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 to B for the blocks to move together will be

Calculation:
Given: Block A has a mass of 4 kg, and block B has a mass of 5 kg. Block B rests on a smooth horizontal table, and the minimum force required to move both blocks together is 12 N.

Let the coefficient of static friction between block A and block B be μ. For the blocks to move together, the frictional force between the two blocks must provide the necessary force to move both blocks as a single unit.

The frictional force, f, between block A and block B is given by:

f = μ × N, where N is the normal force between the two blocks. Since block A is placed on block B, the normal force is equal to the weight of block A, which is:

N = m_A × g = 4 × 10 = 40 N

Therefore, the maximum frictional force is:

f = μ × 40 N

Now, the force required to move both blocks together is:

F = (m_A + m_B) × a, where a is the acceleration of the blocks.

We know that the minimum force required to move the blocks together is 12 N, and this force must overcome the friction between the blocks. Thus, the frictional force must be equal to the applied force for both blocks to move together:

12 N = μ × 40 N

Solving for μ:

μ = 12 / 40 = 0.3

Now, the maximum force that can be applied to block B without sliding is:

F_max = (m_A + m_B) × g × μ = (4 + 5) × 10 × 0.3

F_max = 9 × 10 × 0.3 = 27 N

Thus, the correct option is Option 3: 27 N