A thin uniform wire AB of length 50 cm and resistance `1 Omega` is connected to be terminals of a battery of emf `epsilon_(1) = 2.2 V` and internal resistance `0.1 Omega`. If the terminals of another cell (assume ideal) are connected to two points 25 cm apart on the wire AB without altering the current in the wire AB, the emf `epsilon_(2)` of cell in volts is :

Options

A. 0.5 V
B. 1 V
C. 1.2 V
D. 0.8 V

Answer

B

Explanation

current `u = (epsilon_(1))/(r + R)= (2.2)/(0.1 +1) = 2A`
resistance of 25 cm wire
so emf `epsilon_(2)=2A xx(1)/(2)=1V`.