Physics-
General
Easy

Question

A uniform rope is moving with a constant acceleration ‘a’ on a smooth horizontal surface. The ratio of the tension in the rope at its mid point to the applied force is,

  1. 1:1    
  2. 1:2    
  3. 0    
  4. 2:1    

The correct answer is: 1:2


    Let total mass of the rope be M and total length L. Now let us find the tension at a point at a distance l from the end where the force F (let) is applied.
    From F.B.D. of front part shown

    N2 = fraction numerator M over denominator L end fraction. l g (1)
    F - T = fraction numerator M over denominator L end fraction. l a (2)

    From F.B.D. of rear part shown
    N1 = open parentheses M minus fraction numerator M over denominator L end fraction. l close parentheses g (3)
    T = open parentheses M minus fraction numerator M over denominator L end fraction. l close parentheses a (4)

    From (2) and (4),
    F = Ma (5)
    Therefore, a = fraction numerator F over denominator M end fraction (6)
    From (4) and (6),
    T = M open parentheses 1 minus fraction numerator l over denominator L end fraction close parentheses  .   fraction numerator F over denominator M end fraction
    Therefore, fraction numerator T over denominator F end fraction   equals   1 minus fraction numerator l over denominator L end fraction (7)
    At the mid point, fraction numerator l over denominator L end fraction   equals   fraction numerator 1 over denominator 2 end fraction, therefore fraction numerator T over denominator F end fraction   equals   fraction numerator 1 over denominator 2 end fraction
    2.If the mass of the rope is not accounted then fraction numerator T over denominator F end fraction equals fraction numerator 1 over denominator 1 end fraction
    Therefore (B)

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