INPHO

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.

Note by Aryan Goyat
4 years, 5 months ago

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  Easy Math Editor

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Comments

here is another question!!

Natural uranium consists of mainly two isotopes, U238U^{238} and U235U^{235},whereas the relative concentration of the latter is 0.7%.Uranium is "enriched"(ie., concentration of U235U^{235} is increased) by implementing a multi-stage process,where at each stage,evaporated chemical compound UF6UF_{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 - 4 years, 5 months ago

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110 stages ??

aryan goyat - 4 years, 5 months ago

iska to bta do

aryan goyat - 4 years, 5 months ago

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you are off by 53 stages

Rohith M.Athreya - 4 years, 5 months ago

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

aryan goyat - 4 years, 5 months ago

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

Rohith M.Athreya - 4 years, 5 months ago

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

take ratio of masses of uf6

Rohith M.Athreya - 4 years, 5 months ago

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

The mass density along the rod varies as σ=σ0(1+xnan)\sigma=\sigma_{0}(1+\frac{x^{n}}{a^{n}}) where nn 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., (a2,0)(\frac{a}{2},0)

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

We note that the actual xx co-ordinate is a2+δX\frac{a}{2}+\delta X where XX is the actual co-ordinate of the COM.

Define 100δXX100\frac{\delta X}{X} as percentage error

Approximately what maximum percentage error will we have made in assigning the COM to rods from n=1n=1 to n=20n=20 ie.,10020n=1n=20δXnXn\displaystyle \frac{100}{20} \sum_{n=1}^{n=20} \frac{\delta X_{n}}{X_{n}} where XnX_{n} is the COM when nn is the exponent of xx in the surface density expression

Details and Assumptions

the 21st21^{st} harmonic number H21H_{21} = 3.645

1/23+1/24+1/25=0.125141/23 + 1/24 + 1/25 = 0.12514

give your answer to the nearest integer

Rohith M.Athreya - 4 years, 4 months ago

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@Rohith M.Athreya an APHO type problem

Rohith M.Athreya - 4 years, 4 months ago

@Rohith M.Athreya 7

A Former Brilliant Member - 4 years, 4 months ago

Rajdeep Dhingra - 4 years, 5 months ago

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

Sumanth R Hegde - 4 years, 5 months ago

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I am also getting 2xg/H.

Archit Agrawal - 4 years, 5 months ago

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@Archit Agrawal Wrong

Rajdeep Dhingra - 4 years, 5 months ago

is it 4xg/H ?

aryan goyat - 4 years, 5 months ago

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@Aryan Goyat Correct

Rajdeep Dhingra - 4 years, 5 months ago

@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 - 4 years, 5 months ago

No.

Rajdeep Dhingra - 4 years, 5 months ago

but it isn't INPHO level, is it ?

A Former Brilliant Member - 4 years, 5 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 - 4 years, 4 months ago

Another such seemingly simple problem simple trick

Rohith M.Athreya - 4 years, 4 months ago

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Nice problem it is .

aryan goyat - 4 years, 4 months ago

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 - 4 years, 4 months ago

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

aryan goyat - 4 years, 4 months ago

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how did u solve it??

Rohith M.Athreya - 4 years, 4 months 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 - 4 years, 4 months ago

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

so how to achieve 0.4mgsin@??

Rohith M.Athreya - 4 years, 4 months 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 - 4 years, 4 months ago

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@Aryan Goyat see my comment too...

A Former Brilliant Member - 4 years, 4 months ago

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

aryan goyat - 4 years, 4 months ago

@Aryan Goyat Oh shit i didnt noticed that😂

aryan goyat - 4 years, 4 months ago

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@Aryan Goyat yes that is why i am stuck

Rohith M.Athreya - 4 years, 4 months ago

@Aryan Goyat it cannot pure roll

Rohith M.Athreya - 4 years, 4 months ago

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@Rohith M.Athreya

aryan goyat - 4 years, 4 months ago

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

A Former Brilliant Member - 4 years, 4 months ago

@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

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member Ok

aryan goyat - 4 years, 4 months ago

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

Rohith M.Athreya - 4 years, 4 months ago

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

Spandan Senapati - 4 years, 4 months ago

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

aryan goyat - 4 years, 4 months ago

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@Aryan Goyat lol, mahabharat

A Former Brilliant Member - 4 years, 4 months ago

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

Spandan Senapati - 4 years, 4 months ago

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

aryan goyat - 4 years, 4 months ago

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

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member vaise tu kaun se chocolate dega :p

aryan goyat - 4 years, 4 months ago

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@Aryan Goyat tujhe follow kar lunga :P

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member ohk

aryan goyat - 4 years, 4 months ago

@Spandan Senapati in which class are u?

aryan goyat - 4 years, 4 months ago

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@Aryan Goyat Eleven.you???

