# Physics & Maths JEE Challanges : Help

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

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)-$$=\cfrac { -x\sec { x } }{ x\sin { x } +n\cos { x } } +\tan { x } +C$$

Note by Karan Shekhawat
3 years ago

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Comment deleted May 11, 2016

Really this hearts... Cbse is really asshole ... They always Give wrong answers... Really ashole... We have to pay1000 Rs for their mistakes not our's ....

Are you planning to report ?

- 3 years ago

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

- 3 years 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.

- 3 years ago

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

- 3 years ago

i dont think paying 3000 rupees for 15 marks makes sense, but it leads to a big difference in score,

I hope reputed institutes like FIITJEE would challenge on behalf for their students, also we have to challenge within 24 hours only

- 3 years ago

Is it only four 24 hrs or 4 days i doubt??????

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

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=b inwhich case it is $$\pi$$ given 'a' and 'b' are integers

now 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 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

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

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 )

- 3 years ago

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 :)

- 3 years ago

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)

- 3 years ago

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

- 3 years ago

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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)-$$=\cfrac { -x\sec { x } }{ x\sin { x } +n\cos { x } } +\tan { x } +C$$

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

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 $$f(x)=nsinx-xcosx$$.

- 3 years ago

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

- 3 years ago

I did

- 3 years ago

Comment deleted Apr 17, 2015

Sorry ... But No It is not correct

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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 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 $$P$$(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.

- 3 years ago

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) = $$\beta \gamma \theta$$

- 3 years ago

@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)

think of it as surface tension

(shall i post my solution?)

- 3 years ago

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 $$P \pi 2 R t = r^2 \pi \beta \gamma \theta$$ 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

- 3 years 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. And @Karan Shekhawat, these are really good problems, thank you for posting them! Please continue doing so!

- 3 years ago

Do solve an easy one Like this

- 3 years ago

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

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

- 3 years 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 :P

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

- 3 years 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)

- 3 years ago

Okey dokey :)

- 3 years ago

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

- 3 years ago

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 $$m$$ and radius $$R$$ is rotating about it's COM with angular velocity $$\omega$$(neglect gravity). Find the tension acting inside the ring. Alternatively, find the tensile stress if the area of cross section is given to be $$A$$.

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.

- 3 years ago

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

- 3 years ago

Can you please tell the trick, if you want to ?

- 3 years ago

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

- 3 years ago

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

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 ?

- 3 years ago

@Karan Shekhawat . Here is the full solution.

- 3 years 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 ....

- 3 years ago

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

- 3 years ago