Question

The length of a potentiometer wire is 10m. The distance between the null points on its wire corresponding to two cells comes out to be 60 cm. If the difference of emf's of the cells is 0.4 volt then the potential gradient on potentiometer wire will be

• A. 0.67 V/m
• B. 0.5 V/m
• C. 2.5 V/m
• D. 0 V/m

Answer 1

Answer: (A)

0.67 V/m

Let, $E_1$ and $E_2$ be the emf's of two cells, $l_1$ and $l_2$ be the balancing lengths obtained for two cells respectively.
Given that,

$E_1-E_2=0.4V$
and

$l_1-l_2=60cm=0.6m$

Now, the potential is the fall of potential per unit length of potentiometer wire,

hence

$k=\frac{E_1-E_2}{l_1-l_2}$

$k=\frac{0.4}{0.6}$

$k=0.67V{m}^{-1}$