frac{1}{n^{2}}\right)^{-\frac{1}{2}}=\frac{n}{\sqrt{n^{2}-1}} .
$$
Thus the particle must have a kinetic energy greater than
$$
\begin{aligned}
T &=(\gamma-1) m_{0} c^{2} \\
&=\left[\frac{n}{\sqrt{(n+1)(n-1)}}-1\right] m_{0} c^{2} \\
& \approx\left(\frac{1}{\sqrt{2 \times 1.35 \times 10^{-4}}}-1\right) \times 0.5 \\
& \approx 29.93 \mathrm{MeV} .
\end{aligned}
$$For a relativistic particle of momentum $P \gg m_{0} c$,
$$
\gamma=\frac{E}{m_{0} c^{2}}=\frac{\sqrt{P^{2} c^{2}+m_{0}^{2} c^{4}}}{m_{0} c^{2}} \approx \frac{P}{m_{0} c} .
$$
With $P$ fixed we have
$$
d \gamma=-\frac{P}{c} \cdot \frac{d m_{0}}{m_{0}^{2}} .
$$
Now $\beta=\frac{\gamma \beta}{\gamma}=\sqrt{\frac{\gamma^{2}-1}{\gamma^{2}}}=\sqrt{1-\frac{1}{\gamma^{2}}} \approx 1-\frac{1}{2 \gamma^{2}}$ for $\gamma \gg 1$, so that
$$
d \beta=\frac{d \gamma}{\gamma^{3}} .
$$
For the Čerenkov radiation emitted by the particle, we have
$$
\cos \theta=\frac{1}{\beta n},
$$
or
$$
d \beta=n \beta^{2} \sin \theta d \theta .
$$
Combining the above we have
$$
\left|\frac{d m_{0}}{m_{0}}\right|=\frac{m_{0} c}{P} d \gamma \approx \frac{d \gamma}{\gamma}=\gamma^{2} \beta=n \beta^{2} \gamma^{2} \sin \theta d \theta=\beta \gamma^{2} \tan \theta d \theta .
$$
With $\gamma=\frac{P c}{m_{0} c^{2}}=\frac{100}{1}=100, n=1+1.35 \times 10^{-4}$, we have
$$
\beta \approx 1-\frac{1}{2 \times 10^{4}}=1-5 \times 10^{-5},
$$
$\cos \theta=\frac{1}{\beta n}=\left(1-5 \times 10^{-5}\right)^{-1}\left(1+1.35 \times 10^{-4}\right)^{-1} \approx 1-8.5 \times 10^{-5}$,
$$
\begin{aligned}
\tan \theta=\sqrt{\frac{1}{\cos ^{2} \theta}-1} &=\sqrt{\left(1-8.5 \times 10^{-5}\right)^{-2}-1} \\
& \approx \sqrt{1.7 \times 10^{-4}} \approx 1.3 \times 10^{-2},
\end{aligned}
$$
and hence
$$
\left|\frac{d m_{0}}{m_{0}}\right|=\left(1-5 \times 10^{-5}\right) \times 10^{4} \times 1.3 \times 10^{-2} \times 10^{-3}=0.13
$$ |
At room temperature, $k_{\mathrm{B}} T / e=26 \mathrm{mV}$. A sample of cadmium sulfide displays a mobile carrier density of $10^{16} \mathrm{~cm}^{-3}$ and a mobility coefficient $\mu=10^{2} \mathrm{~cm}^{2} /$ volt sec.
If the charge carriers have an effective mass equal to $0.1$ times the mass of a free electron, what is the average time between successive scatterings? | The electrical conductivity is given by $\sigma=n e \mu$. With $n=10^{22} \mathrm{~m}^{-3}$, $e=1.6 \times 10^{-19} \mathrm{C}, \mu=10^{-2} \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$, we have for the material $\sigma=$ $16 \Omega^{-1} \mathrm{~m}^{-1}$.The law of equipartition of energy
$$
\frac{1}{2} m \bar{v}_{x}^{2}=\frac{1}{2} m \bar{v}_{y}^{2}=\frac{1}{2} m \bar{v}_{z}^{2}=\frac{1}{2} k_{\mathrm{B}} T
$$
gives
$$
\frac{1}{2} m \bar{v}^{2}=\frac{3}{2} k_{\mathrm{B}} T,
$$
or
$$
\bar{v}^{2}=\sqrt{\frac{3 k_{\mathrm{B}} T}{m}} .
$$
The rms distance $l$ between successive trappings is given by
$$
l^{2}=\bar{v}^{2} t^{2} .
$$
Hence
$$
l=\sqrt{\frac{3 k_{\mathrm{B}} T}{m}} t=\sqrt{3\left(\frac{k_{\mathrm{B}} T}{e}\right) \frac{e}{m}} t .
$$
With $\frac{k_{\mathrm{B}} T}{e}=26 \times 10^{-3} \mathrm{~V}, \frac{e}{m}=1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}, t=10^{-5} \mathrm{~s}$, we have $l=1.17 \mathrm{~m}$.The free electron model of metals gives
$$
\sigma=\frac{n e^{2}\langle\tau\rangle}{m^{*}},
$$
where $m^{*}$ is the effective mass of an electron. Then the average time between successive scatterings is
$$
\langle\tau\rangle=\frac{0.1 \sigma}{n e}\left(\frac{m}{e}\right)=5.7 \times 10^{-15} \mathrm{~s} .
$$ |
For a rock salt $(\mathrm{NaCl})$ crystal placed in front of the tube, calculate the Bragg angle for a first order reflection maximum at $\lambda=0.5 \AA$. $\left(\rho_{\mathrm{NaCl}}=\right.$ $\left.2.165 \mathrm{~g} / \mathrm{cm}^{3}\right)$ | When a high voltage is applied to the picture tube, electrons emitted from the negative electrode will be accelerated by the electric field to strike the screen target. If the energy of the electrons exceeds a certain value, they can knock off inner-shell electrons in the target atoms and make holes in the inner shells. Then as the outer-shell electrons fall in to fill up these holes, X-rays are emitted.
The maximum energy of the X-ray photons produced, $h \nu_{\max }$, is equal to the energy $\mathrm{eV}$ of the incident electrons. Hence the minimum wavelength of the X-rays is
$$
\lambda_{\min }=\frac{h c}{e V}=\frac{12000}{V}=\frac{12000}{25 \times 10^{3}}=0.48 \AA .
$$Bragg's law
$$
2 d \sin \theta=n \lambda
$$
gives the angle $\theta$ for the first order diffraction $(n=1)$ maximum:
$$
\sin \theta=\frac{\lambda}{2 d},
$$
where $d$ is the distance between two neighboring ions in the $\mathrm{NaCl}$ crystal.
As $\mathrm{NaCl}$ crystal has a simple cubic structure with $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions arranged alternately, there are $N_{0} \mathrm{Na}^{+}$ions and $N_{0} \mathrm{Cl}^{-}$ions in a mole of $\mathrm{NaCl}$, where $N_{0}$ is Avogadro's number. As $\mathrm{NaCl}$ has a molar weight $M=58.45 \mathrm{~g} / \mathrm{mol}$ and a density $\rho=2.165 \mathrm{~g} / \mathrm{cm}^{3}$, its crystal has
$$
d=\left(\frac{\frac{M}{\rho}}{2 N_{0}}\right)^{1 / 3}=\left(\frac{\frac{58.45}{2.165 \times 10^{6}}}{2 \times 6.02 \times 10^{23}}\right)^{1 / 3}=2.82 \AA .
$$
This gives
$$
\sin \theta=\frac{0.5}{2 \times 2.82}=0.088
$$
and hence
$$
\theta=5^{\circ} \text {. }
$$ |
A charge $q=2 \mu \mathrm{C}$ is placed at $a=10 \mathrm{~cm}$ from an infinite grounded conducting plane sheet. Find the force on the charge $q$. | The method of images requires that an image charge $-q$ is placed symmetrically with respect to the plane sheet. This means that the total induced charge on the surface of the conductor is $-q$.The force acting on $+q$ is
$$
F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{(2 a)^{2}}=9 \times 10^{9} \times \frac{\left(2 \times 10^{-6}\right)^{2}}{0.2^{2}}=0.9 \mathrm{~N}
$$
where we have used $\varepsilon_{0}=\frac{1}{4 \pi \times 9 \times 10^{9}} C^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$. |
Consider an opaque screen with 5 equally spaced narrow slits (spacing $d$) and with monochromatic plane-wave light (wavelength $\lambda$) incident normally. Assume $d / \lambda=10$. Approximately what is the ratio in intensity of the least intense to the most intense peak? | For multiple-slit interference,
$$
I \propto\left(\frac{\sin \frac{N \delta}{2}}{\sin \frac{\delta}{2}}\right)^{2}=\left(\frac{\sin \frac{5 \pi d \sin \theta}{\lambda}}{\sin \frac{\pi d \ln \theta}{\lambda}}\right)^{2}=\left(\frac{\sin (50 \pi \sin \theta)}{\sin (10 \pi \sin \theta)}\right)^{2} .
$$
For $\theta=0$ to $\frac{1}{5}$ rad we may take the approximation $\sin \theta \approx \theta$. Thus
$$
I \propto\left(\frac{\sin (50 \pi \theta)}{\sin (10 \pi \theta)}\right)^{2} .
$$
Therefore, intensity maxima occur when $10 \pi \theta=m \pi$, i.e., $\theta=m / 10$, where $m=0, \pm 1, \pm 2, \ldots$. Intensity minima occur when $50 \pi \theta=n \pi$, i.e., $\theta=n / 50$ where $n$ is an integer such that $n \neq 0, \pm 5, \pm 10, \ldots$
Fig. 2.54 Figure $2.54$ shows the intensity distribution. The least intense peak is midway between two adjacent most intense peaks, for which
$$
\theta=\frac{1}{2} \times \frac{1}{10}=0.05 \mathrm{rad} .
$$
The ratio in intensity of the least to the most intense peak is
$$
\begin{aligned}
\frac{I(\theta=0.05)}{I(\theta=0)} &=\left(\frac{\sin (50 \pi \times 0.05)}{\sin (10 \pi \times 0.05)}\right)^{2} /\left(\frac{\sin (50 \pi \times 0)}{\sin (10 \pi \times 0)}\right)^{2} \\
&=\left[\frac{\sin (0.5 \pi)}{\sin (0.5 \pi)}\right]^{2} /\left[\lim _{\theta \rightarrow 0} \frac{\sin (50 \pi \varepsilon)}{\sin (10 \pi \varepsilon)}\right]^{2}=\frac{1}{25} .
\end{aligned}
$$
The sketch shows the angular distance of the first subsidiary peak from $\theta=0$ to be
$$
\frac{3}{2} \times \frac{1}{50}=0.03 \mathrm{rad} .
$$ |
What is the smallest possible time necessary to freeze $2 \mathrm{~kg}$ of water at $0^{\circ} \mathrm{C}$ if a 50 watt motor is available and the outside air (hot reservoir) is at $27^{\circ} \mathrm{C}$ ? | When $2 \mathrm{~kg}$ of water at $0^{\circ} \mathrm{C}$ becomes ice, the heat released is
$$
Q_{2}=1.44 \times 2 \times 10^{3} / 18=1.6 \times 10^{2} \mathrm{kcal} .
$$
The highest efficiency of the motor is
$$
\varepsilon=\frac{T_{2}}{T_{1}-T_{2}}=\frac{Q_{2}}{W_{\min }} .
$$
Thus,
$$
W_{\min }=Q_{2} \frac{T_{1}-T_{2}}{T_{2}} .
$$
If we use the motor of $P=50 \mathrm{~W}$, the smallest necessary time is
$$
\tau=\frac{W_{\min }}{P}=\frac{Q_{2}}{P} \cdot \frac{T_{1}-T_{2}}{T_{2}} .
$$
With $T_{1}=300 \mathrm{~K}, T_{2}=273 \mathrm{~K}$, we find
$$
\tau=1.3 \times 10^{3} \mathrm{~s} .
$$ |
A gas of molecular hydrogen $\mathrm{H}_{2}$, is originally in equilibrium at a temperature of $1,000 \mathrm{~K}$. It is cooled to $20 \mathrm{~K}$ so quickly that the nuclear spin states of the molecules do not change, although the translational and rotational degrees of freedom do readjust through collisions. What is the approximate internal energy per molecule in terms of temperature units $\mathrm{K}$ ?
Note that the rotational part of the energy for a diatomic molecule is $A l(l+1)$ where $l$ is the rotational quantum number and $A \sim 90 \mathrm{~K}$ for $\mathrm{H}_{2}$. Vibrational motion can be neglected. | Originally the temperature is high and the para- and orthohydrogen molecules are in equilibrium in a ratio of about $1: 3$. When the system is quickly cooled, for a rather long period the nuclear spin states remain the same. The ratio of parahydrogen to orthohydrogen is still $1: 3$. Now the para- and orthohydrogen are no longer in equilibrium but, through collisions, each component is in equilibrium by itself. At the low temperature of $20 \mathrm{~K}$, $\exp (-\beta A) \sim \exp (-90 / 20) \ll 1$, so that each is in its ground state. Thus $\bar{E}_{r, \mathrm{p}}=0, \bar{E}_{\mathrm{r}, \mathrm{o}}=A(1+1) \cdot 1=2 A=180 \mathrm{~K}$, giving
$$
\bar{E}_{\mathrm{r}}=\frac{1}{4} \bar{E}_{\mathrm{r}, \mathrm{p}}+\frac{3}{4} \bar{E}_{\mathrm{r}, \mathrm{O}}=135 \mathrm{~K} .
$$
From equipartition of energy, we have
$$
\bar{E}_{\mathrm{t}}=\frac{3}{2} k T=30 \mathrm{~K} .
$$
The average energy of a molecule is
$$
\bar{E}=\bar{E}_{\mathrm{t}}+\bar{E}_{\mathrm{r}}=165 \mathrm{~K} .
$$ |
Čerenkov radiation is emitted by a high energy charged particle which moves through a medium with a velocity greater than the velocity of electromagnetic wave propagation in the medium.
Hydrogen gas at one atmosphere and at $20^{\circ} \mathrm{C}$ has an index of refraction $n=1+1.35 \times 10^{-4}$. What is the minimum kinetic energy in $\mathrm{MeV}$ which an electron (of rest mass $0.5 \mathrm{MeV} / \mathrm{c}^{2}$) would need in order to emit Čerenkov radiation in traversing a medium of hydrogen gas at $20^{\circ} \mathrm{C}$ and one atmosphere? | As shown in Fig. 5.13, the radiation emitted by the charge at $Q^{\prime}$ at time $t^{\prime}$ arrives at $P$ at time $t$ when the charge is at Q. As the radiation propagates at the speed $c / n$ and the particle has speed $v$ where $v>c / n$, we have
$$
\mathrm{Q}^{\prime} \mathrm{P}=\frac{c}{n}\left(t-t^{\prime}\right), \quad \mathrm{Q}^{\prime} \mathrm{Q}=v\left(t-t^{\prime}\right),
$$
or
$$
\frac{\mathrm{Q}^{\prime} \mathrm{P}}{\mathrm{Q}^{\prime} \mathrm{Q}}=\cos \theta=\frac{c}{v n}=\frac{1}{\beta n},
$$
where $\beta=\frac{v}{c}$. At all the points intermediate between $Q^{\prime}$ and $Q$ the radiation emitted will arrive at the line QP at time $t$. Hence QP forms the wavefront of all radiation emitted prior to $t$.
Fig. $5.13$
As $|\cos \theta| \leq 1$, we require $\beta \geq \frac{1}{n}$ for emission of Čerenkov radiation. Hence we require
$$
\gamma=\frac{1}{\sqrt{1-\beta^{2}}} \geq\left(1-\frac{1}{n^{2}}\right)^{-\frac{1}{2}}=\frac{n}{\sqrt{n^{2}-1}} .
$$
Thus the particle must have a kinetic energy greater than
$$
\begin{aligned}
T &=(\gamma-1) m_{0} c^{2} \\
&=\left[\frac{n}{\sqrt{(n+1)(n-1)}}-1\right] m_{0} c^{2} \\
& \approx\left(\frac{1}{\sqrt{2 \times 1.35 \times 10^{-4}}}-1\right) \times 0.5 \\
& \approx 29.93 \mathrm{MeV} .
\end{aligned}
$$ |
A beam of electrons is fired in free space over a distance of $10^{4} \mathrm{~km}$. If the transversal width of the initial packet is $1 \mu \mathrm{m}$, what will be the minimum spread upon arrival if the kinetic energy of the electrons is $13.6 \mathrm{eV}$ ? | The initial wave packet of an electron in the beam has uncertainties $\Delta p_{x}, \Delta x$ in a transverse direction related by the uncertainty relation
$$
\Delta p_{x} \cdot \Delta x \geq \frac{\hbar}{2} .
$$
If it has momentum $p$, the angle of divergence is
$$
\theta=\frac{\Delta p_{x}}{p} \gtrsim \frac{1}{2 \Delta x} \frac{\hbar}{p}=\frac{1}{2 \Delta x} \frac{\hbar c}{p c} .
$$
As
$$
\begin{gathered}
p=\sqrt{2 m T}=\frac{1}{c} \sqrt{2 m c^{2} T}=\frac{1}{c} \sqrt{2 \times 0.51 \times 10^{6} \times 13.6}=3.72 \times 10^{3} \mathrm{eV} / \mathrm{c}, \\
\Delta x \approx 10^{-6} \mathrm{~m},
\end{gathered}
$$
we have
$$
\theta \gtrsim \frac{6.6 \times 10^{-16} \times 3 \times 10^{8}}{2 \times 10^{-6} \times 3.72 \times 10^{3}}=2.66 \times 10^{-5} \mathrm{radian}
$$
and the minimum transverse spread is
$$
L \theta_{\min }=266 \mathrm{~m} .
$$
|
Assume that the earth's magnetic field is caused by a small current loop located at the center of the earth. Given that the field near the pole is $0.8$ gauss, that the radius of the earth is $R=6 \times 10^{6} \mathrm{~m}$, and that $\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$, use the Biot-Savart law to calculate the strength of the mangnetic moment of the small current loop. | Assume that the axis of the current loop, of small radius a, coincides with the axis of rotation of the earth, which is taken to be the $z$-axis as shown in Fig. 2.8. The contribution of a current element $I d I$ to the magnetic induction $B$ at an axial point $z$ is, according to the Biot-Savart law,
$$
d \mathrm{~B}=\frac{\mu_{0}}{4 \pi} \frac{I d \rrbracket \times \mathrm{r}}{r^{3}} .
$$
$d \mathrm{~B}$ is in the plane containing the $z$-axis and $\mathbf{r}$ and is perpendicular to r. Summing up the contributions of all current elements of the loop, by symmetry the resultant $B$ will be along the $z$-axis, i.e.,
$$
\begin{aligned}
&B=B_{z} \mathbf{e}_{z}, \text { or } \\
&d B_{z}=d B \cdot \frac{a}{r} .
