Zai geen, sayonara, goodbye (for now at least).

With the start of school looming so close, I fear that I won't have enough time to update this blog daily. Thus, I have decided to take a break on posting questions 'til winter vacation comes. I'll still be in the background gathering questions, but they won't be released to the public until December.

Thanks for all the comments, you guys really taught me a lot. I hope I was able to help all of you too (even if it was just the slightest bit).

- Paul

PS: The title of this post is from a song my friend wrote. His alias is LittleColumbus. This text is in white so that I can properly attribute him while not turning this post into a bad advertisement.
PPS: It would be awesome if you guys spread the word about this blog. I'd love to get more people commenting so we can have more discussions.

Math GRE - #34

$f$ is the function defined by: \[ f(x)=\begin{cases} xe^{-x^{2}-x^{-2}} & \text{if } x\neq0\\ 0 & \text{otherwise.}\end{cases}\] At how many values of $x$ does $f$ have a horizontal tangent line?

1. None
2. One
3. Two
4. Three
5. Four

Solution :

Choice 4 is the answer.

$f$ has a horizontal tangent line only when the derivative of $x$ is 0. The derivative of $f$ is:
\[f^\prime(x) = e^{-x^2-x^{-2}}+(-2x+2x^{-3})xe^{-x^2-x^{-2}}=(1-2x^2+2x^{-2})e^{-x^2-x^{-2}}=0.\] This is only zero in two cases: in the limit $\displaystyle\lim_{x\rightarrow 0} f^\prime(x)$ and when $1-2x^2+2x^{-2}=0$.

The quadratic-like equation above can be solved by solving $x^2-2x^4+2=0.$ This equation has two non-zero solutions. These solutions, along with the with the solution obtained in the limit above, gives us three horizontal tangent lines in total.

Physics GRE - # 34

Eigenfunctions for a rigid dumbbell rotating about its centre have a $\phi$ dependence of the form $\psi(\phi)=Ae^{im\phi}$, where $m$ is a quantum number and $A$ is a constant. Which of the following values of $A$ will properly normalize the eigenfunction?

1. $\sqrt{2\pi}$


2. $2\pi$


3. $(2\pi)^2$


4. $\dfrac{1}{\sqrt{2\pi}}$


5. $\dfrac{1}{2\pi}$

Solution :

Choice 4 is the answer.

For $\psi(\phi)$ to be normalized on $a$ to $b$, recall the normalization condition: \[\int_a^b{\left|\psi(\phi)\right|^2}d\phi=1.\] Since $\phi$ ranges from $0$ to $2\pi$, we have: \[\int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1\] where the second equality holds as $|e^{im\phi}|=1$ for any $m$ and any $\phi$.

Thus, we know that: \[\int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.\]

Math GRE - #33

How many integers from 1 to 1000 are divisible by 30 by not by 16?

1. 29
2. 31
3. 32
4. 33
5. 38

Solution :

Choice 1 is the answer.

There are 33 multiples of 30 under 1000 (1000/30 = 33.33..). So we only need to focus our attention on choices 1 to 4.

Note that the factorization of 30 is: $2\cdot 3\cdot 5.$
Now lets consider multiples of 30 in the form: $30n$. For $30n$ to be divisible by 16, $n$ must be divisible by 8 (since we will have a factor of 2 from 30 and another factor of 8 from $n$ to make a factor of 16).
Since $n$ can only go from 1 to 33 (as there are only 33 multiples of 30 under 1000), we only need to find numbers under 33 that are divisible by 8.

There are four such numbers: 8, 16, 24, 32. These numbers correspond to the multiples: 240, 480, 720, 960.

All the multiples remaining are not divisible by 16. Thus there are 33 - 4 = 29 multiples of 30 under 1000 that are divisible by 30 but not 16.

Physics GRE - #33

A radioactive nucleus decays, with the activity shown in the plot below. The half-life of the nucleus is:

1. 2 min
2. 7 min
3. 11 min
4. 18 min
5. 23 min

Solution :

Choice 2 is the answer.

The initial activity is $6\cdot 10^3$ (yes, 6 since the line marked $10^3$ is really $1\cdot 10^3$ and you start counting up from there). At $3\cdot 10^3,$ the time is around 5 to 10 minutes. Thus the only possible answer is choice 2.

Math GRE - #32

Let $\mathbb{R}$ be the set of real numbers and let $f$ and $g$ be functions from $\mathbb{R}$ into $\mathbb{R}$. Which of the following is the negation of the statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

1. For each $s$ in $\mathbb{R}$, there does not exist an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$.
2. For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
3. There exists an $s$ in $\mathbb{R}$ such that for each $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
4. There exists an $s$ in $\mathbb{R}$ and there exists an $r$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.
5. For each $r$ in $\mathbb{R}$, there exists an $s$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.

