Question bank/JEE Main/Physics

The length of a potentiometer wire is 120 cm and it carries a
current of 60 mA. For a cell of emf 5V 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$60\Omega$B$120\Omega$C$100\Omega$D$80\Omega$

Show answer and solution▾

[IMAGE] Potential gradient
$= \frac{5}{1000} = \frac{V_{p}}{1200}$
$V_{P} = 6V$
And $R_{P} = \frac{V_{P}}{I} = \frac{6}{60 \times 10^{- 3}} = 100\Omega$

Source: Grass-G / P-20 Physcis Paper 12 Dec.docx