InterviewSolution
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InterviewSolution
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A small bell starts moving from `A` over `a` fixed track as shown in the figure . Surface `AB` is friction from `A` to `B` the bell roll ralis without sipping `BC` is friction , `K_(A)K_(B)` and` K_(C)` are kinetic energy of the bell at `A, B` and `C` respacecvely.Then A. `h_(A) gt h_(C) , K_(B) gt k_(c)`B. `h_(A) gt h_(C) , K_(c) gt k_(A)`C. `h _(A) = h_(C) , K_(B) gt k_(c)`D. `h_(A) lt h_(C) , K_(B) gt k_(c)` |
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Answer» Correct Answer - A::B At point `A` , potential energy of the ball `= mgh_(A)` At point `C` , potential energy of the ball `= mgh_(C)` Total energy at point `A,E_(A) = K_(A) + mgh_(A)` Total energy at point `B,E_(A) = K_(A) + mgh_(A)` Total energy at point `C,E_(C) = K_(C) + mgh_(C)` According to the law of conserveation of energy `E_(A) = E_(B) = E_(C) ` ....(i) E_(A) = E_(B) rArr E_(C)gt k_(C)` ....(ii) E_(A) = E_(C)` K_(A) + mgh_(A) = K_(B) + mgh_(C) ` ...(iii) `rArr b_(A) gt b_(C) , K_(C) gt K_(A) ` ...(iv) Option (b) is correct force (i) (ii) and (iv) , we get h_(A) ,K_(B) gt K_(C)` Option (a) is correct . |
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