A rope of length L and mass M is being pulled on a rough horizontal floor by a constant horizontal force F=Mg. The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between rope and floor is 1/2 . Then, the tension at the midpoint of the rope is

(a) Mg/4

(b) 2Mg/5

(c) Mg/8

(d) Mg/2

Ans: (d)

Sol: For First half part

$ \displaystyle F- T -\frac{\mu M g}{2} = \frac{M}{2}a $ …(i)

For other half

$ \displaystyle T -\frac{\mu M g}{2} = \frac{M}{2}a $ …(ii)

subtracting (i) & (ii)

F – 2T = 0

T = F/2 = Mg/2