Question

A potentiometer is connected across $A$ and $B$ and a balance is obtained at $64.0\text{cm}$. When the potentiometer is connected across $A$ and $C$, a balance is found at $8.0\text{cm}$. If the potentiometer is now connected across $B$ and $C$, a balance will be found at,

• A. $8.0\text{cm}$
• B. $56.0\text{cm}$
• C. $64.0\text{cm}$
• D. $72.0\text{cm}$

Answer 1

The correct option is B $56.0\text{cm}$

Let the emfs of the cells connected across $AB$ and $AC$ are $E_1$ and $E_2$ respectively and $l_1$ and $l_2$ are their respective balancing lengths, when connected individually
So, $l_1=64\text{cm}$ and $l_1-l_2=8\text{cm}$

For a potentiometer, $V\propto l$

Across points $A$ and $B$, $E_1\propto l_1$

Across points $A$ and $C$, $E_1-E_2\propto l_2$

$\therefore \frac{E_1}{E_1-E_2}=\frac{l_1}{l_1-l_2}=\frac{64}{8}=8$

$\Rightarrow l_1=8l_1-8l_2$

$\Rightarrow l_2=\frac{7l_1}{8}=\frac{64\times 7}{8}=56\text{cm}$

Hence, $\left(B\right)$ is the correct answer.

Key concept- Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.