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We don't have even 15 days for preparing .Hence i think we should prefer just discussing only few important topics via questions.

----Note

---1. Pls ensure that ques deals more with physics and less with maths(though i love such questions)

---2. Before posting Ensure that ques is really of INPHO level.

---3.Ques which involve dealing with major Topics are Welcomed.

---4.The one who solves any particular q needs to post the next q within 3 hrs.

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## Comments

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TopNewesthere is another question!!

Natural uranium consists of mainly two isotopes, \(U^{238}\) and \(U^{235}\),whereas the relative concentration of the latter is 0.7%.Uranium is "enriched"(ie., concentration of \(U^{235}\) is increased) by implementing a multi-stage process,where at each stage,evaporated chemical compound \(UF_{6}\) is led through a porous wall.the porous wall can be considered to be a thin film having microscopic holes in it.How many stages are nedded to double the concentration

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110 stages ??

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iska to bta do

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you are off by 53 stages

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u are wrong coz u probably took ratio of masses as 238/235

take ratio of masses of uf6

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Take a look at the thin rod(of length \(a\)) above.

The mass density along the rod varies as \(\sigma=\sigma_{0}(1+\frac{x^{n}}{a^{n}})\) where \(n\) is a positive integer.

While calculating the co-ordinates of the centre of mass of the rod, we assume(wrongly) the COM to be at the geometric centre ie., \((\frac{a}{2},0)\)

Now, we know that's wrong and find the actual co-ordinate.

We note that the actual \(x\) co-ordinate is \(\frac{a}{2}+\delta X\) where \(X\) is the actual co-ordinate of the COM.

Define \(100\frac{\delta X}{X}\) as percentage error

Approximately what maximum percentage error will we have made in assigning the COM to rods from \(n=1\) to \(n=20\) ie.,\(\displaystyle \frac{100}{20} \sum_{n=1}^{n=20} \frac{\delta X_{n}}{X_{n}}\) where \(X_{n}\) is the COM when \(n\) is the exponent of \(x\) in the surface density expression

Details and Assumptionsthe \(21^{st}\) harmonic number \(H_{21}\) = 3.645

\(1/23 + 1/24 + 1/25 = 0.12514\)

give your answer to the nearest integer

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Isn't it \(\frac{2xg}{H}\) ?

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I am also getting 2xg/H.

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is it 4xg/H ?

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No.

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but it isn't INPHO level, is it ?

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No it is as it is not that much tricky Though by the name it may suggest it uses a trick but actually it is a widely used technique which may be asked in inpho.

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Another such seemingly simple problem simple trick

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Nice problem it is .

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Hey try this. Asphere of radius r is projected up an inclined plane of @inclination with initial speed u and omega In a direction in which it will roll up.the coefficient of friction is tan@/7.and v>omega*r.find the total time of motion of the sphere up the plane before it stops.as this is quite a old q I don't have any source to match my ans.post yours so that we can check.....

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(123vi-18wr)/(147gsin@)

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how did u solve it??

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so how to achieve 0.4mgsin@??

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I got my ans as (17v+4wr)/18gsin@.

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And this is a typical inpho q.not many will notice the fact that friction reverses its direction.I hope you liked it.thanx.

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OK guys here is my first question

1--- A small ball moves at a constant velocity v along a horizontal surface and at a point A falls into a vertical cylindrical well of depth H and radius r.The velocity v of the ball forms an angle alpha with the diameter of the well drawn through point A. Determine relation b/w v,r,H and alpha so that ball gets out of well after elastic impact.(take all collision to be elastic +friction=0)

(u don't need to solve any ques to post your ques if you have one please post it and share this note with as much people as you can)

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OK I have solved this.we can take the projection of the motion on a horizontal plane.the problem then reduces that the velocities get reflected along a normal to a point at which the ball stikes in the plane.so we can conclude mathematically that in a circle I take a line with @ angle to the horizontal and every time it strikes the circle it gets reflected.then the length of that segment is 2rcos@ and of I keep on doing so at some time I must return to the original point.the total length is then 2nrcos@.and this must be v

k2√2h/g.as v in the projected plane is constant.and the constant 2with √2h/g for the cyclic processLog in to reply

Krotov question I guess??

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Yes

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can you attach a a pic ?because you say the velocity of ball is parallel to hori. and the well is vertical so in this case alpha is 0 , right ? by the way in this case i get answer as r=v*root(2h/g) is it right ?if alpha=0 ?

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It means that the balls velocity was initially in a horizontal plane with an angle alpha to the actual horizontal line.

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nrcos(alpha)=kv√(2H/g) where n and k belong to integers.

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yes , that's what i get if alpha is 0, so i guess u r right :)

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@Archit Agrawal @Aniket Sanghi @Prakhar Bindal @Rajdeep Dhingra @Harsh Shrivastava @Swapnil Das @Ankit Dhindhwal @puneet mangla and ......

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@shubham dhull @Spandan Senapati @Gauri shankar Mishra @Akarsh Singh @Samarth Agarwal @CH Nikhil @neelesh vij @Nihar Mahajan @Abhineet Nayyar @Abhisek Mohanty @avi solanki @Aditya Chauhan @Kishore S Shenoy @Somyaneel Sinha @Jyotisman Das @Svatejas Shivakumar @Ayush Agarwal @Sarthak Singhal and ...........

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Let L1 be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) L1 is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon’s orbit is circular and the Moon can be taken as a point. iii) The Earth’s axis of rotation and the Moon’s axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of E I , the moment of inertia of the Earth; ωE1 , the present angular frequency of the Earth’s rotation; M1 I , the present moment of inertia of the Moon with respect to the Earth´s axis; and ω M1 , the present angular frequency of the Moon’s orbit

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Isn't it L1 = (E I)

(wE1) + (M1 I)(w M1) ?Log in to reply

Ya that's quite apparent.you can access the full prob.it gets more interesting after.ipho 2009/10(most prob)

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i think that must be so .pls post the next q.

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isn't anybody participating ?

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the line shown in fig. is string not spring ( understood what i mean ?) k is given , mass is given , find whether they do s.h.m or not, if yes find the time period . if not , state under what condition can they do so . they refer to he circles made which in the figure represent cylinders assume pure rolling ( you may argue about the decreasing normal reaction, assume infinite c. of friction . the angles in figure are 30,30,120 which you can assign by just looking at the figure :) this is original .

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the wedge is fixed .

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try with spring too ..:)

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the bottom one depicts equilibrium position

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isn't anybody participating ?

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if u could add what kind of displacement are u giving

are you shifting it kepping string inclined at same angle as that of equillibrium position

or are you displacing only one mass kepping the other non disturbed

or ?

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both down

or one up other down

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nobody dares to do my problem ? :P

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that connecting moves through the wedge??

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Ok finally its over. Pls post the QUES of inpho and provide link here

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quit pho , cbse is more imporant , lol !

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We will add it on 31st. How was your paper ? Solved Q5 and Q6 ? I was getting -ve refractive index for that X-ray

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can you uplod the questions ?

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yep did q5 and q6 a,b part

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what value u got N and \(\omega_{0}\)

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https://brilliant.org/profile/aryan-2xcg8y/sets/that-is-why-i-love-physics/462021/problem/inpho/

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N kya hai bhai :(

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I got -ve value for refractive index

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what value u got for N??

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omega = 4.57 * (10^15)

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i think i wont make it to OCSC

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pls mention here the marks u know in inpho here are some expected marks which i know 46 37 [44,49] [42,51] 60 65 pls also do mention nos u know

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i am getting around 48-50

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What are your INPhO marks ?

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43 :(

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u?

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