A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is l. A conducting massless rodof resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass m, tied to the other end of the string, hangs vertically. A constant magnetic field B exists perpendicular to the table. If the system is released from rest, calculate the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity (in m/s2).

If v is the velocity of the rod at any instant, the emf induced in it will be Bvl. And as the resistance of the rod is R, the induced current will beI=eR=BvlRDue to this induced current a force will act on the rod which will oppose its motion.FM=iBl=B2l2vR   ....(i)So if T is the tension in the string, equation of motion of rod and mass m will beT−FM=0×a and mg−T=maSo eliminating T between thesemg−FM=ma i.e., a=g−FMmand on substituting the value of FM from Eq. (i)a=g−B2l2mRv i.e.,  a=g−vt with τ=mR(Bl)2    .......(ii)So the rod will achieve terminal velocity vT when a= 0, i.e.,  g−vTτ=0 i.e.,  vT=τg=mgR(Bl)2   .......(iii) and so for v=12vT=12τg, from Eq. (ii), a=g−1τ12τg=12g=5m/s2