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A solid uniform ball having volume $V$ and density $\rho$ floats at the interface of two immiscible liquids as shown in Fig. $8.34$. The densities of the upper and the lower liquids are $\rho_{1}$ and $\rho_{2}$, respectively, such that $\rho_{1}<$ $\rho<\rho_{2}$. The fraction of the volume of the ball in the lower liquid is (A) $\frac{\rho-\rho_{2}}{\rho_{1}-\rho_{2}}$ (B) $\frac{\rho_{1}}{\rho_{1}-\rho_{2}}$ (C) $\frac{\rho_{1}-\rho}{\rho_{1}-\rho_{2}}$ (D) $\frac{\rho_{1}-\rho_{2}}{\rho_{2}}$

Hello. Streams left out our discussion. You are very family here with the form Love Dan. City dead is my US by volume No, for a particle instead. If this then city is constant for a material, then this mask will be directly proportional toe pull you. And if brought a graph mass, worse is volume and this will provide you a straight line. Okay, If you plot a graph between mass and volume, this will provide. You are straight line, but only at Constant, then city. Okay, No support. If you have provided a doctor with the I am eager and masses No, you have to blow the graph between math and the volume. Okay, so first you have to convert this data in for you. I suppose she took 175 diameter. No, you are increasing the side or fume activity one 1.54 2.16 2.54 Okay, this isn't sentiment. This is the diameter of an object. We are increasing gradually now because the size is directly proportional to Mars. So my house is also increased. This is 1.81 decision, Gramps. So you for nine fight 15 point date. Not yet. Born six. 68. Point Doing okay? No. With off by the formal off four upon three My r Q. This is the diameter for stricken. What? This diameter Ready I And then we will find the volume by this formula. Centimeter Cube. So this will comes out. Toby General Point do one. Gino Moin five to 1.89 5.24 And it won 52 Now you have to plot the graph between mass and volume. Okay. Suppose okay. Ah, we take ah must in but access and value in x axis. Okay, so just you done 20 tarty 40 50 and 60. And in XX is one tool three for fine. Six sound. This is the volume in ST Demeter Que and this is the masses. Ingram's okay, so no. With this given data, we can easily brought a craft. Ah, in one point, a general 0.1 volume, we have the mass 1.81 This is I approximately here, and yeah, see, boy, you don't want five. We have 3.95 ground. This relied nearly here. Okay? On day at 1.89 grams, we have 15.8 drum. This will lighten years. Lee here and at 5.24 gram, we have did useful light nearly here on. And actually it won 52 If I used this access this is here comes or could be it. Is this silent on that 8152 We have 68 approximately here. No, if you draw a straight line, this is the straight line. Okay, so the points I do you use this This This This Okay, Now you have to find the slope off this graph. We can easily find the slope of bigger by drinking late a b by BC. So at pointed, the point is, why going to 41 to 38.6 at point B. The point is, Goemon 15 point did, and we'll see the point is 5.24 coma No, 1.89 Goma, 15 point it. Okay. You know what these points we can easily find are the slow that is, that is law a B upon B c. This will be 38.6 minus 15 point it and 5.2 full minus 1.8 night. Okay, so after soldering this, you will get 6.8 gram for sentiment. Acu, this is the slow Well, all right, graph. Okay, well, this is all for me for this video. I hope you will like the video. Thank you.

Okay. See the problem three times for a variance of it. So you have a she with positive charges on it, And then you have, ah, charge to spend it on a string somehow. Ah, uh, I guess this is a good model for, like, pith balls, which I've seen in lab him anyway. And then, you know that this angle is 30 degrees, and then the goal is to get the charge density on the plates goes question mark, I'll go ahead and start it by writing out what's given. So, um, the masses. Um, so it's one milligram, and so I need to convert that to kilograms. And so that's one times 10 to the minus six in the charge is I think it's two ton of Muncie. Let me double check two point. Oh, times 10 to the minus six. Okay, that's also one point. Oh, um, I can write it. Okay, great. So do this. You want to just do a free body diagram on the so you have the weight down, which I'm gonna do no force of gravity. You have attention force. So call that have some tea, and then you have the repulsive electric force was gonna double check the signs of all these? Yes, positive. She and positive charge. And then the charge isn't moving. And so you know about the total for us zero. So, um, so then, yeah, you can say that the forces in either direction balance each other, so I'm gonna label this angle is 60 degrees that this one's 30 because you could make a 30 60 90 triangle. So in the ex direction, that means that the force from the electric field has to be balanced by the horizontal component of the tension forced. So that's f t cose 60. And then for why, you know that the, um, the vertical tension for us has to be balanced by gravity. So you can say f g equals f t times this sign of 60. And, um, basically, you have you two things you don't have are the electric force and the tension. But you can do some algebra to get the, uh, the eliminate tension because the problem doesn't ask about tension, though it is an unknown. And if you do that substitution, then hopefully you got what I got, which is f g over tan 60 equals f e. And then again, I used that they use is Trig. So signed 60 over CO 60 is 10 60. So now we it's time to plug in the gravitational force and the electric force in terms of, uh, the things that were given. So I'll go ahead and do that. And, um, after years, MGs lives mg over tan 60 equals a sigma over to Absalon night. So that's the electric forests. And then our excuse me, the electric field and then forces charged times. Um, Times Field. And so So you got that. And so algebraic Lee isolating sigma. I'll do that on the next page. It's probably not necessary. It feels more complete, though, if I do so it's to Absalon, not mg. I feel like I'm cutting out all the trig and you need help with that. Um then it's probably like, I don't know, this is a bit more basic than than all the trig, but yeah, it feels a lot more complete to write this out this way on DDE. Um, so if I plug in everything So, um JIA JIA's nine pointy If, uh, probably didn't forget that somehow it's a memorable number, I think. Um, yes. When I put that into my calculator, I got five point. Looks like a six 5.0. It was like a lag here. 5.0 times. 10 to the minus nine Newton per pool. Meter squared. I'm just gonna double check the calculation. Okay. Great. It looks good to me.

