@JohnRennie assuming the spiral in the question to be a disc of inner radius a and outer radius b won't change anything, right? It gets the right answer but I was wondering if there would be any issues in practical scenario

@Swan Hi :-)

Hey :)

@Swan Yes, you just treat the spiral as a series of infinitesimal loops.

In real life how good an approximation this is will depend on the thickness of the wire, but it would probably be pretty good.

Thanks!! But if we do take a disc instead of such a spiral in real life then I'm thinking taking a disc in real life might reduce the resistance considerably and thus there would be more current in the disc, am I thinking it wrong?

Am I even making sense sir @JohnRennie?

I guess a solid disk of the same thickness of the wire would have more copper in it because the gaps between the wires would all be filled. To be a fair comparison you'd need to reduce the thickness of the disk so both the disk and the spiral had the same weight of copper.

And you'd need some way to induce a circular current in the disk.

Practically it would actually be quite difficult to compare the two.

Got it!

>And you'd need some way to induce a circular current in the disk.

Is there any known way to do it?

Like something simple enough that's roughly within the bounds of my syllabus

Yes, I think a magnetic field field varying in the appropriate way could induce circular currents.

I'd have to sit down and think about exactly how to do this but I think it's possible.

Thanks again ☺️!

You're welcome :-)

@JohnRennie hello sir

Hi :-)

Are you free right now?

Yes :-)

Is there tranfer of heat between state like if there is water at 0 degree Celsius and ice at 0 degree Celsius will the heat flow from water to ice?

Are you thinking about ice melting to water at 0°? Or the reverse i.e. water freezing to ice?

I was thinking ice melting to water

We normally consider an idealised reversible process where the two phases are always in equilibrium. That means both the water and the ice have to be exactly at the melting point, and the transfer of heat from the water to the ice is infinitely slow.

In real life the water would have to be slightly above the melting point so there is a temperature difference between the water and the ice. Then heat flows from the water to the ice.

Is this what you are asking about?

how can there be transfer of heat if there is no temperature difference that is what i was actually asking

It's because we are considering a limit not a realistic system.

Does this make sense so far?

Yes

But the non-zero ΔT means the ice and water are not in equilibrium and for a reversible process we need the system to be in equilibrium.

So what we do is consider the limit of ΔT ⟶ 0 in which case dQ/dt ⟶ 0

This isn't physically realistic as it would take in infinite time for the ice to melt.

But we use the limit just for the calculation.

In practice we just need ΔT to be small - it doesn't really have to be zero.

Does this help?

This means that if there was no temperature difference the heat will not flow irrespective of state of object

Right?

Yes

Ok thank you sir that helped :)

OK :-)

One more question if a system of ice and water was to be equilibrium then this would mean both of them will be at 0 degree Celsius right?

Yes

Ok thank you I'll ask again if I face any problems

@JohnRennie Hello sir

@RonaldBecker Hi :-)

toppr.com/ask/question/…

@JohnRennie This question

How far did you get with this?

I got temperature 55 degree Celsius :(

In liquid state

I was subjective question in my book

How do you want to do this? Do you want to go through your calculation?

Or do you want me to show how I would do it?

Should I tell what I tried and you can point out the mistake ?

OK, go ahead ...

We know heat lost by steam is equal to heat gained by ice

Yes, so you start by calculating the heat lost by the steam. Yes?

So heat lost by steam while turning into water is 27000 Cal as done in solution + heat lost in lowering in temperature to T is 50*1*(100-T)

Yes

Which is equal to heat gained by ice that is 36000 Cal + 450*1*(T-0)

Remember that the ice starts at 253K i.e. -20°C

You need to include the heat needed to heat the ice from 253K to 273K before it can start melting.

Yes I forgot to add that

Let me do the calculation again

OK :-)

Ah now T is Comming -ve which is not possible as it cannot drop below 0 degree Celsius when I'm doing this calculation

T -ve would mean Ice never melted :(

@JohnRennie Does this mean they will be in water ice equilibrium?

@RonaldBecker What you've found is that condensing the steam then cooling the water to 0°C does not produce enough heat to melt all the ice. Yes?

Yes

So you are indeed going to end up with ice and water in equilibrium at 273K.

With problems like this it's hard to know in advance what the final state will be i.e. whether condensing then cooling the steam will produce enough heat to melt all the ice or not.

You just have to try the calculation to see what happens.

And not miss heat required to bring ice at 273 K while calculating lol

@JohnRennie Thank you sir

You're welcome :-)

imgur.com/a/SaAwMXk | In this question, I can't do the third subpart. It says to find the variation of the refractive index required wrt time, but I think to get central maxima at O we only need one with a specific constant refractive index.

I have attached my work, and as all this variation with time in the refractive index of the medium is doing is linearly contracting the fringes, we need to get central maxima at O, which will not be affected by contractions.

specifically, $\mu_p = \frac{d}{p} sin\phi + 1$

The book's answer is $\mu_p = \mu_o + kt + \frac{d}{p}sin\phi$

@JohnRennie ?

Hi sir please look to this when you're free.

Q: A solid sphere is in rotation about its polar axis ( z- axis) with angular accn= k(angular displacement) where when angular displacement=0, angular velocity= 0 at t=0 and motion was started by giving a very small push to the periphery of the sphere. A point p on the sphere is at 37 degree latitude from the center, what is the unit vector in the direction if acceleration of point P at the instant 'P' completes its first rotation?

I tried : P's radius for its circular motion in x-y plane is 4R/5. angular velocity * d/d(ang displacement) of angular velocity = k * angular displacement, so I integrated this on the RHL from 0 to 2pi and on the LHL from 0 to final angular velocity, I got Angular velocity final = (√k) * 2π

So acceleration vector = centripetal acceleration (I) + tangential acceleration (j) and substituted. The answer though has the tangential acceleration in the negative direction I.e out of the page, how is the direction realised ? The question doesnt mention if its clockwise or anticlockwise....?

To be precise the answer I get is (2π (î) + j)/√(4π²+1). The answer given is: (2π (î) - j)/ √(4π²+1).