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Match the List-I with List-II (O is the point object shown in diagram) |
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Answer» `{:(P,Q,R,S),(3,1,4,2):}` (Q) Let angle of incidence is `theta` Velocity of object `v_(0)=0` velocity of mirror `v_(m) =-v_(i)` velocity of object w.r.t. mirror `v_(om)=v_(o)-v_(m)=+vi` `v_(IM) = v_(I)-v_(m)` or `v_(I) =v_(IM)+v_(m)` `V_(IM) = -vcos2theta HAT(i) - vsin 2thet hat(j)` `v_(I)=(-vcos2theta-v)hat(i)-v sin 2theta hat(j)` `theta =30^(0),v =sqrt(3)m//s` `|v_(I)|=3 m//s` (R) Let angle of incidence is `theta` `r_(I)=(-xcos2theta)hat(i)+(-x sin 2 theta)hat(j)` `(d)/(dt)(r_(I))=V_(I),(d theta)/(dt)=omega` `v_(1)=[-x xx(-SIN2 theta)xx2.(d theta)/(dt)hat(i)]-x.cos2 theta xx2.(d theta)/(dt)hat(j)` `V_(I)=2omegax sin 2 theta hat(i) - 2 omega x cos 2 theta hat(j) ` Put the values. `|V_(I)|=2m//s` (S) Let angle of incidence is `theta` Position vector of image `=r_(I)=(-x cos 2theta hat(i))+(-x sin 2 theta)hat(j)` `(d)/(dt)(r_(I))=v_(I),(DX)/(dt)=v, (d theta)/(dt)=omega` `(d)/(dt)(r_(I))=[(-dx)/(dt). cos2 theta-x(-sin2 theta).2(d theta)/(dt)]hat(i)` `+[((-dx)/(dt))sin 2 theta-x.cos2 theta xx2.(d theta)/(dt)]hat(j)` `V_(I)=[-Vcos2 theta+2omegax sin 2 theta]hat(i)+[-V sin 2 theta-2omega x cos 2 theta]hat(j)` Put the values of `theta, omega` & v `|V_(I)|=a=(31)/(9)cm//s` |
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