Spandan Senapati - 4 years, 4 months ago

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@Spandan Senapati 12

aryan goyat - 4 years, 4 months ago

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@Aryan Goyat Are you appearing for INAO tomorrow.

Spandan Senapati - 4 years, 4 months ago

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@Spandan Senapati No

aryan goyat - 4 years, 4 months 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.

Spandan Senapati - 4 years, 4 months ago

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

aryan goyat - 4 years, 4 months 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 - 4 years, 4 months ago

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@Spandan Senapati ohk

aryan goyat - 4 years, 4 months ago

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 - 4 years, 5 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 - 4 years, 5 months ago

Krotov question I guess??

Spandan Senapati - 4 years, 5 months ago

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Yes

Archit Agrawal - 4 years, 5 months 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 ?

A Former Brilliant Member - 4 years, 5 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 - 4 years, 5 months ago

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

Archit Agrawal - 4 years, 5 months ago

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

A Former Brilliant Member - 4 years, 5 months ago

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

A Former Brilliant Member - 4 years, 5 months ago

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

Rajdeep Dhingra - 4 years, 5 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 - 4 years, 5 months ago

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

aryan goyat - 4 years, 5 months ago

isn't anybody participating ?

A Former Brilliant Member - 4 years, 5 months ago

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 .

A Former Brilliant Member - 4 years, 5 months ago

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the wedge is fixed .

A Former Brilliant Member - 4 years, 5 months ago

try with spring too ..:)

A Former Brilliant Member - 4 years, 5 months ago

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aryan goyat - 4 years, 5 months ago

the bottom one depicts equilibrium position

aryan goyat - 4 years, 5 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 . ;)

A Former Brilliant Member - 4 years, 5 months ago

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

A Former Brilliant Member - 4 years, 5 months ago

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@A Former Brilliant Member yahi to puch raha tha dhanyvad :p

aryan goyat - 4 years, 5 months ago

isn't anybody participating ?

A Former Brilliant Member - 4 years, 5 months 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 - 4 years, 5 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 .

A Former Brilliant Member - 4 years, 5 months ago

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@A Former Brilliant Member no abt equilibrium but how abt equilibrium

both down

or one up other down

aryan goyat - 4 years, 5 months ago

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

A Former Brilliant Member - 4 years, 5 months ago

nobody dares to do my problem ? :P

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member i didnt quite understand the setup fully

that connecting moves through the wedge??

Rohith M.Athreya - 4 years, 4 months ago

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

aryan goyat - 4 years, 4 months ago

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quit pho , cbse is more imporant , lol !

A Former Brilliant Member - 4 years, 4 months ago

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 - 4 years, 4 months ago

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can you uplod the questions ?

A Former Brilliant Member - 4 years, 4 months ago

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

Rajdeep Dhingra - 4 years, 4 months 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)

A Former Brilliant Member - 4 years, 4 months ago

@Rajdeep Dhingra from us too

aryan goyat - 4 years, 4 months ago

yep did q5 and q6 a,b part

aryan goyat - 4 years, 4 months ago

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

Rajdeep Dhingra - 4 years, 4 months ago

what value u got N and ω0\omega_{0}

Rohith M.Athreya - 4 years, 4 months ago

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

aryan goyat - 4 years, 4 months ago

N kya hai bhai :(

aryan goyat - 4 years, 4 months ago

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@Aryan Goyat ohh! anybody did??

Rohith M.Athreya - 4 years, 4 months ago

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@Rohith M.Athreya N kya hai bhai

aryan goyat - 4 years, 4 months ago

I got -ve value for refractive index

Rajdeep Dhingra - 4 years, 4 months 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 - 4 years, 4 months 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 - 4 years, 4 months ago

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@Aryan Goyat ques not clr !

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member about which point w is given ?

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member nahi diya u think

aryan goyat - 4 years, 4 months ago

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@Aryan Goyat atleast give the direction of B

A Former Brilliant Member - 4 years, 4 months ago

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@A Former Brilliant Member take it into the plane

aryan goyat - 4 years, 4 months ago

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

A Former Brilliant Member - 4 years, 4 months ago

@Rajdeep Dhingra vo q nahi padha time up ho gya

aryan goyat - 4 years, 4 months ago

omega = 4.57 * (10^15)

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra oh! i got around 104510^{45}

i think i wont make it to OCSC

Rohith M.Athreya - 4 years, 4 months ago

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 - 4 years, 4 months ago

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i am getting around 48-50

Rohith M.Athreya - 4 years, 4 months ago

What are your INPhO marks ?

Rajdeep Dhingra - 4 years, 4 months ago

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43 :(

Rohith M.Athreya - 4 years, 4 months ago

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u?

Rohith M.Athreya - 4 years, 4 months ago
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