Question

The length of a potentiometer wire is $1200\, cm$ and it carries a current of $60\, mA$. For a cell of emf $5\, V$ and internal resistance of $20 \, \Omega$, the null point on it is found to be at $1000\, cm$. The resistance of whole wire is :

• A. 120 $\Omega$
• B. 80 $\Omega$
• C. 60 $\Omega$
• D. 100 $\Omega$

Answer 1

Answer: (D)

100 $\Omega$

Answer 2

$5=\lambda\ell$ where $\lambda$ is potential gradient $\& L$ is total length of wire. $5=\frac{\Delta V}{L}\ell$ $\Delta V=\frac{5\times L}{\ell}=5\times\frac{12}{10}=6\,V=60\,mA\times R$ $R=100\,\Omega$