For the arrangement shown in figure the coefficient of friction between the two blocks is\[\mu \]. If both the block are identical, then the acceleration of each block is-
A)\[\frac{F}{2m}-2\mu g\]
B) \[\frac{F}{2m}\]
C)\[\frac{F}{2m}-\mu g\]
D)zero
Correct Answer:
C
Solution :
In fig, direction of friction is shown, \[F-T-\mu mg=ma\] \[T-\mu mg=ma\] Solving the equations, \[a\,\,=\,\,\frac{F-2\mu mg}{2m}\]