Neel Chatterjee 12th Sample

PCM here too, can confirm the adrenaline before a chess tournament. My heart rate goes crazy even before I touch a piece. Now I know it's the medulla's fault.

4d ago

Harsh, this is SN1, not SN2, so crowding at the carbon doesn't matter. What matters is the stability of the carbocation intermediate: $3^\circ > 2^\circ > 1^\circ$. The tertiary carbocation is stabilised by three alkyl groups (inductive effect and hyperconjugation), so it forms fastest and the reaction goes immediately. The primary carbocation is so unstable that the reaction barely goes at room temperature. Your SN2 intuition would be correct for the Williamson ether synthesis, which is the next question.

4d ago

Coefficient of restitution $e$ is the ratio of the relative speed of separation to the relative speed of approach: $$ e=\frac{v_2-v_1}{u_1-u_2} $$ - $e=1$: elastic (relative speed unchanged) - $e=0$: perfectly inelastic (bodies move together after collision, no separation) - $0<e<1$: ordinary inelastic collision Check on your elastic one: $(4-(-2))/(6-0)=1$ ✓. Inelastic: $(2-2)/6=0$ ✓.

5d ago

Start with this and the rest follows. Put your page flat on the desk, $\hat i$ pointing right, $\hat j$ pointing up the page (away from you), then $\hat k$ points **out of the page toward your face**. That's the standard orientation. Now take your fingers, point them along $\hat i$ and curl them toward $\hat j$ (the shorter way, 90 degrees). Your thumb points toward your face, which is $+\hat k$. So $\hat i \times \hat j = +\hat k$, and rotating from $\hat i$ toward $\hat j$ is what? Right to up. That's anticlockwise when you look at the page. So "torque along $+\hat k$" and "anticlockwise" are the same statement. The vector direction is the axis of rotation, and the right-hand rule tells you which sense along that axis goes with which spin.

6d ago

Try this: for each option, write your regret in one sentence. "I picked A and I'm lonely and my branch is boring." "I picked B and I wonder what A would have been." Then ask which sentence you can live with more easily. For me, the second one hurts less than the first, but yours may differ. Chess version: don't count only material, count how many bad moves each position forces on you.

7d ago

As far as I know, no. The architecture entrance tests usually need maths as a subject in class 12. I'd check the official eligibility pages, since there are changes sometimes. Design courses (product, communication, animation) are the more flexible ones, many accept all streams.

7d ago

Architecture is another bridge: drawing plus physics plus maths of structures. Though I should say it needs maths in 12th, so with PCB you'd need to check the eligibility. Several design courses are open to students from any stream.

7d ago

Two different things, Kabir. Rolling resistance and slipping (kinetic friction) do dissipate energy, but ideal pure rolling on a rough surface involves only static friction at a point that does not slide. That's the case NCERT uses. If the body slips, then yes kinetic friction heats the surfaces and the energy method needs a loss term.

8d ago

Quietly: a pendulum is also a good way to measure $g$ locally. Rearranging, $g = 4\pi^2 L / T^2$. With your clip numbers you could get a decent estimate, within a few percent.

8d ago

On filler: most long shows have community-maintained "filler lists" online that mark which episodes skip the main story. Look one up before you start and skip those. Also, if a show has more than ~100 episodes, treat it as a separate decision for later. For now stick with completed, short, well-regarded ones. Your 30 minutes a day fits one 24-minute episode comfortably.

9d ago

Resolve $\vec v$ into $v_\parallel = v\cos\theta$ along $\vec B$ and $v_\perp = v\sin\theta$ perpendicular to $\vec B$. The parallel part feels no force, so the particle keeps drifting along $\vec B$ at $v\cos\theta$. The perpendicular part goes round in a circle of radius $r = \dfrac{mv\sin\theta}{qB}$. Circle plus steady drift = helix. Time for one circle: $T = \dfrac{2\pi r}{v\sin\theta} = \dfrac{2\pi m}{qB}$. The $v$ cancels, which is why the period (and frequency $\dfrac{qB}{2\pi m}$, called cyclotron frequency) doesn't depend on speed. Pitch = distance along B in one period: $$ p = v\cos\theta\cdot T = \frac{2\pi m v\cos\theta}{qB} $$

9d ago

The inverse function bit is my favourite part. One small thing, the rectangle tiling is literal only when the curve is monotone like in your sketch. For a general curve it still holds but you have to count the areas with signs.

9d ago

Good plan. Also a quick check for the case where it's thrown horizontally from height $h$ (θ = 0): $t = \sqrt{2h/g}$ and $R = u\sqrt{2h/g}$. Easy to get wrong if you apply the level-ground $T$ formula, which would give 0.

9d ago

Nah, not in 7x. You'll see Saturn as an oddly oval yellow dot. The rings need a telescope with a lot more magnification.

10d ago

Not faked, just a different detector. In the dark our eyes switch to rod cells, which are very sensitive but basically colourblind. The colour cells (cones) need more light. A camera on a long exposure keeps collecting light for seconds or minutes, so it picks up the faint red glow from hydrogen gas that your eye can't register. Some people do report a faint grey-green tint in the nebula by eye, but nothing like the photos.

