ipht physics problems - A compilation of undergraduate...

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A compilation of undergraduate physics problems Thomas Garel Service de Physique Th´ eorique, CEA/DSM/SPhT Unit´ e de recherche associ´ ee au CNRS 91191 Gif-sur-Yvette cedex, France This compilation of physics exercises and comments may be useful for undergraduates. It contains seven chapters (Introduction, Mechanics, Central forces, Vibrations, Optics, Electromagnetism, Thermodynamics), and does not aim at being exhaustive or original. Many exercises can be found in specialized Journals (Am. J. Phys., Eur. J. Phys., .... ), or books. Some post-exercises comments suggest further studies for the interested student.
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2 Introduction I1 Diameter of a fuse A fuse is a cylindrical wire, designed to melt and disconnect an electric circuit, if the current exceeds a certain value I c . For a given material, what is the dependence of I c on the length L and on the diameter D of the fuse? The fuse melts when the Joule heat cannot be evacuated through its surface, so that its temperature exceeds the melting point. The Joule power is P J RI 2 , where R 4 ρL πD 2 is the resistance of the fuse. The surface evacuated power is P S f 0 πDL , where f 0 , like the resistivity ρ , depends on the nature of the fuse material. One gets P J P S A 0 I 2 D 3 where A 0 4 ρ π 2 f 0 only depends on the material. This shows that the critical current I c (i) does not depend on the fuse length L (ii) depends on the fuse diameter as I c D 3 / 2 . I2 Some results on the Earth (2-a) Small rocks are not round, but large planets, such as Jupiter, are. Compare the gravitational and electrostatic energies of cohesion for the case of the Earth. One expects that the gravitational energy E G of a spherical body (mass M , radius R ) is given by E G ∼ − β GM 2 R , where β = O (1). To calculate the electrostatic energy E el of this body, we imagine that it contains 2 N charged particles ( N with charge q and N with charge q ), on a lattice of periodicity a . We get E el ∼ − α Nq 2 4 πǫ 0 a , with an Ewald constant α = O (1). One has the relation R N 1 / 3 a , up to numerical prefactors, leading to E G E el ( N N c ) 2 / 3 where N c ( q 2 / 4 πǫ 0 Gm 2 ) 3 / 2 , with m the mass of the particles. One can then show that an iron-made earth is not completely dominated by gravitation. Jupiter, on the contrary has N N c . Note that both forces are long- ranged, but charge neutrality has no equivalent in gravitation. The gravitational energy is therefore not extensive: ( E G N 5 / 3 ), contrarily to the electrostatic energy E el N . Hence the existence of a critical N c . (2-b) A celestial body has a rotation period of 1 10 second. Find its composition. Suppose that the cohesive energy of the rotating body is due to gravitation. This requires m ( ω 2 r g ( r )) < 0, where g ( r ) is the gravitational field at distance r from the center. Using Gauss theorem leads to ω 2 < 4 π 3 ρG , where ρ is the specific density of the star. One gets ρ> ω 2 G , showing that this star has the density of a nucleus.
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