Question

In a potentiometer, auniform potential gradient of $0.8 \,V \,m^{-1}$ is maintained across its $10 \,m$ wire. The potential difference across two points on the wire located at $0 \,m \,65\, cm$ and $2 \,m\, 45 \,cm$ is

• A. $1.44 \,V$
• B. $0.144 \,V$
• C. $1.96\, V$
• D. $0.196\, V$

Answer 1

Answer: (A)

$1.44 \,V$

Answer 2

Here,
Length of the wire $= 0.8\, m$
Potential gradient of the wire, $k=0.8\,V\,m^{-1}$
The potential at point $A$ located at $0\,m$ $65\,cm$
$(= 0.65 m)$ on the wire is
$V_{A}=kl=(0.8\,V\,m^{-1})(0.65\,m)=0.52\,V$
and that at point $B$ located at $2\, m$ $45\, cm (=2\,m+0.45\,m=2.45\,m)$ on the wire is
$V_{B}=(0.8\,V\,m^{-1})(2.45\,m)=1.96\,V$
$\therefore$ The potential difference between across $A$ and $B$ is
$\Delta\,V=V_{B}-V_{A}=1.96\,V-0.52\,V$
$=1.44\,V$