Open to all !

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.

## 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 – Rohith M.Athreya · 2 months, 1 week ago

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– Aryan Goyat · 2 months, 1 week 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 · 2 months ago

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– Shubham Dhull · 1 month, 3 weeks ago

7Log in to reply

– Rohith M.Athreya · 2 months ago

an APHO type problemLog in to reply

take ratio of masses of uf6 – Rohith M.Athreya · 2 months, 1 week ago

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– Rohith M.Athreya · 2 months, 1 week ago

you are off by 53 stagesLog in to reply

– Aryan Goyat · 2 months, 1 week ago

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

– Rohith M.Athreya · 2 months, 1 week ago

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

– Aryan Goyat · 2 months, 1 week ago

110 stages ??Log in to reply

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..... – Spandan Senapati · 1 month, 4 weeks ago

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– Aryan Goyat · 1 month, 4 weeks ago

(123vi-18wr)/(147gsin@)Log in to reply

– Spandan Senapati · 1 month, 4 weeks ago

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

– Aryan Goyat · 1 month, 4 weeks ago

oh yech, it was a good one.Log in to reply

– Spandan Senapati · 1 month, 4 weeks ago

You guys want to try another one.fro inpho 1999.if you want I am gonna post it.Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

ohkLog in to reply

– Spandan Senapati · 1 month, 4 weeks ago

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

– Aryan Goyat · 1 month, 4 weeks ago

in which class are u?Log in to reply

– Spandan Senapati · 1 month, 4 weeks ago

Eleven.you???Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

12Log in to reply

– Spandan Senapati · 1 month, 4 weeks ago

Are you appearing for INAO tomorrow.Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

NoLog in to reply

– Aryan Goyat · 1 month, 4 weeks ago

bhai niche itte mahabharat likhe hai vo bhi padh le.Log in to reply

– Shubham Dhull · 1 month, 4 weeks ago

see my chem ques , sorry for advertising here :P https://brilliant.org/problems/can-you-get-the-chocolate/Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

vaise tu kaun se chocolate dega :pLog in to reply

– Shubham Dhull · 1 month, 4 weeks ago

tujhe follow kar lunga :PLog in to reply

– Aryan Goyat · 1 month, 4 weeks ago

ohkLog in to reply

– Spandan Senapati · 1 month, 4 weeks ago

I had not opened my tab since yesterday.so I was unaware..looks like someone computed something wrong.Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

ok bhai ye ans sahi hai tera i have verified that :PLog in to reply

– Shubham Dhull · 1 month, 4 weeks ago

lol, mahabharatLog in to reply

– Rohith M.Athreya · 1 month, 4 weeks ago

how did u solve it??Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

First find out time t1 time taken for pure rolling kepping f=uN Then find t2 time to stop after starting of pure rolling Here f=2mgsin(@)/5 Since here not maximum f acts but that value of f acts for which it is pure rolling which comes to be as above. Add t1 +t2Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

Oh shit i didnt noticed that😂Log in to reply

– Rohith M.Athreya · 1 month, 4 weeks ago

it cannot pure rollLog in to reply

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– Rohith M.Athreya · 1 month, 3 weeks ago

yeah thanks i got the same as well :)Log in to reply

– Shubham Dhull · 1 month, 4 weeks ago

my answer is same as yours ;) i just took the direction wrong at first when i asked you that but realized my mistake later , so i'm getting now the same answer , lol i was just making links for my pics :PLog in to reply

– Aryan Goyat · 1 month, 4 weeks ago

OkLog in to reply

– Shubham Dhull · 1 month, 4 weeks ago

wait a min. i'm gonna upload too :)Log in to reply

– Rohith M.Athreya · 1 month, 4 weeks ago

yes that is why i am stuckLog in to reply

so how to achieve 0.4mgsin@?? – Rohith M.Athreya · 1 month, 4 weeks ago

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– Aryan Goyat · 1 month, 4 weeks ago

