Arjun Nair 12th Sample

COP isn't an efficiency, it's the ratio of what you want (heat removed from inside, $Q_2$) to what you pay (work $W$). For an engine, the useful output (work) can't exceed the input heat, so efficiency < 1. For a fridge, the fridge isn't *creating* the heat it moves, it is just **moving** it from cold to hot using a little work. So $Q_2$ can be much larger than $W$, and $\alpha = 9$ means for every 1 J of work, 9 J of heat is pumped out of the inside. Note that the smaller the temperature difference between inside and outside, the larger the COP. As $T_1 \to T_2$, $\alpha \to \infty$, and that's why a fridge set to a less cold temperature is cheaper to run.

3d ago

The "large numbers" part is very much a statistics point. A 3:1 ratio is the expectation, not what you will get every time. If you only have 8 offspring you could easily get 5:3. With hundreds, the observed ratio gets close to 3:1. Mendel's sample was big enough for the pattern to show up. (PCM person here, but this is probability, so it's mine too.)

4d ago

Think of it as a game with an opponent. The opponent says: "I want $f(x)$ to be within $\varepsilon$ of $L$", where $\varepsilon$ is any tolerance they like, as tiny as they want. You reply: "Fine. If $x$ is within $\delta$ of $a$, I guarantee it." The limit exists if you can *always* reply, whatever $\varepsilon$ they pick. So the opponent chooses $\varepsilon$, you choose $\delta$, and $\delta$ will usually depend on $\varepsilon$. For $3x+1$: $|f(x)-7| = |3x-6| = 3|x-2|$. To make that less than $\varepsilon$ you need $|x-2|<\varepsilon/3$. So $\delta=\varepsilon/3$. And yes, working backwards is exactly how you *find* $\delta$ (scratch work). The proof is the same steps written forwards: "Given $\varepsilon>0$, take $\delta=\varepsilon/3$. If $0<|x-2|<\delta$, then $|3x+1-7| = 3|x-2| < \varepsilon$." Done.

4d ago

A note on hack 6, since it's a Physics point. Momentum conservation requires no *net external force* on the system, as you wrote. Strictly, the total momentum is conserved along any direction in which the net external force component is zero. So a block sliding down a frictionless incline on a movable wedge conserves horizontal momentum but not vertical. That's the exact nuance that students lose marks on. Otherwise, excellent post. The teddy is a better teacher than most.

5d ago

One more thing on the metronome: it trains your ear more than your hand. Typing on my keyboard I learned the same, rhythm first, speed later. 60 feels slow only until you try to be exactly on the click.

6d ago

the regret sentences thing is a good idea, writing them down tonight. A sentence: "I picked A, I'm lonely, and the branch is boring" makes me feel something, so maybe the branch matters more than I admit.

7d ago

1. In pure rolling the point of contact is **instantaneously at rest** relative to the surface. Friction is static, acts at that point, and the point doesn't move, so the work done by friction is $f\cdot v_{contact}\,dt=0$. Friction does not dissipate energy here. What it does do is give a torque that makes the body spin, and it reduces the translational acceleration compared to a frictionless slide ($g\sin\theta$) by a fraction. Energy conservation holds: $mgh=\tfrac12mv^2(1+k^2/R^2)$.

8d ago

Also practicals count and you can't cram them. And the board marks matter for quite a few college and eligibility things. Dry take: ignoring them is a bad trade.

9d ago

Small detail for people who want to try this: the period is independent of mass only because the weight cancels out. Force on the bob along the arc is $mg\sin\theta$ and acceleration is that divided by $m$, so $m$ drops out. And the formula is valid only for small angles (a rule of thumb is under about 10 degrees), where $\sin\theta \approx \theta$. For big swings, the period gets longer than the formula says. Also the standing-on-the-swing thing is cool. You're right that you lower/raise the centre of mass and so change the effective length.

9d ago

Velocity is a vector. A force perpendicular to $\vec v$ changes the direction of $\vec v$, not its magnitude. KE depends on $v^2$ (speed), not on direction. So the speed stays constant, KE constant, but the velocity vector rotates. Work done is $\vec F\cdot d\vec r = 0$ since $\vec F\perp d\vec r$. So the force changes momentum (direction) but not energy. That's why a magnetic field can steer a particle but cannot speed it up.

9d ago

Right, signed areas. I left it out to keep the post short, but yes, that is the careful version.

9d ago

No, they are two different speeds that happen to have the same symbol. The wave speed $v = \omega/k = 20$ m/s is how fast the *pattern* moves along $x$. The particle just goes up and down about its mean position, so its speed is $\partial y/\partial t = 0.02\times100\cos(100t-5x)$, which peaks at $A\omega = 2$ m/s. Here 20 and 2 are different. Don't equate them. (For a wave, they would be equal only by coincidence.)

10d ago

Yes exactly the same idea. Area under the v-t graph is $\int v\,dt$, thin strips of height $v$ and width $dt$. Here the only new thing is that you also cut the rectangle the other way, with horizontal strips, and the two cuts have to add up to the whole.

10d ago

45°. At 45°, $\sin 2\theta = \sin 90° = 1$, which is the biggest value sine can take, so that's the maximum range. 30° and 60° are equally far from 45° on either side, that's why they tie.

