logical deduction - Raven's Inception

Raven's Inception

Fluff (no clues):

The horde is chasing you. You run for your life along the main street of a small nondescript town you wish you had never set foot in. You turn a hard right, put your entire body behind you and crash into the door of a building. You burst into a small office in a explosion of wooden splinters. Disoriented and with pain throbbing everywhere you look around. An ominous staircase leads to a row of heavy ornate doors numbered 1 to 4.

At the bottom of the staircase sits Leonardo DiCaprio playing with a dreidel. He wears a severed raven's head as a hat. The wings of the poor animal are shoddily tied to his arms. They barely cover his forearms and looks scrawny. It does not look hygienic. As you approach him he KaKwaa's twice and hands you a scroll and points at the doors.

You look at the scroll, you hear noise outside, the horde has caught up...

Tips:

http://www.highiqpro.com/solve-matrices-iq-problems Drew inspiration from other places as well, I don't know how official the rules stated on that site are.

Hint:

In the big grid there are 3 patterns, you need atleast 2 to to find the final answer.
1: Color of the correct answer; this pattern is observed in the solved state of the smaller puzzles

2: Something to do with a specific recurring shape in the puzzles;this pattern is observed in the unsolved state of the smaller puzzles

3: Something to do with the same shape as in 2; this pattern is observed in the solved state of the smaller puzzles

electrostatics - Force on dielectric when inserted in charged capacitor

When we insert dielectric in a charged capacitor then, dielectric is attracted by the capacitor which makes dielectric being "sucked up" by the capacitor. Now, in order to derive the formula for force on dielectric, we apply energy conservation which I feel to be wrong as heat will be lost and hence energy conservation should not be applied, furthermore I feel that the energy term in the derivation corresponding to work done by force is actually heat released in the process.Where am I going wrong, also see [ Heat produced when dielectric inserted in a capacitor ] for more clarification of the doubt. Thanks

Answer

This problem is equivalent to releasing a mass which is at the end of an unextended vertical spring.

The mass loses gravitational potential energy whilst at the same time gains elastic potential energy and kinetic energy.

When the mass reaches the static equilibrium position it has kinetic energy and so overshoots that static equilibrium position to eventually stop when the loss of gravitational potential energy is equal to the gain in elastic potential energy.

At that maximum downward excursion there is a net upward force on the mass so it starts to move upwards losing elastic potential energy whilst at the same time gaining gravitational potential energy and kinetic energy.

The mass passes through the static equilibrium position and carries on until it reaches its initial position with the spring unextended and the mass at rest.
The process repeats itself the the mass oscillates about the static equilibrium position.

In the real world friction is present and so the mass undergoes damped harmonic motion eventually stopping at the static equilibrium position.

The spring-mass system now has less mechanical energy than it started with (it is actually half as much) because of the negative work done on the sysyem by the frictional forces - heat is generated.

Now with your capacitor and dielectric a very similar thing happens.
Start with the dielectric (same size as one of the plates of the capacitor) just outside the capacitor and release the dielectric.
There is a force on the dielectric which pulls it into the capacitor and electric potential energy is converted into kinetic energy of the dielectric.

When the dielectric reaches the static equilibrium position it has kinetic energy and will overshoot the static equilibrium position and eventually end up at rest jest outside the capacitor on the other side from where it started.
Although at rest there is a force on the dielectric pulling it into the capacitor and so the dielectric now starts moving in the opposite direction finishing up at rest at its starting point.

If one does not ignore dissipative process then the dielectric will undergo a motion which will result in it finally being at rest in the capacitor - the static equilibrium position.
Overall the dielectric-capacitor (and possibly battery) would have lost electrical energy and generated heat.

When you work out the force on a dielectric you can do it by ignoring dissipative processes and allow the dielectric to move from its current position a small distance .$dx$.
What you do next depends on whether the capacitor is isolated (constant charge $Q$ or connected to a battery (constant potential difference $V$).

If the capacitor has a constant charge then the change in energy stored $dU$ in the capacitor $C$ when the dielectric is moved a distance $dx$ in a direction parallel to the sides of the plates is $dU = -F\,dx$ where $F$ is the force on the dielectric.

So $F = -\frac{dU}{dx} $ with $U= \frac 1 2 \frac {Q^2}{C}$ which gives $ F = \frac 1 2 \frac {Q^2}{C^2}\frac{dC}{dx}$.

Knowing an expression for $C$ in terms of $x$ the force $F$ can be found.

Conservation of energy has been used in the absence of any dissipative processes, there are no frictional forces and no losses within the dielectric when its polarisation changes.

electromagnetism - Permeability constant in Ampère's circuital law must be vacuum permeability $mu_0$?

Just to be sure, is the permeability constant in Ampère's circuital law always equal to $\mu_0$, regardless of which medium the Amperian loop is placed in? That is, $\oint {\bf B} \cdot d {\bf \ell} = \mu_0 I$ and never equal to $\mu I$.

