\(Q-1\) [

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

\(Q-2\)- [

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

\(Q-3\)- [

Maths]\(\displaystyle{I(n)=\int { \cfrac { { x }^{ 2 }+n(n-1) }{ { \left( x\sin { x } +n\cos { x } \right) }^{ 2 } } } dx\\ I(n)=?}\)

Please Help , me ! Urgent

PLease Post Solutions , Instead of answers....

Thanks alot......

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

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

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

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TopNewest1) \[I_{m}=2\int_{0}^{\pi}\prod_{r=1}^{m}cos(rx)\mathrm{d}x= 2\int_{0}^{\pi}\prod_{r=1}^{m}cos(r(\pi-x))\mathrm{d}x \]

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

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

\[I_{m}=0 \Rightarrow I_{m}=-I_{m} \]

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

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

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

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

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, \(2\pi RtS=\pi R^2 p\).

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

Differentiate and compare this with the given expression to get \(f(x)=nsinx-xcosx\). – Shashwat Shukla · 1 year, 9 months ago

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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 · 1 year, 9 months ago

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– Raghav Vaidyanathan · 1 year, 9 months ago

I didLog in to reply

@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 \(2 \pi \) )

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

it is easy to show that this is 0 unless and untill

a=binwhich case it is \(\pi\) given 'a' and 'b' are integersnow consider \(\int { cos(ax)cos(bx) } cos(cx)\)

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

in general it can be stated that

\(\int { \prod _{ r=1 }^{ m }{ cos({ a }_{ r }x) } } =0\\ \\ unless\quad { a }_{ r }=|{ a }_{ 1 }\pm { a }_{ 2 }\pm { a }_{ 3 }......\pm { a }_{ m-1 }|\)

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

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

which is again equivalen to asking whether

the terms \(t^m\)or \(t^{-m}\)exists in the expansion of

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

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

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

in our expansion whose result is

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

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 · 1 year, 9 months ago

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@Mvs Saketh – Shashwat Shukla · 1 year, 9 months ago

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@Karan Shekhawat - post a solution for Q1 incase you can find one (from your friend) – Mvs Saketh · 1 year, 9 months ago

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Using Euler's formula's ... – Karan Shekhawat · 1 year, 9 months ago

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– Mvs Saketh · 1 year, 9 months ago

please post a solution :)Log in to reply

– Shashwat Shukla · 1 year, 9 months ago

That is precisely what I was also wondering! I hope there is an easier way.Log in to reply

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

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

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

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

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

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

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

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

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

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

Now we equate \(F_1=F_2\) and put \(\alpha \to 0\) to get:

\(S=\frac {RP} {2t}\)

Now all that is left to do is find the value of \(P\). From the definition of bulk modulus:

\(\beta =-V \frac {dP} {dV}\)

Where \(V\) is original volume, \(dV\) is change in volume, and \(dP=P\). According to me, they have committed a mistake in following steps:

Let initial volume be \(V_0\)

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

The rest is left as an exercise. – Raghav Vaidyanathan · 1 year, 9 months ago

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@Karan Shekhawat . Here is the full solution. – Raghav Vaidyanathan · 1 year, 9 months ago

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But Please Tell what is blunder in this ...

Stress=Pressure(P) = \(\beta \gamma \theta \) – Karan Shekhawat · 1 year, 9 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 · 1 year, 9 months ago

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

then we have \( P \pi 2 R t = r^2 \pi \beta \gamma \theta \) 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 · 1 year, 9 months ago

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– Karan Shekhawat · 1 year, 9 months ago

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 :PLog in to reply

– Shashwat Shukla · 1 year, 9 months ago

Haha :). Thanks for the undue praise... And yes, I also made the same mistake as I misinterpreted the word 'tensile'.Log in to reply

– Mvs Saketh · 1 year, 9 months ago

yes, its a common error, thanks for posting, please post more such problems when you want, these were pretty good quality problemsLog in to reply

@Mvs Saketh – Shashwat Shukla · 1 year, 9 months ago

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– Mvs Saketh · 1 year, 9 months ago

