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Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance 'a' from the center P (as shown in figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is : |
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Answer» `(F)/(2m)(a)/(sqrt(a^(2)-x^(2))`  Fig. (ii) 2T sin theta=F…(i) Also for the mass A, Fig. ..(ii) `R+T sin theta= mg...(ii)` ` T cos theta= mf...(iii)` where .f is the acceleration with which each ball moves. From (1) and (iii) Acceleration, `f =(F cos theta)/(2m)=(F)/(2m)xx(x)/sqrt(a^(2)x^(2))` HENCE (b) is the correct choice |
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