Two block A and B of masses 4 kg and 12 kg are placed on a smooth horizontal plane, a force F of 16N is applied on A, as shown, then the contact force between the block is

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Both blocks move together (smooth plane, contact maintained), so the common acceleration is

$a = \frac{F}{m _{A} + m _{B}} = \frac{16}{4 + 12} = 1 m / s^{2}$

The contact force N is the only horizontal force on block B, giving it this acceleration:

$N = m_{B} a = 12 \times 1 = 12 N$

Correct option: (C).

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