Question

A standard cell of 1.08 V is balanced by the p.d. across 90 cms of a meter long wire supplied by a cell of emf 2V through a series resistor of resistance 2 Ω. If the internal resistance of cell in primary circuit is zero then the resistance per unit length of potentiometer wire is :

• A. 3 ohm/cm
• B. 0.3 ohm/cm
• C. 3 ohm/m
• D. 3 ohm/mm

Answer 1

Answer: (C)

3 ohm/m

Answer 2

E = $\left( \frac{E_{o}}{L} \times \frac{R_{w}}{R_{w} + r + R_{h}} \right)$× l (Q r = 0, R_h = 2 Ω)

∴ 1.08 = $\left( \frac{2}{1m} \times \frac{R_{w}}{R_{w} + 2} \right)$ × 0.90 m

$\frac{1.08}{2 \times 0.90}$ = $\frac{R_{w}}{R_{w} + 2}$

Hence : 0.6 = $\frac{R_{w}}{R_{w} + 2}$

0.6 R_w + 1.2 = R_w ⇒ 1.2 = 0.4 R_w ⇒ 3 = R_w

∴ Resistance per unit length is 3Ω /meter