Solution :

Choice 3 is the answer.

If you've got an innate talent for logic, you could probably tell that choice 3 is the answer after a bit of thinking. 

Another way is to use quantifiers to translate the statement into formal logic.

The statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

can be translated as: \[\forall s\in\mathbb{R},\;\exists r\in\mathbb{R},\; f(r)>0\rightarrow g(s)>0.\]

We can negate this statement (imagine the negation as an arrow, flipping each quantifier as it flies past it) to be: \[\exists s\in\mathbb{R},\;\forall r\in\mathbb{R},\; f(r)>0\wedge g(s)\leq0.\]

This is choice 3 precisely.

Physics GRE - #32

As shown below, a coaxial cable having radii $a$, $b$, and $c$ carries equal and opposite currents of magnitude $i$ on the inner and outer conductors. What is the magnitude of the magnetic induction at a point $P$ outside the cable at a distance $r$ from the axis?

1. $0$
2. $\dfrac{\mu_0 ir}{2\pi a^2}$
3. $\dfrac{\mu_0 i}{2\pi r}$
4. $\dfrac{\mu_0 i}{2\pi r}\dfrac{c^2-r^2}{c^2-b^2}$
5. $\dfrac{\mu_0 i}{2\pi r}\dfrac{r^2-b^2}{c^2-b^2}$

Solution :

Choice 1 is the answer.

The currents cancel out the magnetic fields created by each other, resulting in a net field of zero. More formally, this is an application of Ampere's Law.

Math GRE - #31

Let $\text{F}$ be a constant force that is given by the vector $\left(\begin{array}{c}-1\\0\\1\end{array}\right)$. What is the work done by $\text{F}$ on a particle that moves along the path given by $\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)$ between time $t=0$ and $t=1$?

1. $-\dfrac{1}{4}$


2. $-\dfrac{1}{4\sqrt{2}}$


3. $0$


4. $\sqrt{2}$


5. $3\sqrt{2}$

Solution :

Choice 3 is the answer.

Recall that work can be calculated according to the line integral: \[W=\int_C{\text{F}\cdot d\text{r}}\] along a curve $C$ with force $\text{F}$ and position vector $\text{r}$. In this case, we have \[\text{r}=\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)\implies d\text{r}=\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)dt.\] The curve goes from 0 to 1, so our integral is: \[W=\int_0^1{\left(\begin{array}{c}-1\\0\\1\end{array}\right)\cdot \left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)}dt=\int_0^1{(-1+0+3t^2)}dt=0\] which gives zero work as we wanted.

Physics GRE - #31

Which of the following is nearly the mass of the Earth? The radius of the Earth is about $6.4\cdot10^6\,\text{m}$.

1. $6\cdot10^{24}\,\text{kg}$
2. $6\cdot10^{27}\,\text{kg}$
3. $6\cdot10^{30}\,\text{kg}$
4. $6\cdot10^{33}\,\text{kg}$
5. $6\cdot10^{36}\,\text{kg}$

Solution :

Choice 1 is the answer.

If you don't remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]

Math GRE - #30

What is \[\int_{-3}^3{|x+1|\,dx}?\]

1. 0
2. 5
3. 10
4. 15
5. 20

Solution :

Choice 3 is the answer.

This is a very basic question. The usual method of dealing with absolute values in integrals is to split it into a sum of two integrals and remove the absolute values. Since $|x+1|$ is negative on $[-3, 1)$ and positive on $[-1, 3]\,\,\,$ , we can split the integral into: \[\int_{-3}^3{|x+1|\,dx}=-\int_{-3}^{-1}{(x+1)\,dx}+\int_{-1}^3{(x+1)\,dx}=10.\] Another way of doing this problem is to imagine the graph of $|x+1|$ and realize that the integral is the sum of two 45-45-90 triangles (one with base 2 and height 2,  the other with base 4 and height 4). Simply add the area of triangles up and we're done.

Bonus question: Given that $c$ is a constant, what is \[\int_{-\infty}^{\infty}{e^{-|x+c|}\,dx}?\]

Physics GRE - #30

The energy levels of the hydrogen atom are given in terms of the principal quantum number $n$ and a positive constant $A$ by the expression:

1. $A\left(n+\frac{1}{2}\right)$
2. $A(1-n^2)$
3. $A\left(\frac{1}{n^2}-\frac{1}{4}\right)$
4. $An^2$
5. $-\frac{A}{n^2}$

Solution :

Choice 5 is the answer.

Recall that the energy levels of the hydrogen atom is given by \[E_n = -\frac{13.6\,\text{eV}}{n^2}.\] It's just one of those things that you've got to remember, as it takes time you don't have to derive it from first principals (I only ever remember that $E_n\propto \frac{1}{n^2}\,$ since the constant of 13.6 eV is usually unimportant).

