In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs ?

CONCEPT:

As we know in a potentiometer EMF has two cells in which \({\ell_1}\) and \({\ell_2}\) be the null points and\({\ E_1}\) and \({\ E_2}\) be the EMFs,  therefore we can write it as,

\(\frac{{{E_1}}}{{{E_2}}} = \frac{{{\ell_1}}}{{{\ell_2}}}\)

CALCULATION:

Given:

\(\ \begin{array}{*{20}{l}} {{E_1} = 1.5V} \\ {{E_2} = 2.5V} \\ {{l_1} = 36{\text{ cm}}} \\ {{l_2} = ?} \end{array} \ \)

Using \(\frac{{{E_1}}}{{{E_2}}} = \frac{{{\ell_1}}}{{{\ell_2}}}\) we get; 

 ⇒  \(\frac{{{1.5}}}{{{2.5}}} = \frac{{{\ 36}}}{{{\ell_2}}}\)

 ⇒ \(\ {l_2} = \frac{{36 \times 2.5}}{{1.5}}\ \)

 ⇒ \({l_2} = 60{\text{ cm}}\)

Hence, option B) is the correct answer.