CRACKNEETPhysics

Did you know :

Tension vs weight tells you power direction.
In a rope lift: if the rope’s pull is upward and the load moves up, the machine is doing positive power on the load. If the load moves down while the rope still pulls up (you’re lowering gently), the load does negative power on the rope—energy flows back to the machine or is dissipated.

Two everyday pictures

Elevator:

Going up → motor delivers power to the car.

Coming down gently → motor can take power back (modern lifts feed it to the building’s electrical system), or waste it in a brake if regen isn’t available.

Lowering a bucket by hand: your hands feel the rope pulling up while the bucket moves down → your muscles are absorbing energy (they heat up). That’s negative power done by the rope on the bucket and positive power done by the bucket on you.


Question ;-

You rise through the same height H in three different ways. Start and finish at rest in all cases; ignore air drag and don’t hold the hand-rail.


Ride an upward escalator, standing still on a step.

Climb a stationary staircase at a steady pace.

Walk up an upward escalator so that your speed relative to the ground is the same as in case 2 (you just reach the top sooner).

Let “contact forces” mean only the forces from the steps/floor on you (normal + any friction). Over the ascent, which statement about the net work done by these contact forces on you is correct?

A. Same in all three cases: zero, because normal forces do no work.
B. Case 1 is positive and equals your gain in gravitational energy; Case 2 is zero; Case 3 is greater than Case 1.
C. Case 1 is positive and equals your gain in gravitational energy; Case 2 is zero; Case 3 is equal to Case 1.
D. Case 1 is zero; Case 2 is positive and equals your gain in gravitational energy; Case 3 is zero.

Watch Work Power Energy Playlist :- www.youtube.com/playlist?list...

4 hours ago | [YT] | 0

CRACKNEETPhysics

A sealed bulb contains a mono atomic gas X at 300 K . Placed in a weak non-uniform magnetic field, the bulb shows no measurable drift. When UV light ionizes a tiny fraction ( <1% ) of atoms to X+ (temperature unchanged), the bulb now drifts toward the stronger field. When the UV is switched off and the gas fully recombines, the drift disappears again.

Which ground-state valence configuration of X best explains this behavior?

A) ns2 np6


B) ns2 np1


C) ns2 np3


D) ns2 np5

Solution :- https://youtu.be/XuUK26FUvUg

12 hours ago | [YT] | 2

CRACKNEETPhysics

Did you know?

Curie Temperature ( TC ). It’s the “forgetting point” of a magnet: above TC , a ferromagnet (or ferrimagnet) loses spontaneous magnetization and behaves paramagnetically (no hysteresis, weak attraction only).

What actually changes is order, not the atoms: individual atomic moments don’t vanish; thermal agitation disorders them so domains can’t stay aligned.

Typical TC values: Iron ≈ 1043 K (770 °C), Nickel ≈ 631 K (358 °C), Cobalt ≈ 1393 K (1120 °C), Magnetite (Fe₃O₄) ≈ 858 K (585 °C).

Exam traps:

TC is material-specific; not about sample size or domain count.

Above TC , the B–H loop collapses (no remanence/coercivity).

Don’t confuse Curie temperature with the Curie constant in χ=C/T (paramagnetism) or with stronger magnet = higher TC —not always true.


Four unknown solids P,Q,R,S are tested at 300K.

1.In a weak non-uniform magnetic field, P is repelled while Q,R,S are attracted.

2. Quasi-static B – H loops: Q and S show hysteresis; P and R do not.

3. Heating Q to T≈1040 K makes its hysteresis loop vanish; on cooling back to 300 K in zero field, it has no net magnetization until re-magnetized.