10d ago

Yes. Don't hold them free in the air. Rest your elbows on the parapet or a wall, or sit and brace your elbows on your knees. The best one for me is lying back on a folding chair and resting the barrels on my brow bones, then it's your head that holds them, not your arms. Also breathe out slowly before you look, the shake drops a lot.

10d ago

Faraday's law is $\varepsilon = -N\,\dfrac{d\Phi}{dt}$. It cares about how fast flux *changes*, not how big it is. For a coil rotating at constant ω with normal along B at t = 0: $\Phi = BA\cos\omega t$. So $$\varepsilon = -N\frac{d\Phi}{dt} = NBA\omega\sin\omega t$$ Cos is at its peak when sin is zero. Flux max means the graph of Φ against t is flat at the top for a moment, slope zero, so emf zero. Flux is zero when the Φ curve is steepest, so emf is max there. Easy picture: height of a ball thrown up vs its speed. At the top the height is max but the velocity is zero.

10d ago

It had a lens cap once. Its whereabouts are an unsolved family mystery.

10d ago

Ishaan, careful, it's for overtaking: the **rear** of B (faster) must go past the **front** of A, so B has to gain the full $100+150=250$ m on A. Relative speed $10$ m/s so $t=25$ s. Sara got it right. For "crossing" or "overtaking" both cases the relative distance is the sum of lengths.

10d ago

Small correction. 256 is the *number of different values* one byte can hold, but they start from 0, so the biggest value is 255. Eight ones in a row, 11111111, is 128+64+32+16+8+4+2+1 = 255. Same as how two decimal digits give 100 different values, 00 to 99, but the biggest is 99, not 100.

13d ago

Good chess parallel here. Players don't improve by playing more games alone, they improve by studying their lost games and doing endgame drills, which is the most boring and most effective part. Same as your mistake-analysis block. Also "a defeat isn't a verdict, it's a to-do list" is going in my notebook.

13d ago

In the idealised case (no air resistance, ball lands at the same height it was kicked from), $45^\circ$ gives the farthest range. If the ball lands lower than it started, say you throw from a cliff, the best angle is slightly less than $45^\circ$. And with air drag it's also lower in practice. For a football kicked on level ground it's something like 30 to 40 degrees in reality, but I haven't measured that, so take it as "less than 45".

15d ago

Dry as promised. Good. I'd add one asymmetric point: the board paper rewards partial credit. JEE (especially MCQ) rewards you only for the final answer. That changes how you handle half-solved problems.

15d ago

The "body double" idea is quietly brilliant. I play chess online and it's the same thing, I focus better when someone's on the other side of the board even if we don't talk. May try the silent study call with a friend.

15d ago

Dev, no, molality is moles of solute per kg of **solvent**, and 'solvent' means only the water (100 g), not the glucose plus water. Kavya's mistake is exactly this one. $$m = \frac{18/180}{0.100\ \text{kg}} = \frac{0.1}{0.1} = 1.0\ m$$ Mole fraction: moles of water $= 100/18 = 5.56$. $$x_{glucose} = \frac{0.1}{0.1+5.56} = \frac{0.1}{5.66} \approx 0.0177$$ So molality 1.0 m and $x \approx 0.0177$.

16d ago

Another way to see the "bring from infinity" bit. Bring the test charge in along the equatorial line from infinity. At every point the force from the dipole is perpendicular to the equatorial line... wait, along the line itself the force has no component along the path? Yes. Force from the field is perpendicular to the path direction, so each small bit of work is $\vec{F}\cdot d\vec{l}=0$. Total work is zero, so $V=0$. The geometry of the two charges makes this happen.

16d ago

Sensible. For plank form, a common mistake is lifting the hips too high so the abs aren't working. Aim for a straight line and squeeze the glutes. You'll get shorter holds at first but they'll be better holds.

16d ago

Blind contour is the best exercise for observation I know of. It's similar to how in chess you train yourself to look at the position rather than at what you expect. Same discipline: see first, then act.

17d ago

small fix, Friends has 10 seasons, not 12. (Which does not make your 6 rewatches any less of an issue 😄) To Ananya: I think it's fine. A show you know is low-risk. Your brain doesn't have to track new characters or fear a cliffhanger, which is probably why you couldn't sleep with the new one. That's actually a reason to keep the old one for nights. Just treat it like a tool for winding down, with a start and an end, not a place to hide.

18d ago

Your guess for 2 is correct. Combine the two terms into one sine wave: $$3\sin\omega t + 4\cos\omega t = 5\sin(\omega t + \phi)$$ because amplitude is $\sqrt{3^2 + 4^2} = 5$. Then $I_{rms} = 5/\sqrt2 \approx 3.5$ A. If you want to avoid the phase trick, square it: the cross term $2\cdot3\cdot4\sin\omega t\cos\omega t = 12\sin\omega t\cos\omega t$ averages to zero over a cycle. So $\langle i^2\rangle = (9+16)/2 = 12.5$, and the root is the same $\sqrt{12.5}$.

18d ago