In the arrangement shown in figure all surfaces are smooth. Select the correct alternative $\left(s\right)$:

$N_A=m_{A}g\cos \theta \_\_\_\_\_\_\_\_\_\boxed{1}$

$m_{A}g\sin \theta +N=m_A{a}_A\_\_\_\_\_\_\boxed{2}$

$\Rightarrow a_A=g\sin \theta +\frac{N}{m_A}$

FBD of B

$N_B=m_{B}g\cos \theta \_\_\_\_\_\_\_\_\_\boxed{1}$

$m_{B}g\sin \theta -N=m_B{a}_B\_\_\_\_\_\_\boxed{2}$

$\Rightarrow a_B=g\sin \theta -\frac{N}{m_B}$

For normal to be not zero, 

$a_A<a_B$

Which is not possible

Therefore normal force between the blocks is zero. Consecutively accelerations of the blocks will be same