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 centre P (as shown in the 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) /(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

The acceleration of mass m is due to the force `T cos theta`
`:.` `T cos theta =maimplies a=(Tcos theta)/(m)` …(i)
also, `F=2Tsintheta implies T=(F)/(2sintheta)` …(ii)
From (i) and (ii)
`a=((F)/(2 sin theta))(cos theta)/(m)`
`=(F)/(2m tan theta)=(F)/(2m)(x)/(sqrt(a^2-x^2)`
`[ :' tan theta (sqrt(a^2-x^2))/(x)]`