\end{aligned}
$$
At the pole, $z=R$. As $R \gg a, r=\sqrt{R^{2}+a^{2}} \approx R$ and
$$
\begin{aligned}
B_{z} &=\frac{\mu_{0}}{4 \pi} \frac{I a}{R^{3}} \oint d l=\frac{\mu_{0}}{4 \pi} \frac{I a}{R^{3}} \cdot 2 \pi a \\
&=\frac{\mu_{0}}{2 \pi} \frac{I S}{R^{3}},
\end{aligned}
$$
where $S=\pi a^{2}$ is the area of the current loop.
Fig. $2.8$
The magnetic moment of the loop is $\mathbf{m}=I S \mathbf{e}_{z}$, thus
$$
m=\frac{2 \pi R^{3}}{\mu_{0}} B_{z} .
$$
Using the given data $R=6 \times 10^{6} \mathrm{~m}, B_{z}=0.8 \mathrm{Gs}$, we obtain
$$
m \approx 8.64 \times 10^{-26} \mathrm{Am}^{2} \text {. }
$$ |
Consider simple models for the earth's atmosphere. Neglect winds, convection, etc, and neglect variation in gravity.
Assume that the atmosphere is isothermal $\left(\right.$ at $\left.0^{\circ} \mathrm{C}\right)$. By using an expression for the distribution of molecules with height, estimate roughly the height below which half the molecules lie. | The molecular number density at height $h$ is denoted by $n(h)$. From the condition of mechanical equilibrium $d p=-n m g d h$ and the equation of state $p=n k T$, we find
$$
\frac{1}{p} d p=-\frac{m g}{k T} d h .
$$
Thus $n(h)=n_{0} \exp (-m g h / k T)$. Let $\int_{0}^{H} n(h) d h / \int_{0}^{\infty} n(h) d h=\frac{1}{2}$, then
$$
H=\frac{k T}{m g} \ln 2=\frac{R T}{N_{0} m g} \ln 2 .
$$
The average molecular weight of the atmosphere is 30 . We have
$$
H=\frac{8.31 \times 10^{7} \times 273}{30 \times 980} \times \ln 2 \approx 8 \times 10^{5} \mathrm{~cm}=8 \mathrm{~km} .
$$ |
A parabolic mirror of small relative aperture ($10 \mathrm{~cm}$ dia, $500 \mathrm{~cm}$ focal length) is used to photograph stars. Roughly what is the diameter of the image "blob" for a star on axis for visible light (take $\lambda=5000 \AA$)? | For photographing stars, the dominant limitation of resolution is the Fraunhofer diffraction by the aperture of the telescope. The diameter of the image "blob" (the Airy disk) of a star on axis is
$$
d=\frac{1.22 \lambda f}{D} .
$$
With $D=10 \mathrm{~cm}, f=500 \mathrm{~cm}$ and $\lambda=5000 \AA$ we have
$$
d=0.3 \times 10^{-2} \mathrm{~cm} .
$$
For a atar off axis, the ray makes an angle $\theta$ with the axis of the telescope. So the effective diameter of the aperture reduces to $D \cos \theta$, and the diameter of the Airy disk becomes
$$
d^{\prime}=\frac{1.22 \lambda f}{D \cos \theta} .
$$ |
A defective satellite of mass $950 \mathrm{~kg}$ is being towed by a spaceship in empty space. The two vessels are connected by a uniform $50 \mathrm{~m}$ rope whose mass per unit length is $1 \mathrm{~kg} / \mathrm{m}$. The spaceship is accelerating in a straight line with acceleration $5 \mathrm{~m} / \mathrm{sec}^{2}$.
What is the force exerted by the spaceship on the rope? | $$
\begin{aligned}
F &=\left(m_{\text {rope }}+m_{\text {gatellite }}\right) \cdot a \\
&=(950+50) \times 5=5 \times 10^{3} \mathrm{~N} .
\end{aligned}
$$ |
Suppose an electron is in a state described by the wave function
$$
\psi=\frac{1}{\sqrt{4 \pi}}\left(e^{i \phi} \sin \theta+\cos \theta\right) g(r)
$$
where
$$
\int_{0}^{\infty}|g(r)|^{2} r^{2} d r=1
$$
and $\phi, \theta$ are the azimuth and polar angles respectively.
What is the expectation value of $L_{z}$ ? | As
$$
Y_{10}=\sqrt{\frac{3}{4 \pi}} \cos \theta, Y_{1, \pm 1}=\mp \sqrt{\frac{3}{8 \pi}} \sin \theta e^{\pm i \phi},
$$
the wave function can be written as
$$
\psi=\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right) g(r)
$$
Hence the possible values of $L_{z}$ are $+\hbar, 0$.
Since
$$
\begin{aligned}
\int|\psi|^{2} d r &=\frac{1}{4 \pi} \int_{0}^{\infty}|g(r)|^{2} r^{2} d r \int_{0}^{\pi} d \theta \int_{0}^{2 \pi}(1+\cos \phi \sin 2 \theta) \sin \theta d \phi \\
&=\frac{1}{2} \int_{0}^{\pi} \sin \theta d \theta=1
\end{aligned}
$$
the given wave function is normalized. The probability density is then given by $P=|\psi|^{2}$. Thus the probability of $L_{z}=+\hbar$ is $\left(\sqrt{\frac{2}{3}}\right)^{2}$ or $2 / 3$ and that of $L_{z}=0$ is $\left(\frac{1}{\sqrt{3}}\right)^{2}$ or $1 / 3$.
$$
\begin{aligned}
\int \psi^{*} L_{z} \psi r^{2} \sin \theta d \theta d \phi d r=& \int\left[\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right)\right]^{*} \\
& \times \hat{L}_{z}\left[\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right)\right] \\
& \times|g(r)|^{2} r^{2} d r \sin \theta d \theta d \phi \\
=& \frac{2}{3} \hbar \int_{0}^{\pi} d \theta \int_{0}^{2 \pi} Y_{11}^{2} d \phi=\frac{2}{3} \hbar
\end{aligned}
$$ |
It is perhaps too easy to think of mathematical truths as objective entities existing in the universe, simply waiting to be "discovered" by brilliant minds such as Pythagoras, Newton, and Descartes. Indeed, such a mentality may be cemented in Western minds raised in the Platonic tradition of a world of ideals that exists outside of the material world we reside in.
But new research in the fields of cognitive science and developmental psychology may be challenging this long-held belief in mathematics as a truth outside of human experience, and recasting mathematics as a product of the brain, an entity that is not discovered by man but rather created by him.
Such a radical paradigm shift has predictably met with stiff resistance, but the evidence in favor of a created mathematics rather than a discovered mathematics is compelling. Study after study has shown that all people possess an innate arithmetic. Babies as young as three or four days old can differentiate between groups of two and three items. And by the age of four months or so, an infant can see that one plus one is two and two minus one is one. Researchers discovered this startling fact by means of a simple experiment, following what is known as the violation of expectation model in developmental psychology. An infant was presented with a scenario in which a researcher held a puppet before the baby's eyes. Then a screen was moved in front of the puppet and another researcher placed a second puppet behind it. The screen was removed and if there were two puppets now visible, the infant registered no surprise (as measured by both the direction of the child's gaze and the duration of her stare). But if the screen was removed and only one puppet appeared, the infant registered perplexity at the situation. This experiment and others like it strongly suggest that all individuals are born with certain mathematical concepts hardwired into their brains. The ability to quickly and accurately count a small number of items, called subitizing by psychologists, is also found in animals. Researchers using experiments similar to the violation of expectation setup described previously have found an innate mathematical ability not just in primates, our closest evolutionary cousins, but in raccoons, rats, parrots, and pigeons. These findings are consistent with the belief that mathematical thinking is a function of the structures in the brain and not a product of the outside world.
Anomalous cases involving brain injuries and disorders also support the idea that mathematical thinking arises from the organization of the brain. There is a documented case of a subject with a $\mathrm{PhD}$ in chemistry who suffers from acalculia, the inability to perform basic arithmetic functions. Strangely, this patient is unable to perform simple calculations such as five plus three or eight minus two, but has no problem manipulating abstract algebraic operations. This curious fact has led psychologists to conclude that the part of the brain that handles abstract algebraic operations must be different from the part of the brain that works with more concrete arithmetic functions.
The author supports the passage's main thesis with all of the following types of evidence EXCEPT:
Options:
A. experiments
B. case studies
C. testimonials
D. comparisons | To answer this question, you must assess evidence. Since this is an EXCEPT question, you're looking for the choice that is not present in the passage. The passage does discuss experiments, choice $\mathrm{A}$, in the section about violation of expectation tests. The part about the $\mathrm{PhD}$ in chemistry who suffered from acalculia is a case study (choice B). Finally, the author presents a comparison (choice D) when showing how animals and humans have similar abilities. That leaves choice $\mathrm{C}$ as the correct answer. The correct answer is C. |
"I acknowledge Shakespeare to be the world's greatest dramatic poet, but regret that no parent could place the uncorrected book in the hands of his daughter, and therefore I have prepared the Family Shakespeare."
Thus did Thomas Bowdler, a self-appointed editor trained as a physician, explain his creation of a children's edition of Shakespeare that omitted some characters completely, toned down language he considered objectionable, and euphemized such shocking situations as Ophelia's suicide-an accident in Bowdler's version.
Bowdler was hardly the first to tone down the Bard. Poet laureate Nahum Tate rewrote King Lear, banishing the Fool entirely and giving the play a happy ending. His version was staged regularly from the 1680 s through the 18 th century.
Thomas Bowdler was, from all the evidence, a less likely editor. He was born in 1754 near Bath, England, to a wealthy family. He studied medicine but never really practiced, preferring to exercise philanthropy and to play chess, at which he was a master. In his retirement, he decided to try his hand at editorial work. Shakespeare's dramas were his first project. He added nothing to the texts, but by cutting and paraphrasing strove to remove anything that "could give just offense to the religious and virtuous mind."
The result was a 10-volume expurgated version that was criticized widely, although hardly universally. The famed British poet Algernon Swinburne remarked that "no man ever did better service to Shakespeare than the man who made it possible to put him into the hands of intelligent and imaginative children."
There is some indication that Bowdler's sister Harriet did the actual editing of Shakespeare's text. She was a poet and editor in her own right, and clearly more qualified than her brother to lay hands on the Bard of Avon. She may have published the expurgated edition anonymously before allowing her brother to take over the rights. If this is so, it is unsurprising that Harriet would not have wanted her name on the book. If the original Shakespeare were truly objectionable, then it would have been doubly so for a well-bred, unmarried Englishwoman of the day.
Bowdler went on to create children's versions of the Old Testament and of Edward Gibbons's History of the Decline and Fall of the Roman Empire, although neither achieved either the success or ridicule of his Shakespeare. Today, of course, we know Bowdler not for his good works, but instead for the eponym derived from his good name. To bowdlerize is to censor or amend a written work, often with a connotation of prudishness. The process for which Bowdler is known is one that continues to the present day, as texts deemed too difficult for today's high schoolers are "dumbed-down," library editions of The Adventures of Huckleberry Finn have the "n-word" blacked out, and middle-school productions of popular Broadway plays have all expletives deleted. We would be hard-pressed to say that we live in a prudish era, but some of the same impulses that drove Thomas and Harriet Bowdler still exist today.
The author's claim that Bowdler's Shakespeare was not unanimously criticized is supported by:
Options:
A. a quotation from Algernon Swinburne
B. reference to later children's works by Bowdler
C. the definition of the eponym formed from his name
D. the example of Ophelia's revised "suicide" | Although choices B, C, and D appear in the passage, none of them supports the claim that Bowdler's revised Shakespeare was not universally despised. The quote from Swinburne (choice A), however, does-Swinburne praises Bowdler for creating a Shakespeare that can be put in the hands of children. The correct answer is A. |
The Shawnee people of the Ohio and Cumberland Valleys gave the American elk its other name, wapiti. Yet when we think of elk today, we rarely envision them as eastern animals. If we want to see elk in their natural habitat, we travel to Colorado or Montana, where they still roam the mountainous terrain. Elk were, in fact, once found in the East, from Georgia north to New York and Connecticut. By the time of the Civil War, hunting and habitat destruction had caused their extinction in most eastern states. All of the eastern subspecies are now extinct. Elk County, Pennsylvania, and Elk County, Kansas, were without elk for more than a century. At the beginning of the 20th century, herds of elk in the Rocky Mountains faced death by starvation as encroaching farms depleted their winter feeding and finally the government decided to intercede. They gathered up elk from Yellowstone National Park and shipped 50 of them to Pennsylvania. The elk were to be protected, with no hunting of them allowed until eight years had passed. At that early date, 1913, there was little understanding of the kind of acclimatization required when moving large animals from one habitat to another. The elk were released from cattle cars and chased into the wild to fend for themselves. Two years later, 95 more elk were moved from Yellowstone to Pennsylvania. Elk are large animals with large ranges, and the damage these elk caused to cornfields was substantial. The elk tended to move toward farming areas because that was where the food was. Farmers and poachers took some elk illegally, but the herds still began to grow. During the 1920 s, several hunting seasons went by with harvests of one or two dozen elk, but then the numbers declined dramatically, and the elk were again put under protected status. No real census was taken until 50 years later, in 1971. According to the passage, where might you see elk today?
Options:
A. the Tennessee Valley
B. Georgia
C. the Adirondacks
D. Arkansas | "At present," says the passage, "new herds are established in Arkansas, Kentucky, Michigan, and Wisconsin in addition to Pennsylvania." There is talk of moving herds to Tennessee (choiceA) and the Adirondacks (choice C) in the future. Georgia (choice B) is mentioned only as the original southernmost range of eastern elk; it is not a place where they might be seen today. The correct answer is D. |
The Shawnee people of the Ohio and Cumberland Valleys gave the American elk its other name, wapiti. Yet when we think of elk today, we rarely envision them as eastern animals. If we want to see elk in their natural habitat, we travel to Colorado or Montana, where they still roam the mountainous terrain. Elk were, in fact, once found in the East, from Georgia north to New York and Connecticut. By the time of the Civil War, hunting and habitat destruction had caused their extinction in most eastern states. All of the eastern subspecies are now extinct. Elk County, Pennsylvania, and Elk County, Kansas, were without elk for more than a century. At the beginning of the 20th century, herds of elk in the Rocky Mountains faced death by starvation as encroaching farms depleted their winter feeding and finally the government decided to intercede. They gathered up elk from Yellowstone National Park and shipped 50 of them to Pennsylvania. The elk were to be protected, with no hunting of them allowed until eight years had passed. At that early date, 1913, there was little understanding of the kind of acclimatization required when moving large animals from one habitat to another. The elk were released from cattle cars and chased into the wild to fend for themselves. Two years later, 95 more elk were moved from Yellowstone to Pennsylvania. Elk are large animals with large ranges, and the damage these elk caused to cornfields was substantial. The elk tended to move toward farming areas because that was where the food was. Farmers and poachers took some elk illegally, but the herds still began to grow. During the 1920 s, several hunting seasons went by with harvests of one or two dozen elk, but then the numbers declined dramatically, and the elk were again put under protected status. No real census was taken until 50 years later, in 1971. According to the passage, what is true of the elk in Pennsylvania today?
Options:
A. They are the same as the elk who lived there 200 years ago.
B. They are a different subspecies from the old Pennsylvania elk.
C. They are a different subspecies from the elk found in the Rockies.
D. They are only distantly related to the elk found in Yellowstone National Park. | As the passage states clearly, "All of the eastern subspecies are now extinct." The reintroduced subspecies is the western subspecies, the one found in Yellowstone. It is a different subspecies from the original Pennsylvania elk. The correct answer is B. |
\section{Passage $\mathrm{V}$}
In linguistics, metathesis refers to the reversal of phonemes in a word. This can come about by accident, as in the common mispronunciation "aks" for ask or the common (and correct) pronunciation of iron as "i-orn." It may come about on purpose, as in various language games.
Accidental metathesis may be a process that takes place in the evolution of languages. Over time, for example, the Latin words miraculum, periculum, and parabola evolved into the Spanish words milagro, peligro, and palabra. Linguists posit that such changes usually occur when two consonants are of the same type, with similar mouth formations needed to create each one.
The Reverend Archibald Spooner, an Oxford dean, was known for his unintentional transpositions and gave his name to the particular metathesis he represented: Spoonerisms. Most famous Spoonerisms once attributed to Spooner are now believed to be apocryphal, but they are nevertheless amusing; for example, his supposed advice to a substandard student: "You have deliberately tasted two worms and will leave Oxford by the next town drain." Spoonerisms are funny when the metathesis involved changes one word into another, and they seem to lend themselves particularly well to off-color jokes.
Everyone is familiar with Pig Latin, in which initial consonant sounds are moved to the end of a word and followed with the syllable "ay." This is not pure metathesis, but it begins with the kind of metathesis known as backslang, in which syllables or letters in a word are reversed to produce a private code. Pig Latin may have begun as a language spoken in prisoner-of-war camps; it was certainly used in such camps during World War II and in Vietnam.
In Pig Latin, the phrase "Over the river and through the woods" becomes "Overay uhthay iverray anday oughthray uhthay oodsway." In other variations of Pig Latin, it might instead be "Overway uhthay iverray andway oughthray uhthay oodsway." There are computer programs that translate from English to Pig Latin and back again, and there is even a version of the Bible written entirely in Pig Latin. For the most part, however, outside of prisoner-of-war camps, Pig Latin is a funny code used by children.
