Question

The e.m.f of the driver cell of a potentiometer is $2\text{V}$ and its internal resistance is negligible. The length of the potentiometer wire is $100\text{cm}$ and resistance is $5\Omega$. How much resistance is to be connected in series with the potentiometer wire to have a potential gradient of $0.05\text{mV/cm}$ ?

• A. $1990\Omega$
• B. $2000\Omega$
• C. $1995\Omega$
• D. None of these

Answer 1

The correct option is C $1995\Omega$

Given,
$E=2\text{V};R=5\Omega ;l=100\text{cm}$
$K=0.05\text{mV/cm}=0.05\times {10}^{-3}\text{V/cm}$


If $R^{\prime }$ is the external resistance connected with the battery of zero internal resistance and a potentiometer wire having resistance $R$ then,

$i=\frac{E}{R+R^{\prime }}=\frac{Kl}{R}$

After putting the value of given data

$\frac{2}{5+R^{\prime }}=\frac{(0.05\times {10}^{-3})\times 100}{5}$

$10000=25+5R^{\prime }\Rightarrow 5R^{\prime }=9975$

$\therefore R=1995\Omega$

Hence, option $\left(c\right)$ is correct.