Physics & Maths JEE Challanges : Help

Q1Q-1 [MATHS] Let Im=02πcosxcos2xcos3x............cosmxmtimesdx\displaystyle{{ I }_{ m }=\int _{ 0 }^{ 2\pi }{ \underbrace { \cos { x } \cos { 2x } \cos { 3x } ............\cos { mx } }_{ m-times } } dx} Find Integral and condition for mm for which (i)Im0(ii)Im=0\displaystyle{(i)\quad { I }_{ m }\neq 0\\ (ii)\quad { I }_{ m }=0}


Q2Q-2- [Physics] A liquid of voulmetric thermal coff. γ\gamma and bulk modulus β\beta is filled in a spherical tank of negligible coff. of thermal exapanion. It's Radius is R and thickness of wall is t , t<<Rt<<R . when temprature of the liquid is raised by θ\theta . Find Tensile stress develope in walls of tank ?


Q3Q-3- [Maths]I(n)=x2+n(n1)(xsinx+ncosx)2dxI(n)=?\displaystyle{I(n)=\int { \cfrac { { x }^{ 2 }+n(n-1) }{ { \left( x\sin { x } +n\cos { x } \right) }^{ 2 } } } dx\\ I(n)=?}


Please Help , me ! Urgent

PLease Post Solutions , Instead of answers....

Thanks alot......

Answers-1)- Im0:m(m+1)2=even:Im=2π2m1Im=0:m(m+1)2=odd:Im=0{ I }_{ m }\neq 0\quad :\quad \cfrac { m(m+1) }{ 2 } =even\quad :\quad { I }_{ m }=\cfrac { 2\pi }{ { 2 }^{ m-1 } } \\ { I }_{ m }=0\quad :\quad \cfrac { m(m+1) }{ 2 } =odd\quad :\quad { I }_{ m }=0\quad

Answer-2)-βγθR2t\cfrac { \beta \gamma \theta R }{ 2t }

Ans-3)-=xsecxxsinx+ncosx+tanx+C=\cfrac { -x\sec { x } }{ x\sin { x } +n\cos { x } } +\tan { x } +C

Note by Karan Shekhawat
4 years, 4 months ago

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1) Im=20πr=1mcos(rx)dx=20πr=1mcos(r(πx))dxI_{m}=2\int_{0}^{\pi}\prod_{r=1}^{m}cos(rx)\mathrm{d}x= 2\int_{0}^{\pi}\prod_{r=1}^{m}cos(r(\pi-x))\mathrm{d}x

Now, cos(r(πx))=cosrxcos(r(\pi-x))=cosrx if r is even.

And if r is odd, cos(r(πx))=cosrxcos(r(\pi-x))=-cosrx .

Im=0Im=ImI_{m}=0 \Rightarrow I_{m}=-I_{m}

Now, as already mentioned, odd values of r change the sign of ImI_{m} while the even values of r do not.

Thus, we need odd values of r to appear an oddnumberoftimesodd \quad number \quad of \quad times.

This happens when m=1,2,5,6,9,10 etc.

i.e Im=0m=4n+1,4n+2n0 I_{m}=0 \quad \Rightarrow m=4n+1, 4n+2 \quad \forall n \geqslant0

2) See Raghav's explanation. However, I would like to suggest one small simplification: Instead of taking a small part of the surface and working with solid angles and using lots of trignometry, the same condition can also obtained by considering the equilibrium of the two halves of the spherical shell.

Directly, 2πRtS=πR2p2\pi RtS=\pi R^2 p.

3)The (xsinx+ncosx)2(xsinx+ncosx)^2 in the denominator leads us to deduce that the anti-derivative, g(x)g(x) is of the form f(x)(xsinx+ncosx)\frac{f(x)}{(xsinx+ncosx)}

Differentiate and compare this with the given expression to get f(x)=nsinxxcosxf(x)=nsinx-xcosx.

Shashwat Shukla - 4 years, 4 months ago

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Thank you very much again... :) You always help me..!

I understood Q-1 and Q-3 ... Thank you all of you..!

But Please Post sol of Q-2 (Physics) (Please Explain it ... Since I Did not getting answer in terms of R and t) @Shashwat Shukla @Karan Siwach @Raghav Vaidyanathan

Karan Shekhawat - 4 years, 4 months ago

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I did

Raghav Vaidyanathan - 4 years, 4 months ago

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I don't know if I'm right. For question Q2, is this the answer:

Stress=Rβγθ2t(1+γθ)Stress=\frac{R \beta \gamma \theta} {2t(1+\gamma \theta)}

What happens here is that the liquid inside the sphere will tend to expand, but it is restricted from doing so by the walls of the sphere(which are rigid and do not change size). This restricted expansion creates a stress on the fluid, which is equal and opposite to the stress on the walls of the container. Note that this stress is physically manifested as pressure on the walls of the container. Our task is to find the tensile stress in the walls of the container. The tensile stress in the walls act to nullify the effect of pressure exerted by liquid due to expansion.

