Two blocks are in contact on a frictionless table. One has mass m and the other 2m a force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of the force of contact between the two blocks will be:

(A) same
(B) $1:2$
(C) $2:1$
(D) $1:3$

Hint Due to the force of contact both the masses will move together with the same acceleration. Force of contact acting between the two blocks will be the driving force for the movement for block m (when force is acting from the left side) and 2m (when force is acting from the right side).
Formula used: $F = ma$ where F is force, m is mass and a is acceleration

Complete Step-by-step solution
Let, case A be the case in which force is acting on the left side i.e. on block 2m and case B be the case in which force is acting on the right side i.e. on block m.
Due to the force of contact both the masses will move together with the same acceleration.
Hence,
$F = ma$ where F is force, m is mass and a is acceleration
Here, considering both the blocks as one system,
$F = 3ma$
$a = \dfrac{F}{{3m}}$
In case A force of contact will cause movement of the block of mass m.
In case B force of contact will cause movement of a block of mass 2m.
Let ${N_1}$ be the force of contact in case A and ${N_2}$ be the force of contact in case B.
We know that$F = ma$
So,
  
$ \Rightarrow {N_1} = m \times \dfrac{F}{{3m}}$
$ \Rightarrow {N_1} = \dfrac{F}{3}$
So,
  
 $ \Rightarrow {N_2} = 2m \times \dfrac{F}{{3m}}$
$ \Rightarrow {N_2} = \dfrac{{2F}}{3}$
In the question they are asking ratio of ${N_1}$ to ${N_2}$ i.e. ${N_1}:{N_2}$ that is,
\[ \Rightarrow \dfrac{F}{3}:\dfrac{{2F}}{3}\]
$ \therefore 1:2$

$\therefore $ The correct answer is $(B) 1:2$

Note Above we solved considering the free body diagram of the body on which force of contact is acting but we can also solve it by considering the free body diagram of other body by balancing force on it and following $F = ma$where F is force, m is mass and a is acceleration