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A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is p. An electron (charge e, mass m) is released inside the cavity from point P as shown in figure. The centre of sphere and centre of cavity are separated by a distance a. The time after which the electron again touches the sphere is |
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Answer» `sqrt((6sqrt(2)repsilon_(0)m)/(epa))` `L(B_(0)IL)=3(lambdaL)g((L)/(2)+(L)/(6))=3lambdaLg((2L)/(3))` `B_(0)=(2lambdag)/(I)` No EXTERNAL force, so COM cannot displace initial coordinate of `COM=(3(lambdaL)("zero")+2lambdaL((L)/2)+lambdaL(L))/(6lambdag)=(L)/(3)` Final coordinate of `COM=(L)/(3)("same")` But COM DISPLACES with respect to QT by `(2L)/(3)`. So displacement of `QPUT=(2L)/(3)`. Initial magnetic dipole moment M makes an angle of `(pi)/(4)` ACW from +ive x-axis and finally it makes angle `(3pi)/(4)` ACW from `+"ive x-axis" (M=sqrt(2)IL^(2))` So change in `PE=-MB(cos theta_(2)-costheta_(1))=MB_(0)[cos135^(@)-cos45^(@)]=2B_(0)IL^(2)="Gain in KE"=(1)/(2)I omega ^(2)` `I("about QT")=2[(2lambdaL^(3))/(3)+lambdaL^(3)]=(10lambdaL^(3))/(3),2B_(0)IL^(2)=(1)/(2)(Iomega^(2))=(1)/(2)((10lambdaL^(3))/(3))omega^(2),omega^(2)=(6B_(0)I)/(5lambdaL)` |
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