Now, let the pressure exerted by liquid be PP(we will calculate this later). Take a small area element on the sphere whose respective cone has a half angle α\alpha(hope you are familiar with solid angle concept). The area of this element is 2πR2(1cos(α))2\pi R^2(1-\cos {(\alpha)}). Thus, the force acting on this element due to said pressure is:

F1=P×2πR2(1cos(α))F_1=P\times 2\pi R^2(1-\cos {(\alpha)})(radially outwards)

This force is counteracted by the tensile stress in the walls of the container, let the tensile stress be SS. The force due to tesile stress is given by:

F2=S×A×sin(α)F_2=S\times A \times \sin {(\alpha)}

Where AA is the area of the interface of the small element with the rest of the sphere and sin(α) \sin {(\alpha)} is due to the fact that the force due to SS acts in an oblique direction w.r.t the element under consideration.

Obviously: A=2πRsin(α)tA=2 \pi R \sin {(\alpha)} t

F2=S×2πRt(sin(α))2\Rightarrow F_2=S\times 2 \pi R t (\sin {(\alpha)})^2(radially inwards)

Now we equate F1=F2F_1=F_2 and put α0\alpha \to 0 to get:

S=RP2tS=\frac {RP} {2t}

Now all that is left to do is find the value of PP. From the definition of bulk modulus:

β=VdPdV\beta =-V \frac {dP} {dV}

Where VV is original volume, dVdV is change in volume, and dP=PdP=P. According to me, they have committed a mistake in following steps:

Let initial volume be V0V_0

When temperature is increased by θ\theta, the volume is supposed to become V0(1+γθ)V_0(1+\gamma \theta). But it is restricted and it stays as V0V_0. Hence, here change in volume is: dV=V0γθdV=-V_0 \gamma \theta. But what is the original volume? According to me, V=V0(1+γθ)V=V_0(1+\gamma \theta), but they have taken V=V0V=V_0, which is the reason for the difference between my answer and theirs.

The rest is left as an exercise.

Raghav Vaidyanathan - 4 years, 4 months ago

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@Karan Shekhawat . Here is the full solution.

Raghav Vaidyanathan - 4 years, 4 months ago

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First. thanks for replying... But sorry I have Answer , as βγθR2t\displaystyle{\cfrac { \beta \gamma \theta R }{ 2t } }.... Still can you please tell ur method ....

Karan Shekhawat - 4 years, 4 months ago

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Hmm, I don't think that is right... but we will see. I will edit my comment so as to include solution

Raghav Vaidyanathan - 4 years, 4 months ago

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Thanks a lot for Explaining ... But still I need some time to anylise your solution by my own on study table....So that I understand it better ...

But Please Tell what is blunder in this ...

Stress=Pressure(P) = βγθ\beta \gamma \theta

Karan Shekhawat - 4 years, 4 months ago

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@Karan Shekhawat - the blunder is that the pressure you derived is the pressure exerted by the liquid upon sphere normally,

The tensile stress is tension developed in the sphere's walls which support the liquid from expanding and it is along the walls of the container and not normal to it)

(example, if a rope is rotating (circular loop ) , then each element experiences a force of dm w^2 r radially outward, but the tension is not that)

and yes your answer is correct (Raghav i think your answer is wrong)

think of it as surface tension

(shall i post my solution?)

Mvs Saketh - 4 years, 4 months ago

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@Mvs Saketh What do you think about my solution and arguments, are they right? My answer does not match with the one given.

Raghav Vaidyanathan - 4 years, 4 months ago

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@Mvs Saketh Consider, one half of the sphere,

Let the pressure inside the walls be P and directed tangentially at the rim (like surface tension)

then we have Pπ2Rt=r2πβγθ P \pi 2 R t = r^2 \pi \beta \gamma \theta so P is the answer which you gave @Karan Shekhawat

@Raghav Vaidyanathan - the change is linear and hence small, so neglecting the volume change term in denominator would make no difference, that is the reason of difference as you have already realised

Mvs Saketh - 4 years, 4 months ago

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@Mvs Saketh Awesome...... I got It completly .... Yes I realise My mistake ... Example of Circular disc is best , Thanks to you both guys for realising me my blunder.... Even an Genius known as Shashwat shukla ji .. also made this mistake intially .. he posted it and then deleted it ... :P :P

Karan Shekhawat - 4 years, 4 months ago

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@Karan Shekhawat yes, its a common error, thanks for posting, please post more such problems when you want, these were pretty good quality problems

Mvs Saketh - 4 years, 4 months ago

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@Mvs Saketh Karan has posted a closed form for question 1(scroll down to see it)....What would your approach to arrive at the same, be?