French children and teenagers use a similar code. Called verlan (a reversal of l'envers, or "reversal"), it is trickier than Pig Latin because it relies much more on phonetics than on spelling. In verlan, the word café becomes féca, but the word tomber ("to fall") becomes béton. Verlan becomes even more complex in words that end in silent $e$ or which have only one syllable. While just a few words from Pig Latin have entered the vernacular (for example, amscray and ixnay), many verlan words are common lingo in France.
Historically, for reasons that are not quite clear, butchers have been frequent users of metathesis, creating their own backslang to converse without being understood by patrons. In France, this is called louchébem (boucher, or "butcher," with the first consonant moved to the end and the letter $l$ placed in front). In Australia, it is known as rechtub klat. That name, of course, represents the ultimate in metathesis-the reversal of letters instead of merely syllables.
The discussion of prisoner-of-war camps shows primarily that:
Options:
A. Pig Latin can be entertaining even in difficult circumstances.
B. Metathesis developed during the mid-20th century.
C. Prisoners may revert to childish games to pass the time.
D. Metathesis is useful and practical as a secret code. | Although the origin of Pig Latin appears to be unknown, the author refers to its use in the prisoner-of-war camps of World War II and Vietnam. The implication is that it was used to communicate between prisoners in a language their captors could not understand. The correct answer is D. |
\section{Passage $\mathrm{V}$}
In linguistics, metathesis refers to the reversal of phonemes in a word. This can come about by accident, as in the common mispronunciation "aks" for ask or the common (and correct) pronunciation of iron as "i-orn." It may come about on purpose, as in various language games.
Accidental metathesis may be a process that takes place in the evolution of languages. Over time, for example, the Latin words miraculum, periculum, and parabola evolved into the Spanish words milagro, peligro, and palabra. Linguists posit that such changes usually occur when two consonants are of the same type, with similar mouth formations needed to create each one.
The Reverend Archibald Spooner, an Oxford dean, was known for his unintentional transpositions and gave his name to the particular metathesis he represented: Spoonerisms. Most famous Spoonerisms once attributed to Spooner are now believed to be apocryphal, but they are nevertheless amusing; for example, his supposed advice to a substandard student: "You have deliberately tasted two worms and will leave Oxford by the next town drain." Spoonerisms are funny when the metathesis involved changes one word into another, and they seem to lend themselves particularly well to off-color jokes.
Everyone is familiar with Pig Latin, in which initial consonant sounds are moved to the end of a word and followed with the syllable "ay." This is not pure metathesis, but it begins with the kind of metathesis known as backslang, in which syllables or letters in a word are reversed to produce a private code. Pig Latin may have begun as a language spoken in prisoner-of-war camps; it was certainly used in such camps during World War II and in Vietnam.
In Pig Latin, the phrase "Over the river and through the woods" becomes "Overay uhthay iverray anday oughthray uhthay oodsway." In other variations of Pig Latin, it might instead be "Overway uhthay iverray andway oughthray uhthay oodsway." There are computer programs that translate from English to Pig Latin and back again, and there is even a version of the Bible written entirely in Pig Latin. For the most part, however, outside of prisoner-of-war camps, Pig Latin is a funny code used by children.
French children and teenagers use a similar code. Called verlan (a reversal of l'envers, or "reversal"), it is trickier than Pig Latin because it relies much more on phonetics than on spelling. In verlan, the word café becomes féca, but the word tomber ("to fall") becomes béton. Verlan becomes even more complex in words that end in silent $e$ or which have only one syllable. While just a few words from Pig Latin have entered the vernacular (for example, amscray and ixnay), many verlan words are common lingo in France.
Historically, for reasons that are not quite clear, butchers have been frequent users of metathesis, creating their own backslang to converse without being understood by patrons. In France, this is called louchébem (boucher, or "butcher," with the first consonant moved to the end and the letter $l$ placed in front). In Australia, it is known as rechtub klat. That name, of course, represents the ultimate in metathesis-the reversal of letters instead of merely syllables.
Which of the following examples is presented as evidence that butchers are frequent users of metathesis?
I. Spoonerisms
II. louchébem
III. Pig Latin
Options:
A. I only
B. I and II only
C. II only
D. II and III only | The final paragraph discusses two forms of metathesis that derive from the butchers' trade. One is louchébem (II), and the other is rechtub klat. Since only II is correct, the answer is choice C. The correct answer is C. |
\section{Passage $\mathrm{V}$}
Engineers and computer scientists are intrigued by the potential power of nanocomputing. Nanocomputers will use atoms and molecules to perform their functions instead of the standard integrated circuits now employed. Theorists believe that the amount of information a nanocomputer could handle is staggering.
A professor at Massachusetts Institute of Technology (MIT) has attempted to calculate the computational limits of a computer with a weight of 1 kilogram and a volume of 1 liter. According to the laws of physics, the potential amount of computational power is a function of the available energy. Basically, each atom and subatomic particle in the computer has an amount of energy attached to it. Furthermore, the energy of each particle or atom is increased by the frequency of its movement. Thus the power of a computer that uses nanotechnology is bounded by the energy available from its atoms.
Specifically, the relationship between the energy of an object and its computation potential is a proportionate one. As Einstein has famously calculated, the energy of an object is equal to its mass times the speed of light squared. Thus, a theoretical computer weighing a mere kilogram has a huge amount of potential energy. To find the total computational power, the minimum amount of energy required to perform an operation is divided by the total amount of energy.
of energy possessed by a 1-kilogram computer by Planck's constant yields a tremendously large number, roughly $5 \times 10^{50}$ operations per second. Using even the most conservative estimates of the computing power of the human brain, such a computer would have the computational power of five trillion trillion human civilizations. The computer would also have a memory capacity, calculated by determining the degrees of freedom allowed by the state of all the particles comprising it, of $10^{31} \mathrm{bits}$.
These numbers are purely theoretical, however. Were the computer to convert all of its mass to energy, it would be the equivalent of a thermonuclear explosion. And it is unreasonable to expect human technology to ever achieve abilities even close to these limits. However, a project at the University of Oklahoma has succeeded in storing 50 bits of data per atom, albeit on only a limited number of atoms. Given that there are $10^{25}$ atoms in 1 kilogram of material, it may be possible to fit up to $10^{27}$ bits of information in the computer. And if scientists are able to exploit the many properties of atoms to store information, including the position, spin, and quantum state of all its particles, it may be possible to exceed even that number.
One interesting consequence of such staggering increases in computing power is that each advance could provide the basis for further evolution. Once technology can achieve, for instance, a level of computation equal to $10^{40}$ operations per second, it can use that massive power to help bring the theoretical limit ever closer.
The central thesis of the passage is:
Options:
A. Computing power is limited only by the laws of physics.
B. New advances in computer technology allow for staggering levels of memory and computational ability.
C. It may not be possible to achieve the theoretical limits of computing power.
D. Computers using nanotechnology have the potential to tap vast quantities of power. | This question asks you to find the main idea. This passage Passage $\mathrm{V}$ explains that theoretically, a massive amount of computational power is available. However, practical considerations make it unlikely that this limit would ever be reached. Thus choice $\mathrm{C}$ is the best answer. Choice $\mathrm{A}$ is a detail provided in the passage, but it is not the main idea. Choice $\mathrm{B}$ is incorrect, because the advances discussed are still primarily theoretical. Choice $\mathrm{D}$ is also only partly correct, because it doesn't deal with the practical limits mentioned in the passage. The correct answer is C. |
\section{Passage $\mathrm{V}$}
The Nobel Peace Prize has often engendered controversy. The Red Cross, yes, but Henry Kissinger? Mother Teresa, perhaps, but Yasser Arafat? Surprisingly, a loud and ongoing controversy surrounds a choice that at first look seems most benign-that of the Guatemalan freedom fighter, Rigoberta Menchú Tum.
Rigoberta Menchú was born in 1959 to a poor Mayan family in Guatemala. At the time of her birth, General Ydígoras Fuentes had just seized power, following the assassination of Colonel Castillo Armas. The year she turned one, there was a failed coup by a military group, many of whom fled to the countryside and would return to lead the rebellion that would wax and wane for the next 36 years.
Guatemala was divided into factions for that entire period, as military governments controlled the nation and guerilla groups controlled the countryside. At the same time, right-wing vigilantes led a campaign of torture and assassination, eliminating students and peasants they deemed to be allied with the guerillas.
In the countryside where Rigoberta lived, the battles were largely over control of farmland. Rigoberta's father was taken into custody and tortured when the army believed he had assisted in the assassination of a plantation owner. Following a rout of the guerillas by the army, the guerillas regrouped and began to take their fight to the capital, beginning a long series of assassinations of government figures.
Rigoberta, her father, and some of her siblings joined the Committee of the Peasant Union (CUC). In rapid succession, Rigoberta lost her brother, father, and mother to the army's own assassination squads. Rigoberta, though only in her early 20s, became a key figure in the resistance, traveling from village to village to educate the Mayan peasants in overcoming oppression, and leading demonstrations in Guatemala City.
Soon Rigoberta was herself a target. She fled to Mexico and continued her life as an organizer, concentrating her focus on peasants' rights. She participated in the founding of the United Representation of the Guatemalan Opposition (RUOG). This period was the most violent of the entire Guatemalan civil war. Under the new president, General Efrain Ríos Montt, massacres of civilians became an everyday occurrence. The controversy is all about the book. It began when anthropologist David Stoll began independent research on the same era about which Rigoberta dictated and discovered discrepancies in her recall of events. Massacres she described were not remembered by the locals; one of her brothers died by shooting rather than by fire, dates were incorrect, and so on. Stoll's discoveries were enough to roil up conservative historians and columnists, who declared Rigoberta's book the grossest propaganda and her career, therefore, unworthy of the Nobel Prize. Not surprisingly, this brought forth charges of racism, sexism, and cultural ignorance from the other side.
The Nobel Committee maintained a stiff upper lip throughout the political backand-forth and still insists that Rigoberta's prize had to do with her documented good works and not with her personal recorded history. Rigoberta's book is still widely taught, and her place in the history of Central American peasant revolutions seems assured.
In the context of the passage, the word custody means:
Options:
A. fortification
B. safekeeping
C. incarceration
D. supervision | For vocabulary-in-context questions like this one, always locate the word in question because the distracters are likely to be alternate meanings of that word. Here, the passage states, "Rigoberta's father was taken into custody and tortured when the army believed he had assisted in the assassination of a plantation owner." The only answer that makes sense in this context is choice $\mathrm{C}$. The correct answer is C. |
\section{
If our knowledge of the world occurs through the weaving of narratives, as postmodernists would have us believe, then we must judge the truth of each narrative by comparing it with what we value and what we already accept as true. Any metanarrative, or overarching "big story," must be rejected because it attempts to explain away too many individual narratives.
history through that metanarrative, postmodernists would argue, does not allow us to consider the contributions of other groups to the country we live in today. So constructing a metanarrative can often be exclusionary.
Of course, constructing a series of smaller narratives is just as exclusionary because it frames experience in a limited way. It is easy to see this occurring if you look at any university course list today. How is it possible for American History 4111, "Imperialism and Amerindians, 1600-1840" to coexist alongside American History 4546, "American Military History and Policy"? Does English 340X, "Survey of Women's Literature," subsume or incorporate elements of English 342X, "American Indian Women Writers"? Perhaps we should pity the undergraduate student today. Lacking any overarching metanarrative to provide perspective, how can the poor student wade through the often contradictory mininarratives being thrown his or her way?
There is no question that the old way was wrongheaded and elitist. Survey courses that emphasized the white male perspective were removed from the curriculum in the 1970s, '80s, and '90s, and for good reason. But replacing them with a smorgasbord of mininarratives risks eliminating any sense of a big picture. Can students connect the dots to define a coherent reality, or must their reality be a series of frames with no links among them?
Revising the canon was about ridding our perspective of those racist, sexist, or classist notions that insisted on a single Truth with a capital $T$. Of course there is no overriding Truth that applies to everyone. For everyone who believes that Thomas Jefferson was a brilliant reformer, there is someone who thinks he was a two-faced slaveholder. Yet, where does it leave us if everyone we know is approaching history, science, literature, or what have you from a slightly different angle? It's bad enough when partisan politics divides us into red states and blue states. How much worse is it to imagine ourselves clad in thousands upon thousands of shades of purple?
The clearest sign that we are not ready to abandon our metanarratives comes in the current and ongoing clash between those who accept evolutionary theory and those who accept the Bible as the written word of God. The latter grant the Earth 6000 years of life, the former give it several million more, and never the twain shall meet. Each of these viewpoints is a metanarrative, a big story that governs people's understanding of the world. Like many metanarratives, each of these completely negates the other.
So on the one hand, metanarratives don't work well because they are too exclusionary. And on the other hand, mininarratives don't work well because they are too narrow. It will be fascinating to watch the canon evolve over the next few decades and to see whether this dichotomy can ever be resolved.
The passage suggests that the author would MOST likely believe that:
Options:
A. Colleges should offer more courses in American history.
B. Revising the canon in the late 20th century was a bad idea.
C. Students today need a more global perspective than their coursework provides.
D. Divisiveness in politics is preferable to conformity or unilateralism. | You must make this prediction based on what the author has said in the passage. Two examples of courses in American history are given to support the notion that some mininarratives are incompatible, but there is no indication that offering more courses would help (choice A). The author mentions revising the canon (choice B) in paragraphs 4 and 5 but implies that it was a good and necessary idea. Red states and blue states are mentioned in passing, but there is little else to indicate the author's feelings about politics (choice D). The best answer is choice $C$; the author is concerned that students today find it hard to see a "big picture" due to their immersion in competing mininarratives. The correct answer is C. |
It is perhaps too easy to think of mathematical truths as objective entities existing in the universe, simply waiting to be "discovered" by brilliant minds such as Pythagoras, Newton, and Descartes. Indeed, such a mentality may be cemented in Western minds raised in the Platonic tradition of a world of ideals that exists outside of the material world we reside in.
But new research in the fields of cognitive science and developmental psychology may be challenging this long-held belief in mathematics as a truth outside of human experience, and recasting mathematics as a product of the brain, an entity that is not discovered by man but rather created by him.
Such a radical paradigm shift has predictably met with stiff resistance, but the evidence in favor of a created mathematics rather than a discovered mathematics is compelling. Study after study has shown that all people possess an innate arithmetic. Babies as young as three or four days old can differentiate between groups of two and three items. And by the age of four months or so, an infant can see that one plus one is two and two minus one is one. Researchers discovered this startling fact by means of a simple experiment, following what is known as the violation of expectation model in developmental psychology. An infant was presented with a scenario in which a researcher held a puppet before the baby's eyes. Then a screen was moved in front of the puppet and another researcher placed a second puppet behind it. The screen was removed and if there were two puppets now visible, the infant registered no surprise (as measured by both the direction of the child's gaze and the duration of her stare). But if the screen was removed and only one puppet appeared, the infant registered perplexity at the situation. This experiment and others like it strongly suggest that all individuals are born with certain mathematical concepts hardwired into their brains. The ability to quickly and accurately count a small number of items, called subitizing by psychologists, is also found in animals. Researchers using experiments similar to the violation of expectation setup described previously have found an innate mathematical ability not just in primates, our closest evolutionary cousins, but in raccoons, rats, parrots, and pigeons. These findings are consistent with the belief that mathematical thinking is a function of the structures in the brain and not a product of the outside world.
Anomalous cases involving brain injuries and disorders also support the idea that mathematical thinking arises from the organization of the brain. There is a documented case of a subject with a $\mathrm{PhD}$ in chemistry who suffers from acalculia, the inability to perform basic arithmetic functions. Strangely, this patient is unable to perform simple calculations such as five plus three or eight minus two, but has no problem manipulating abstract algebraic operations. This curious fact has led psychologists to conclude that the part of the brain that handles abstract algebraic operations must be different from the part of the brain that works with more concrete arithmetic functions.
Suppose a team of anthropologists discovered an isolated community of individuals who possessed absolutely no mathematical ability whatsoever. This discovery would have what effect on the author's argument?
Options:
A. It would refute it.
B. It would support it.
C. It would neither support nor refute it.
D. It would be necessary to determine the cause of the individuals' lack of mathematical ability in order to say what effect it would have on the argument. | This question asks you to apply new evidence. This question is very difficult because you must understand both the nature of the author's argument and the effect of new evidence on it. The passage argues that mathematical thinking is a product of the innate structures of the brain and mathematical "truths" are created by people. Finding a group of people that lacks mathematical ability might seem to refute the argument as choice A indicates, but upon closer examination it is not the best answer. The author does allow that certain changes in the brain may destroy the ability to perform mathematical operations. Thus, the community of math-deficient people may have a genetic mutation that prevents them from doing math. If so, it would not harm the author's argument. If, however, the group had no brain abnormalities and possessed no mathematical abilities, it would challenge the author's belief that mathematical thought is hardwired into the brain. Thus, choice $\mathrm{D}$ is the best answer. Answers and Explanations The correct answer is D. |
\section{Passage $\mathrm{V}$}
Engineers and computer scientists are intrigued by the potential power of nanocomputing. Nanocomputers will use atoms and molecules to perform their functions instead of the standard integrated circuits now employed. Theorists believe that the amount of information a nanocomputer could handle is staggering.
A professor at Massachusetts Institute of Technology (MIT) has attempted to calculate the computational limits of a computer with a weight of 1 kilogram and a volume of 1 liter. According to the laws of physics, the potential amount of computational power is a function of the available energy. Basically, each atom and subatomic particle in the computer has an amount of energy attached to it. Furthermore, the energy of each particle or atom is increased by the frequency of its movement. Thus the power of a computer that uses nanotechnology is bounded by the energy available from its atoms.