My reasoning is that if $\mu \neq \mu_0$, then using Stoke's theorem, \begin{align*} \oint {\bf B} \cdot d {\bf \ell} &= \mu I \\ \int (\nabla \times {\bf B}) \cdot d{\bf S} &= \mu \int {\bf J \cdot} d{\bf S} \\ \nabla \times {\bf B} &= \mu {\bf J} \\ &= \mu ({\bf J}_f + {\bf J}_b)\\ &= \mu (\nabla \times {\bf H} + \nabla \times {\bf M}) \\ &= \mu [ \nabla \times ({\bf H} + {\bf M}) ] \\ {\bf B} &= \mu({\bf H} + {\bf M}) \end{align*} But this contradicts with ${\bf B} = \mu_0({\bf H} + {\bf M})$.

Therefore, the $\mu$ must be equal to $\mu_0$.

Corollary:
Some books define Ampère's circuital law as $\oint {\bf H} \cdot d {\bf \ell} = I$. This is true if we are dealing with ${\bf B}$ in free space (or if $I=I_f$, see comment below). That is, we place the Amperian loop in free space such that \begin{align*} \oint {\bf B} \cdot d {\bf \ell} &= \mu_0 I\\ \oint \frac{\bf B}{\mu_0} \cdot d{\bf \ell}&= I \\ \oint {\bf H}\cdot d{\bf \ell}&= I \end{align*} If ${\bf B}$ is not in free space then $\frac{\bf B}{\mu_0} \neq {\bf H}$ and thus $\oint {\bf H}\cdot d{\bf \ell} \neq I$ (unless $I=I_f$, see comment below).

Answer

Yes, one of the correct forms of the law is $\oint {\bf B} \cdot d {\bf \ell} = \mu_0 I $, (not$ \mu$)and $I$ is all the currents included in the loop (free current in conductors, bound current due to spins, current due to orbital motion of electrons, everything).

Inside a linear medium (with permeability being constant inside it), one can use the formula $\oint {\bf B} \cdot d {\bf \ell} = \mu I_{free}$, where $ \mu = \mu_r \mu_0$, but this may not work if the loop of integration passes through multiple mediums.

Note that the above can be derived from $\oint {\bf H} \cdot d {\bf \ell} = I_{free}$ (in your question you have written $I$, but it should be $I_{free}$, then everything fits together and it is valid in any medium), which is another form of Ampere's law. In this form only free currents are included.

homework and exercises - Ball flying towards me or me flying towards ball

Suppose a ball is flying towards me at a speed of 10m/s and that, on impact, I feel "x" amount of pain.

If, instead, it was me flying towards the ball at the same speed, with all other conditions being the same, would I feel the same amount of pain?

Answer

Look at it this way:

Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball at 10 m/s (= your second experiment). Since it is the same experiment, you will feel exactly the same pain.

visible light - Does any material glow, under appropriate conditions?

Given a high enough temperature, or a high enough electrical field, can we make every material emit visible light?

homework and exercises - Ascertain the height an object has fallen from given force exerted and mass

An object of a given mass falls from an unknown height. If the force exerted by the object on contact with the ground is known, how would you ascertain the height from which the object fell?

Answer

The answer is a definite maybe!

Ignoring air resistance the velocity a falling object hits the ground can be calculated using the appropriate SUVAT equation or by equating potential energy lost with kinetic energy gained. However this only tells you the speed the object hits the ground, and how hard it hits the ground depends on how fast it decelerates.

If you can measure the force as a function of time during the collision you can calculate the total impulse, and since this is equal to the momentum change you can calculate the original momentum and hence the original velocity. Just calculating the peak force doesn't help because it could be a hard slow moving object decelerating suddenly or a soft fast moving object decelerating slowly.

I haven't gone into specifics, since this looks rather like a homework question, but this should direct you to the physics you need to answer your problem.

probability - Social Engineering (The Birthrate Problem)

The distant country of Toomalia has a birthrate of 67 boys for every 33 girls. In an effort to restore a 50:50 gender ratio in the country, Toomalian policymakers institute a mandatory new social policy: Couples seeking to have children may continue to do so as long as they bear only girls, but must stop as soon as they bear one boy.

Assuming the Toomalian people i) rigorously adhere to the policy, and ii) do not selectively abort their offspring, what will the net effect on the boy:girl birth ratio be when the new policy is put into effect?

Note that you may assume the gender of each conceived child is independent from the gender(s) of its siblings, but you should not assume that every couple wishes to conceive as many children as possible. Many couples will voluntarily stop having children after having 2 or 3, for example.

Answer

The ratio will stay the same. I'm not even sure how to prove that, since it seems so self-evident.

One simple approach:

Since the gender of a child is independent from their siblings, we can ignore who the parents are and ask under this procedure, what will the gender of the next child born to any couple in the country be? Well it'll still have a 67% chance of boy, 33% chance of girl. And since the parents/siblings don't matter in that number, those chances will be constant, so over time as more children continue to be born, it will be in the same ratio.