2:27 AM
@Abcd Writing ice today, gotta finish that first for some reason.

1 hour later…
3:37 AM
Hey @Dante , what are the answers of the 3rd question?

2 hours later…
5:21 AM
hello sir :-) @JohnRennie

@user8718165 hi. I'm working I'm afraid. I won't be around much ths morning.

@JohnRennie That's just fine...no problem...:-)

6:05 AM
@user8718165 I'm free now for about an hour if you want to ask a question

@JohnRennie I've got something to ask you if I can catch you before you're busy working as well.
Just let me know.

@kylecampbell ask now. No-one else seems to be around :-)

okie doke
A satellite of mass m moves in a circular orbit around an asteroid of mass M (M>>m). If the asteroid’s mass was suddenly reduced to one-half its former value, what would happen to the satellite? Describe its new orbit.
I am thinking of using the equation $e = \sqrt{1 + \frac{2E{L}^2}{m_{reduced}{\alpha}^2}}$ and figuring it out in terms of energy. But I'm not so sure it's working out.

@kylecampbell The orbital speed for a circular orbit of radius $r$ is $v = \sqrt{GM/r}$. If the mass is suddenly halved then the velocity is too great for a circular orbit by a factor of $\sqrt{2}$. This means the orbiting object is now at the perigee of an elliptical orbit.

So it's now an elliptical orbit?

6:13 AM
Elliptical orbits are surprisingly complicated for what is on the face of it quite a simple orbit. The simplest way to approach this is probably to use the vis viva equation.

Wait a minute... It's not necessarily elliptical
okay

@kylecampbell the only other option is that after halving the mass the orbiting object has a speed greater than the escape velocity. I'd have to calculate the new total energy to check that.

I was thinking it could be hyperbolic or parabolic.
More likely elliptical or parabolic.
How do we know that it's elliptical?

Ah, yes, it's parabolic.

How did you figure that out?

6:19 AM
You just need to calculate the total energy. In the circular orbit $T = GMm/(2r)$ and $V = -GMm/r$ so $E = -GMm/(2r)$. The total energy is negative so it's a bound orbit.
If we halve $M$ without changing the KE then $E = GMm/(2r) - GMm/(2r) = 0$
The total energy is now zero i.e. the orbit is parabolic.

6:37 AM
@AjayMishra Hello, sorry I was writing a test
Ans is B,D

6:54 AM
@JohnRennie Wouldn't $v_{new} = \sqrt{\frac{\frac{1}{2}GM}{r}} = \frac{1}{\sqrt{2}}\sqrt{\frac{GM}{r}}$?

If I understand the question correctly the idea is that half the mass of the asteroid is suddenly somehow removed without changing the speed of the orbiting body. So just before half the mass is removed the orbital velocity is $v = \sqrt{GM/r}$ and immediately afterwards this speed remains unchanged.
The KE of the orbiting body remains constant while the PE is suddenly halved.

oh I see
right, yeah

@JohnRennie good morning. Are you free

@Scáthach briefly. I need to go in about 15 minutes, then I'll be gone for two-three hours. But if you ask now I'll answer if there's time.

7:10 AM
@JohnRennie OK. Just 1 question

@Scáthach yes?

Q7
Do I add the self induxtion

I'm not sure. I don't think I've ever done a problem with the solenoid core only partially filled ...

Ohkay. See u later then

I wonder if the book has solutions as well.

7:20 AM
looks like it's 7.87 henries

7:39 AM
@Dante , It is inconsistent.
Statement B is general statement, i.e. Whether the charge is specified on the specific capacitor, the statement implies that the potential cannot be determined.
but the second statement says that it can be determined, in the specific condition.
One can reach to a clear relation between $V_o$ and $V_B$ using nodal analysis.
@kylecampbell how did you reach to that? Did you treated the solenoid as a combination of two?
@Scáthach , Is C the answer?

@AjayMishra I'm noob in capacitors. Could you tell the procedure to find out potentials using charge on C2?

8:00 AM
In the middle segment, the net charge is zero, right?

Yes
Sorry (was afk)

Assuming, this is segment of a network in static state, there wont be any potential difference across the resistor.
Take the potential of the middle segment as V

ok,

From symmetry, charge on capacitor $C_1$ and $C_3$ would cancel out?

You mean, it will be same?

8:06 AM
In magnitude.
And opposite in sign.
Is that not clear?

This was where I had got confused, how can we tell they're same (in magnitude)?

Because It is given that capacitance is same, and we know that the magnitude of potential difference across them is same, and It is known that only on these things magnitude of charge depends on.

How can we say potential across them is same?

Tell me the potential difference across $C_1$ , Take potential at O to be V

V+10

8:16 AM
And Across C3?

10-V
I'm telling the magnitudes

Wow! I can only wonder how did I figure out that. Still no problem. Do you agree that $C(V+ 10) + C(V-10) + C(V-V_B ) + C(V-0) = 0$

yes

Then $4CV = CV_B$?

8:31 AM
Yes

Clearly, $V$ and $V_B$ are related by an expression.
Can you find out the potential, if the $Q_2$ is given?

Yeah, right!
Thanks a lot!

Dont you think the B is negation of D?
Not on same level, but a level more general than the level of D?

Charge is not specified in the question. D gives us a special case,
So I don't think it's the negation.

Okay.

8:49 AM
@AjayMishra yes. C is the answer

@AjayMishra I'm just curious, how are you kind enough to clear our doubts while most of the smart JEE aspirants like you busy preparing vigorously for advanced not willing to spare a second?

