Question

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40$\Omega$ maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V gives a balance point at 67.3 cm length of the wire. The standard cell is then replaced by a cell of unknown emf E and the balance point, similarly turns out to be at 82.3 cm, length of the wire. Then the E is

• A. 1 V
• B. 1.25 V
• C. 1.5 V
• D. 1.75 V

Answer 1

The correct option is B 1.25 V

The balance condition for a potentiometer : $\frac{E_1}{E_2}=\frac{l_1}{l_2}$

Here $E_1=1.02V,l_1=67.3cm,l_2=82.3cm$ and $E_2=?$

so, $\frac{1.02}{E_2}=\frac{67.3}{82.3}$ or $E_2=1.25V$