Specifically, the relationship between the energy of an object and its computation potential is a proportionate one. As Einstein has famously calculated, the energy of an object is equal to its mass times the speed of light squared. Thus, a theoretical computer weighing a mere kilogram has a huge amount of potential energy. To find the total computational power, the minimum amount of energy required to perform an operation is divided by the total amount of energy.
of energy possessed by a 1-kilogram computer by Planck's constant yields a tremendously large number, roughly $5 \times 10^{50}$ operations per second. Using even the most conservative estimates of the computing power of the human brain, such a computer would have the computational power of five trillion trillion human civilizations. The computer would also have a memory capacity, calculated by determining the degrees of freedom allowed by the state of all the particles comprising it, of $10^{31} \mathrm{bits}$.
These numbers are purely theoretical, however. Were the computer to convert all of its mass to energy, it would be the equivalent of a thermonuclear explosion. And it is unreasonable to expect human technology to ever achieve abilities even close to these limits. However, a project at the University of Oklahoma has succeeded in storing 50 bits of data per atom, albeit on only a limited number of atoms. Given that there are $10^{25}$ atoms in 1 kilogram of material, it may be possible to fit up to $10^{27}$ bits of information in the computer. And if scientists are able to exploit the many properties of atoms to store information, including the position, spin, and quantum state of all its particles, it may be possible to exceed even that number.
One interesting consequence of such staggering increases in computing power is that each advance could provide the basis for further evolution. Once technology can achieve, for instance, a level of computation equal to $10^{40}$ operations per second, it can use that massive power to help bring the theoretical limit ever closer.
According to the passage, why does the author believe that the theoretical limit of computational power may be approached?
Options:
A. Scientists can exploit many different properties of atoms.
B. Technological advances engender more advances.
C. Computers only require a minimal amount of energy.
D. Recent advances have shown the technology exists to reach the limit. | This question asks about supporting details and evidence. Go back to the passage and find why the author believes technology may reach the theoretical limit discussed. In the last paragraph, the author indicates that each technological advance provides the basis for further evolution, as choice B states. Choice A refers to the author's discussion of memory capacity, not computational power, and indicates a way technicians may exceed the theoretical limits described. Choice $C$ has nothing to do with reaching the theoretical limits of computational power. The technology to reach the limits does not yet exist, as choice D seems to indicate. The correct answer is B. |
\section{Passage $\mathrm{V}$}
The Nobel Peace Prize has often engendered controversy. The Red Cross, yes, but Henry Kissinger? Mother Teresa, perhaps, but Yasser Arafat? Surprisingly, a loud and ongoing controversy surrounds a choice that at first look seems most benign-that of the Guatemalan freedom fighter, Rigoberta Menchú Tum.
Rigoberta Menchú was born in 1959 to a poor Mayan family in Guatemala. At the time of her birth, General Ydígoras Fuentes had just seized power, following the assassination of Colonel Castillo Armas. The year she turned one, there was a failed coup by a military group, many of whom fled to the countryside and would return to lead the rebellion that would wax and wane for the next 36 years.
Guatemala was divided into factions for that entire period, as military governments controlled the nation and guerilla groups controlled the countryside. At the same time, right-wing vigilantes led a campaign of torture and assassination, eliminating students and peasants they deemed to be allied with the guerillas.
In the countryside where Rigoberta lived, the battles were largely over control of farmland. Rigoberta's father was taken into custody and tortured when the army believed he had assisted in the assassination of a plantation owner. Following a rout of the guerillas by the army, the guerillas regrouped and began to take their fight to the capital, beginning a long series of assassinations of government figures.
Rigoberta, her father, and some of her siblings joined the Committee of the Peasant Union (CUC). In rapid succession, Rigoberta lost her brother, father, and mother to the army's own assassination squads. Rigoberta, though only in her early 20s, became a key figure in the resistance, traveling from village to village to educate the Mayan peasants in overcoming oppression, and leading demonstrations in Guatemala City.
Soon Rigoberta was herself a target. She fled to Mexico and continued her life as an organizer, concentrating her focus on peasants' rights. She participated in the founding of the United Representation of the Guatemalan Opposition (RUOG). This period was the most violent of the entire Guatemalan civil war. Under the new president, General Efrain Ríos Montt, massacres of civilians became an everyday occurrence. The controversy is all about the book. It began when anthropologist David Stoll began independent research on the same era about which Rigoberta dictated and discovered discrepancies in her recall of events. Massacres she described were not remembered by the locals; one of her brothers died by shooting rather than by fire, dates were incorrect, and so on. Stoll's discoveries were enough to roil up conservative historians and columnists, who declared Rigoberta's book the grossest propaganda and her career, therefore, unworthy of the Nobel Prize. Not surprisingly, this brought forth charges of racism, sexism, and cultural ignorance from the other side.
The Nobel Committee maintained a stiff upper lip throughout the political backand-forth and still insists that Rigoberta's prize had to do with her documented good works and not with her personal recorded history. Rigoberta's book is still widely taught, and her place in the history of Central American peasant revolutions seems assured.
Which choice(s) by the Nobel Committee does the author seem to indicate was undeserved?
I. Rigoberta Menchú
II. Henry Kissinger
III. The Red Cross
Options:
A. I only
B. II only
C. I and III only
D. II and III only | “The Red Cross, yes, but Henry Kissinger?" asks the author. The author finds the controversy over Rigoberta "surprising" and states, "her place in the history of Central American peasant revolutions seems assured." That indicates that as far as the author is concerned, the committee's choice of Rigoberta is not undeserved (I), but the choice of Kissinger may well be (II). Since the author gives the Red Cross (III) a thumbs-up, only choice B fits the passage. The correct answer is B. |
\section{Passage $\mathrm{V}$}
The Nobel Peace Prize has often engendered controversy. The Red Cross, yes, but Henry Kissinger? Mother Teresa, perhaps, but Yasser Arafat? Surprisingly, a loud and ongoing controversy surrounds a choice that at first look seems most benign-that of the Guatemalan freedom fighter, Rigoberta Menchú Tum.
Rigoberta Menchú was born in 1959 to a poor Mayan family in Guatemala. At the time of her birth, General Ydígoras Fuentes had just seized power, following the assassination of Colonel Castillo Armas. The year she turned one, there was a failed coup by a military group, many of whom fled to the countryside and would return to lead the rebellion that would wax and wane for the next 36 years.
Guatemala was divided into factions for that entire period, as military governments controlled the nation and guerilla groups controlled the countryside. At the same time, right-wing vigilantes led a campaign of torture and assassination, eliminating students and peasants they deemed to be allied with the guerillas.
In the countryside where Rigoberta lived, the battles were largely over control of farmland. Rigoberta's father was taken into custody and tortured when the army believed he had assisted in the assassination of a plantation owner. Following a rout of the guerillas by the army, the guerillas regrouped and began to take their fight to the capital, beginning a long series of assassinations of government figures.
Rigoberta, her father, and some of her siblings joined the Committee of the Peasant Union (CUC). In rapid succession, Rigoberta lost her brother, father, and mother to the army's own assassination squads. Rigoberta, though only in her early 20s, became a key figure in the resistance, traveling from village to village to educate the Mayan peasants in overcoming oppression, and leading demonstrations in Guatemala City.
Soon Rigoberta was herself a target. She fled to Mexico and continued her life as an organizer, concentrating her focus on peasants' rights. She participated in the founding of the United Representation of the Guatemalan Opposition (RUOG). This period was the most violent of the entire Guatemalan civil war. Under the new president, General Efrain Ríos Montt, massacres of civilians became an everyday occurrence. The controversy is all about the book. It began when anthropologist David Stoll began independent research on the same era about which Rigoberta dictated and discovered discrepancies in her recall of events. Massacres she described were not remembered by the locals; one of her brothers died by shooting rather than by fire, dates were incorrect, and so on. Stoll's discoveries were enough to roil up conservative historians and columnists, who declared Rigoberta's book the grossest propaganda and her career, therefore, unworthy of the Nobel Prize. Not surprisingly, this brought forth charges of racism, sexism, and cultural ignorance from the other side.
The Nobel Committee maintained a stiff upper lip throughout the political backand-forth and still insists that Rigoberta's prize had to do with her documented good works and not with her personal recorded history. Rigoberta's book is still widely taught, and her place in the history of Central American peasant revolutions seems assured.
Which of the following statements is/are NOT presented as evidence that Rigoberta may have falsified her autobiography?
I. Massacres she described were not remembered by the locals.
II. One of her brothers died by shooting rather than by fire.
III. Rigoberta dictated her life story to anthropologist Debray.
Options:
A. I only
B. I and II only
C. III only
D. II and III only | The question asks you to find the one or more statements that do NOT support the premise that Rigoberta falsified her book. To do that, you must find the statements that do support that premise and eliminate those. Statements that do support the premise will indicate some kind of disconnect between what she said and what was true. Those statements include I and II. Statement III, on the other hand, is simply a statement of fact. It does not support the premise that she falsified the book. The answer is choice C. The correct answer is C. |
\section{
If our knowledge of the world occurs through the weaving of narratives, as postmodernists would have us believe, then we must judge the truth of each narrative by comparing it with what we value and what we already accept as true. Any metanarrative, or overarching "big story," must be rejected because it attempts to explain away too many individual narratives.
history through that metanarrative, postmodernists would argue, does not allow us to consider the contributions of other groups to the country we live in today. So constructing a metanarrative can often be exclusionary.
Of course, constructing a series of smaller narratives is just as exclusionary because it frames experience in a limited way. It is easy to see this occurring if you look at any university course list today. How is it possible for American History 4111, "Imperialism and Amerindians, 1600-1840" to coexist alongside American History 4546, "American Military History and Policy"? Does English 340X, "Survey of Women's Literature," subsume or incorporate elements of English 342X, "American Indian Women Writers"? Perhaps we should pity the undergraduate student today. Lacking any overarching metanarrative to provide perspective, how can the poor student wade through the often contradictory mininarratives being thrown his or her way?
There is no question that the old way was wrongheaded and elitist. Survey courses that emphasized the white male perspective were removed from the curriculum in the 1970s, '80s, and '90s, and for good reason. But replacing them with a smorgasbord of mininarratives risks eliminating any sense of a big picture. Can students connect the dots to define a coherent reality, or must their reality be a series of frames with no links among them?
Revising the canon was about ridding our perspective of those racist, sexist, or classist notions that insisted on a single Truth with a capital $T$. Of course there is no overriding Truth that applies to everyone. For everyone who believes that Thomas Jefferson was a brilliant reformer, there is someone who thinks he was a two-faced slaveholder. Yet, where does it leave us if everyone we know is approaching history, science, literature, or what have you from a slightly different angle? It's bad enough when partisan politics divides us into red states and blue states. How much worse is it to imagine ourselves clad in thousands upon thousands of shades of purple?
The clearest sign that we are not ready to abandon our metanarratives comes in the current and ongoing clash between those who accept evolutionary theory and those who accept the Bible as the written word of God. The latter grant the Earth 6000 years of life, the former give it several million more, and never the twain shall meet. Each of these viewpoints is a metanarrative, a big story that governs people's understanding of the world. Like many metanarratives, each of these completely negates the other.
So on the one hand, metanarratives don't work well because they are too exclusionary. And on the other hand, mininarratives don't work well because they are too narrow. It will be fascinating to watch the canon evolve over the next few decades and to see whether this dichotomy can ever be resolved.
The author states that metanarratives have been replaced on campuses by a "smorgasbord" of mininarratives. The word smorgasbord is used to indicate:
Options:
A. freshness
B. choice
C. flavor
D. foreignness | Smorgasbord is a Swedish word meaning "a buffet meal featuring a variety of dishes." It is this variety, or choice, to which the author refers. Even if you do not know the word, you should be able to guess this from context. The correct answer is B. |
\section{
If our knowledge of the world occurs through the weaving of narratives, as postmodernists would have us believe, then we must judge the truth of each narrative by comparing it with what we value and what we already accept as true. Any metanarrative, or overarching "big story," must be rejected because it attempts to explain away too many individual narratives.
history through that metanarrative, postmodernists would argue, does not allow us to consider the contributions of other groups to the country we live in today. So constructing a metanarrative can often be exclusionary.
Of course, constructing a series of smaller narratives is just as exclusionary because it frames experience in a limited way. It is easy to see this occurring if you look at any university course list today. How is it possible for American History 4111, "Imperialism and Amerindians, 1600-1840" to coexist alongside American History 4546, "American Military History and Policy"? Does English 340X, "Survey of Women's Literature," subsume or incorporate elements of English 342X, "American Indian Women Writers"? Perhaps we should pity the undergraduate student today. Lacking any overarching metanarrative to provide perspective, how can the poor student wade through the often contradictory mininarratives being thrown his or her way?
There is no question that the old way was wrongheaded and elitist. Survey courses that emphasized the white male perspective were removed from the curriculum in the 1970s, '80s, and '90s, and for good reason. But replacing them with a smorgasbord of mininarratives risks eliminating any sense of a big picture. Can students connect the dots to define a coherent reality, or must their reality be a series of frames with no links among them?
Revising the canon was about ridding our perspective of those racist, sexist, or classist notions that insisted on a single Truth with a capital $T$. Of course there is no overriding Truth that applies to everyone. For everyone who believes that Thomas Jefferson was a brilliant reformer, there is someone who thinks he was a two-faced slaveholder. Yet, where does it leave us if everyone we know is approaching history, science, literature, or what have you from a slightly different angle? It's bad enough when partisan politics divides us into red states and blue states. How much worse is it to imagine ourselves clad in thousands upon thousands of shades of purple?
The clearest sign that we are not ready to abandon our metanarratives comes in the current and ongoing clash between those who accept evolutionary theory and those who accept the Bible as the written word of God. The latter grant the Earth 6000 years of life, the former give it several million more, and never the twain shall meet. Each of these viewpoints is a metanarrative, a big story that governs people's understanding of the world. Like many metanarratives, each of these completely negates the other.
So on the one hand, metanarratives don't work well because they are too exclusionary. And on the other hand, mininarratives don't work well because they are too narrow. It will be fascinating to watch the canon evolve over the next few decades and to see whether this dichotomy can ever be resolved.
Which of the following assertions does the author support with an example?
I. Constructing mininarratives frames experience in a limited way.
II. We are not ready to abandon our metanarratives.
III. Constructing a metanarrative may be exclusionary.
Options:
A. I only
B. III only
C. I and II only
D. I, II, and III | Supposition I is supported by the examples of competing American history courses. Supposition II is supported by the clash between evolutionary theory and strict Bible interpretation. Supposition III is supported by the example of a Eurocentric look at American history (see paragraph 2). Since all three suppositions are supported by examples, choice $\mathrm{D}$ is the best answer. The correct answer is D. |
\section{
Language consists of three main parts. The first is simply vocabulary, the lexicon of words available to the speaker to describe particular concepts, actions, and states. The second part is morphology, the set of rules that dictates how words and parts of words may be combined to form other words. The final part of language is syntax, the rules that combine words into phrases and sentences. It is morphology that we shall turn our attention to first.
Morphology has two major subdivisions. They are derivation and inflection. Derivation refers to those rules by which new words may be formed out of preexisting words. Inflection, sometimes referred to as conjugation and declension, states how words are changed to indicate their function within a sentence.
Although common mythology holds that English is one of the most difficult languages to learn, many linguists would beg to differ. In fact, when it comes to inflection,
the conditional, the subjunctive, the imperative, and others. Languages from outside the Indo-European family tree may be even more complex. The Bantu language of Kivunjo has roughly half a million combinations of prefixes and suffixes for a verb. English, however, has a mere four combinations. A verb such as to walk is limited to the following forms: walk, walks, walked, and walking.
Although there are only four basic verb forms, the English language does have more than 13 possible inflections. It just makes do with these four forms for the various functions. English also differs significantly from many other Indo-European languages in that in English, the stem of the word can be pronounced and is in fact part of the language. In Spanish, the stem of the verb to walk is camin-, but this is not a word in the language. In order to pronounce this word, it must be conjugated and combined with a suffix. English uses the stem form of the verb for four inflections: the present tense (except for the third-person singular), the infinitive, the imperative, and the subjunctive. The verb form ending in $-s$ is used in only one case, the third-person singular.
The two possible verb suffixes, -ing and $-e d$, are put to use in a variety of ways as well. The -ing ending appears in the progressive participle, the present participle, the gerund form, and the verbal adjective form. That leaves the $-e d$ form to handle the remaining inflections, including the past tense, the perfect participle, the passive participle, and some verbal adjectives.
Why are there so many different inflections for verb forms that basically sound the same? Simply put, the meanings implied by the various moods are different, even if the words used are fairly similar. Compare the meaning conveyed by the simple past tense (he walked) to the meaning inherent in the perfect participle (he has walked).
The passage suggests that the author MOST likely believes that:
Options:
A. The English language does not suffer from its lack of inflections.
B. English is the only language in which the verb stem is a word.
C. English is not as hard to learn as many think.
D. At one time the English language had more inflections. | This application question asks you to draw conclusions. The author states that the English language makes do with fewer verb forms than other languages, but it still can convey different meanings. This makes choice A correct. Choices B and D are not supported by the passage, and choice C reflects what some linguists believe. The correct answer is A. |
\section{Passage $\mathrm{V}$}
The Nobel Peace Prize has often engendered controversy. The Red Cross, yes, but Henry Kissinger? Mother Teresa, perhaps, but Yasser Arafat? Surprisingly, a loud and ongoing controversy surrounds a choice that at first look seems most benign-that of the Guatemalan freedom fighter, Rigoberta Menchú Tum.
Rigoberta Menchú was born in 1959 to a poor Mayan family in Guatemala. At the time of her birth, General Ydígoras Fuentes had just seized power, following the assassination of Colonel Castillo Armas. The year she turned one, there was a failed coup by a military group, many of whom fled to the countryside and would return to lead the rebellion that would wax and wane for the next 36 years.
Guatemala was divided into factions for that entire period, as military governments controlled the nation and guerilla groups controlled the countryside. At the same time, right-wing vigilantes led a campaign of torture and assassination, eliminating students and peasants they deemed to be allied with the guerillas.
In the countryside where Rigoberta lived, the battles were largely over control of farmland. Rigoberta's father was taken into custody and tortured when the army believed he had assisted in the assassination of a plantation owner. Following a rout of the guerillas by the army, the guerillas regrouped and began to take their fight to the capital, beginning a long series of assassinations of government figures.