9:06 AM
That was deep sarcasm, I guess.
@Scáthach , Just treat them as combination of two solenoid/inductors.
And law of combination of inductor is same as that resistors.

@AjayMishra What? o.0 Seriously? No dude....
Nvm if you found anything wrong in it (which I still can't, after reading it thrice), sorry.

I am not preparing for advanced, My chemistry is just terrible, I dont think I can even score 10-20 in JEE advanced.

Hmm okay. That gives me the answer

in chemistry

Oh, I didn't know that.

9:12 AM
@Scáthach , If that does not sound reasonable you can break any solenoid into parts, and verify that that is right by combining them.

Has assumed it must be good, since you're getting a good score in main.

Leave it.

1 hour later…
10:37 AM
@JohnRennie Hi .Are you free

@Scáthach hi, yes I'm around for a couple of hours or so.

Q14
Shouldn't the capacitance be zero here?

This exact question was there in MyPat Bitsat yesterday :-0

@tatan My pat bitsat? do u have to pay for the mock tests

Not if you already have an account

10:47 AM
@Scáthach I just sketched a phasor diagram and we get $R = 10$ and $\omega L = 10$ so just with those two the phase angle is 45°.
So yes, I get a power factor of $1/\sqrt{2}$ without any capacitor.

@JohnRennie Hw did u sketch

But if you add a capacitor with $1/\omega C = 20$ you also get a phase angle of 45°, so I'd guess the answer is (3).
@Scáthach that's without the capacitor. With the 500uF capacitor I get:

@JohnRennie it's correct but i didn't get how you drew
@JohnRennie Okay but the angle comes 135?

@Scáthach there
The $\omega L$ and $1/\omega C$ point in opposite directions.
With a 500uF capacitor I get $1/\omega C = 20$ downwards, so when you add the $\omega L = 10$ upwards you get a resultant of $Z = 10$ downwards.

@JohnRennie Got it

10:59 AM
So that gives a phase angle of -45°
$C=0$ is a valid solution, but isn't one of the options.

11:10 AM
@JohnRennie Hello

@Scáthach hi

What is balancing the reaction force of hinge along the rod
I don't see any force

@Scáthach There are two forces on the hinge. If the rod is rotating with angular velocity $\omega$ then there is a centripetal force that the hinge supplies. With the rod vertical this force is straight down so this contributes to $F_y$.
You can calculate $\omega$ using conservation of energy.

okay.That explains it .Let me try
@JohnRennie conservation of energy about com of rod?
or the hinge

Yes. The COM of the rod falls a distance $L/2$ so the PE change is $MgL/2$
And the KE is $\tfrac{1}{2}I\omega^2$

11:19 AM
@JohnRennie I guess I is about hinge?

Yes, $I = \tfrac{1}{3}ML^2$

@JohnRennie How to calculate Fy

$F_y$ is equal to the weight of the rod, $Mg$, plus the centripetal force.
I have to go now. Possibly back later or failing that tomorrow morning.

12:09 PM
@JohnRennie Hi, could you explain me what pressure energy actually means in hydrodynamics?

12:53 PM
17

While deriving Bernoulli's Theorem, our teacher said that the sum of KE, PE and Pressure Energy per unit volume remains constant at any two points. $$P + \rho g h + \frac{\rho v^2}{2} = \text{Constant}$$ In this, he stated that the first term is Pressure Energy per unit Volume. What exactly is ...

Thanks

3 hours later…
3:38 PM
@Dante hi

3:55 PM
@tatan Hi

@Dante Can you help with a HCV problem?

Yeah, I can try.

Check Q.no. 49 pg 135

part?

If the rod is not massless and has a mass "m" . Then how to solve it?
Part 1
Work Energy chapter

4:02 PM
@tatan

yes
you got it the question?

It just completes vertical circle, that means at highest point it's velocity tends to zero.

yes

Kinetic energy at lowest point is completely converted to potential energy at highest point.

Yes...

4:03 PM
$0.5mv^2=mg(2l)$

Yes thats the original problem

But I have changed it a bit... let the rod has a mass "m" and is not massless
@Dante I have changed the situation a bit

Then consider change in potential energy of the COM of rod also.

4:05 PM
Yeah, You've to consider that also, along with KE of the block.

So can you write the equation... i want to match with what I got

$0.5(mv^2+\dfrac{Iv^2}{l^2})=2mgl+Mgl$

ok... matched

where $I=\dfrac{Ml^2}{3}$
Okay

yep

4:19 PM
@Dante or anyone else available

A?

yes
how?

If loops was complete, net force was zero.

oh.. thanks

Understood?

4:25 PM
yes

Okay

4:37 PM
@Dante 1 more
All I know is $$\Delta T(t) = \Delta T_0 e^{-t/A}$$
But when I plug in the values the thing is becoming unsolvable ;-(

A ?
@tatan

yes

Use approximated simplified form of newton's cooling law.
Second formula

I just can't understand where to use approximation where not to
;-(

Look at the data given given to you.

4:45 PM
Apr 5 at 6:36, by tatan
A body cools from 67◦C to 37◦C. If this takes time t when the surrounding temperature is 27◦C,
what will be the time taken if the surrounding temperature is 7◦C?
This one got solved without approximation

I think using approximation is always better.
When they've given initial and final temperatures, examiner wants us to use that formula.
@AdvilSell The above formula is an approximation only right?

@Dante yeah