4. For R , the susceptibility obeys χ≈C/T and is small and positive over 300−900K.

5. S shows a clear hysteresis loop at 300 K , but even in a very strong field its magnetization saturates at a small value—as if many tiny magnets inside are opposing each other and only a small ‘leftover’ remains

Identify P,Q,R,S :

A) P : diamagnetic, Q : ferromagnetic, R : paramagnetic, S : ferrimagnetic
B) P: diamagnetic, Q : ferrimagnetic, R : paramagnetic, S: ferromagnetic
C) P paramagnetic, Q : ferromagnetic, R : diamagnetic, S : ferrimagnetic
D) P: diamagnetic, Q : ferromagnetic, R : ferrimagnetic, S: paramagnetic

Watch Full PlayList :- www.youtube.com/playlist?list...

Topics Covered : Para-, dia- and ferromagnetic substances with examples, effect of temperature on magnetic properties

Hystersis Theory/Curve :-

https://youtu.be/wZJNqMCwg2Q?si=J8g_3...

https://youtu.be/SycsmlJCvs8?si=MAsRn...

https://youtu.be/T1CIg3k-TyE?si=raPaq...

https://youtu.be/bxntO5tTxrk?si=CIXAu...

23 hours ago | [YT] | 1

CRACKNEETPhysics

Most schools and Coaching Institutes would not teach you like this. Watch this Video if you think you have Doubts on Newtons second law o f Motion. Very Good Theory, Concept practise and MCQs. Step by Step .

youtube.com/live/YATrLiPFD78



You stand on a spring scale in a lift. The scale reading becomes more than your weight for a short time and then returns to normal while the lift keeps moving upward. Which situation fits this record?


A. Lift starts upward and keeps speeding up
B. Lift moves upward at constant speed
C. Lift moving upward comes to a stop at an upper floor
D. Lift is moving downward and speeds up

3 days ago | [YT] | 4

CRACKNEETPhysics

Two teams play tug-of-war on dry level ground. The rope is light and taut. At some instant the right team begins to slide left while the left team’s feet do not slip.

Which statement is correct?

A) The left team pulls the rope more than the rope pulls the left team.
B) The rope pulls both teams with equal magnitude, but the right team loses because the friction from the ground on the right team is smaller than that on the left team.
C) Since the right team accelerates left, the rope’s pull on the right team must be greater than the rope’s pull on the left team.
D) The net external horizontal force on the “two teams + rope” system is zero, so that system cannot accelerate

3 days ago | [YT] | 4

CRACKNEETPhysics

A straight conducting bar is made by joining two long segments in series, X and Y, each of the same length and the same cross-sectional area. A steady current flows from left to right when an ideal battery is connected. The free-electron density is the same in X and Y, but their resistivities are different (Y is the more resistive material). Temperature is constant.

Which one of the following statements is incorrect?

A) The current is the same through X and Y in steady state.

B) The electric field inside Y is stronger than inside X (since Y is more resistive) even though their cross-sections are the same.


C) The drift speed of electrons is larger in Y than in X because Y is more resistive, even when the cross-sections are equal.


D) A non-uniform surface charge distribution appears along the bar and at the X–Y junction to steer the field; reversing battery polarity flips the sign of these surface charges.

Did school/coaching leave your NEET Physics doubts unresolved? Subscribe here for crystal-clear concepts and exam-level practice questions—no formula cramming, just understanding.

youtube.com/live/BnPaG87ku1c

4 days ago | [YT] | 8

CRACKNEETPhysics

A solid metal bar AB is made of a single material with uniform resistivity. Its cross-section decreases smoothly from end A (thick) to end B (thin). The ends are connected to an ideal battery so that a steady current flows. Temperature remains constant.

Consider these statements about the steady state inside the bar:

A. The current is the same through every cross-section; the drift speed of electrons is larger near B than near A.

B.The electric field inside the bar has the same magnitude everywhere; only drift speed changes.

C.The potential falls uniformly from A to B (equal drop per unit length).

