A problem proposal for IPhO 2019 participants

All the best to all the 5 participants will be selected to represent India at IPhO 2019, to be held at Tel Aviv, Israel.

Here is a doable problem for you guys :P

A Dielectric slab of thickness tt, relative permittivity ϵr\epsilon_r is between two fixed metallic parallel plates. Faces of the slab and the plates are parallel. Distance between the plates is dd. Find the minimum voltage applied between the plates sufficient to rupture the slab, if the breaking stress of slab is ω\omega. Consider a uniform string of length ll, tension TT, mass per unit length ρ\rho that is stretched between two immovable walls. Show that the total energy of the string, which is the sum of its kinetic and potential energies E=1201[ρ(yt)+T(yx)2]dxE=\frac {1}{2}\int_0^1 [\rho (\frac {\partial y}{\partial t})+T (\frac {\partial y}{\partial x})^2]dx. Where y(x,t)y (x,t) is the string's transverse displacement relatively small. The general motion of the string can be represented as a linear superposition of the normal modes: that is y(x,t)=n=1,(Sin(nπxl))Cos(nπvtlϕn)y (x,t)=\sum_{n=1,\infty}(Sin(n\pi\frac {x}{l}))Cos (n\pi\frac {vt}{l}-\phi_n). Here v=Tρv={\sqrt {\frac{T}{\rho}}}. Demonstrate that E=n=1,EnE=\sum_{n=1,\infty}E_n, where En=14mωn2An2E_n=\frac {1}{4}m\omega_n^2A_n^2 is the energy of nthn^{th} normal mode. Here, m=ρlm=\rho l is the mass of the string, and ωn=nπvl\omega_n=n\pi\frac {v}{l} is the angular frequency of the nthn^{th} normal place.

Note by Pawan Goyal
1 month, 4 weeks ago

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Ya, best of luck. @Pawan Goyal, do you know them? How?

Mr. India - 1 month, 3 weeks ago

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@Pawan Goyal @Rajdeep Dhingra @Archit Boobna who all are in Indian team this year?

Kushal Thaman - 1 week, 4 days ago

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@Archit Boobna @Rajdeep Dhingra best of luck for the TSTs of IPhO 2019.

Pawan Goyal - 1 month, 3 weeks ago

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