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INPHO

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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.

Note by Aryan Goyat
11 months ago

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here 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 - 11 months ago

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iska to bta do

Aryan Goyat - 11 months ago

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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 Assumptions

the \(21^{st}\) harmonic number \(H_{21}\) = 3.645

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

give your answer to the nearest integer

Rohith M.Athreya - 10 months, 3 weeks ago

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@Rohith M.Athreya 7

Brilliant Member - 10 months, 2 weeks ago

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@Rohith M.Athreya an APHO type problem

Rohith M.Athreya - 10 months, 3 weeks ago

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

take ratio of masses of uf6

Rohith M.Athreya - 11 months ago

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you are off by 53 stages

Rohith M.Athreya - 11 months ago

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@Rohith M.Athreya yep now i am getting 163 .(really what a mess i did)

Aryan Goyat - 11 months ago

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@Aryan Goyat :) I realised Ur mistake because I did the same mistake first :p

Rohith M.Athreya - 11 months ago

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110 stages ??

Aryan Goyat - 11 months ago

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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.....

Spandan Senapati - 10 months, 2 weeks ago

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

Aryan Goyat - 10 months, 2 weeks ago

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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.

Spandan Senapati - 10 months, 2 weeks ago

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@Spandan Senapati oh yech, it was a good one.

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat You guys want to try another one.fro inpho 1999.if you want I am gonna post it.

Spandan Senapati - 10 months, 2 weeks ago

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@Spandan Senapati ohk

Aryan Goyat - 10 months, 2 weeks ago

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

Spandan Senapati - 10 months, 2 weeks ago

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@Spandan Senapati in which class are u?

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat Eleven.you???

Spandan Senapati - 10 months, 2 weeks ago

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@Spandan Senapati 12

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat Are you appearing for INAO tomorrow.

Spandan Senapati - 10 months, 2 weeks ago

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@Spandan Senapati No

Aryan Goyat - 10 months, 2 weeks ago

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@Spandan Senapati bhai niche itte mahabharat likhe hai vo bhi padh le.

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat see my chem ques , sorry for advertising here :P https://brilliant.org/problems/can-you-get-the-chocolate/

Brilliant Member - 10 months, 2 weeks ago

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@Brilliant Member vaise tu kaun se chocolate dega :p

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat tujhe follow kar lunga :P

Brilliant Member - 10 months, 2 weeks ago

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@Brilliant Member ohk

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat I had not opened my tab since yesterday.so I was unaware..looks like someone computed something wrong.

Spandan Senapati - 10 months, 2 weeks ago

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@Spandan Senapati ok bhai ye ans sahi hai tera i have verified that :P

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat lol, mahabharat

Brilliant Member - 10 months, 2 weeks ago

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how did u solve it??

Rohith M.Athreya - 10 months, 2 weeks ago

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@Rohith M.Athreya 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 +t2

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat Oh shit i didnt noticed that😂

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat it cannot pure roll

Rohith M.Athreya - 10 months, 2 weeks ago

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@Rohith M.Athreya

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat yeah thanks i got the same as well :)

Rohith M.Athreya - 10 months, 2 weeks ago

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@Aryan Goyat 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 :P

Brilliant Member - 10 months, 2 weeks ago

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@Brilliant Member Ok

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat wait a min. i'm gonna upload too :)

Brilliant Member - 10 months, 2 weeks ago

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@Aryan Goyat yes that is why i am stuck

Rohith M.Athreya - 10 months, 2 weeks ago

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@Aryan Goyat max friction is 0.1428mgsin@ right??

so how to achieve 0.4mgsin@??

Rohith M.Athreya - 10 months, 2 weeks ago

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@Rohith M.Athreya Getting (17v+4wr)/18gsin@

Aryan Goyat - 10 months, 2 weeks ago

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@Rohith M.Athreya 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.

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat see my comment too...

Brilliant Member - 10 months, 2 weeks ago

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Comment deleted 10 months ago

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@Brilliant Member Ae bhai its initially not pure rolling but after some time f and g component make him do so.

Aryan Goyat - 10 months, 2 weeks ago

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Rajdeep Dhingra - 11 months ago

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Another such seemingly simple problem simple trick

Rohith M.Athreya - 10 months, 3 weeks ago

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Nice problem it is .

Aryan Goyat - 10 months, 3 weeks ago

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but it isn't INPHO level, is it ?

Brilliant Member - 11 months ago

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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.

Aryan Goyat - 10 months, 3 weeks ago

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

Sumanth R Hegde - 11 months ago

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No.

Rajdeep Dhingra - 11 months ago

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is it 4xg/H ?

Aryan Goyat - 11 months ago

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@Aryan Goyat yes! it must be 4xg/h because when u push it in, liquid moves up so that volume of wood submerged is more

Rohith M.Athreya - 11 months ago

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@Aryan Goyat Correct

Rajdeep Dhingra - 11 months ago

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I am also getting 2xg/H.

Archit Agrawal - 11 months ago

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@Archit Agrawal Wrong

Rajdeep Dhingra - 11 months ago

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What are your INPhO marks ?

Rajdeep Dhingra - 10 months ago

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43 :(

Rohith M.Athreya - 10 months ago

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u?