Getting (17v+4wr)/18gsin@Log in to reply

– Aryan Goyat · 1 month, 4 weeks ago

No problem calculate t2 with f =umgsin@ only because and off course up the plane since we know for always pure rolling f must me 0.4mgsin@ and if f factor decreases then v of lowest point due to both translational and rotational become more effective in down the incline so f only acts up at its max and hence we get t2.Log in to reply

– Shubham Dhull · 1 month, 4 weeks ago

see my comment too...Log in to reply

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– Aryan Goyat · 1 month, 4 weeks ago

Ae bhai its initially not pure rolling but after some time f and g component make him do so.Log in to reply

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simple trick – Rohith M.Athreya · 2 months ago

Another such seemingly simple problemLog in to reply

– Aryan Goyat · 2 months ago

Nice problem it is .Log in to reply

– Shubham Dhull · 2 months, 1 week ago

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

– Aryan Goyat · 2 months 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 · 2 months, 1 week ago

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

– Rajdeep Dhingra · 2 months, 1 week ago

No.Log in to reply

– Aryan Goyat · 2 months, 1 week ago

is it 4xg/H ?Log in to reply

– Rohith M.Athreya · 2 months, 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 · 2 months, 1 week ago

CorrectLog in to reply

– Archit Agrawal · 2 months, 1 week ago

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

– Rajdeep Dhingra · 2 months, 1 week ago

WrongLog in to reply

What are your INPhO marks ? – Rajdeep Dhingra · 1 month, 1 week ago

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– Rohith M.Athreya · 1 month, 1 week ago

43 :(Log in to reply

– Rohith M.Athreya · 1 month, 1 week ago

u?Log in to reply

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 – Aryan Goyat · 1 month, 3 weeks ago

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– Rohith M.Athreya · 1 month, 3 weeks ago

i am getting around 48-50Log in to reply

Ok finally its over. Pls post the QUES of inpho and provide link here – Aryan Goyat · 1 month, 3 weeks ago

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– Rohith M.Athreya · 1 month, 3 weeks ago

what value u got N and \(\omega_{0}\)Log in to reply

– Rajdeep Dhingra · 1 month, 3 weeks ago

omega = 4.57 * (10^15)Log in to reply

i think i wont make it to OCSC – Rohith M.Athreya · 1 month, 3 weeks ago

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– Rajdeep Dhingra · 1 month, 3 weeks ago

I got -ve value for refractive indexLog in to reply

– Aryan Goyat · 1 month, 3 weeks ago

vo q nahi padha time up ho gyaLog in to reply

what value u got for N?? – Rohith M.Athreya · 1 month, 3 weeks ago

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– Aryan Goyat · 1 month, 3 weeks ago

https://brilliant.org/profile/aryan-2xcg8y/sets/that-is-why-i-love-physics/462021/problem/inpho/ ye w sahi hai kyaLog in to reply

– Shubham Dhull · 1 month, 3 weeks ago

ques not clr !Log in to reply

– Shubham Dhull · 1 month, 3 weeks ago

about which point w is given ?Log in to reply

– Aryan Goyat · 1 month, 3 weeks ago

nahi diya u thinkLog in to reply

– Shubham Dhull · 1 month, 3 weeks ago

atleast give the direction of BLog in to reply

– Aryan Goyat · 1 month, 3 weeks ago

take it into the planeLog in to reply

– Shubham Dhull · 1 month, 3 weeks ago

i was on verge of solving by taking it out of plane lol :PLog in to reply

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– Aryan Goyat · 1 month, 3 weeks ago

