# CAP 2015 Solutions

This post will cover the multiple choice section of the 2015 CAP exam.

# CAP 2015 Question 1

As shown in the picture, a ball is attached to a ceiling and a wall with massless ropes $A$ and $B$. Rope $A$ is at angle $\theta$ from the vertical direction, and rope $B$ is horizontal. The system is static. If rope $B$ is cut, what is the ratio of the tension in rope $A$ immediately after it is cut to the tension in $A$ before it is cut?

$\displaystyle a) \, \, 1$

$\displaystyle b) \, \, \cos{\theta}$

$\displaystyle c) \, \, \frac{1}{\cos{\theta} }$

$\displaystyle d) \, \, \cos^{2}\theta$

$\displaystyle e) \, \, \cos{\theta}\sin{\theta}$

##### Solution

The answer is $d)$

Let the ball have a mass $M$ and the initial tension in $A$ be $T_A$. Then balancing forces vertically, $T_A \cos{\theta} = Mg$.

Once $B$ is cut, what happens to $A$? There’s one thing you know for sure. The length of the string remains constant. Therefore the acceleration in the radial direction is zero. Mathematically, if the new tension is $T_A'$, then

$T_A' = Mg \cos{\theta}$. Dividing our two equations, we get

$T_A'/T_A = \cos^{2} \theta$

# CAP 2015 Question 2

Consider a circuit made of a wire with uniform resistance in a shape of a circle as shown in the picture. The circle is connected diagonally from point $A$ to point $B$ with the same type of wire. If the current passing through the circuit is $i_0$, what is the current passing through the wire $AB$ as a function of angle $\theta$?

$\displaystyle a) \, \, 0$

$\displaystyle b) \, \, \frac{\theta}{\pi - \theta} i_0$

$\displaystyle c) \, \, \frac{\pi - 2 \theta}{\pi + 2}i_0$

$\displaystyle d) \, \, \frac{\pi-2 \theta}{\pi+4}i_0$

##### Solution

The answer is $d)$

Whenever you see a problem like this on a multiple choice test, the first thing you should do is look at a special case of $\theta$. What’s the most obvious choice? $\theta = 0$. The possibilities when $\theta = 0$ are

$\displaystyle a) \, \, 0$

$\displaystyle b) \, \, 0$

$\displaystyle c) \, \, \frac{\pi }{\pi + 2}i_0$

$\displaystyle d) \, \, \frac{\pi}{\pi+4}i_0$

Notice that the answer is definitely not $0$ because certainly some current will pass through $AB$! But the other two options are different! So if we solve the case when $\theta = 0$, we can unambiguously determine the answer. It turns out this is very easy. It’s just three resistors in parallel! But what are their resistances? Let the resistance of $AB$ be $R$. From geometry, the circular arcs are $\frac{\pi}{2}$ times longer. And since resistance is proportional to length (we are assuming a constant resistivity), their resistance is $\frac{\pi}{2}$ times higher. Therefore the total resistance is

$\displaystyle \frac{1}{R_{tot}} = \frac{1}{R} + \frac{1}{\frac{\pi}{2} R} + \frac{1}{\frac{\pi}{2} R}$

$\displaystyle \Rightarrow \frac{R_{tot}}{R} = \frac{\pi}{4+\pi}$

You may be able to immediately see that the current through $AB$ is $\frac{R_{tot}}{R} i_0$. If not, here’s an explanation of why. The voltage drop over the resistors is just $R_{tot}i_0$. But this must be the voltage drop over $AB$ as well! If the current in $AB$ is $i_{AB}$, then

$\displaystyle i_{AB} R = R_{tot}i_0 \Rightarrow i_{AB} = \frac{R_{tot}}{R} i_0 = \frac{\pi}{4+\pi} i_0$

This is the answer you would have gotten using $d)$.

# cap 2015 Question 3

The above graph represents the speed of a car as a function of time. We know that as the car speeds up there is a friction force with air that can be approximately considered to be proportional to the speed of the car. Which of the following graphs can be the force of the engine as a function of time?

$a)$

$b)$

$c)$

$d)$

$e)$

##### Solution

The answer is $c)$.

We are given $F \propto v$. If you’re unfamiliar with this notation, it just means that $F$ is proportional to $v$. Therefore since $v$ is accelerating uniformly, $v \propto t \Rightarrow F \propto t$. So $F$ must increase in a straight line in the beginning. This means the answer must be $b)$ or $c)$. What’s different about the two? In $b)$, the final force is equal to the initial force. But this can’t possibly be true because there’s now a strong force from air resistance! Therefore $c)$ must be correct.

# cap 2015 Question 4

If electromagnetic radiation with intensity $I$ (power per unit area) is absorbed by a surface, it exerts a pressure on the surface that is given by $I \cos{\theta}$, where $c$ is the speed of light and $\theta$ is the angle the light rays hit the surface relative to a direction perpendicular to the surface. The Sun has a radiation power of $P = 3.9 \cdot 10^{26} \, W$. The absorption of sunlight by Earth causes a force on Earth that pushes it away from the Sun. Which of the following is closest to the magnitude of this force?

