Two particles of mass 'm' each are tied at eh ends of a light string of length '2a' The whole system is kept on frictionless horizontal surface with the string held tight so that each mass is at a distance 'a' from the centre 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:

Options

A. `F/(2m)a/sqrt(a^(2)-x^(2))`
B. `F/(2m)x/sqrt(a^(2)-x^(2))`
C. `F/(2m)x/a`
D. `F/(2m)sqrt((a^(2)-x^(2))/x)`

Answer

B

Explanation

From the figure

`F=2T sin theta` ….(i)
`mA=T cos theta` ….(ii)
Dividing eqn (ii) by eqn (i)
`(cos theta)/(2sin theta)=(mA)/F`
`A=F/(2m)cottheta=F/(2m)x/sqrt(a^(2)-x^(2))`