heavy rod ko itna hi likha tha:PLog in to reply

– Aryan Goyat · 1 month, 3 weeks ago

N kya hai bhai :(Log in to reply

– Rohith M.Athreya · 1 month, 3 weeks ago

ohh! anybody did??Log in to reply

– Aryan Goyat · 1 month, 3 weeks ago

N kya hai bhaiLog in to reply

– Aryan Goyat · 1 month, 3 weeks ago

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

– Rajdeep Dhingra · 1 month, 3 weeks ago

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

– Aryan Goyat · 1 month, 3 weeks ago

yep did q5 and q6 a,b partLog in to reply

– Rajdeep Dhingra · 1 month, 3 weeks ago

Can you write a brief solution to Q5 ?Log in to reply

– Shubham Dhull · 1 month, 3 weeks ago

can you uplod the questions ?Log in to reply

– Rajdeep Dhingra · 1 month, 3 weeks ago

They away the QP from us. I will try to post the questions by evening.Log in to reply

– Aryan Goyat · 1 month, 3 weeks ago

from us tooLog in to reply

– Shubham Dhull · 1 month, 3 weeks ago

oh , ok ! i will be happy to solve them and write a solution(if i am able to :P)Log in to reply

– Shubham Dhull · 1 month, 3 weeks ago

quit pho , cbse is more imporant , lol !Log in to reply

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– Shubham Dhull · 2 months ago

can we discuss ur problem it has certain ambiguities :| , if the motor was detached , what is the role of fringes , light and stefan's constant ? if not, write it there , also you din't tell the initial temp. as the light source was not playing any role at that time , i think the problem needs slight changes .Log in to reply

its far too convoluted there are far too many ambiguities to bother correcting – Rohith M.Athreya · 2 months ago

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– Shubham Dhull · 2 months ago

see my report thereLog in to reply

– Shubham Dhull · 2 months ago

oh , you deleted it , nice :P it ws relly wrong :)Log in to reply

– Shubham Dhull · 2 months, 1 week ago

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

for example, 1m is a unimodular constant – Rohith M.Athreya · 2 months, 1 week ago

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– Shubham Dhull · 2 months, 1 week ago

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

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– Shubham Dhull · 2 months, 1 week ago

isn't anybody participating ?Log in to reply

– Shubham Dhull · 2 months ago

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

that connecting moves through the wedge?? – Rohith M.Athreya · 2 months 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 · 2 months, 1 week ago

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– Shubham Dhull · 2 months, 1 week 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 · 2 months, 1 week ago

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– Shubham Dhull · 2 months, 1 week ago

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

– Shubham Dhull · 2 months, 1 week ago

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

– Aryan Goyat · 2 months, 1 week ago

the bottom one depicts equilibrium positionLog in to reply

– Shubham Dhull · 2 months, 1 week 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 · 2 months, 1 week ago

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

– Aryan Goyat · 2 months, 1 week ago

yahi to puch raha tha dhanyvad :pLog in to reply

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– Shubham Dhull · 2 months, 1 week ago

the wedge is fixed .Log in to reply

isn't anybody participating ? – Shubham Dhull · 2 months, 1 week 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 · 2 months, 1 week ago

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(wE1) + (M1 I)(w M1) ? – Rajdeep Dhingra · 2 months, 1 week agoLog in to reply

– Aryan Goyat · 2 months, 1 week ago

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

– Spandan Senapati · 2 months, 1 week 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 · 2 months, 1 week ago

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@Archit Agrawal @Aniket Sanghi @Prakhar Bindal @Rajdeep Dhingra @Harsh Shrivastava @Swapnil Das @Ankit Dhindhwal @puneet mangla and ...... – Aryan Goyat · 2 months, 1 week 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 · 2 months, 1 week 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 · 2 months, 1 week agoLog in to reply

– Archit Agrawal · 2 months, 1 week ago

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

– Shubham Dhull · 2 months, 1 week ago

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

– Shubham Dhull · 2 months, 1 week 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 · 2 months, 1 week 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 · 2 months, 1 week ago

Krotov question I guess??Log in to reply

– Archit Agrawal · 2 months, 1 week ago

YesLog in to reply