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)Log in to reply

– Shashwat Shukla · 1 year, 9 months ago

Okey dokey :)Log in to reply

@Karan Shekhawat, these are really good problems, thank you for posting them! Please continue doing so! – Raghav Vaidyanathan · 1 year, 9 months ago

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. AndLog in to reply

Like this – Karan Shekhawat · 1 year, 9 months ago

Do solve an easy oneLog in to reply

here V=intial vol. Since it is neglibly changed ... @Raghav Vaidyanathan @Mvs Saketh – Karan Shekhawat · 1 year, 9 months ago

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– Raghav Vaidyanathan · 1 year, 9 months ago

What do you think about my solution and arguments, are they right? My answer does not match with the one given.Log in to reply

acts on the inner wallsof the container. What is required is the pressure(tensile stress) which is presentinside the walls. There is a difference. For example consider this problem:Try the above problem. Remember \(T \ne m(\omega )^2R\).

You will understand the second possible blunder when you fully read and understand my solution. – Raghav Vaidyanathan · 1 year, 9 months ago

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– Krishna Sharma · 1 year, 9 months ago

Do you know the short trick for finding tension in problems like the one you gave? (just asking Iknow trick)Log in to reply

– Karan Siwach · 1 year, 9 months ago

Can you please tell the trick, if you want to ?Log in to reply

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

Rcm = 2R/pi (for half ring )

Am I right @Krishna Sharma – Karan Shekhawat · 1 year, 9 months ago

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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 · 1 year, 9 months ago

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– Karan Shekhawat · 1 year, 9 months ago

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

– Raghav Vaidyanathan · 1 year, 9 months ago

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

3) (nsinx - xcosx)/(ncosx + xsinx) – Karan Siwach · 1 year, 9 months ago

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– Karan Shekhawat · 1 year, 9 months ago

Yest it is correct .... Please Post a solution also ,..... ThanksLog in to reply

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Are you planning to report ? – Karan Shekhawat · 1 year, 9 months ago

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– Shashwat Shukla · 1 year, 9 months ago

WTF is right. This sucks...Is anyone planning to challenge?Log in to reply

– Raghav Vaidyanathan · 1 year, 9 months ago

We will do something, each will contribute 100 bucks and the highest scorer in our peer group will challenge on behalf of the rest. It's better than doing nothing.Log in to reply

– Karan Shekhawat · 1 year, 9 months ago

Raghav Vaidyanathan Shashwat Shukla Mvs Saketh I had posted an Note for this... So that we Can gather all our other friends ...Log in to reply

I hope reputed institutes like FIITJEE would challenge on behalf for their students, also we have to challenge within 24 hours only – Mvs Saketh · 1 year, 9 months ago

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– Abdulmuttalib Lokhandwala · 1 year, 9 months ago

Is it only four 24 hrs or 4 days i doubt??????Log in to reply

@Ronak Agarwal ; Is there any method to do Q.3 rather than inspection ? – Karan Siwach · 1 year, 9 months ago

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Okay I'am Posting Answers::

– Karan Shekhawat · 1 year, 9 months agoLog in to reply

– Shashwat Shukla · 1 year, 9 months ago

Out of curiosity, where did you get these questions from?Log in to reply

– Karan Shekhawat · 1 year, 9 months ago

It is taken from JEE advance module of my one of friend...Log in to reply

– Shashwat Shukla · 1 year, 9 months ago

I see. Thanks for posting these questions :)...I really liked all of them.Log in to reply

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– Karan Shekhawat · 1 year, 9 months ago

Sorry ... But No It is not correctLog in to reply

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

Please Reply .... – Karan Shekhawat · 1 year, 9 months ago

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– Shashwat Shukla · 1 year, 9 months ago

I made a mistake in my calculations. Will repost soon.Log in to reply

@Shashwat Shukla @Sudeep Salgia @Raghav Vaidyanathan @Mvs Saketh @Ronak Agarwal – Karan Shekhawat · 1 year, 9 months ago

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