$\displaystyle a) \, \,10^7 \, N$

$\displaystyle b) \, \, 10^9 \, N$

$\displaystyle b) \, \, 10^{11} \, N$

$\displaystyle b) \, \, 10^{13} \, N$

##### Solution

The answer is $b).$

Let $R_E$ be the radius of the Earth and $R_{ES}$ be the distance between the Earth and the Sun. From the data sheet, we know that  $R_E = 6.371 \cdot 10^6 \, m$ and $R_{ES} = 1.49598 \cdot 10^{11} \, m$. We also know that $c = 3.00 \cdot 10^8 \, \frac{m}{s}$. What’s the intensity $I$ absorbed by the earth? Well power is proportional to area, and the area the sunlight is spread over is the area of a sphere with radius $R_{ES}$. This is

$A_S = 4 \pi R_{ES}^2$

But the total power of the sun’s light on (not in) this sphere is just the power of the sun! So the intensity is

$\displaystyle I = \frac{P}{A_S} = \frac{3.9 \cdot 10^{26} \, W}{4 \pi (1.49598 \cdot 10^{11} \, m)^2 } = 1.4 \cdot 10^{3} \, \frac{W}{m^2}$

Now we need the area of the Earth perpendicular to the sun’s rays. Why? Because the sunlight only cares about the surface perpendicular to itself. Think about this a little bit.

Hopefully you got $A_E = \pi R_E^2$. Just imagine looking at the silhouette of the Earth against the backdrop of the sun! All you see is a circle of radius $R_E$! The force is now

$\displaystyle F = \frac{I A_E}{c} = \frac{1.4 \cdot 10^{3} \, \frac{W}{m^2} \cdot \pi (6.371\cdot 10^6 \, m)^2}{3.00 \cdot 10^8 \, \frac{m}{s}} = 6.0 \cdot 10^8 \, N \sim 10^9 \, N$

# cap 2015 Question 5

As shown in the picture a cylinder with volumetric thermal expansion coefficient of $\beta_{c} = 3 \cdot 10^{-6} \, / {}^{\circ} C$ is completely submerged and floating in a fluid at the temperature $50^{\circ} C$. The fluid has volumetric thermal expansion coefficient of $\beta_{f} = 8 \cdot 10^{-5} \, / {}^{\circ} C$. If we cool down the cylinder and the fluid to $0^{\circ} C$ what percentage of the cylinder’s height will be out of the fluid?

$\displaystyle a) \, \, 0.39 \%$

$\displaystyle b) \, \, 0.42 \%$

$\displaystyle c) \, \, 0.78 \%$

$\displaystyle d) \, \, 0.84 \%$

##### Solution

The answer is $a)$.

First notice that if you have an object with density $\rho_{obj}$ in a liquid of density $\rho_{l}$ such that $\rho_{l} >= \rho_{obj}$, the submerged fraction is $\alpha = \frac{\rho_{obj}}{\rho_{l}}$. This is a useful equation to remember. If you’re unfamiliar with it, here’s a quick derivation. Let the object’s volume be $V$. Then the buoyant force is

$F_{B} = \alpha V \rho_{l} g$

This must balance the total weight of the object. Therefore

$\alpha V \rho_{l} g= V \rho_{obj} g \Rightarrow \alpha = \frac{\rho_{obj}}{\rho_{l}}$

OK. Let’s use this equation. Initially, the submerged fraction is exactly $1$. Therefore without even using the equation, we know that $\rho_{obj} = \rho_{l} = \rho$. However, after a change in temperature, the volumes change. The total volume of the object is multiplied by a factor $1-\beta_{c} \Delta T$. Since the mass stays the same, the density is multiplied by a factor of $\frac{1}{1-\beta_{c} \Delta T }$. It’s the same for the liquid except the coefficient of expansion is $\beta_{f}$. Therefore the ratio of densities is now

$\displaystyle \alpha = \frac{ \frac{\rho}{1-\beta_{c} \Delta T} }{\frac{\rho}{1-\beta_{f} \Delta T} } = \frac{1-\beta_{f} \Delta T}{1-\beta_{c} \Delta T}$

Plugging in values,

$\displaystyle \alpha = \frac{1-8 \cdot 10^{-5} \, / {}^{\circ} C \cdot 50^{\circ} C }{1-3 \cdot 10^{-6} \, / {}^{\circ} C \cdot 50^{\circ} C} = 0.996$

Therefore the fraction outside the fluid is $1-\alpha = 3.9 \cdot 10^{-3} = 0.39 \%$

# cap 2015 Question 6

The amount of sunlight absorbed by the Earth’s atmosphere is approximately proportional to the length of air through which it travels to reach to the Earth. Which of the following is closest to the ratio of the absorption of sunlight during sunset relative to the absorption when the Sun is exactly at the center of the sky? The effective height of the atmosphere is about 10 km.

##### Solution

The answer is $b)$.

