Physics & Maths JEE Challanges : Help

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 $f(x)=nsinx-xcosx$.

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.

@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=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) (nsinx - xcosx)/(ncosx + xsinx)

@Shashwat Shukla @Sudeep Salgia @Raghav Vaidyanathan @Mvs Saketh @Ronak Agarwal

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

Please Reply ....

Okay I'am Posting Answers::

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

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