Introduction:
In this scenario, there are two blocks, A and B, of masses m1 and m2 respectively, placed in contact on a frictionless table. The experimenter applies a force on block A, causing it to accelerate. As a result, block A exerts a force on block B. We need to determine the force exerted by the experimenter on block A.

Explanation:
To understand the force exerted by the experimenter on block A, we need to consider Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In this case, the force exerted by block A on block B is the action, and the force exerted by block B on block A (and hence the experimenter) is the reaction.

Force exerted by block A on block B:
When block A is pushed by the experimenter, it experiences an applied force. Due to this force, block A accelerates. According to Newton's second law of motion, the force exerted on an object is equal to its mass multiplied by its acceleration (F = ma). Here, the force exerted by block A on block B is equal to mass m1 of block A multiplied by its acceleration.

Force exerted by block B on block A:
As per Newton's third law, the force exerted by block B on block A is equal in magnitude but opposite in direction to the force exerted by block A on block B. Therefore, the force exerted by block B on block A is also equal to mass m1 of block A multiplied by its acceleration.

Force exerted by the experimenter on block A:
Since block A and the experimenter are in contact and form an interacting system, the force exerted by block B on block A is the force exerted by the experimenter on block A. Therefore, the force exerted by the experimenter on block A is equal to mass m1 of block A multiplied by its acceleration.

Conclusion:
In summary, when the experimenter pushes block A causing it to accelerate, the force exerted by the experimenter on block A is equal to mass m1 of block A multiplied by its acceleration. This force is equal in magnitude but opposite in direction to the force exerted by block A on block B, according to Newton's third law of motion.

Ans:(m1+m2)a