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A long chain of mass of mass `M` length `L` is being pulled with constant velocity on a rough incline with coefficient of friction `mu`. What rate frictional force on chain is increasing A. `(3)/(10)mu(M)/(L)gv`B. `(5)/(6)mu(M)/(L)gv`C. zeroD. `mu(M)/(L)gv` |
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Answer» Correct Answer - A In a small time let `dx` length crossing from lower incline to upper incline `df=mu(M(dx))/(L)g cos 37^(@)-mu (M)/(L)dx cos 60^(@)` `df=mu(M)/(L)g (cos 37-cos 60^(@))dx` `(df)/(dt)=(muM)/(L)g((4)/(5)-(1)/(2))(dx)/(dt)=(3)/(10)(muM)/(L)gv` |
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