Q60 with all steps pls n emf balanced against 600 cm length is
e = kl =0.0002 x 600 = 0.12 V
le 60 In the circuit diagram given below, AB is
a uniform wire of resistance 15 ohm and length one
metre. It is connected to a series arrangement of cell €1
of emf 2.0 V and negligible internal resistance and a
resistor R. Terminal A is also connected to an electro-
chemical cell €2 of emf75 mV and a galvanometer G. In
this set-up, a balancing point is obtained at 30 cmmark
from A. Calculate the resistance of R. If €2 were to have
an emf of 300 mV, where will you expect the balancing
D 990
point to be ?
a
s
€1
€2
Figure 3.62
Solution. Current through the potentio-
meter wire,
2