Math GRE - #29

If $z=e^{2\pi i/5},$ then \[1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=\]

1. $0$
2. $4e^{3\pi i/5}$
3. $5e^{4\pi i/5}$
4. $-4e^{2\pi i/5}$
5. $-5e^{3\pi i/5}$

Solution :

Choice 5 is the answer.

Recall that the sum of all the $n-th$ roots of unity for $n>1$ ($n=5$ in this case) is 0. For this particular question, this means that \[1+z+z^2+z^3+z^4 = 0.\] This motivates us to write the sum in the question as: \[\begin{eqnarray*}
(1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9 & = & 0+4z^4\cdot0+5z^9 \\
& = & 5z^9.
\end{eqnarray*}\] Now we can easily evaluate the sum: \[\begin{eqnarray*}
5z^9=5\left(e^{2\pi i/5}\right)^9 & = & 5\left(e^{2\pi i/5}\right)^4\cdot\left(e^{2\pi i/5}\right)^5 \\
& = & 5\left(e^{2\pi i/5}\right)^4 \\
& = &5e^{8\pi i/5}=-5e^{3\pi i/5}.
\end{eqnarray*}\]

Physics GRE - #29

Solid argon is held together by which of the following bonding mechanisms?

1. Ionic bond only.
2. Covalent bond only.
3. Partly covalent and partly ionic.
4. Metallic bond.
5. Van der Waals bond

Solution :

Choice 5 is the answer.

Argon is not charged and stable as a monoatomic element (due to it's full orbitals). Thus, because it is stable alone, it does not ionically bond with itself since argon does not exist naturally as an ion. This eliminates choice 1.

Because covalent bonding occurs with electron sharing and the stability from filling unfilled orbitals, Argon does not covalently bond with itself since it already has full orbitals. This eliminates choice 2.

It cannot be partly ionic and partly covalent as we just stated above that it is to stable to do either. This eliminates choice 3.

Argon is not a metal and so it does not participate in metallic bonding (metallic bonding occurs in metals due to massive electron sharing among many metal atoms, but again, due to argon's completely filled orbitals, it is too stable to do this). This eliminates choice 4.

Van der Waals bonding occurs in all chemicals and compounds as they are mostly caused by instantaneously inducted dipoles due to movement of electrons in the atom. Argon is no exception to this rule. Hence choice 5 is the answer. Note that although Van der Waals bonding is always present, it is by no means the strongest bonding mechanism. If this question was applied to other elements, Van der Waals may not be the answer since it could be negligible compared to other bonding mechanisms.

Math GRE - #28

What is the volume of the solid formed by revolving about the x-axis the region in the first quadrant of the xy-plane bounded by the coordinate axes and the graph of the equation: \[y=\frac{1}{\sqrt{1+x^2}}?\]

1. $\dfrac{\pi}{2}$
2. $\pi$
3. $\dfrac{\pi^2}{4}$
4. $\dfrac{\pi^2}{2}$
5. $\infty$

Solution :

Choice 4 is the correct answer.

Recall that the volume of revolution around the x-axis is given by the formula \[V=\int{\pi y(x)^2}\,dx.\] In this case, we can substitute for $y$ and plug in the limits of integration to obtain: \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx.\]
Almost everyone with some calculus experience has had the great displeasure of memorizing: \[\int{\frac{1}{1+x^2}}\,dx = \arctan x + C.\] Now we can put it to good use! We can see that the answer is (if you'll pardon my poor notation): \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx=\pi\left[\arctan\infty-\arctan0\right]=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}.\]

Interesting tidbit: Here's a funny little thing you'll often see in first year calculus. It's called Gabriel's Horn. It's got finite volume but INFINITE surface area. Just something interesting I wanted to share.

Physics GRE - #28

lalalala test element $^A_ZX$ decays by natural radioactivity in two stages to $^{A-4}_{Z-1}Y$. The two stages would most likely be which of the following?

	First stage	Second stage
1.	$\beta^-$ emission with an antineutrino	$\alpha$ emission
2.	$\beta^-$ emission	$\alpha$ emission with a neutrino
3.	$\beta^-$ emission	$\gamma$ emission
4.	Emission of a deuteron	Emission of two neutrons
5.	$\alpha$ emission	$\gamma$ emission

Solution :

Choice 1 is the answer.

For choice 1, we have \[^A_ZX\rightarrow ^A_{Z+1}X+e^-+\bar{v}_e\] for the $\beta^-$ emission, and then \[^A_{Z+1}X \rightarrow ^{A-4}_{Z-1}Y+^4_2\text{He}\] for the $\alpha$ emission. Everything works out, charge/lepton/baryon numbers are conserved as they should be.