Rohith M.Athreya - 10 months ago

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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

Aryan Goyat - 10 months, 2 weeks ago

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i am getting around 48-50

Rohith M.Athreya - 10 months, 2 weeks ago

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

Aryan Goyat - 10 months, 2 weeks ago

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

Rohith M.Athreya - 10 months, 2 weeks ago

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omega = 4.57 * (10^15)

Rajdeep Dhingra - 10 months, 2 weeks ago

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@Rajdeep Dhingra oh! i got around \(10^{45}\)

i think i wont make it to OCSC

Rohith M.Athreya - 10 months, 2 weeks ago

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I got -ve value for refractive index

Rajdeep Dhingra - 10 months, 2 weeks ago

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@Rajdeep Dhingra vo q nahi padha time up ho gya

Aryan Goyat - 10 months, 2 weeks ago

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@Rajdeep Dhingra i didnt have time for the last two parts :(

what value u got for N??

Rohith M.Athreya - 10 months, 2 weeks ago

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@Rohith M.Athreya https://brilliant.org/profile/aryan-2xcg8y/sets/that-is-why-i-love-physics/462021/problem/inpho/ ye w sahi hai kya

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat ques not clr !

Brilliant Member - 10 months, 2 weeks ago

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@Brilliant Member about which point w is given ?

Brilliant Member - 10 months, 2 weeks ago

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@Brilliant Member nahi diya u think

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat atleast give the direction of B

Brilliant Member - 10 months, 2 weeks ago

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@Brilliant Member take it into the plane

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat i was on verge of solving by taking it out of plane lol :P

Brilliant Member - 10 months, 2 weeks ago

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Comment deleted 10 months ago

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@Brilliant Member heavy rod ko itna hi likha tha:P

Aryan Goyat - 10 months, 2 weeks ago

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N kya hai bhai :(

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat ohh! anybody did??

Rohith M.Athreya - 10 months, 2 weeks ago

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@Rohith M.Athreya N kya hai bhai

Aryan Goyat - 10 months, 2 weeks ago

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

Aryan Goyat - 10 months, 2 weeks ago

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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

Rajdeep Dhingra - 10 months, 2 weeks ago

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yep did q5 and q6 a,b part

Aryan Goyat - 10 months, 2 weeks ago

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@Aryan Goyat Can you write a brief solution to Q5 ?

Rajdeep Dhingra - 10 months, 2 weeks ago

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can you uplod the questions ?

Brilliant Member - 10 months, 2 weeks ago

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@Brilliant Member They away the QP from us. I will try to post the questions by evening.

Rajdeep Dhingra - 10 months, 2 weeks ago

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@Rajdeep Dhingra from us too

Aryan Goyat - 10 months, 2 weeks ago

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@Rajdeep Dhingra oh , ok ! i will be happy to solve them and write a solution(if i am able to :P)

Brilliant Member - 10 months, 2 weeks ago

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quit pho , cbse is more imporant , lol !

Brilliant Member - 10 months, 2 weeks ago

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Comment deleted 10 months ago

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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 .

Brilliant Member - 10 months, 3 weeks ago

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no leave it

its far too convoluted there are far too many ambiguities to bother correcting

Rohith M.Athreya - 10 months, 3 weeks ago

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@Rohith M.Athreya see my report there

Brilliant Member - 10 months, 3 weeks ago

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@Brilliant Member oh , you deleted it , nice :P it ws relly wrong :)

Brilliant Member - 10 months, 3 weeks ago

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i dunno unimodulr constnt would you tell bout it ?

Brilliant Member - 10 months, 4 weeks ago

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a unimodular constant means a quantity with magnitude1

for example, 1m is a unimodular constant

Rohith M.Athreya - 10 months, 4 weeks ago

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@Rohith M.Athreya lol , i did't knew it :P

Brilliant Member - 10 months, 4 weeks ago

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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 .

Brilliant Member - 11 months ago

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isn't anybody participating ?

Brilliant Member - 11 months ago

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nobody dares to do my problem ? :P

Brilliant Member - 10 months, 3 weeks ago

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@Brilliant Member i didnt quite understand the setup fully

that connecting moves through the wedge??

Rohith M.Athreya - 10 months, 3 weeks ago

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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 ?

Aryan Goyat - 11 months ago

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@Aryan Goyat no, displacing both about euil. , you can do that too when only the 10 kg one is moved .

Brilliant Member - 11 months ago

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@Brilliant Member no abt equilibrium but how abt equilibrium

both down

or one up other down

Aryan Goyat - 11 months ago

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@Aryan Goyat i didn't get what u r saying .

Brilliant Member - 11 months ago

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try with spring too ..:)

Brilliant Member - 11 months ago

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the bottom one depicts equilibrium position

Aryan Goyat - 11 months ago

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@Aryan Goyat 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 . ;)

Brilliant Member - 11 months ago

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@Brilliant Member they r both dragged down by same force || to the /\ planes ;)

Brilliant Member - 11 months ago

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@Brilliant Member yahi to puch raha tha dhanyvad :p

Aryan Goyat - 11 months ago

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Aryan Goyat - 11 months ago

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the wedge is fixed .

Brilliant Member - 11 months ago

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isn't anybody participating ?

Brilliant Member - 11 months 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

Brilliant Member - 11 months ago

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Isn't it L1 = (E I)(wE1) + (M1 I)(w M1) ?

Rajdeep Dhingra - 11 months ago

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

Aryan Goyat - 11 months ago

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

Spandan Senapati - 11 months 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 - 11 months ago

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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 vk2√2h/g.as v in the projected plane is constant.and the constant 2with √2h/g for the cyclic process

Spandan Senapati - 11 months ago

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

Archit Agrawal - 11 months ago

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

Brilliant Member - 11 months ago

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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 ?

Brilliant Member - 11 months ago

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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.

Spandan Senapati - 11 months ago

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Krotov question I guess??

Spandan Senapati - 11 months ago

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Yes

Archit Agrawal - 11 months ago

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