A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied on B for the blocks to move together will be

Idea: the friction between the blocks is the same ceiling in both cases. Whichever block is pushed, that friction must drive the other one, so the two limiting forces are in the ratio of the masses they have to move.

Take the limiting friction between A and B as the 12 N the stem supplies. When F is applied to B, the pair accelerates together at

a=F/(m_A+m_B)=F/9,

and A is carried along by friction alone:

f=m_Aa=(4F)/9.

Setting f to its ceiling,

(4F)/9=12, so F=(12×9)/4=27 N.

Two cautions about this question. The stem says 'minimum' where the physics gives a maximum — on a smooth table the blocks move together for small forces and slide apart for large ones. And the printed answer only follows if the 12 N is the limiting friction itself; reading it as the largest force applicable to A would give a limiting friction of 12×5/9=6.7 N and an answer of 15 N, which is not among the options.