This is just a geometry question. Let the Earth’s radius be $R$ and the atmosphere’s thickness be $d$. From the data table, $R = 6371 \, km$ and from the question $d = 10^4 \, m$. Look at the diagram below.

When the sun is directly overhead, it simply travels a distance $d$ from $B$ to $A$. However, at sunset the light travels along path $CA$. Notice that $OA \perp CA$ so we can apply Pythagorean’s Theorem on $\triangle OAC$.

$OA = R$,

$OC = d+R$, so

$AC = \sqrt{(OC)^2-(OA)^2} = \sqrt{(R+d)^2-R^2}$

Therefore since sunlight absorption is proportional to distance through the atmosphere, the ratio is

$\displaystyle \gamma = \frac{\sqrt{(R+d)^2-R^2}}{d} =\frac{ \sqrt{(6371 \, km + 10 \, km)^2 - (6371 \, km)^2}} {10 \, km} = 36$

# cap 2015 Question 7

A beam of red light is made up of a stream of photons as shown above. The size of dots represents the photon energy, and the spacing represents the spatial distribution between photons.:

If we use photons of half the wavelength of red light, keeping the intensity constant, what should the stream look like?

$a)$

$b)$

$c)$

$d)$

$e)$

##### Solution

The answer is $c)$.

First notice that halving the wavelength doubles the energy, since $E = \frac{hc}{\lambda}$. Because the size represents the energy, doubling the energy should increase the size. Therefore we can immediately rule out $d)$ and $e)$. Next notice that since the intensity is constant and the energy of each photon is doubled, the number of photons per unit time should be halved. In other words, there is twice as much space between each photon, which corresponds to $c)$.

# cap 2015 Question 8

To paint the surface of a solid metal sculpture we need 100 buckets of paint. If we melt the sculpture and create 1000 smaller but otherwise identical sculptures, how many buckets of paint are needed to paint the surface of all 1000 small sculptures? Assume that the original and the smaller sculptures are all solid with no cavity inside.

$\displaystyle a) \, \, 10$

$\displaystyle b) \, \, 100$

$\displaystyle c) \, \, 1000$

$\displaystyle d) \, \, 10000$

##### Solution

The answer is $c)$.

The volume of each small sculpture is a thousandth of the original sculpture. Since volume goes down by the cube of the length, we know that the length of the small sculpture is a tenth of the original sculpture ($10^3 = 1000$). Because area goes down by the square of the length, the surface area of each small sculpture is a hundredth of the original. But since there are a thousand of these sculptures, there combined surface area is $1000 \cdot \frac{1}{100} = 10$ times the original surface area. Since the number of buckets is proportional to the area, we need $10 \cdot 100 = 1000$ buckets.

# cap 2015 Question 9

A person is standing on a platform that is sliding down the hill as shown in the picture. The slope of the hill is $30^{\circ}$ with respect to the horizontal. The person is standing on a scale positioned on the surface of the platform and the scale is reading only $150 lbs$ while his real weight is $160 lbs$. Which of the following is the coefficient of kinetic friction between the platform and the slope?

$\displaystyle a) \, \, \frac{\sqrt{3}}{2}$

$\displaystyle b) \, \, \frac{\sqrt{3}}{4}$

$\displaystyle c) \, \, \frac{\sqrt{3}}{3}$

$\displaystyle d) \, \, \frac{\sqrt{2}}{3}$

$\displaystyle e) \, \,$ there is not enough information.

##### Solution

The answer is $b)$.

We can immediately find the acceleration of the person in the direction of gravity. If his mass is $m_p$, then the normal force from the scale is $\frac{150}{160} mg = \frac{15}{16} mg$. Therefore his net acceleration downwards is $a_{y} = g - \frac{15}{16}g = \frac{1}{16}g$. Since the acceleration of the block must be along the ramp, if the total acceleration is $a$, then $a_{y} = a \sin{30^{\circ}} = \frac{a}{2} \Rightarrow a = \frac{g}{8}$

We’re done with the person now. Let’s just treat the platform and the person as one composite object of mass $M$. This is just a standard block down the ramp! Call the normal force from the hill onto the platform $N$ and the coefficient of kinetic friction $\mu$. By balancing forces perpendicular to the ramp, we see that the normal force is

$N = Mg \cos{30^{\circ}} = Mg \frac{\sqrt{3}}{2}$

Using Newton’s Law along the ramp, we get

$Mg \sin{30^{\circ}} - \mu N = \frac{Mg}{2} - \mu Mg \frac{\sqrt{3}}{2} = Ma$

Simplifying,

$\mu = \frac{\sqrt{3}}{4}$

# cap 2015 Question 10

A constant amount of ideal gas, at the temperature $T_0$, undergoes a process that changes its pressure from $P_0$ to $2P_0$. Then its volume is increased from $V_0$ to $3V_0$ at a constant pressure, as shown on the $P-V$ diagram:

Which of the $P-T$ diagrams below correctly reflects this process?

$a)$

$b)$

$c)$

$d)$

##### Solution

The answer is $b)$.