Correct Answer - B
Here, Mass of block A, `m_(A)=10kg`
gt Mass of the block B, `m_(B)=15kg`
Coefficient of friction between the blocks and the surface, u=0.2
Applied force=200N
Force of friction of `a=muN_(A)g=0.2xx10xx10=20N`
`"Force of friction on B"=muN_(B)=mum_(B)g=0.2xx15xx10=30N`
Taking two blocks forming one system, therefore net force acting on the blocksis F=200-20=30=150N
Let a be common acceleration of the system
`therefore a=(F)/(m_(A)+m_(B))=(150N)/((10+15))kg=6ms^(-2)`