A thin semicircular conducting ring of radius R is falling with its...

The induced motional emf in an arbitrarily shaped moving conductor in a uniform magnetic field is equivalent to the emf induced in a straight conductor connecting its two ends.

The effective length of the semicircular ring is the straight line distance between its ends M and Q.

Effective length $l = 2R$

The magnitude of the motional emf developed is given by $\varepsilon = Bvl$

Substituting the values, we get $\varepsilon = B \cdot V \cdot 2R = 2RBV$

To determine the polarity, we consider the magnetic force on a positive charge carrier, given by $\vec{F} = q(\vec{V} \times \vec{B})$

Since the velocity vector $\vec{V}$ is downwards and the magnetic field vector $\vec{B}$ is into the plane of the paper, the cross product $\vec{V} \times \vec{B}$ points towards the right.

Positive charges are pushed towards end Q, making Q at a higher potential than end M.

Answer: $2 RBV$ and Q is at higher potential