Rigoberta, her father, and some of her siblings joined the Committee of the Peasant Union (CUC). In rapid succession, Rigoberta lost her brother, father, and mother to the army's own assassination squads. Rigoberta, though only in her early 20s, became a key figure in the resistance, traveling from village to village to educate the Mayan peasants in overcoming oppression, and leading demonstrations in Guatemala City.
Soon Rigoberta was herself a target. She fled to Mexico and continued her life as an organizer, concentrating her focus on peasants' rights. She participated in the founding of the United Representation of the Guatemalan Opposition (RUOG). This period was the most violent of the entire Guatemalan civil war. Under the new president, General Efrain Ríos Montt, massacres of civilians became an everyday occurrence. The controversy is all about the book. It began when anthropologist David Stoll began independent research on the same era about which Rigoberta dictated and discovered discrepancies in her recall of events. Massacres she described were not remembered by the locals; one of her brothers died by shooting rather than by fire, dates were incorrect, and so on. Stoll's discoveries were enough to roil up conservative historians and columnists, who declared Rigoberta's book the grossest propaganda and her career, therefore, unworthy of the Nobel Prize. Not surprisingly, this brought forth charges of racism, sexism, and cultural ignorance from the other side.
The Nobel Committee maintained a stiff upper lip throughout the political backand-forth and still insists that Rigoberta's prize had to do with her documented good works and not with her personal recorded history. Rigoberta's book is still widely taught, and her place in the history of Central American peasant revolutions seems assured.
Which new information, if true, would BEST support conservatives' claims that Rigoberta was a fraud?
Options:
A. the discovery that Rigoberta was really born in early 1960
B. findings that show Rigoberta educated thousands of Mayan peasants
C. the discovery that neither Rigoberta nor her parents were part of CUC
D. findings that show Ríos Montt was overthrown by a peasant army | You are asked here to assess the relative relevance of evidence. Falsifying a birth date (choice A) might be evidence of a minor deception, but it would not be enough to support charges of fraud. Falsifying a critical part of her history as it relates to her reputation, as choice C would do, would certainly indicate a desire to deceive. The correct answer is C. |
\section{
If our knowledge of the world occurs through the weaving of narratives, as postmodernists would have us believe, then we must judge the truth of each narrative by comparing it with what we value and what we already accept as true. Any metanarrative, or overarching "big story," must be rejected because it attempts to explain away too many individual narratives.
history through that metanarrative, postmodernists would argue, does not allow us to consider the contributions of other groups to the country we live in today. So constructing a metanarrative can often be exclusionary.
Of course, constructing a series of smaller narratives is just as exclusionary because it frames experience in a limited way. It is easy to see this occurring if you look at any university course list today. How is it possible for American History 4111, "Imperialism and Amerindians, 1600-1840" to coexist alongside American History 4546, "American Military History and Policy"? Does English 340X, "Survey of Women's Literature," subsume or incorporate elements of English 342X, "American Indian Women Writers"? Perhaps we should pity the undergraduate student today. Lacking any overarching metanarrative to provide perspective, how can the poor student wade through the often contradictory mininarratives being thrown his or her way?
There is no question that the old way was wrongheaded and elitist. Survey courses that emphasized the white male perspective were removed from the curriculum in the 1970s, '80s, and '90s, and for good reason. But replacing them with a smorgasbord of mininarratives risks eliminating any sense of a big picture. Can students connect the dots to define a coherent reality, or must their reality be a series of frames with no links among them?
Revising the canon was about ridding our perspective of those racist, sexist, or classist notions that insisted on a single Truth with a capital $T$. Of course there is no overriding Truth that applies to everyone. For everyone who believes that Thomas Jefferson was a brilliant reformer, there is someone who thinks he was a two-faced slaveholder. Yet, where does it leave us if everyone we know is approaching history, science, literature, or what have you from a slightly different angle? It's bad enough when partisan politics divides us into red states and blue states. How much worse is it to imagine ourselves clad in thousands upon thousands of shades of purple?
The clearest sign that we are not ready to abandon our metanarratives comes in the current and ongoing clash between those who accept evolutionary theory and those who accept the Bible as the written word of God. The latter grant the Earth 6000 years of life, the former give it several million more, and never the twain shall meet. Each of these viewpoints is a metanarrative, a big story that governs people's understanding of the world. Like many metanarratives, each of these completely negates the other.
So on the one hand, metanarratives don't work well because they are too exclusionary. And on the other hand, mininarratives don't work well because they are too narrow. It will be fascinating to watch the canon evolve over the next few decades and to see whether this dichotomy can ever be resolved.
Based on the information in the passage, which of these might be considered a metanarrative?
I. Marxism
II. rationalism
III. libertarianism
Options:
A. I only
B. II only
C. II and III only
D. I, II, and III
| All three philosophies are all-encompassing ways to look at the world, so all three could be considered metanarratives. Since all three are valid, the answer is choice $\mathrm{D}$. The correct answer is D. |
It is perhaps too easy to think of mathematical truths as objective entities existing in the universe, simply waiting to be "discovered" by brilliant minds such as Pythagoras, Newton, and Descartes. Indeed, such a mentality may be cemented in Western minds raised in the Platonic tradition of a world of ideals that exists outside of the material world we reside in.
But new research in the fields of cognitive science and developmental psychology may be challenging this long-held belief in mathematics as a truth outside of human experience, and recasting mathematics as a product of the brain, an entity that is not discovered by man but rather created by him.
Such a radical paradigm shift has predictably met with stiff resistance, but the evidence in favor of a created mathematics rather than a discovered mathematics is compelling. Study after study has shown that all people possess an innate arithmetic. Babies as young as three or four days old can differentiate between groups of two and three items. And by the age of four months or so, an infant can see that one plus one is two and two minus one is one. Researchers discovered this startling fact by means of a simple experiment, following what is known as the violation of expectation model in developmental psychology. An infant was presented with a scenario in which a researcher held a puppet before the baby's eyes. Then a screen was moved in front of the puppet and another researcher placed a second puppet behind it. The screen was removed and if there were two puppets now visible, the infant registered no surprise (as measured by both the direction of the child's gaze and the duration of her stare). But if the screen was removed and only one puppet appeared, the infant registered perplexity at the situation. This experiment and others like it strongly suggest that all individuals are born with certain mathematical concepts hardwired into their brains. The ability to quickly and accurately count a small number of items, called subitizing by psychologists, is also found in animals. Researchers using experiments similar to the violation of expectation setup described previously have found an innate mathematical ability not just in primates, our closest evolutionary cousins, but in raccoons, rats, parrots, and pigeons. These findings are consistent with the belief that mathematical thinking is a function of the structures in the brain and not a product of the outside world.
Anomalous cases involving brain injuries and disorders also support the idea that mathematical thinking arises from the organization of the brain. There is a documented case of a subject with a $\mathrm{PhD}$ in chemistry who suffers from acalculia, the inability to perform basic arithmetic functions. Strangely, this patient is unable to perform simple calculations such as five plus three or eight minus two, but has no problem manipulating abstract algebraic operations. This curious fact has led psychologists to conclude that the part of the brain that handles abstract algebraic operations must be different from the part of the brain that works with more concrete arithmetic functions.
In the discussion of the violation of expectation experiment, the author assumes that:
Options:
A. Certain observable behaviors accurately indicate inner mental states.
B. The infants in the study did not know what happened to the puppets behind the screen.
C. Researchers used the same types of puppets in all of the experiments.
D. Animals and infants performed at similar levels of competency in the experiments. | In order to answer this question, you must be able to judge credibility. The author cites the violation of expectation experiment as support for the author's claim that mathematics is an innate part of human cognition. In order for these experiments to be credible, the author is assuming that the researchers can accurately determine that the infants are in fact surprised when the number of puppets changes. The only evidence for this is the direction and duration of the stares of the subjects. Accepting these results as evidence indicates the author believes outward signs accurately indicate inner emotional states; otherwise, the evidence would not support the argument. The correct answer is A. |
In sharp contrast to the modern, confessional poetry of the 20th century, the oeuvre of Henry Wadsworth Longfellow seems quaint and primly Victorian. During his lifetime, however, he was the most celebrated poet in the country. A closer look at the history of American poetry reveals that, despite his eminence, Longfellow wrote in a mold of both form and content that was already being challenged during his lifetime. But why, a century later, do the artistic works of many of his contemporaries continue to be enjoyed and studied while Longfellow languishes in the tomb of cultural artifacts?
One answer lies in the radical shift that began to take place in poetry in the mid19th century. Longfellow's themes and steadfast rhymes (and those of John Greenleaf Whittier, Oliver Wendell Holmes, and James Russell Lowell) gave way gradually to confessional verse, whose subjects were more personal and rhymes were looser and less conventional. But to understand this shift, one must first understand the nature of Longfellow's work and his standing in the American literary scene of his time.
Longfellow took as his subject his country's historical imagination, writing on an epic scale about Paul Revere, the Indian Hiawatha, and the pilgrim Miles Standish. He bestowed a mythic dimension on these figures, giving American readers iconic images that helped form a collective consciousness in the new country (indeed, Longfellow was a part of the first generation to be born in the United States). But Longfellow's content went beyond nationalistic pride-it was highly accessible and incredibly popular. Its accessibility is explained by his obvious themes that could be easily understood and embraced by the public. Longfellow did not challenge his readers, but appealed to their desire for stories that expounded an optimistic, sentimental, and moralistic worldview. Those themes were explored in rhyme that allowed readers to commit the poems to memory, much like songs. In 1857, The Song of Hiawatha, arguably his bestknown poem, sold 50,000 copies, an astounding number at the time. The next year, The Courtship of Miles Standish sold 25,000 copies in two months and in London sold 10,000 copies in one day. His success allowed him to give up a professorship at Harvard and focus full time on his writing.
Walt Whitman, Longfellow's contemporary, wrote poetry similar to that of Longfellow-romantic and sentimental, with conventional rhyme and meter. But in the 1850s, indeed two years before The Song of Hiawatha, he wrote and published Leaves of Grass; a more radical departure from his previous work could not have been imagined. The 12 unnamed poems comprising Leaves of Grass are written in free verse-that is, without conventional rhyme and meter. Yet, like Longfellow, he was determined to explore the subject of America and his love for his country.
Whitman looked to the writings of Ralph Waldo Emerson for inspiration. Emerson wrote "America is a poem in our eyes; its ample geography dazzles the imagination, and it will not wait long for metres." Indeed, Whitman paraphrased Emerson in his preface. Whitman's groundbreaking free verse changed the trajectory of American poetry. The next generation of poets, including Ezra Pound, Hart Crane, Sherwood Anderson, and William Carlos Williams, celebrated their debt to Whitman. Decades later, the influence of Whitman's work on Allen Ginsberg and Langston Hughes, among many others, continues his legacy.
Which of the following statements is NOT presented as evidence that Whitman is responsible for the radical shift in American poetry that occurred in the 19th century?
Options:
A. He used iconic American figures as his subjects.
B. His poetry was more personal and intimate than that of his predecessors or contemporaries.
C. He wrote about the common man and commonplace events.
D. He began writing in free verse rather than conventional rhymes. | This question asks you to assess evidence. To answer it, first note the hypothesis as stated in the question stem: Whitman was responsible for the radical shift in American poetry. Then determine which statements support it and which one does not. The one that does not is the correct answer. The fact that his poetry was personal and intimate (choice B) supports the hypothesis, as does his use of free verse (choice D). Choice C, choosing as his subjects the common man and commonplace events, also supports the hypothesis. Nowhere in the passage does it state that Whitman used iconic American figures as his subjects; in fact, that is what Longfellow did, and his poetry was what Whitman's shifted away from. That means choice A is correct. The correct answer is A. |
\section{Passage $\mathrm{V}$}
Engineers and computer scientists are intrigued by the potential power of nanocomputing. Nanocomputers will use atoms and molecules to perform their functions instead of the standard integrated circuits now employed. Theorists believe that the amount of information a nanocomputer could handle is staggering.
A professor at Massachusetts Institute of Technology (MIT) has attempted to calculate the computational limits of a computer with a weight of 1 kilogram and a volume of 1 liter. According to the laws of physics, the potential amount of computational power is a function of the available energy. Basically, each atom and subatomic particle in the computer has an amount of energy attached to it. Furthermore, the energy of each particle or atom is increased by the frequency of its movement. Thus the power of a computer that uses nanotechnology is bounded by the energy available from its atoms.
Specifically, the relationship between the energy of an object and its computation potential is a proportionate one. As Einstein has famously calculated, the energy of an object is equal to its mass times the speed of light squared. Thus, a theoretical computer weighing a mere kilogram has a huge amount of potential energy. To find the total computational power, the minimum amount of energy required to perform an operation is divided by the total amount of energy.
of energy possessed by a 1-kilogram computer by Planck's constant yields a tremendously large number, roughly $5 \times 10^{50}$ operations per second. Using even the most conservative estimates of the computing power of the human brain, such a computer would have the computational power of five trillion trillion human civilizations. The computer would also have a memory capacity, calculated by determining the degrees of freedom allowed by the state of all the particles comprising it, of $10^{31} \mathrm{bits}$.
These numbers are purely theoretical, however. Were the computer to convert all of its mass to energy, it would be the equivalent of a thermonuclear explosion. And it is unreasonable to expect human technology to ever achieve abilities even close to these limits. However, a project at the University of Oklahoma has succeeded in storing 50 bits of data per atom, albeit on only a limited number of atoms. Given that there are $10^{25}$ atoms in 1 kilogram of material, it may be possible to fit up to $10^{27}$ bits of information in the computer. And if scientists are able to exploit the many properties of atoms to store information, including the position, spin, and quantum state of all its particles, it may be possible to exceed even that number.
One interesting consequence of such staggering increases in computing power is that each advance could provide the basis for further evolution. Once technology can achieve, for instance, a level of computation equal to $10^{40}$ operations per second, it can use that massive power to help bring the theoretical limit ever closer.
According to the information in the passage, which of the following could increase the computational power of the theoretical computer?
Options:
A. increasing the amount of energy it takes to perform an operation
B. decreasing the volume of the computer
C. decreasing the amount of energy it takes to perform an operation
D. decreasing the mass of the computer | The question asks you to solve a problem. To answer it, you must first understand how computational power is calculated. The passage states that the power is found by dividing the total energy by the energy required for an operation. Decreasing the amount of energy, as in choice $\mathrm{C}$, would yield a larger quotient. Choice A would do the opposite, and choices B and D would decrease the total energy available. The correct answer is C. |
"I acknowledge Shakespeare to be the world's greatest dramatic poet, but regret that no parent could place the uncorrected book in the hands of his daughter, and therefore I have prepared the Family Shakespeare."
Thus did Thomas Bowdler, a self-appointed editor trained as a physician, explain his creation of a children's edition of Shakespeare that omitted some characters completely, toned down language he considered objectionable, and euphemized such shocking situations as Ophelia's suicide-an accident in Bowdler's version.
Bowdler was hardly the first to tone down the Bard. Poet laureate Nahum Tate rewrote King Lear, banishing the Fool entirely and giving the play a happy ending. His version was staged regularly from the 1680 s through the 18 th century.
Thomas Bowdler was, from all the evidence, a less likely editor. He was born in 1754 near Bath, England, to a wealthy family. He studied medicine but never really practiced, preferring to exercise philanthropy and to play chess, at which he was a master. In his retirement, he decided to try his hand at editorial work. Shakespeare's dramas were his first project. He added nothing to the texts, but by cutting and paraphrasing strove to remove anything that "could give just offense to the religious and virtuous mind."
The result was a 10-volume expurgated version that was criticized widely, although hardly universally. The famed British poet Algernon Swinburne remarked that "no man ever did better service to Shakespeare than the man who made it possible to put him into the hands of intelligent and imaginative children."
There is some indication that Bowdler's sister Harriet did the actual editing of Shakespeare's text. She was a poet and editor in her own right, and clearly more qualified than her brother to lay hands on the Bard of Avon. She may have published the expurgated edition anonymously before allowing her brother to take over the rights. If this is so, it is unsurprising that Harriet would not have wanted her name on the book. If the original Shakespeare were truly objectionable, then it would have been doubly so for a well-bred, unmarried Englishwoman of the day.
Bowdler went on to create children's versions of the Old Testament and of Edward Gibbons's History of the Decline and Fall of the Roman Empire, although neither achieved either the success or ridicule of his Shakespeare. Today, of course, we know Bowdler not for his good works, but instead for the eponym derived from his good name. To bowdlerize is to censor or amend a written work, often with a connotation of prudishness. The process for which Bowdler is known is one that continues to the present day, as texts deemed too difficult for today's high schoolers are "dumbed-down," library editions of The Adventures of Huckleberry Finn have the "n-word" blacked out, and middle-school productions of popular Broadway plays have all expletives deleted. We would be hard-pressed to say that we live in a prudish era, but some of the same impulses that drove Thomas and Harriet Bowdler still exist today.
The passage implies that Bowdler's sister was a more likely editor than he because:
Options:
A. She would have been truly offended by Shakespeare's plots.
B. Unlike her brother, she was a published writer and editor.
C. Bowdler was known for his misuse of the English language.
D. Women of the time were more likely to read Shakespeare. | Bowdler himself was trained as a doctor; the passage makes clear that he was not an editor. His sister, however, was a poet and editor, and therefore was more likely than he to have made the editorial changes that led to the revised Shakespearean texts. There is no evidence in the passage to support choice $\mathrm{C}$ or $\mathrm{D}$, and although there is some implication that choice $\mathrm{A}$ is true, it does not explain why she would have been a more likely editor than he. The correct answer is B. |
\section{
Language consists of three main parts. The first is simply vocabulary, the lexicon of words available to the speaker to describe particular concepts, actions, and states. The second part is morphology, the set of rules that dictates how words and parts of words may be combined to form other words. The final part of language is syntax, the rules that combine words into phrases and sentences. It is morphology that we shall turn our attention to first.
Morphology has two major subdivisions. They are derivation and inflection. Derivation refers to those rules by which new words may be formed out of preexisting words. Inflection, sometimes referred to as conjugation and declension, states how words are changed to indicate their function within a sentence.
