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MR^2/2
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great....thanks a lot mam...can u also solve my question
yes... I'll try
okkk....i am posting it...plzz see it
see it mam...i posted...plzz
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Here mass of the wire=pL.As the wire is bent to the form of circular loop.Its radius will be given as =pL/(2π).So,Moment of Inertia of the loop about the vertical z axis
=MR²=(pL){(pL)/(2π)}²=p³L³/(4π²).As the loop is symmetrical about both x and y axes along its plane,hence perpendicular axis theorem brings 2(Ix)=p³L³/(4π²)=Ix=p³L³/(8π²)
=MR²=(pL){(pL)/(2π)}²=p³L³/(4π²).As the loop is symmetrical about both x and y axes along its plane,hence perpendicular axis theorem brings 2(Ix)=p³L³/(4π²)=Ix=p³L³/(8π²)
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