A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleration α, the torque on the child is τ. The child moves to a position half way between the center and edge of the merry-go-round, and the angular acceleration increases to 2α. The torque on the child is now

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of  the mass of the child times the square of the distance to the center.When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

       

So, τ = 3/2*m*r²*α (2)When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

       

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

       τ = 3/4*m*r² * (2α) = 3/2*m*r²

        same result than in (2), so the torque remains the same.

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The correct answer is c. hope this