Question

A potnetiometer wire, 10 m long, has a resistance of 40 $\Omega$. It is connected in series with a resistance box and a 2 V storage cell. If the potential gradient along the wire is 0.1 m V/cm, the resistance unplugged in the box is:

• A. 260 $\Omega$
• B. 760 $\Omega$
• C. 960 $\Omega$
• D. 1060 $\Omega$

Answer 1

The correct option is B 760 $\Omega$

Potential gradient along wire $=\frac{\text{potential difference along wire}}{\text{length of wire}}$
or $0.1mV/cm=\frac{I\left(40\right)}{10}$ where $I=$ current in wire.
or $\frac{0.1\times {10}^{-3}}{{10}^{-2}}V/m=\frac{I\left(40\right)}{10}$
or $I=\frac{1}{400}A$
Let $R$ be the resistance unplugged in the box . when it is connected in series ,the current in the wire is
$I^{\prime }=\frac{2}{R+40}$
As the potentiometer is connected in series with R so same current will be flow in wire.
thus, $I^{\prime }=I\Rightarrow \frac{2}{R+40}=\frac{1}{400}$
or $R+40=800$
$R=800-40=760\Omega$