Although common mythology holds that English is one of the most difficult languages to learn, many linguists would beg to differ. In fact, when it comes to inflection,
the conditional, the subjunctive, the imperative, and others. Languages from outside the Indo-European family tree may be even more complex. The Bantu language of Kivunjo has roughly half a million combinations of prefixes and suffixes for a verb. English, however, has a mere four combinations. A verb such as to walk is limited to the following forms: walk, walks, walked, and walking.
Although there are only four basic verb forms, the English language does have more than 13 possible inflections. It just makes do with these four forms for the various functions. English also differs significantly from many other Indo-European languages in that in English, the stem of the word can be pronounced and is in fact part of the language. In Spanish, the stem of the verb to walk is camin-, but this is not a word in the language. In order to pronounce this word, it must be conjugated and combined with a suffix. English uses the stem form of the verb for four inflections: the present tense (except for the third-person singular), the infinitive, the imperative, and the subjunctive. The verb form ending in $-s$ is used in only one case, the third-person singular.
The two possible verb suffixes, -ing and $-e d$, are put to use in a variety of ways as well. The -ing ending appears in the progressive participle, the present participle, the gerund form, and the verbal adjective form. That leaves the $-e d$ form to handle the remaining inflections, including the past tense, the perfect participle, the passive participle, and some verbal adjectives.
Why are there so many different inflections for verb forms that basically sound the same? Simply put, the meanings implied by the various moods are different, even if the words used are fairly similar. Compare the meaning conveyed by the simple past tense (he walked) to the meaning inherent in the perfect participle (he has walked).
According to the passage, which English inflection can be expressed by two different verb suffixes?
I. the present tense
II. the verbal adjective
III. the present participle
Options:
A. I only
B. I and II only
C. III only
D. I, II, and III | This question asks about supporting details. The passage states that the present tense (statement I) is usually expressed using the stem form of the verb, but can also be expressed in the third-person singular by using the $-s$ suffix. Similarly, verbal adjectives (statement II) are sometimes expressed using the -ing suffix and sometimes with the -ed suffix. The correct answer is B. |
The man whom Franklin Delano Roosevelt christened “The Happy Warrior" in a nominating speech would later become a thorn in Roosevelt's side. Some thought the switch in loyalties was sour grapes, but others saw Alfred E. Smith as the epitome of William Wordsworth's "happy warrior" and therefore a man who "makes his moral being his prime care"-one who never made a move without consulting his conscience.
Alfred E. Smith was both a successful politician and an unsuccessful one. A fourterm governor of New York State, he seemed a sure candidate for president, and indeed he ran three times for that position, only to lose each time.
He had several strikes against him. He was the Catholic son of Irish and ItalianGerman immigrants, making him anathema to nativists, the xenophobes who underwent a resurgence in the 1920s. He was from New York City, viewed even in the early twentieth century as disconnected from the national character. He was a progressive, which made conservatives of all stripes nervous. And he favored the repeal of Prohibition, a position that lost him the backing of many party leaders.
Who was this unlikely candidate? Born Alfred Emanuel Smith in 1873, Smith grew up on the Lower East Side of Manhattan. His father died when Smith was young, and the boy dropped out of school to take care of the family. At age 21, Smith supported a losing candidate in a local race and came to the attention of New York politicos, who took him under their wing. Nine years later, he ran successfully for the New York State Assembly, where he rapidly rose in the ranks to become Majority Leader and finally Speaker. He played a pivotal role in the revamping of New York's constitution, was elected sheriff of New York County, and then ran for governor in 1918, winning handily. Although he lost a re-election bid two years later, he surged back in 1922 and would remain in the governor's seat for six more years. His terms were marked by unparalleled improvements in labor laws and laws protecting civil liberties, for Smith's goal was to support those he saw as most in need of government's assistance.
In this goal, he was allied with Franklin Roosevelt, and the two were close enough that Roosevelt nominated Smith for president in 1924. The Drys, or Prohibitionists, backed William McAdoo, a son-in-law of former President Woodrow Wilson. Smith's supporters and McAdoo's supporters were so far apart that finally a compromise candidate, John Davis, was nominated and promptly lost the general election to Calvin Coolidge.
In 1928, Smith received his party's nomination on the second ballot, but along with his anti-Prohibition leanings, his religion became a major issue during the campaign, paving the way for Herbert Hoover to win the general election. Meanwhile, Smith had arranged for the nomination of his New York colleague, Franklin Roosevelt, to be governor of New York. Although Smith lost his bid, Roosevelt did not. Then came the Depression and the election of 1932. Backroom dealings ensured that Franklin Roosevelt won the nominating process, with another would-be presidential candidate in the vice-presidential spot on the ticket. Smith was left out in the cold on his third try for the presidency. Despite his former regard for Roosevelt, he began to work hard, if unsuccessfully, to defeat the New Deal. He supported two Republicans rather than Roosevelt in the 1936 and 1940 elections, and he died in 1944, a man who had never reached his desired goal, yet had inspired many of the changes for which his rival is now known.
If Smith were to run for office today on a platform similar to his original plan, which of the following outcomes would MOST likely occur?
Options:
A. He would be supported by Progressives and union leaders.
B. He would be opposed by civil libertarians.
C. He would be supported by those who favor immigration reform.
D. He would be opposed by those who support social welfare programs. | Smith was a progressive born of immigrant parents, and he supported both civil liberties and unions, as well as "those he saw as most in need of government's assistance." Of the answers given, only choice A is likely. This question requires you to combine information. The correct answer is A. |
\section{
Language consists of three main parts. The first is simply vocabulary, the lexicon of words available to the speaker to describe particular concepts, actions, and states. The second part is morphology, the set of rules that dictates how words and parts of words may be combined to form other words. The final part of language is syntax, the rules that combine words into phrases and sentences. It is morphology that we shall turn our attention to first.
Morphology has two major subdivisions. They are derivation and inflection. Derivation refers to those rules by which new words may be formed out of preexisting words. Inflection, sometimes referred to as conjugation and declension, states how words are changed to indicate their function within a sentence.
Although common mythology holds that English is one of the most difficult languages to learn, many linguists would beg to differ. In fact, when it comes to inflection,
the conditional, the subjunctive, the imperative, and others. Languages from outside the Indo-European family tree may be even more complex. The Bantu language of Kivunjo has roughly half a million combinations of prefixes and suffixes for a verb. English, however, has a mere four combinations. A verb such as to walk is limited to the following forms: walk, walks, walked, and walking.
Although there are only four basic verb forms, the English language does have more than 13 possible inflections. It just makes do with these four forms for the various functions. English also differs significantly from many other Indo-European languages in that in English, the stem of the word can be pronounced and is in fact part of the language. In Spanish, the stem of the verb to walk is camin-, but this is not a word in the language. In order to pronounce this word, it must be conjugated and combined with a suffix. English uses the stem form of the verb for four inflections: the present tense (except for the third-person singular), the infinitive, the imperative, and the subjunctive. The verb form ending in $-s$ is used in only one case, the third-person singular.
The two possible verb suffixes, -ing and $-e d$, are put to use in a variety of ways as well. The -ing ending appears in the progressive participle, the present participle, the gerund form, and the verbal adjective form. That leaves the $-e d$ form to handle the remaining inflections, including the past tense, the perfect participle, the passive participle, and some verbal adjectives.
Why are there so many different inflections for verb forms that basically sound the same? Simply put, the meanings implied by the various moods are different, even if the words used are fairly similar. Compare the meaning conveyed by the simple past tense (he walked) to the meaning inherent in the perfect participle (he has walked).
Which of the following assertions does the author support with an example?
I. The English language is relatively simple in its inflections.
II. The English language is not as hard to learn as some believe.
III. Inflections in the English language can convey a variety of meanings.
Options:
A. I only
B. III only
C. I and III only
D. II and III only | This question involves analyzing an argument. Statement $\mathrm{I}$ is supported by the examples of the Spanish language and the Kivunjo language, which have far more forms than the English language. Statement II is unsupported by a specific example. Statement III is supported by the example in the final paragraph of the different meanings of similar-sounding verb forms, which makes choice $\mathrm{C}$ the best answer. The correct answer is C. |
"I acknowledge Shakespeare to be the world's greatest dramatic poet, but regret that no parent could place the uncorrected book in the hands of his daughter, and therefore I have prepared the Family Shakespeare."
Thus did Thomas Bowdler, a self-appointed editor trained as a physician, explain his creation of a children's edition of Shakespeare that omitted some characters completely, toned down language he considered objectionable, and euphemized such shocking situations as Ophelia's suicide-an accident in Bowdler's version.
Bowdler was hardly the first to tone down the Bard. Poet laureate Nahum Tate rewrote King Lear, banishing the Fool entirely and giving the play a happy ending. His version was staged regularly from the 1680 s through the 18 th century.
Thomas Bowdler was, from all the evidence, a less likely editor. He was born in 1754 near Bath, England, to a wealthy family. He studied medicine but never really practiced, preferring to exercise philanthropy and to play chess, at which he was a master. In his retirement, he decided to try his hand at editorial work. Shakespeare's dramas were his first project. He added nothing to the texts, but by cutting and paraphrasing strove to remove anything that "could give just offense to the religious and virtuous mind."
The result was a 10-volume expurgated version that was criticized widely, although hardly universally. The famed British poet Algernon Swinburne remarked that "no man ever did better service to Shakespeare than the man who made it possible to put him into the hands of intelligent and imaginative children."
There is some indication that Bowdler's sister Harriet did the actual editing of Shakespeare's text. She was a poet and editor in her own right, and clearly more qualified than her brother to lay hands on the Bard of Avon. She may have published the expurgated edition anonymously before allowing her brother to take over the rights. If this is so, it is unsurprising that Harriet would not have wanted her name on the book. If the original Shakespeare were truly objectionable, then it would have been doubly so for a well-bred, unmarried Englishwoman of the day.
Bowdler went on to create children's versions of the Old Testament and of Edward Gibbons's History of the Decline and Fall of the Roman Empire, although neither achieved either the success or ridicule of his Shakespeare. Today, of course, we know Bowdler not for his good works, but instead for the eponym derived from his good name. To bowdlerize is to censor or amend a written work, often with a connotation of prudishness. The process for which Bowdler is known is one that continues to the present day, as texts deemed too difficult for today's high schoolers are "dumbed-down," library editions of The Adventures of Huckleberry Finn have the "n-word" blacked out, and middle-school productions of popular Broadway plays have all expletives deleted. We would be hard-pressed to say that we live in a prudish era, but some of the same impulses that drove Thomas and Harriet Bowdler still exist today.
The mention of Nahum Tate shows primarily that:
Options:
A. Few dared to change Shakespeare's works during his lifetime.
B. Actors and directors often changed Shakespeare's words to suit their talents.
C. Different versions of King Lear were used depending on the audience.
D. The Bowdlers were not the first to alter Shakespeare's texts. | Look for the answer that is supported by the text. Although choices A, B, and C may in fact be true, none of them receives support in the passage. The author's reason for including mention of Nehum Tate is to show that the Bowdlers were not the first to edit Shakespeare. The correct answer is D. |
The Shawnee people of the Ohio and Cumberland Valleys gave the American elk its other name, wapiti. Yet when we think of elk today, we rarely envision them as eastern animals. If we want to see elk in their natural habitat, we travel to Colorado or Montana, where they still roam the mountainous terrain. Elk were, in fact, once found in the East, from Georgia north to New York and Connecticut. By the time of the Civil War, hunting and habitat destruction had caused their extinction in most eastern states. All of the eastern subspecies are now extinct. Elk County, Pennsylvania, and Elk County, Kansas, were without elk for more than a century. At the beginning of the 20th century, herds of elk in the Rocky Mountains faced death by starvation as encroaching farms depleted their winter feeding and finally the government decided to intercede. They gathered up elk from Yellowstone National Park and shipped 50 of them to Pennsylvania. The elk were to be protected, with no hunting of them allowed until eight years had passed. At that early date, 1913, there was little understanding of the kind of acclimatization required when moving large animals from one habitat to another. The elk were released from cattle cars and chased into the wild to fend for themselves. Two years later, 95 more elk were moved from Yellowstone to Pennsylvania. Elk are large animals with large ranges, and the damage these elk caused to cornfields was substantial. The elk tended to move toward farming areas because that was where the food was. Farmers and poachers took some elk illegally, but the herds still began to grow. During the 1920 s, several hunting seasons went by with harvests of one or two dozen elk, but then the numbers declined dramatically, and the elk were again put under protected status. No real census was taken until 50 years later, in 1971. Which new information, if true, would most CHALLENGE the claim that elk reintroduction benefits other species as well?
Options:
A. record sightings of wild turkey, pheasant, and grouse on reclaimed land in Pennsylvania
B. surprising numbers of multiple births in moose from central and western Maine and New Hampshire
C. a census showing that whitetail deer were migrating northward from Pennsylvania into New York
D. an increase in temperature of one or two degrees in the headwaters of the Yellowstone River | The claim is that elk reintroduction benefits other species. Your answer must contradict this contention. Choice A shows a clear benefit to other populations, so it does not contradict the claim. Choice B shows a change in a species that does not live where elk live. Choice D does not refer to other species at all. Only choice $\mathrm{C}$ indicates a possible problem; the fact that deer are migrating northward may mean that elk are encroaching on their habitats. The correct answer is C. |
\section{Passage $\mathrm{V}$}
In linguistics, metathesis refers to the reversal of phonemes in a word. This can come about by accident, as in the common mispronunciation "aks" for ask or the common (and correct) pronunciation of iron as "i-orn." It may come about on purpose, as in various language games.
Accidental metathesis may be a process that takes place in the evolution of languages. Over time, for example, the Latin words miraculum, periculum, and parabola evolved into the Spanish words milagro, peligro, and palabra. Linguists posit that such changes usually occur when two consonants are of the same type, with similar mouth formations needed to create each one.
The Reverend Archibald Spooner, an Oxford dean, was known for his unintentional transpositions and gave his name to the particular metathesis he represented: Spoonerisms. Most famous Spoonerisms once attributed to Spooner are now believed to be apocryphal, but they are nevertheless amusing; for example, his supposed advice to a substandard student: "You have deliberately tasted two worms and will leave Oxford by the next town drain." Spoonerisms are funny when the metathesis involved changes one word into another, and they seem to lend themselves particularly well to off-color jokes.
Everyone is familiar with Pig Latin, in which initial consonant sounds are moved to the end of a word and followed with the syllable "ay." This is not pure metathesis, but it begins with the kind of metathesis known as backslang, in which syllables or letters in a word are reversed to produce a private code. Pig Latin may have begun as a language spoken in prisoner-of-war camps; it was certainly used in such camps during World War II and in Vietnam.
In Pig Latin, the phrase "Over the river and through the woods" becomes "Overay uhthay iverray anday oughthray uhthay oodsway." In other variations of Pig Latin, it might instead be "Overway uhthay iverray andway oughthray uhthay oodsway." There are computer programs that translate from English to Pig Latin and back again, and there is even a version of the Bible written entirely in Pig Latin. For the most part, however, outside of prisoner-of-war camps, Pig Latin is a funny code used by children.
French children and teenagers use a similar code. Called verlan (a reversal of l'envers, or "reversal"), it is trickier than Pig Latin because it relies much more on phonetics than on spelling. In verlan, the word café becomes féca, but the word tomber ("to fall") becomes béton. Verlan becomes even more complex in words that end in silent $e$ or which have only one syllable. While just a few words from Pig Latin have entered the vernacular (for example, amscray and ixnay), many verlan words are common lingo in France.
Historically, for reasons that are not quite clear, butchers have been frequent users of metathesis, creating their own backslang to converse without being understood by patrons. In France, this is called louchébem (boucher, or "butcher," with the first consonant moved to the end and the letter $l$ placed in front). In Australia, it is known as rechtub klat. That name, of course, represents the ultimate in metathesis-the reversal of letters instead of merely syllables.
In the context of the passage, the word transposition refers to a:
Options:
A. reordering of letters or syllables
B. movement of terms in an equation
C. rendering into another language
D. musical performance in a new key | Spooner was known for his transpositions, of which "You have deliberately tasted two worms" (wasted two terms) is an example. Although transposition has a variety of possible meanings, here it refers specifically to the exchange of letters or word parts. The correct answer is A. |
Biologists use the term "living fossil" to designate a species that maintains many of the features of its ancient ancestors. These species have evidenced very little evolutionary change, indicating not that they are primitive creatures as might be popularly believed, but instead that they are exquisitely suited for their biological and ecological niches and thus have not had any selection pressures to respond to.
The four most celebrated living fossils are the lungfish, the horseshoe crab, the lampshell, and the coelacanth. The horseshoe crab is the relative youngster of the group, belonging to the Jurassic period and having shown little structural changes in its 200-million-year lifespan. The lampshell, a member of the phylum Brachiopoda, is part of an even older lineage, resembling ancestors that lived nearly 400 million years ago.
The oldest members of the living fossil family are the lungfish and the coelacanth, both species that are remarkably similar to their forebears that roamed the Earth almost 425 million years ago. The lungfish, of which there are six surviving species, has the same basic structure and fleshy lobe fins of its progenitor, the sarcopterygian. Of course, the lungfish has long interested biologists not just because of its connections to the distant past but because of the bridge it represents between aquatic species and land dwellers. In fact, the African and South American species of lungfish live part of their lives entirely on land, burrowing into mud and respiring through a tiny breathing hole in their earthy homes. Although the lungfish has remained evolutionarily stable for close to 250 million years, prior to that the species did experience rapid and dynamic evolutionary change. However, by the end of the Permian period, the changes ground to a halt, and the lungfish has persevered happily ever since.
The singular story of the coelacanth stretches all the way from the Silurian period, some 425 million years ago, to the modern day. The coelacanths were well-known from the fossil record and dating technologies placed them in a time when plants were just beginning to encroach on the land. However, biologists thought the species went extinct long before the dinosaurs did. Imagine the surprise when in 1938 a South American trawler captured a strange-looking fish, some five feet in length and a pale blue and silver in color. By an astounding coincidence, the captain of the boat had a relationship with the curator of the East London Museum and had made a habit of sending any interesting finds along to her. Upon seeing this strange specimen, the curator contacted a leading ichthyologist, who, after examining the creature, said he "would not have been more surprised if I had seen a dinosaur walking down the street." But despite all protestations of common sense and amid allegations of forgery and fakery, upon further study, the creature was indubitably a coelacanth. Since that day, explorers have found many more members of the species and have even turned up a second species in the Indian Ocean.
