Question

A potentiometer wire of length $10m$ is connected in series with a battery. The e.m.f. of a cell balances against $250cm$ length of wire. If length of potentiometer wire is increased by $1m$, the new balancing length of wire will be

• A. $2.00m$
• B. $2.25m$
• C. $2.50m$
• D. $2.75m$

Answer 1

The correct option is D $2.75m$

$l=10m$

Balancing length$=250cm=2.5m$

Lets, voltage drop across $10m=VVolt$

$⟹$ Voltage drop across $1m=\frac{V}{10}=0.v$

$⟹$ Voltage drop across $2.5m=0.25V$

Now, we know the e.m.f of cell=$0.25V$

Case=$II$ When $l=11m$

Lets, voltage drop across $11m=VVolt$

To drop V potential we use $11m$ long wire.

To drop $1V$ of potential we will find the balancing length at $=\frac{}{V}m$

To drop $0.25V$ balancing length will be $=\frac{1}{1}V\times 0.25V$

$=2.75m$