And this problem. We're told that we we want to analyze a bowling ball that has a, uh, core that is different from the outer. Um um, the outside shell. So the whole boiling Wallace said the way 3.2 kg and have a diameter of 2.21 point six centimeters. We're told that this inner core has a diameter of 19.6 centimeters and it weighs half of the total weight of the bowling ball. So the we can just we can add up the, uh, moments of inertia for the for the interior sphere and the outer shell. So for the interior sphere, we get to fifth M r. Squared M is the mass of the inner the inner sphere, and R is the radius of that half of this diameter. And then we have on for the outer shell. Um, again, we're assuming that it's thin here, and so we have a two thirds m on M two is the same as it has. The same mass is the core. So this is also M two times are three squared where r three is the average diameter here. So we're gonna take the average of this diameter in this diameter, so that will give us diameter from the middle to the middle. And again, we'll take divided by two to get the radius. So plugging in numbers, we get that, um, the inner core has a moment of inertia of 6.15 times 10 to the minus 3 kg per meter squared. And the outer shell has a, uh, moment of inertia of 11.3 times 10 to the minus 3 kg Meters squared. So in total, it has a moment of inertia of 17 point four times, 10 to the minus 3 kg per meter squared. Or I guess this would be 17.4 g grand meters squared. I don't, um, So then I asked us if we just had a uniforms, Um, a uniformed ball that had this total mass so that we didn't have extra mass. Um, kind of extra mass towards the outside ring here on debt was just uniformed throughout what would be the moment of inertia? Well, that's just to fifth times the total mass times the total radius, which is to fifth times 3.2 kg. Um, times there this diameter half of that squared and in the end we plug all that in and we get 14.9 times 10 to the minus three kilograms meters squared, so we can see that, in fact, that we have same mass. But we have a larger moment of inertia when we have this course shell structure, and that's what we want in a bowling ball. We want Thio, have a lot of have a large moment of inertia and a low mass. So again, this helps having this shell and or the shell around the core.

Hi friends here it is given two charged historical words each of radius art with the economy in approach it charge dentistry. Let's go and minus room. The center of the boats are yeah out Plus here by two and uh minus a way to respectively. So the equation of their surfaces are are vector minus A by two vectors say equal to capital up. Yeah. Oh ar minus a fight course of Tito equal to our And are less AY2 or substitute critical to our taking. It will be very small. The distance between the two surfaces. Uh huh. In radial direction add angle cheater. See that this is that minus director. This is fighting. That is the distance between the two, surfacing rated direction at angle theta you will get equals of theater and does not depend on as if you tell him get it is seen from the diagram that. Mhm. Uh huh. Sigma not physical to go into it. Sorry, my notice got to ruin to a and surface dancing effective are faced instability, Sigma not or South Dakota inside and uniformly charged spherical ball. Just a Spiritual one. The electric field is really and has the magnitude according to go stop four by our square into Egypt It's Cartoon 45 x three. Arturo diverted by absolutely not. So he's cut to roll into our upon three. Absolutely not in that presentation. It can be utilized. He back to risk all to rule upon three. Absoluteal not AR -8 x two. Back to minus row three. Absolutely not harbor scripless Hey by two, that's all. Thanks for watching it. Sorry. You can further simplify it. I'm extremely sorry. You can simplify it. A row a upon three. Absoluteal not. Or a role not upon three. Absolutely not came up. That's all. Thanks for.


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