In contrast to the other three living fossils, the horseshoe crab:
Options:
A. was alive at the same time the dinosaurs were
B. has experienced few structural changes
C. is no longer evolutionarily stable
D. evolved well after land-dwelling plants did | This question asks you to compare and contrast information found in the passage. We know from the second paragraph that the horseshoe crab is the most recent living fossil at 200 million years old. The other three all date from around 400 million years ago. The final paragraph says that at that time, land-dwelling plants were just beginning to evolve. Thus, choice D is the best answer because the land-dwelling plants had been evolving for about 200 million years by the time the horseshoe crab appeared on the scene. Choices A and B indicate aspects the crab shares with some of the other living fossils. There is no evidence of choice C. The correct answer is D. |
\section{Passage $\mathrm{V}$}
In linguistics, metathesis refers to the reversal of phonemes in a word. This can come about by accident, as in the common mispronunciation "aks" for ask or the common (and correct) pronunciation of iron as "i-orn." It may come about on purpose, as in various language games.
Accidental metathesis may be a process that takes place in the evolution of languages. Over time, for example, the Latin words miraculum, periculum, and parabola evolved into the Spanish words milagro, peligro, and palabra. Linguists posit that such changes usually occur when two consonants are of the same type, with similar mouth formations needed to create each one.
The Reverend Archibald Spooner, an Oxford dean, was known for his unintentional transpositions and gave his name to the particular metathesis he represented: Spoonerisms. Most famous Spoonerisms once attributed to Spooner are now believed to be apocryphal, but they are nevertheless amusing; for example, his supposed advice to a substandard student: "You have deliberately tasted two worms and will leave Oxford by the next town drain." Spoonerisms are funny when the metathesis involved changes one word into another, and they seem to lend themselves particularly well to off-color jokes.
Everyone is familiar with Pig Latin, in which initial consonant sounds are moved to the end of a word and followed with the syllable "ay." This is not pure metathesis, but it begins with the kind of metathesis known as backslang, in which syllables or letters in a word are reversed to produce a private code. Pig Latin may have begun as a language spoken in prisoner-of-war camps; it was certainly used in such camps during World War II and in Vietnam.
In Pig Latin, the phrase "Over the river and through the woods" becomes "Overay uhthay iverray anday oughthray uhthay oodsway." In other variations of Pig Latin, it might instead be "Overway uhthay iverray andway oughthray uhthay oodsway." There are computer programs that translate from English to Pig Latin and back again, and there is even a version of the Bible written entirely in Pig Latin. For the most part, however, outside of prisoner-of-war camps, Pig Latin is a funny code used by children.
French children and teenagers use a similar code. Called verlan (a reversal of l'envers, or "reversal"), it is trickier than Pig Latin because it relies much more on phonetics than on spelling. In verlan, the word café becomes féca, but the word tomber ("to fall") becomes béton. Verlan becomes even more complex in words that end in silent $e$ or which have only one syllable. While just a few words from Pig Latin have entered the vernacular (for example, amscray and ixnay), many verlan words are common lingo in France.
Historically, for reasons that are not quite clear, butchers have been frequent users of metathesis, creating their own backslang to converse without being understood by patrons. In France, this is called louchébem (boucher, or "butcher," with the first consonant moved to the end and the letter $l$ placed in front). In Australia, it is known as rechtub klat. That name, of course, represents the ultimate in metathesis-the reversal of letters instead of merely syllables.
The author's claim that most Spoonerisms are apocryphal could BEST be supported by the inclusion of:
Options:
A. a list of transpositions attributed to the Reverend Spooner
B. examples of modern-day transpositions based on famous Spoonerisms
C. interviews with Spooner's contemporaries denying their authenticity
D. Spooner's own letters to relatives, colleagues, and friends | If most Spoonerisms are apocryphal, or fictional, that would not be proved by a list of attributed transpositions (choice A), which would simply seem to support their authenticity. Examples of modern-day transpositions (choice B) would do nothing to support or deny the authenticity of Spoonerisms, and Spooner's letters (choice D) would not help because his transpositions, if they existed, were oral, not written. Interviews with people who knew Spooner (choice C) would be the best way to settle the issue one way or the other. The correct answer is C. |
The man whom Franklin Delano Roosevelt christened “The Happy Warrior" in a nominating speech would later become a thorn in Roosevelt's side. Some thought the switch in loyalties was sour grapes, but others saw Alfred E. Smith as the epitome of William Wordsworth's "happy warrior" and therefore a man who "makes his moral being his prime care"-one who never made a move without consulting his conscience.
Alfred E. Smith was both a successful politician and an unsuccessful one. A fourterm governor of New York State, he seemed a sure candidate for president, and indeed he ran three times for that position, only to lose each time.
He had several strikes against him. He was the Catholic son of Irish and ItalianGerman immigrants, making him anathema to nativists, the xenophobes who underwent a resurgence in the 1920s. He was from New York City, viewed even in the early twentieth century as disconnected from the national character. He was a progressive, which made conservatives of all stripes nervous. And he favored the repeal of Prohibition, a position that lost him the backing of many party leaders.
Who was this unlikely candidate? Born Alfred Emanuel Smith in 1873, Smith grew up on the Lower East Side of Manhattan. His father died when Smith was young, and the boy dropped out of school to take care of the family. At age 21, Smith supported a losing candidate in a local race and came to the attention of New York politicos, who took him under their wing. Nine years later, he ran successfully for the New York State Assembly, where he rapidly rose in the ranks to become Majority Leader and finally Speaker. He played a pivotal role in the revamping of New York's constitution, was elected sheriff of New York County, and then ran for governor in 1918, winning handily. Although he lost a re-election bid two years later, he surged back in 1922 and would remain in the governor's seat for six more years. His terms were marked by unparalleled improvements in labor laws and laws protecting civil liberties, for Smith's goal was to support those he saw as most in need of government's assistance.
In this goal, he was allied with Franklin Roosevelt, and the two were close enough that Roosevelt nominated Smith for president in 1924. The Drys, or Prohibitionists, backed William McAdoo, a son-in-law of former President Woodrow Wilson. Smith's supporters and McAdoo's supporters were so far apart that finally a compromise candidate, John Davis, was nominated and promptly lost the general election to Calvin Coolidge.
In 1928, Smith received his party's nomination on the second ballot, but along with his anti-Prohibition leanings, his religion became a major issue during the campaign, paving the way for Herbert Hoover to win the general election. Meanwhile, Smith had arranged for the nomination of his New York colleague, Franklin Roosevelt, to be governor of New York. Although Smith lost his bid, Roosevelt did not. Then came the Depression and the election of 1932. Backroom dealings ensured that Franklin Roosevelt won the nominating process, with another would-be presidential candidate in the vice-presidential spot on the ticket. Smith was left out in the cold on his third try for the presidency. Despite his former regard for Roosevelt, he began to work hard, if unsuccessfully, to defeat the New Deal. He supported two Republicans rather than Roosevelt in the 1936 and 1940 elections, and he died in 1944, a man who had never reached his desired goal, yet had inspired many of the changes for which his rival is now known.
The passage implies that being from New York City is bad for a national candidate because:
Options:
A. Urban candidates have trouble relating to those in the heartland.
B. New York is perceived as too different from the rest of the country.
C. Westerners rarely support or give money to eastern candidates.
D. New York is the center of liberalism, but candidates must be neutral. | Although choice A may be true, only choice B is alluded to in the passage, allowing you to draw conclusions. New York City, says the author, was "viewed even in the early twentieth century as disconnected from the national character." The correct answer is B. |
\section{Passage $\mathrm{V}$}
The Nobel Peace Prize has often engendered controversy. The Red Cross, yes, but Henry Kissinger? Mother Teresa, perhaps, but Yasser Arafat? Surprisingly, a loud and ongoing controversy surrounds a choice that at first look seems most benign-that of the Guatemalan freedom fighter, Rigoberta Menchú Tum.
Rigoberta Menchú was born in 1959 to a poor Mayan family in Guatemala. At the time of her birth, General Ydígoras Fuentes had just seized power, following the assassination of Colonel Castillo Armas. The year she turned one, there was a failed coup by a military group, many of whom fled to the countryside and would return to lead the rebellion that would wax and wane for the next 36 years.
Guatemala was divided into factions for that entire period, as military governments controlled the nation and guerilla groups controlled the countryside. At the same time, right-wing vigilantes led a campaign of torture and assassination, eliminating students and peasants they deemed to be allied with the guerillas.
In the countryside where Rigoberta lived, the battles were largely over control of farmland. Rigoberta's father was taken into custody and tortured when the army believed he had assisted in the assassination of a plantation owner. Following a rout of the guerillas by the army, the guerillas regrouped and began to take their fight to the capital, beginning a long series of assassinations of government figures.
Rigoberta, her father, and some of her siblings joined the Committee of the Peasant Union (CUC). In rapid succession, Rigoberta lost her brother, father, and mother to the army's own assassination squads. Rigoberta, though only in her early 20s, became a key figure in the resistance, traveling from village to village to educate the Mayan peasants in overcoming oppression, and leading demonstrations in Guatemala City.
Soon Rigoberta was herself a target. She fled to Mexico and continued her life as an organizer, concentrating her focus on peasants' rights. She participated in the founding of the United Representation of the Guatemalan Opposition (RUOG). This period was the most violent of the entire Guatemalan civil war. Under the new president, General Efrain Ríos Montt, massacres of civilians became an everyday occurrence. The controversy is all about the book. It began when anthropologist David Stoll began independent research on the same era about which Rigoberta dictated and discovered discrepancies in her recall of events. Massacres she described were not remembered by the locals; one of her brothers died by shooting rather than by fire, dates were incorrect, and so on. Stoll's discoveries were enough to roil up conservative historians and columnists, who declared Rigoberta's book the grossest propaganda and her career, therefore, unworthy of the Nobel Prize. Not surprisingly, this brought forth charges of racism, sexism, and cultural ignorance from the other side.
The Nobel Committee maintained a stiff upper lip throughout the political backand-forth and still insists that Rigoberta's prize had to do with her documented good works and not with her personal recorded history. Rigoberta's book is still widely taught, and her place in the history of Central American peasant revolutions seems assured.
The author's claim that Rigoberta's book is still widely taught could BEST be supported by the inclusion of:
Options:
A. two students' comments on the ongoing controversy
B. data on the sales for her book from publication until the present day
C. a record of bookstores that continue to carry her autobiography
D. a list of universities that include her book in their required reading lists | This question is purely speculative, since the author did not include information to support the statement. If the author had, it would have had to be information that told more about the teaching of Rigoberta's book in schools. Two students' comments (choice A) would not be enough evidence, and choices B and C might show that the book is still sold but not that it is widely taught. Only choice D includes reasonable evidence. The correct answer is D. |
It is perhaps too easy to think of mathematical truths as objective entities existing in the universe, simply waiting to be "discovered" by brilliant minds such as Pythagoras, Newton, and Descartes. Indeed, such a mentality may be cemented in Western minds raised in the Platonic tradition of a world of ideals that exists outside of the material world we reside in.
But new research in the fields of cognitive science and developmental psychology may be challenging this long-held belief in mathematics as a truth outside of human experience, and recasting mathematics as a product of the brain, an entity that is not discovered by man but rather created by him.
Such a radical paradigm shift has predictably met with stiff resistance, but the evidence in favor of a created mathematics rather than a discovered mathematics is compelling. Study after study has shown that all people possess an innate arithmetic. Babies as young as three or four days old can differentiate between groups of two and three items. And by the age of four months or so, an infant can see that one plus one is two and two minus one is one. Researchers discovered this startling fact by means of a simple experiment, following what is known as the violation of expectation model in developmental psychology. An infant was presented with a scenario in which a researcher held a puppet before the baby's eyes. Then a screen was moved in front of the puppet and another researcher placed a second puppet behind it. The screen was removed and if there were two puppets now visible, the infant registered no surprise (as measured by both the direction of the child's gaze and the duration of her stare). But if the screen was removed and only one puppet appeared, the infant registered perplexity at the situation. This experiment and others like it strongly suggest that all individuals are born with certain mathematical concepts hardwired into their brains. The ability to quickly and accurately count a small number of items, called subitizing by psychologists, is also found in animals. Researchers using experiments similar to the violation of expectation setup described previously have found an innate mathematical ability not just in primates, our closest evolutionary cousins, but in raccoons, rats, parrots, and pigeons. These findings are consistent with the belief that mathematical thinking is a function of the structures in the brain and not a product of the outside world.
Anomalous cases involving brain injuries and disorders also support the idea that mathematical thinking arises from the organization of the brain. There is a documented case of a subject with a $\mathrm{PhD}$ in chemistry who suffers from acalculia, the inability to perform basic arithmetic functions. Strangely, this patient is unable to perform simple calculations such as five plus three or eight minus two, but has no problem manipulating abstract algebraic operations. This curious fact has led psychologists to conclude that the part of the brain that handles abstract algebraic operations must be different from the part of the brain that works with more concrete arithmetic functions.
The passage implies that:
Options:
A. Structures for arithmetic calculations are hardwired into the brain, but those for algebraic manipulations are not.
B. Most animals are able to perform simple arithmetic operations.
C. Infants at the age of three or four days can already perform simple calculations.
D. Mathematical concepts are arbitrary. | This question asks you to draw conclusions about an assertion that is not directly stated. The second paragraph states that mathematics is created by man, which implies that mathematical concepts are to some degree arbitrary. Choice A is contradicted by the passage in the final paragraph when the author says that the acalculia patient "has no problem manipulating abstract algebraic operations." Choice B is not supported because the passage only mentions that some animals have mathematical abilities. Choice $\mathrm{C}$ is incorrect; infants at the age of a few months can perform simple calculations. At younger ages they can only differentiate between two and three items. The correct answer is D. |
\section{
If our knowledge of the world occurs through the weaving of narratives, as postmodernists would have us believe, then we must judge the truth of each narrative by comparing it with what we value and what we already accept as true. Any metanarrative, or overarching "big story," must be rejected because it attempts to explain away too many individual narratives.
history through that metanarrative, postmodernists would argue, does not allow us to consider the contributions of other groups to the country we live in today. So constructing a metanarrative can often be exclusionary.
Of course, constructing a series of smaller narratives is just as exclusionary because it frames experience in a limited way. It is easy to see this occurring if you look at any university course list today. How is it possible for American History 4111, "Imperialism and Amerindians, 1600-1840" to coexist alongside American History 4546, "American Military History and Policy"? Does English 340X, "Survey of Women's Literature," subsume or incorporate elements of English 342X, "American Indian Women Writers"? Perhaps we should pity the undergraduate student today. Lacking any overarching metanarrative to provide perspective, how can the poor student wade through the often contradictory mininarratives being thrown his or her way?
There is no question that the old way was wrongheaded and elitist. Survey courses that emphasized the white male perspective were removed from the curriculum in the 1970s, '80s, and '90s, and for good reason. But replacing them with a smorgasbord of mininarratives risks eliminating any sense of a big picture. Can students connect the dots to define a coherent reality, or must their reality be a series of frames with no links among them?
Revising the canon was about ridding our perspective of those racist, sexist, or classist notions that insisted on a single Truth with a capital $T$. Of course there is no overriding Truth that applies to everyone. For everyone who believes that Thomas Jefferson was a brilliant reformer, there is someone who thinks he was a two-faced slaveholder. Yet, where does it leave us if everyone we know is approaching history, science, literature, or what have you from a slightly different angle? It's bad enough when partisan politics divides us into red states and blue states. How much worse is it to imagine ourselves clad in thousands upon thousands of shades of purple?
The clearest sign that we are not ready to abandon our metanarratives comes in the current and ongoing clash between those who accept evolutionary theory and those who accept the Bible as the written word of God. The latter grant the Earth 6000 years of life, the former give it several million more, and never the twain shall meet. Each of these viewpoints is a metanarrative, a big story that governs people's understanding of the world. Like many metanarratives, each of these completely negates the other.
So on the one hand, metanarratives don't work well because they are too exclusionary. And on the other hand, mininarratives don't work well because they are too narrow. It will be fascinating to watch the canon evolve over the next few decades and to see whether this dichotomy can ever be resolved.
The ideas in this passage would be MOST useful to:
Options:
A. authors of novels in the postmodern tradition
B. committees responsible for establishing college curricula
C. students of ancient Roman or Greek philosophy
D. journalists, detectives, and other seekers of truth | The main idea of the passage has to do with the tension between metanarratives and mininarratives, especially as they affect the way we learn things. This would be most useful for people who develop curricula. The correct answer is B. |
Biologists use the term "living fossil" to designate a species that maintains many of the features of its ancient ancestors. These species have evidenced very little evolutionary change, indicating not that they are primitive creatures as might be popularly believed, but instead that they are exquisitely suited for their biological and ecological niches and thus have not had any selection pressures to respond to.
The four most celebrated living fossils are the lungfish, the horseshoe crab, the lampshell, and the coelacanth. The horseshoe crab is the relative youngster of the group, belonging to the Jurassic period and having shown little structural changes in its 200-million-year lifespan. The lampshell, a member of the phylum Brachiopoda, is part of an even older lineage, resembling ancestors that lived nearly 400 million years ago.
The oldest members of the living fossil family are the lungfish and the coelacanth, both species that are remarkably similar to their forebears that roamed the Earth almost 425 million years ago. The lungfish, of which there are six surviving species, has the same basic structure and fleshy lobe fins of its progenitor, the sarcopterygian. Of course, the lungfish has long interested biologists not just because of its connections to the distant past but because of the bridge it represents between aquatic species and land dwellers. In fact, the African and South American species of lungfish live part of their lives entirely on land, burrowing into mud and respiring through a tiny breathing hole in their earthy homes. Although the lungfish has remained evolutionarily stable for close to 250 million years, prior to that the species did experience rapid and dynamic evolutionary change. However, by the end of the Permian period, the changes ground to a halt, and the lungfish has persevered happily ever since.
