# 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
1 year, 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

- 1 year, 11 months ago

iska to bta do

- 1 year, 11 months ago

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

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

- 1 year, 10 months ago

7

- 1 year, 10 months ago

an APHO type problem

- 1 year, 10 months ago

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

take ratio of masses of uf6

- 1 year, 11 months ago

you are off by 53 stages

- 1 year, 11 months ago

yep now i am getting 163 .(really what a mess i did)

- 1 year, 11 months ago

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

- 1 year, 11 months ago

110 stages ??

- 1 year, 11 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.....

- 1 year, 10 months ago

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

- 1 year, 10 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.

- 1 year, 10 months ago

oh yech, it was a good one.

- 1 year, 10 months ago

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

- 1 year, 10 months ago

ohk

- 1 year, 10 months ago

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

- 1 year, 10 months ago

in which class are u?

- 1 year, 10 months ago

Eleven.you???

- 1 year, 10 months ago

12

- 1 year, 10 months ago

Are you appearing for INAO tomorrow.

- 1 year, 10 months ago

No

- 1 year, 10 months ago

bhai niche itte mahabharat likhe hai vo bhi padh le.

- 1 year, 10 months ago

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

- 1 year, 10 months ago

vaise tu kaun se chocolate dega :p

- 1 year, 10 months ago

- 1 year, 10 months ago

ohk

- 1 year, 10 months ago

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

- 1 year, 10 months ago

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

- 1 year, 10 months ago

lol, mahabharat

- 1 year, 10 months ago

how did u solve it??

- 1 year, 10 months 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 +t2

- 1 year, 10 months ago

Oh shit i didnt noticed that😂

- 1 year, 10 months ago

it cannot pure roll

- 1 year, 10 months ago

- 1 year, 10 months ago

yeah thanks i got the same as well :)

- 1 year, 10 months 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 :P

- 1 year, 10 months ago

Ok

- 1 year, 10 months ago

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

- 1 year, 10 months ago

yes that is why i am stuck

- 1 year, 10 months ago

max friction is 0.1428mgsin@ right??

so how to achieve 0.4mgsin@??

- 1 year, 10 months ago

Getting (17v+4wr)/18gsin@

- 1 year, 10 months 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.

- 1 year, 10 months ago

see my comment too...

- 1 year, 10 months ago

Comment deleted Jan 27, 2017

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

- 1 year, 10 months ago

- 1 year, 11 months ago

Another such seemingly simple problem simple trick

- 1 year, 10 months ago

Nice problem it is .

- 1 year, 10 months ago

but it isn't INPHO level, is it ?

- 1 year, 11 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.

- 1 year, 10 months ago

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

- 1 year, 11 months ago

No.

- 1 year, 11 months ago

is it 4xg/H ?

- 1 year, 11 months ago

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

- 1 year, 11 months ago

Correct

- 1 year, 11 months ago

I am also getting 2xg/H.

- 1 year, 11 months ago

Wrong

- 1 year, 11 months ago

What are your INPhO marks ?

- 1 year, 10 months ago

43 :(

- 1 year, 10 months ago

u?

- 1 year, 10 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

- 1 year, 10 months ago

i am getting around 48-50

- 1 year, 10 months ago

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

- 1 year, 10 months ago

what value u got N and $$\omega_{0}$$

- 1 year, 10 months ago

omega = 4.57 * (10^15)

- 1 year, 10 months ago

oh! i got around $$10^{45}$$

i think i wont make it to OCSC

- 1 year, 10 months ago

I got -ve value for refractive index

- 1 year, 10 months ago

vo q nahi padha time up ho gya

- 1 year, 10 months ago

i didnt have time for the last two parts :(

what value u got for N??

- 1 year, 10 months ago

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

- 1 year, 10 months ago

ques not clr !

- 1 year, 10 months ago

about which point w is given ?

- 1 year, 10 months ago

nahi diya u think

- 1 year, 10 months ago

atleast give the direction of B

- 1 year, 10 months ago

take it into the plane

- 1 year, 10 months ago

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

- 1 year, 10 months ago

Comment deleted Jan 29, 2017

heavy rod ko itna hi likha tha:P

- 1 year, 10 months ago

N kya hai bhai :(

- 1 year, 10 months ago

ohh! anybody did??

- 1 year, 10 months ago

N kya hai bhai

- 1 year, 10 months ago

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

- 1 year, 10 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

- 1 year, 10 months ago

yep did q5 and q6 a,b part

- 1 year, 10 months ago

Can you write a brief solution to Q5 ?

- 1 year, 10 months ago

can you uplod the questions ?

- 1 year, 10 months ago

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

- 1 year, 10 months ago

from us too

- 1 year, 10 months ago

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

- 1 year, 10 months ago

quit pho , cbse is more imporant , lol !

- 1 year, 10 months ago

Comment deleted Jan 25, 2017

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 .

- 1 year, 10 months ago

no leave it

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

- 1 year, 10 months ago

see my report there

- 1 year, 10 months ago

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

- 1 year, 10 months ago

i dunno unimodulr constnt would you tell bout it ?

- 1 year, 11 months ago

a unimodular constant means a quantity with magnitude1

for example, 1m is a unimodular constant

- 1 year, 11 months ago

lol , i did't knew it :P

- 1 year, 11 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 .

- 1 year, 11 months ago

isn't anybody participating ?

- 1 year, 11 months ago

nobody dares to do my problem ? :P

- 1 year, 10 months ago

i didnt quite understand the setup fully

that connecting moves through the wedge??

- 1 year, 10 months ago

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 ?

- 1 year, 11 months ago

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

- 1 year, 11 months ago

no abt equilibrium but how abt equilibrium

both down

or one up other down

- 1 year, 11 months ago

i didn't get what u r saying .

- 1 year, 11 months ago

try with spring too ..:)

- 1 year, 11 months ago

the bottom one depicts equilibrium position

- 1 year, 11 months 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 . ;)

- 1 year, 11 months ago

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

- 1 year, 11 months ago

yahi to puch raha tha dhanyvad :p

- 1 year, 11 months ago

- 1 year, 11 months ago

the wedge is fixed .

- 1 year, 11 months ago

isn't anybody participating ?

- 1 year, 11 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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

- 1 year, 11 months ago

- 1 year, 11 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)

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 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 ?

- 1 year, 11 months ago

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

- 1 year, 11 months ago

Krotov question I guess??

- 1 year, 11 months ago

Yes

- 1 year, 11 months ago