10d ago

Nice trick for the projectile one that needs no calculator: for the same speed, launch angles $\theta$ and $90° - \theta$ give the same range, because $\sin 2\theta = \sin(180° - 2\theta)$. So 30° and 60° must give the same answer. If a chatbot gives you different ranges for the two, you've caught it. Free sanity check.

10d ago

LIATE is a fine heuristic, it works most of the time. It just fails when you meet something it has no opinion on. The picture is for those days.

10d ago

Derivation of the magic one: divide $H$ by $R$. $$ \frac{H}{R} = \frac{u^2\sin^2\theta/2g}{u^2\cdot 2\sin\theta\cos\theta/g} = \frac{\sin\theta}{4\cos\theta} = \frac{\tan\theta}{4} $$ So $H = \frac{R\tan\theta}{4}$, or $\tan\theta = \frac{4H}{R}$. Your $R = 4H$ gives $\tan\theta = 1$, so 45°. Correct. The formula holds only for level ground (same launch and landing height).

10d ago

This is the post I wish I'd read before I wasted money on my first scope. 7x50 is honestly a great beginner setup, the wide field means you learn the sky properly instead of just chasing one object. Also the Moon's terminator tip is the thing nobody tells you. Full moon through any instrument is basically a flashlight to the face.

10d ago

On the sign convention: $v_{AB} = v_A - v_B$ means "velocity of A **as seen from** B". Choose a positive direction first, put signs on both velocities, then subtract. Example (a): take B's direction as positive. Then $v_B=+30$ and $v_A=-20$. So $v_{AB}=-20-30=-50$ m/s, A moves at 50 m/s in the negative direction as seen from B. The magnitude is 50, which is what time needs. $v_{BA}=+50$. Opposite sign, same magnitude: $v_{BA} = -v_{AB}$. So it doesn't matter for time questions (magnitude only), but it matters when asked for the **direction** or in graph / sign-based questions.

10d ago

Your numerical is fine. Converting joules to eV by dividing by $1.6\times10^{-19}$ is correct. Shortcut worth memorising: $$E\,(\text{eV}) = \frac{1240}{\lambda\,(\text{nm})}$$ so for 400 nm, $E = 3.1$ eV. $K_{max} = 3.1 - 2.0 = 1.1$ eV, and the stopping potential satisfies $eV_0 = K_{max}$, so $V_0 = 1.1$ V. The threshold wavelength of this metal would be $1240/2.0 = 620$ nm. Light of 700 nm wouldn't eject anything.

12d ago

The $\sqrt2$: in the simple derivation you assume the target molecules are standing still and only your molecule moves with the average speed $\bar v$. But all the molecules are moving. What matters for collisions is the **relative** speed. For two molecules moving in random directions, the average relative speed works out to $\sqrt2\,\bar v$. So the sweeping rate goes up by $\sqrt2$, there are $\sqrt2$ times more collisions per second and the path between collisions shrinks by $\sqrt2$. That's why it sits in the denominator.

12d ago

Also the series view, for completeness: $$0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots = \frac{9/10}{1 - 1/10} = 1$$ It's a geometric series with first term 9/10 and ratio 1/10. And your teacher's subtraction trick is only legitimate *because* that sum has a well-defined value, we are allowed to do arithmetic on it. Good that you're suspicious of the trick, it is an informal version of this.

13d ago

That describes the sequence 0.9, 0.99, 0.999, ... Each of those is less than 1 and they keep getting closer, yes. But "0.999..." with the dots is not any one of those terms. It means the *limit* of that sequence, and the limit is exactly 1. The "approaches but never reaches" language is about the list of numbers, not about the value the dots stand for. Also, in the real numbers we use at school there are no positive numbers smaller than every $1/10^n$, so the "infinitely small gap" doesn't exist there.

13d ago

I skipped part of ours to finish a mock paper review. Didn't enjoy the paper review. Regret the skipping. Reading this tells me what I missed. Not doing that at the next event.

14d ago

You have it right. Axial field is in the direction of the dipole moment p (from -q to +q). Equatorial field is opposite to p. Since both are written as magnitudes in the ratio question, direction doesn't change the ratio of 2.

15d ago

Not quite. A pulley with mass has rotational inertia, so some of the energy goes into spinning it, and the acceleration is smaller. With moment of inertia $I$ and radius $r$: $$a = \frac{(m_1 - m_2)\,g}{m_1 + m_2 + I/r^2}$$ Note the limit check works here too. Let $I \to 0$ (light pulley) and you get Neel's formula back.

15d ago

Do a bit now. One long answer a week in class 11 is enough, so your hand doesn't forget. In class 12 increase. Better to build it slowly than to panic at the end.

15d ago

Agree with all of it. One thing worth adding to the dimensions check: it also tells you what *combinations* are possible. If you know the answer can only depend on $L$ and $g$, dimensions force $T \propto \sqrt{L/g}$ up to a constant. You can get the form of a result without deriving anything. It's a very good trick for unfamiliar problems.

15d ago

Honestly, 4 or 5 full answers a week is enough if you do them properly: write them with the diagram, then compare with the marking scheme or your textbook. Pick the ones you know least. Do not copy, write from memory. It builds the habit of stepwise writing.

15d ago

Temperature: volume changes with temperature (liquids expand), mass does not. So anything with volume in the denominator, **molarity** (and normality), changes with T. **Molality**, **mole fraction**, and mass percent use only mass or moles, so they don't change with temperature. That's why molality is preferred in physical chemistry (e.g., colligative properties) where you heat or cool a solution.

16d ago