A potentiometer with a cell of 2.4 volt and internal resistance of $2 \Omega$ maintains a potential drop across the resistance wire AB of length 2 meters and resistance $10 \Omega$. A standard cell which maintains a constant emf of 'V' volt with internal resistance $0.2 \Omega$ gives a balance point at 1.6 m length of the wire. The value of emf of second (standard) cell (V) is:

1.6 volt

The correct answer is Option (4) → 1.6 volt

Internal Resistance of the cell, $r=2Ω$

R, Resistance of the potentiometer wire = $10Ω$

$R_{total}=R+r=10Ω+2Ω=12Ω$

$I=\frac{\text{EMF of Primary cell}}{\text{Total Resistance}}$

$=\frac{2.4V}{12Ω}=0.2A$

$V_{wire}=I×R=(0.2A)×(10Ω)=2V$

Potential Gradient, $R=\frac{V_{wire}}{\text{Total length of wire}}$

$=\frac{2V}{2m}=1V/m$

V (corresponding to balance length) = $k×l$

$=R×1.6$

$=1.6V$