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A uniform conducting ring of mass `m=2kg`, radius `r=2m` and resistance `8Omega` is kept on smooth, horizontal surface. A time varying magnetic field `vecB=(hati+t^(2)hatj)` Tesla is present in the region, where `t` is time in second and take vertical as `y`-axis. (Take `pi^(2)=10`). ThenA. Time when ring starts toppling is 1 secB. Time when ring starts toppling is `3//4` secC. Heat generated through the ring till the instant when the ring start toppling is `40//3` Joule.D. Heat generated through the ring till the instant when the ring start toppling is `80//3` Joule |
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Answer» Correct Answer - A::D `phi=BA=(t^(2))(pir^(2))` `epsilon=(dphi)/(dt)=2pir^(2)t` `i=(epsilon)/R=(2pir^(2)t)/R` For toppling `tau=|vec(mu)xxvecB|gemgr` `(i.pir^(2)).1gemgr` `(2pir^(2)t)/R.pir^(2) gemgr` `t ge (mgR)/(2pi^(2)r^(3))=1`sec Heat `=int i^(2)Rdt=int((2pir^(2)t)/R)^(2)=dt.R` Heat `=(4pi^(2)r^(4))/R int_(0)^(1) t^(2) dt =80/3` Joule |
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