D. The surface charge along the bar is uniform, because the material is uniform

Did school/coaching leave your NEET Physics doubts unresolved? Subscribe here for crystal-clear concepts and exam-level practice questions—no formula cramming, just understanding. Watch this Video if your school/Coaching did not help you understand the concepts :- youtube.com/live/BnPaG87ku1c

4 days ago | [YT] | 5

CRACKNEETPhysics

A long straight conductor of radius a initially carries a steady current uniformly across its cross-section. A coaxial cylindrical hole of radius a/2 is drilled out (no current in the hole). The total current is kept the same by redistributing it uniformly in the remaining metal (the annular region). Let B(r) be the magnitude of the magnetic field at distance r from the axis.

Which statement about the shape of B(r) is correct?

A) B=0 for r<a/2 it then jumps to a finite value at r=a/2 , rises up to the surface, and falls outside.

B) B=0 for r<a/2 and also remains 0 just outside r=a/2 ; it becomes non-zero only near r=a .

C) B≠0 inside the hole because the surrounding current shell acts like an external wire.

D) B=0 for r<a/2 ; at r=a/2 the value of B is the same from inside and just outside (no jump), but the slope changes: for a/2<r<a it increases with r, then beyond r=a it decreases with distance.

Solution https://www.youtube.com/watch?v=oWMix...

5 days ago | [YT] | 5

CRACKNEETPhysics

Two identical long solenoids share the same axis; one is slid inside the other so their bores coincide. Both carry the same current magnitude.

Case I: Currents in the same direction.

Case II: Currents in opposite directions.

Which statement best describes the resultant field far from the ends?

A) Case I: nearly double inside, almost zero outside; Case II: almost zero everywhere except small end regions.
B) Case I: zero inside; Case II: doubled inside.
C) Both cases: zero inside due to cancellation.
D) Both cases: strong outside, weak inside.

Solution :- https://youtu.be/3V_oric5g_k?si=xmaNP...

https://youtu.be/VnoFW1uu6_c

https://youtu.be/M8v9y7rzn20

youtube.com/live/YEJZIk1MauU?feature=share

5 days ago | [YT] | 3

CRACKNEETPhysics

Gauss’s Law in Magnetism — quick, friendly reminder

What it says (one line):
No magnetic monopoles ⇒ total magnetic flux through any closed surface is zero.
(Magnetic field lines never start or end; they always form closed loops.)

Picture to keep in mind:
Imagine rubber-band loops (field lines) threading space. Draw any closed bubble (sphere, cube, weird potato). For every line that enters the bubble, it must leave somewhere else. Enters = leaves ⇒ net flux = 0.

Bar magnet trap:
Even if your bubble surrounds the “north” end of a bar magnet, lines that enter from the sides/inside also leave—you can’t trap an isolated pole with a closed surface.

Solenoid reminder:
A surface capping the bore has strong inward/outward flux, but the small, spread-out outside flux + edge effects exactly cancel it → net zero.

Time-varying fields:
Current changing, fields changing… still no magnetic charge. Net flux through a closed surface remains zero.


5-second checklist

Closed surface? → 0 net flux

Open surface? → use dot with the area direction and symmetry

Watch end effects and sign conventions

Beware “isolated pole” language

Revisit the theory short notes: field lines = closed loops, no monopoles, net flux through any closed surface = 0. Once this is burned in, most Gauss-law-in-magnetism MCQs collapse to one-step answers.


QUESTION

A closed surface S is drawn in each situation below. Decide which single claim about the net magnetic flux through

S is correct.

A) S cuts a bar magnet so that only the “north” face lies inside while the “south” face is outside. Therefore the net flux through S is positive.

B) S is threaded by a long straight wire carrying steady current. Since field lines circle the wire and pierce
S , the net flux through S is non-zero.

C) S encloses one circular end-cap inside the bore of a very long solenoid, while the rest of S lies outside where the field is tiny. The net flux is zero.

D) In case (B), if the current is ramped with time, “displacement current” acts like magnetic charge, so the net flux through S becomes non-zero.


Watch This Video Fully :- youtube.com/live/lxUiFbAxDuE?si=uaSVq2UZadJY9e9N

5 days ago | [YT] | 4