@Mvs Saketh

Shashwat Shukla - 4 years, 4 months ago

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@Shashwat Shukla i am presently working on it, i think i have found the solution, but there are some problems , i shall make it sufficiently rigorious and post it (let me feel i am right)

Mvs Saketh - 4 years, 4 months ago

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@Mvs Saketh Okey dokey :)

Shashwat Shukla - 4 years, 4 months ago

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@Karan Shekhawat Haha :). Thanks for the undue praise... And yes, I also made the same mistake as I misinterpreted the word 'tensile'.

Shashwat Shukla - 4 years, 4 months ago

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@Mvs Saketh Yes He also correct .. but thing is while writing this Raghav also little mistke ... β=VdpdV\beta =-V\cfrac { dp }{ dV }

here V=intial vol. Since it is neglibly changed ... @Raghav Vaidyanathan @Mvs Saketh

Karan Shekhawat - 4 years, 4 months ago

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@Mvs Saketh Yes, but still, they shouldn't have given the answer like that :/ And kudos on taking half sphere, i never thought of that. Too busy proving them wrong.. xD. And @Karan Shekhawat, these are really good problems, thank you for posting them! Please continue doing so!

Raghav Vaidyanathan - 4 years, 4 months ago

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@Raghav Vaidyanathan Do solve an easy one Like this

Karan Shekhawat - 4 years, 4 months ago

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See, this is what I explained in my solution. What you have found is the pressure which acts on the inner walls of the container. What is required is the pressure(tensile stress) which is present inside the walls. There is a difference. For example consider this problem:

A thin uniform circular solid ring of mass mm and radius RR is rotating about it's COM with angular velocity ω\omega(neglect gravity). Find the tension acting inside the ring. Alternatively, find the tensile stress if the area of cross section is given to be AA.

Try the above problem. Remember Tm(ω)2RT \ne m(\omega )^2R.

You will understand the second possible blunder when you fully read and understand my solution.

Raghav Vaidyanathan - 4 years, 4 months ago

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@Raghav Vaidyanathan Thanks a lot... This is really helpfull for me... I got it ... Answer should be Tmω2R2πS=TAs\displaystyle{T\approx \cfrac { m\omega ^{ 2 }R }{ 2\pi } \\ S=\cfrac { T }{ { A }_{ s } } }

Now I Realise my blunder... Tensile stress is develope due to Pressure , But it is never being equal to pressure applied , instead it is equal to Pressure develope inside the walls ...

Forces are equal but pressure don't .... Am I right ?

Also What is 2 nd Blunder ?

Karan Shekhawat - 4 years, 4 months ago

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@Raghav Vaidyanathan Do you know the short trick for finding tension in problems like the one you gave? (just asking Iknow trick)

Krishna Sharma - 4 years, 4 months ago

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@Krishna Sharma Can you please tell the trick, if you want to ?

Karan Siwach - 4 years, 4 months ago

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@Karan Siwach I guess It should be ... by Calculating Force on Half Ring ....

2T=(m/2)w^2(2R/pi)

Rcm = 2R/pi (for half ring )

Am I right @Krishna Sharma

Karan Shekhawat - 4 years, 4 months ago

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@Shashwat Shukla - i have found a crude method to solve question number 1

please check it out, patiently

consider

(note- all integration limits are 0 to 2π2 \pi )

cos(ax)cos(bx)\int { cos(ax)cos(bx) }

it is easy to show that this is 0 unless and untill a=b inwhich case it is π\pi given 'a' and 'b' are integers

now consider cos(ax)cos(bx)cos(cx)\int { cos(ax)cos(bx) } cos(cx)

cos(ax)cos(bx)cos(cx)=(1/2)cos(ax+bx)cos(cx)+cos(axbx)cos(cx)=0unlessc=a+borc=ab\int { cos(ax)cos(bx) } cos(cx)\\ \\ =(1/2)\int { cos(ax+bx)cos(cx)+cos(ax-bx) } cos(cx)\\ =0\\ \\ unless\quad c=|a+b|\\ or\\ c=|a-b|