For choice 2, lepton number is not conserved as the $\beta^-$ emission does not occur with an antineutrino.

For choice 3, the atomic number remains unchanged under both emissions.

For choice 4, the emissions specified are extremely rare. In other words, this type of emission is impossible for the 'natural radioactivity' specified in the question.

For choice 5, the atomic numbers do not match up (there are fewer protons than needed with this decay path).

Physics GRE - #27

Suppose that there is a very small shaft in the Earth such that a point mass can be placed at a radius of $R/2$ where $R$ is the radius of the Earth. If $F(r)$ is the gravitational force of the Earth on a point mass at a distance $r$, what is: \[\frac{F(R)}{F(2R)}?\]

1. 32
2. 8
3. 4
4. 2
5. 1

Solution :

Choice 3 is the answer.

A very simple question today.

Recall that the gravitational force between two objects is given by \[F(r)=\frac{GMm}{r^2}=\frac{k}{r^2}\] where $k$ is a constant.
Hence we have: \[\frac{F(R)}{F(2R)}=\left.\frac{k}{R^2}\right/\frac{k}{(2R)^2}=4.\]

On the same exam, they also asked for \[\frac{F(R)}{F\left(\frac{R}{2}\right)}.\] Try it if you want some extra practice.

Math GRE - #27

Which of the following is the best approximation of $\sqrt{1.5}(266)^{3/2}$?

1. 1000
2. 2700
3. 3200
4. 4100
5. 5300

Solution :

Choice 5 is the answer.

With some simplification, we note that: \[\begin{eqnarray*}
\sqrt{1.5}(266)^{3/2}=\sqrt{\frac{3}{2}}\cdot(266)^{3/2} & = & 266\sqrt{\frac{3\cdot 266}{2}} \\
& = & 266\sqrt{399}\approx266\cdot20\approx 5300.
\end{eqnarray*}\]
Those of us familiar with Taylor series may try approximations with calculus, however this question teaches us that too much knowledge may be a little bit dangerous if misapplied. And that most of the time, there are simpler solutions than you'd think.

Physics GRE - #26

An engine absorbs heat at a temperature of 727 degrees Celsius and exhausts heat at a temperature of 527 degrees Celcius. If the engine operates at maximum possible efficiency, for 2000 J of heat the amount of work the engine performs is most nearly:

1. 400 J
2. 1450 J
3. 1600 J
4. 2000 J
5. 2760 J

Solution :

Choice 1 is the answer.

Recall that the Carnot Efficiency of a heat engine is: \[\eta=1-\frac{T_C}{T_H}\] where $T_C$ and $T_H$ are in Kelvins. For this problem, the efficiency is: \[\eta\approx1-\frac{527+273}{727+273}=1-\frac{800}{1000}=0.2.\] Thus the maximum amount of work we can get is \[W=\eta\cdot2000\,J=400\,J.\]

Math GRE - #26

A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?

1. $\dfrac{x^2}{9}$
2. $\dfrac{x^2}{8}$
3. $\dfrac{x^2}{4}$
4. $x^2$
5. $2x^2$

Solution :

Choice 2 is the answer.

Let's call one side of the fence $a$ and the other side $b$. Since we are only looking at 3 sides of a rectangular fence (e.g. we're fencing in our back yard and one side is the wall of the house), let's call the side length that we have to fence twice. So we have: \[x = 2a+b.\] The area of the fence is \[A = ab = a(x-2a)\] where the last equality comes from our equation for $x$. To maximize the area, we take the derivative of $A$ with respect to $a$ and set equal to 0 to obtain: \[0 = x-4a\implies a=\frac{x}{4}.\] We know this represents the maximum since the second derivative of $A$ is always negative. Using our equation for the perimeter, we now know that: \[x=2a + b\implies b=\frac{x}{2}.\] Hence the maximum possible area is: \[A=ab=\frac{x}{4}\cdot\frac{x}{2}=\frac{x^2}{8}.\]

Physics GRE - #25

A 2-kilogram box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally until the horizontal force is 10 N. The box is then in equilibrium as shown below.

The angle that the rope makes with the vertical is closest to:

1. $\arctan0.5$
2. $\arcsin0.5$
3. $\arctan2.0$
4. $\arcsin2.0$
5. $45^\circ$

Solution :

Choice 1 is the answer.

Since the block is in equilibrium, we must have $\vec{F}+\vec{F_g}=\vec{T}$ as shown below.

Hence $\theta=\arctan\frac{|\vec{F}|}{|\vec{F_g}|}=\arctan\frac{10}{2g}\approx\arctan\frac{10}{20}=\arctan0.5$.