a block A of mass4 kg is placed on another block B of mass 5kg B rests on smooth table for sliding block A onB a horizontal force of 12 N is required to be applied on A how much max force can be applied on Bso that both A andB move together

1st case when the force is applied on the block A

then the total acceleration of both the block together is = Forces/mass = 12 /( 4+5)= 12/9 m/sec²

the pseudo accleating force on the block b must be balance by the frictional force between the blocks 

therefore fricition for μ m₁g = m₂a = m₂* 12/9

now is and force F ( maximum) is applied on the block B.

then acceleration of both blocks together is = Force / mass = F/ 9 meter per second square 

the pseudo force on A must be balanced by the frictional force between the blocks .

therefore 

μ m₁g = m₁a

μ m₁g = m₁F/9

μ m₁g from earlier case

(12m₂)/9 = m₁F/9

F = 15 N