in general it can be stated that

r=1mcos(arx)=0unlessar=a1±a2±a3......±am1\int { \prod _{ r=1 }^{ m }{ cos({ a }_{ r }x) } } =0\\ \\ unless\quad { a }_{ r }=|{ a }_{ 1 }\pm { a }_{ 2 }\pm { a }_{ 3 }......\pm { a }_{ m-1 }|

now, in the given situation, this is equivalent to asking

whether m=1±2±3......±m1m=|1\pm 2\pm 3......\pm { m-1 }| is possible for some selection of signs (+ and -)

which is again equivalen to asking whether

the terms tmt^mor tmt^{-m}exists in the expansion of

(t+1t)(t2+1t2)......(tm1+1tm1)(t+\frac { 1 }{ t } )({ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } )......({ t }^{ m-1 }+\frac { 1 }{ { t }^{ m-1 } } )

now we have already shown that it is non zero when m(m+1)/2 is even

hence, considering the case when its even, we simply have to prove that there is a unique combination of signs such that the condition is satisfied

because then we will have the term

12m2cos2(mx)dx\\ \frac { 1 }{ { 2 }^{ m-2 } } \int { { cos }^{ 2 } } (mx)dx

in our expansion whose result is

12m2cos2(mx)dx=2π2m1\frac { 1 }{ { 2 }^{ m-2 } } \int { { cos }^{ 2 } } (mx)dx\quad =\quad \frac { 2\pi }{ { 2 }^{ m-1 } }

all the other terms will cancel out to 0 since the condition is not satisfied for them

(presently, i am trying to rigoriously prove that the coefficient of t^m or 1/t^m is 1 and unique )

Mvs Saketh - 4 years, 4 months ago

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Brilliant :D ...You have essentially considered orthogonal functions in n-dimensions and this process is closely linked to a discrete Fourier transform (as you must be well aware)...I tried something of the sort but got muddled up in the details...Hats-off again :)

@Mvs Saketh

Shashwat Shukla - 4 years, 4 months ago

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thanks :) though there are some details to fill, but i dont think i could have solved it in the exam hall, we need to find a slicker way

@Karan Shekhawat - post a solution for Q1 incase you can find one (from your friend)

Mvs Saketh - 4 years, 4 months ago

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@Mvs Saketh That is precisely what I was also wondering! I hope there is an easier way.

Shashwat Shukla - 4 years, 4 months ago

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@Mvs Saketh Now I found one another approach .... By using Complex Numbers.
Using Euler's formula's ...

Karan Shekhawat - 4 years, 4 months ago

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@Karan Shekhawat please post a solution :)

Mvs Saketh - 4 years, 4 months ago

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3) (nsinx - xcosx)/(ncosx + xsinx)

Karan Siwach - 4 years, 4 months ago

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Yest it is correct .... Please Post a solution also ,..... Thanks

Karan Shekhawat - 4 years, 4 months ago

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@Shashwat Shukla why you had deleted your solution ? Pleasse repost it....

Please Reply ....

Karan Shekhawat - 4 years, 4 months ago

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I made a mistake in my calculations. Will repost soon.

Shashwat Shukla - 4 years, 4 months ago

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Okay I'am Posting Answers::

Answers-1)- Im0:m(m+1)2=even:Im=2π2m1Im=0:m(m+1)2=odd:Im=0{ I }_{ m }\neq 0\quad :\quad \cfrac { m(m+1) }{ 2 } =even\quad :\quad { I }_{ m }=\cfrac { 2\pi }{ { 2 }^{ m-1 } } \\ { I }_{ m }=0\quad :\quad \cfrac { m(m+1) }{ 2 } =odd\quad :\quad { I }_{ m }=0\quad

Answer-2)-βγθR2t\cfrac { \beta \gamma \theta R }{ 2t }

Ans-3)-=xsecxxsinx+ncosx+tanx+C=\cfrac { -x\sec { x } }{ x\sin { x } +n\cos { x } } +\tan { x } +C

Karan Shekhawat - 4 years, 4 months ago

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Out of curiosity, where did you get these questions from?

Shashwat Shukla - 4 years, 4 months ago

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It is taken from JEE advance module of my one of friend...

Karan Shekhawat - 4 years, 4 months ago

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@Karan Shekhawat I see. Thanks for posting these questions :)...I really liked all of them.

Shashwat Shukla - 4 years, 4 months ago

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@Ronak Agarwal ; Is there any method to do Q.3 rather than inspection ?

Karan Siwach - 4 years, 4 months ago

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