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Q.

Two identical small masses each of mass m are connected by a light inextensible string on a smooth horizontal floor. A constant force F is applied at the mid point of the string as shown in the figure. The acceleration of each mass towards each other is,

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a

F23m

b

3F2m

c

23Fm

d

None of these

answer is A.

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Detailed Solution

Let the tension in the string be T at any angular position θ, the acceleration of each ball along x and y axes be a and a1 respectively. Writing the equation of motion of m, we obtain ∑Fx=ma⇒ Tcos⁡θ=ma    ......(i)∑Fy=ma1⇒ Tsin⁡θ=ma1    .......(ii)At point P, as it is accelerating with an acceleration a, therefore F−2Tcos⁡θ=mpa  where mp= mass of the string at the point P≅0⇒ F=2Tsos⁡θ      ........(iii)(2)÷(1)⇒ tan⁡θ=a1a⇒ a1=atan⁡θ where a=Tcos⁡θm from (i).  Putting T=F2cos⁡θ, from (iii), we obtain a1=F2mtan⁡θ Putting θ=30∘⇒a1=F23m
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