Question

A potentiometer wire has a length of $10\text{m}$ and resistance $4\Omega {\text{m}}^{-1}$. An accumulator of emf $2\text{V}$ and a resistance box are connected in series with it. Calculate the resistance to be introduced in the box to get a potential gradient of $0.1\text{V/m}$.

• A. $20\Omega$
• B. $30\Omega$
• C. $40\Omega$
• D. $50\Omega$

Answer 1

The correct option is C $40\Omega$

Given:
$l=10\text{m};R/l=4\Omega {\text{m}}^{-1};E=2\text{V}$
$V/l=0.1\text{V/m}$

Resistance of the potentiometer wire,

$R=10\text{m}\times 4\Omega {\text{m}}^{-1}=40\Omega$

Let $r$ be the resistance introduced in the box.

So, the current drawn in the circuit,

$i=\frac{E}{R+r}$

Now, potential difference across potentiometer wire,

$V=iR=\frac{ER}{R+r}$

Thus, potential gradient will be

$\frac{V}{l}=\frac{ER}{(R+r)l}$

$\Rightarrow 0.1=\frac{2\times 40}{(40+r)\times 10}$

$\Rightarrow 40+r=80$

$\therefore r=40\Omega$

Therefore, option $\left(C\right)$ is the right answer.