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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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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 – Rohith M.Athreya · 1 week ago

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– Aryan Goyat · 6 days, 21 hours ago

iska to bta doLog in to reply

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 – Rohith M.Athreya · 1 day, 21 hours ago

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– Rohith M.Athreya · 1 day, 21 hours ago

an APHO type problemLog in to reply

take ratio of masses of uf6 – Rohith M.Athreya · 5 days, 21 hours ago

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– Rohith M.Athreya · 6 days, 21 hours ago

you are off by 53 stagesLog in to reply

– Aryan Goyat · 5 days, 18 hours ago

yep now i am getting 163 .(really what a mess i did)Log in to reply

– Rohith M.Athreya · 5 days, 18 hours ago

:) I realised Ur mistake because I did the same mistake first :pLog in to reply

– Aryan Goyat · 1 week ago

110 stages ??Log in to reply

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simple trick – Rohith M.Athreya · 2 days, 18 hours ago

Another such seemingly simple problemLog in to reply

– Aryan Goyat · 2 days, 17 hours ago

Nice problem it is .Log in to reply

– Shubham Dhull · 1 week ago

but it isn't INPHO level, is it ?Log in to reply

– Aryan Goyat · 1 day, 18 hours ago

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.Log in to reply

– Sumanth R Hegde · 1 week ago

Isn't it \(\frac{2xg}{H}\) ?Log in to reply

– Rajdeep Dhingra · 1 week ago

No.Log in to reply

– Aryan Goyat · 1 week ago

is it 4xg/H ?Log in to reply

– Rohith M.Athreya · 1 week ago

yes! it must be 4xg/h because when u push it in, liquid moves up so that volume of wood submerged is moreLog in to reply

– Rajdeep Dhingra · 1 week ago

CorrectLog in to reply

– Archit Agrawal · 1 week ago

I am also getting 2xg/H.Log in to reply

– Rajdeep Dhingra · 1 week ago

WrongLog in to reply

BTW,u could try my question simplified complications for practice

the numbers sure are weird but share the expressions u get for the radius – Rohith M.Athreya · 5 days, 3 hours ago

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– Shubham Dhull · 5 days ago

i dunno unimodulr constnt would you tell bout it ?Log in to reply

for example, 1m is a unimodular constant – Rohith M.Athreya · 5 days ago

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– Shubham Dhull · 4 days, 23 hours ago

lol , i did't knew it :PLog in to reply

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– Shubham Dhull · 5 days, 23 hours ago

isn't anybody participating ?Log in to reply

– Shubham Dhull · 1 day, 20 hours ago

nobody dares to do my problem ? :PLog in to reply

that connecting moves through the wedge?? – Rohith M.Athreya · 1 day, 20 hours ago

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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 ? – Aryan Goyat · 5 days, 17 hours ago

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– Shubham Dhull · 5 days, 17 hours ago

no, displacing both about euil. , you can do that too when only the 10 kg one is moved .Log in to reply

both down

or one up other down – Aryan Goyat · 5 days, 17 hours ago

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– Shubham Dhull · 5 days, 17 hours ago

i didn't get what u r saying .Log in to reply

– Shubham Dhull · 1 week ago

try with spring too ..:)Log in to reply

– Aryan Goyat · 5 days, 17 hours ago

the bottom one depicts equilibrium positionLog in to reply

– Shubham Dhull · 5 days, 17 hours ago

erh...then......what about it ? what you have drawn is common sense :P , no need to draw , i didn't just understand ur language , i want you to prove it does / or not s.h.m by condition of shm i.e f =kx and find time period . hint : since it is string, it'll apply force only when extended . ;)Log in to reply

– Shubham Dhull · 5 days, 17 hours ago

they r both dragged down by same force || to the /\ planes ;)Log in to reply

– Aryan Goyat · 5 days, 17 hours ago

yahi to puch raha tha dhanyvad :pLog in to reply

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– Shubham Dhull · 1 week ago

the wedge is fixed .Log in to reply

isn't anybody participating ? – Shubham Dhull · 1 week, 1 day ago

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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 – Shubham Dhull · 1 week, 1 day ago

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(wE1) + (M1 I)(w M1) ? – Rajdeep Dhingra · 1 week, 1 day agoLog in to reply

– Aryan Goyat · 1 week ago

i think that must be so .pls post the next q.Log in to reply

– Spandan Senapati · 1 week, 1 day ago

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

@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 ........... – Aryan Goyat · 1 week, 1 day ago

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@Archit Agrawal @Aniket Sanghi @Prakhar Bindal @Rajdeep Dhingra @Harsh Shrivastava @Swapnil Das @Ankit Dhindhwal @puneet mangla and ...... – Aryan Goyat · 1 week, 1 day ago

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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) – Aryan Goyat · 1 week, 1 day ago

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k2√2h/g.as v in the projected plane is constant.and the constant 2with √2h/g for the cyclic process – Spandan Senapati · 1 week, 1 day agoLog in to reply

– Archit Agrawal · 1 week, 1 day ago

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

– Shubham Dhull · 1 week, 1 day ago

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

– Shubham Dhull · 1 week, 1 day ago

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 ?Log in to reply

– Spandan Senapati · 1 week, 1 day ago

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

– Spandan Senapati · 1 week, 1 day ago

Krotov question I guess??Log in to reply

– Archit Agrawal · 1 week, 1 day ago

YesLog in to reply