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A bob of mass m = 100 gm having charge q = 50 mu C connected with a non-conducting string of length l = 1m is whirled in vertical plane with minimum speed such that it can complete a circular path in space where electric field of strength E=2 xx 10^(4)N//Cis switched on as shown in figure. The velocity of bob at the point where tension in string become zero is |
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Answer» `vec(v)=(sqrt(5)hat(i)+sqrt(3)hat(j))` `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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