The singular story of the coelacanth stretches all the way from the Silurian period, some 425 million years ago, to the modern day. The coelacanths were well-known from the fossil record and dating technologies placed them in a time when plants were just beginning to encroach on the land. However, biologists thought the species went extinct long before the dinosaurs did. Imagine the surprise when in 1938 a South American trawler captured a strange-looking fish, some five feet in length and a pale blue and silver in color. By an astounding coincidence, the captain of the boat had a relationship with the curator of the East London Museum and had made a habit of sending any interesting finds along to her. Upon seeing this strange specimen, the curator contacted a leading ichthyologist, who, after examining the creature, said he "would not have been more surprised if I had seen a dinosaur walking down the street." But despite all protestations of common sense and amid allegations of forgery and fakery, upon further study, the creature was indubitably a coelacanth. Since that day, explorers have found many more members of the species and have even turned up a second species in the Indian Ocean.
The primary purpose of the passage is to:
Options:
A. explain how living fossils can remain unchanged by evolution
B. describe a dilemma facing evolutionary biologists
C. detail the interesting story of the coelacanth
D. provide an overview of a biological category | This question requires you to find the main idea. The passage discusses living fossils, a term biologists use to designate a species that has changed little since its inception. It then provides details on four different living fossils. Choice D is the best description of the passage. The passage doesn't primarily explain how these species remain unchanged as choice A states. Nor does it detail a dilemma (choice B) or focus mainly on the coelacanth (choice C). The correct answer is D. |
The Shawnee people of the Ohio and Cumberland Valleys gave the American elk its other name, wapiti. Yet when we think of elk today, we rarely envision them as eastern animals. If we want to see elk in their natural habitat, we travel to Colorado or Montana, where they still roam the mountainous terrain. Elk were, in fact, once found in the East, from Georgia north to New York and Connecticut. By the time of the Civil War, hunting and habitat destruction had caused their extinction in most eastern states. All of the eastern subspecies are now extinct. Elk County, Pennsylvania, and Elk County, Kansas, were without elk for more than a century. At the beginning of the 20th century, herds of elk in the Rocky Mountains faced death by starvation as encroaching farms depleted their winter feeding and finally the government decided to intercede. They gathered up elk from Yellowstone National Park and shipped 50 of them to Pennsylvania. The elk were to be protected, with no hunting of them allowed until eight years had passed. At that early date, 1913, there was little understanding of the kind of acclimatization required when moving large animals from one habitat to another. The elk were released from cattle cars and chased into the wild to fend for themselves. Two years later, 95 more elk were moved from Yellowstone to Pennsylvania. Elk are large animals with large ranges, and the damage these elk caused to cornfields was substantial. The elk tended to move toward farming areas because that was where the food was. Farmers and poachers took some elk illegally, but the herds still began to grow. During the 1920 s, several hunting seasons went by with harvests of one or two dozen elk, but then the numbers declined dramatically, and the elk were again put under protected status. No real census was taken until 50 years later, in 1971. According to this passage, a major early threat to elk populations was:
Options:
A. Wolves and cougars
B. vehicular traffic
C. disease
D. hunting | Hunting and habitat destruction are the only reasons mentioned for the depopulation of elk prior to the 20th century. The correct answer is D. |
\section{Passage $\mathrm{V}$}
Engineers and computer scientists are intrigued by the potential power of nanocomputing. Nanocomputers will use atoms and molecules to perform their functions instead of the standard integrated circuits now employed. Theorists believe that the amount of information a nanocomputer could handle is staggering.
A professor at Massachusetts Institute of Technology (MIT) has attempted to calculate the computational limits of a computer with a weight of 1 kilogram and a volume of 1 liter. According to the laws of physics, the potential amount of computational power is a function of the available energy. Basically, each atom and subatomic particle in the computer has an amount of energy attached to it. Furthermore, the energy of each particle or atom is increased by the frequency of its movement. Thus the power of a computer that uses nanotechnology is bounded by the energy available from its atoms.
Specifically, the relationship between the energy of an object and its computation potential is a proportionate one. As Einstein has famously calculated, the energy of an object is equal to its mass times the speed of light squared. Thus, a theoretical computer weighing a mere kilogram has a huge amount of potential energy. To find the total computational power, the minimum amount of energy required to perform an operation is divided by the total amount of energy.
of energy possessed by a 1-kilogram computer by Planck's constant yields a tremendously large number, roughly $5 \times 10^{50}$ operations per second. Using even the most conservative estimates of the computing power of the human brain, such a computer would have the computational power of five trillion trillion human civilizations. The computer would also have a memory capacity, calculated by determining the degrees of freedom allowed by the state of all the particles comprising it, of $10^{31} \mathrm{bits}$.
These numbers are purely theoretical, however. Were the computer to convert all of its mass to energy, it would be the equivalent of a thermonuclear explosion. And it is unreasonable to expect human technology to ever achieve abilities even close to these limits. However, a project at the University of Oklahoma has succeeded in storing 50 bits of data per atom, albeit on only a limited number of atoms. Given that there are $10^{25}$ atoms in 1 kilogram of material, it may be possible to fit up to $10^{27}$ bits of information in the computer. And if scientists are able to exploit the many properties of atoms to store information, including the position, spin, and quantum state of all its particles, it may be possible to exceed even that number.
One interesting consequence of such staggering increases in computing power is that each advance could provide the basis for further evolution. Once technology can achieve, for instance, a level of computation equal to $10^{40}$ operations per second, it can use that massive power to help bring the theoretical limit ever closer.
The author of the passage indicates that:
Options:
A. The theoretical computer may have even more computational power than described.
B. Technicians have already built computers that can store $10^{27}$ bits of data.
C. $10^{27}$ bits of data is the theoretical limit of memory capacity in a computer.
D. The technology exists to create a computer that can perform $10^{40}$ operations per second. | You are asked to draw a conclusion in order to answer this question. The author indicates that choices $\mathrm{B}$ and $\mathrm{D}$ are still in the realm of the theoretical. The author also states that the memory limit in choice $\mathrm{C}$ may be "exceeded." That leaves choice A, which is supported when the author states that the calculation of the computer's power is based on "the most conservative estimates" of human computational power. Thus the proposed computers can be even more powerful than the author describes. The correct answer is A. |
\section{Passage $\mathrm{V}$}
The Nobel Peace Prize has often engendered controversy. The Red Cross, yes, but Henry Kissinger? Mother Teresa, perhaps, but Yasser Arafat? Surprisingly, a loud and ongoing controversy surrounds a choice that at first look seems most benign-that of the Guatemalan freedom fighter, Rigoberta Menchú Tum.
Rigoberta Menchú was born in 1959 to a poor Mayan family in Guatemala. At the time of her birth, General Ydígoras Fuentes had just seized power, following the assassination of Colonel Castillo Armas. The year she turned one, there was a failed coup by a military group, many of whom fled to the countryside and would return to lead the rebellion that would wax and wane for the next 36 years.
Guatemala was divided into factions for that entire period, as military governments controlled the nation and guerilla groups controlled the countryside. At the same time, right-wing vigilantes led a campaign of torture and assassination, eliminating students and peasants they deemed to be allied with the guerillas.
In the countryside where Rigoberta lived, the battles were largely over control of farmland. Rigoberta's father was taken into custody and tortured when the army believed he had assisted in the assassination of a plantation owner. Following a rout of the guerillas by the army, the guerillas regrouped and began to take their fight to the capital, beginning a long series of assassinations of government figures.
Rigoberta, her father, and some of her siblings joined the Committee of the Peasant Union (CUC). In rapid succession, Rigoberta lost her brother, father, and mother to the army's own assassination squads. Rigoberta, though only in her early 20s, became a key figure in the resistance, traveling from village to village to educate the Mayan peasants in overcoming oppression, and leading demonstrations in Guatemala City.
Soon Rigoberta was herself a target. She fled to Mexico and continued her life as an organizer, concentrating her focus on peasants' rights. She participated in the founding of the United Representation of the Guatemalan Opposition (RUOG). This period was the most violent of the entire Guatemalan civil war. Under the new president, General Efrain Ríos Montt, massacres of civilians became an everyday occurrence. The controversy is all about the book. It began when anthropologist David Stoll began independent research on the same era about which Rigoberta dictated and discovered discrepancies in her recall of events. Massacres she described were not remembered by the locals; one of her brothers died by shooting rather than by fire, dates were incorrect, and so on. Stoll's discoveries were enough to roil up conservative historians and columnists, who declared Rigoberta's book the grossest propaganda and her career, therefore, unworthy of the Nobel Prize. Not surprisingly, this brought forth charges of racism, sexism, and cultural ignorance from the other side.
The Nobel Committee maintained a stiff upper lip throughout the political backand-forth and still insists that Rigoberta's prize had to do with her documented good works and not with her personal recorded history. Rigoberta's book is still widely taught, and her place in the history of Central American peasant revolutions seems assured.
According to the passage, why did Rigoberta leave Guatemala?
Options:
A. She wanted to alert the world to the problems of the peasants.
B. She was under fire from the army's assassination squads.
C. She was invited to Paris to dictate her personal memoirs.
D. She no longer cared to stay after her family was murdered. | This question calls for a cause-and-effect relationship. You must find the cause of Rigoberta's departure. That is not simply something that happened before her departure (choice D), or something that was true but not a direct cause (choice A). The cause appears in the sentence "Soon Rigoberta was herself a target." She left because otherwise she might have been assassinated (choice B). The correct answer is B. |
\section{Passage $\mathrm{V}$}
In linguistics, metathesis refers to the reversal of phonemes in a word. This can come about by accident, as in the common mispronunciation "aks" for ask or the common (and correct) pronunciation of iron as "i-orn." It may come about on purpose, as in various language games.
Accidental metathesis may be a process that takes place in the evolution of languages. Over time, for example, the Latin words miraculum, periculum, and parabola evolved into the Spanish words milagro, peligro, and palabra. Linguists posit that such changes usually occur when two consonants are of the same type, with similar mouth formations needed to create each one.
The Reverend Archibald Spooner, an Oxford dean, was known for his unintentional transpositions and gave his name to the particular metathesis he represented: Spoonerisms. Most famous Spoonerisms once attributed to Spooner are now believed to be apocryphal, but they are nevertheless amusing; for example, his supposed advice to a substandard student: "You have deliberately tasted two worms and will leave Oxford by the next town drain." Spoonerisms are funny when the metathesis involved changes one word into another, and they seem to lend themselves particularly well to off-color jokes.
Everyone is familiar with Pig Latin, in which initial consonant sounds are moved to the end of a word and followed with the syllable "ay." This is not pure metathesis, but it begins with the kind of metathesis known as backslang, in which syllables or letters in a word are reversed to produce a private code. Pig Latin may have begun as a language spoken in prisoner-of-war camps; it was certainly used in such camps during World War II and in Vietnam.
In Pig Latin, the phrase "Over the river and through the woods" becomes "Overay uhthay iverray anday oughthray uhthay oodsway." In other variations of Pig Latin, it might instead be "Overway uhthay iverray andway oughthray uhthay oodsway." There are computer programs that translate from English to Pig Latin and back again, and there is even a version of the Bible written entirely in Pig Latin. For the most part, however, outside of prisoner-of-war camps, Pig Latin is a funny code used by children.
French children and teenagers use a similar code. Called verlan (a reversal of l'envers, or "reversal"), it is trickier than Pig Latin because it relies much more on phonetics than on spelling. In verlan, the word café becomes féca, but the word tomber ("to fall") becomes béton. Verlan becomes even more complex in words that end in silent $e$ or which have only one syllable. While just a few words from Pig Latin have entered the vernacular (for example, amscray and ixnay), many verlan words are common lingo in France.
Historically, for reasons that are not quite clear, butchers have been frequent users of metathesis, creating their own backslang to converse without being understood by patrons. In France, this is called louchébem (boucher, or "butcher," with the first consonant moved to the end and the letter $l$ placed in front). In Australia, it is known as rechtub klat. That name, of course, represents the ultimate in metathesis-the reversal of letters instead of merely syllables.
Which example(s) of metathesis does the author call accidental?
I. verlan and other backslang
II. parabola to palabra
III. the pronunciation of iron
Options:
A. I only
B. II only
C. I and III only
D. II and III only | Verlan and other backslang (I) are word games and therefore purposeful, not accidental. The examples of accidental metathesis include the change between Latin originals and Spanish derivatives (II) and mispronunciations or skewed pronunciations of English words (III). Since II and III are correct, the answer is choice D. Answers and Explanations The correct answer is D. |
It is perhaps too easy to think of mathematical truths as objective entities existing in the universe, simply waiting to be "discovered" by brilliant minds such as Pythagoras, Newton, and Descartes. Indeed, such a mentality may be cemented in Western minds raised in the Platonic tradition of a world of ideals that exists outside of the material world we reside in.
But new research in the fields of cognitive science and developmental psychology may be challenging this long-held belief in mathematics as a truth outside of human experience, and recasting mathematics as a product of the brain, an entity that is not discovered by man but rather created by him.
Such a radical paradigm shift has predictably met with stiff resistance, but the evidence in favor of a created mathematics rather than a discovered mathematics is compelling. Study after study has shown that all people possess an innate arithmetic. Babies as young as three or four days old can differentiate between groups of two and three items. And by the age of four months or so, an infant can see that one plus one is two and two minus one is one. Researchers discovered this startling fact by means of a simple experiment, following what is known as the violation of expectation model in developmental psychology. An infant was presented with a scenario in which a researcher held a puppet before the baby's eyes. Then a screen was moved in front of the puppet and another researcher placed a second puppet behind it. The screen was removed and if there were two puppets now visible, the infant registered no surprise (as measured by both the direction of the child's gaze and the duration of her stare). But if the screen was removed and only one puppet appeared, the infant registered perplexity at the situation. This experiment and others like it strongly suggest that all individuals are born with certain mathematical concepts hardwired into their brains. The ability to quickly and accurately count a small number of items, called subitizing by psychologists, is also found in animals. Researchers using experiments similar to the violation of expectation setup described previously have found an innate mathematical ability not just in primates, our closest evolutionary cousins, but in raccoons, rats, parrots, and pigeons. These findings are consistent with the belief that mathematical thinking is a function of the structures in the brain and not a product of the outside world.
Anomalous cases involving brain injuries and disorders also support the idea that mathematical thinking arises from the organization of the brain. There is a documented case of a subject with a $\mathrm{PhD}$ in chemistry who suffers from acalculia, the inability to perform basic arithmetic functions. Strangely, this patient is unable to perform simple calculations such as five plus three or eight minus two, but has no problem manipulating abstract algebraic operations. This curious fact has led psychologists to conclude that the part of the brain that handles abstract algebraic operations must be different from the part of the brain that works with more concrete arithmetic functions.
The author probably mentions the "Platonic tradition" in order to:
Options:
A. explain why some people may be less accepting of certain conclusions
B. demonstrate the philosophical and historical underpinnings of mathematical studies
C. highlight another influential Western figure in mathematical thought
D. argue that followers of non-Western philosophical traditions are more likely to agree with the author's thesis | This evaluation question asks you to analyze an argument. Specifically, you need to figure out why the author mentioned the Platonic tradition. In the first paragraph, the author states that the prevailing view of mathematics is an "easy" one that may be "cemented" in the minds of people. Later, in the third paragraph, the author states that the new view of mathematics has met with "stiff resistance." Thus, the mention of the Platonic tradition helps to explain why there is resistance to the conclusions of the new paradigm, which is what choice A says. The correct answer is A. |
\section{Passage $\mathrm{V}$}
Engineers and computer scientists are intrigued by the potential power of nanocomputing. Nanocomputers will use atoms and molecules to perform their functions instead of the standard integrated circuits now employed. Theorists believe that the amount of information a nanocomputer could handle is staggering.
A professor at Massachusetts Institute of Technology (MIT) has attempted to calculate the computational limits of a computer with a weight of 1 kilogram and a volume of 1 liter. According to the laws of physics, the potential amount of computational power is a function of the available energy. Basically, each atom and subatomic particle in the computer has an amount of energy attached to it. Furthermore, the energy of each particle or atom is increased by the frequency of its movement. Thus the power of a computer that uses nanotechnology is bounded by the energy available from its atoms.
Specifically, the relationship between the energy of an object and its computation potential is a proportionate one. As Einstein has famously calculated, the energy of an object is equal to its mass times the speed of light squared. Thus, a theoretical computer weighing a mere kilogram has a huge amount of potential energy. To find the total computational power, the minimum amount of energy required to perform an operation is divided by the total amount of energy.
of energy possessed by a 1-kilogram computer by Planck's constant yields a tremendously large number, roughly $5 \times 10^{50}$ operations per second. Using even the most conservative estimates of the computing power of the human brain, such a computer would have the computational power of five trillion trillion human civilizations. The computer would also have a memory capacity, calculated by determining the degrees of freedom allowed by the state of all the particles comprising it, of $10^{31} \mathrm{bits}$.
These numbers are purely theoretical, however. Were the computer to convert all of its mass to energy, it would be the equivalent of a thermonuclear explosion. And it is unreasonable to expect human technology to ever achieve abilities even close to these limits. However, a project at the University of Oklahoma has succeeded in storing 50 bits of data per atom, albeit on only a limited number of atoms. Given that there are $10^{25}$ atoms in 1 kilogram of material, it may be possible to fit up to $10^{27}$ bits of information in the computer. And if scientists are able to exploit the many properties of atoms to store information, including the position, spin, and quantum state of all its particles, it may be possible to exceed even that number.
One interesting consequence of such staggering increases in computing power is that each advance could provide the basis for further evolution. Once technology can achieve, for instance, a level of computation equal to $10^{40}$ operations per second, it can use that massive power to help bring the theoretical limit ever closer.
The ideas in this passage would MOST likely be presented in:
Options:
A. an academic journal
B. a newspaper
C. a tabloid
D. a popular science magazine | This question requires you to combine information. The passage discusses computers, technology, and even physics. However, the information is presented at the level of someone with a casual, rather than expert, knowledge of these fields. Thus choice D would be the best answer. Choice A may be close, but the information in the passage is not written at the level of expertise that a reader would expect in an academic journal. The correct answer is D. |