In a potentiometer experiment two cells of emf E1 and E2 are used in series and in conjunction and the balancing length is found to be 58 cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29 cm. The ratio (E1/E2) of the emfs of the two cells is

Question

In a potentiometer experiment two cells of emf E1 and E2 are used in series and in conjunction and the balancing length is found to be 58 cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29 cm. The ratio (E1/E2) of the emfs of the two cells is (A) 1: 1 (B) 2: 1 (C) 3: 1 (D) 4: 1

Tardigrade Admin · Accepted Answer

E1+E2=k × 58 ldots (i) E1-E2=k × 29 ldots (ii) ∴ (E1+E2/E1-E2)=2 ⇒ (E1/E2)=(3/1)

$1 : 1$

$2 : 1$

$3 : 1$

$4 : 1$

$E_{1} + E_{2} = k \times 58 \dots$ (i)
$E_{1} - E_{2} = k \times 29 \dots$ (ii)
$∴ \frac{E _{1} + E _{2}}{E _{1} - E _{2}} = 2$
$\Rightarrow \frac{E _{1}}{E _{2}} = \frac{3}{1}$

Q. In a potentiometer experiment two cells of emf E1​ and E2​ are used in series and in conjunction and the balancing length is found to be 58cm of the wire. If the polarity of E2​ is reversed, then the balancing length becomes 29cm. The ratio E2​E1​​ of the emfs of the two cells is

1:1

2:1

3:1

4:1

E1​+E2​=k×58… (i) E1​−E2​=k×29… (ii) ∴E1​−E2​E1​+E2​​=2 ⇒E2​E1​​=13​

$1 : 1$

$2 : 1$

$3 : 1$

$4 : 1$

$E_{1} + E_{2} = k \times 58 \dots$ (i)
$E_{1} - E_{2} = k \times 29 \dots$ (ii)
$∴ \frac{E _{1} + E _{2}}{E _{1} - E _{2}} = 2$
$\Rightarrow \frac{E _{1}}{E _{2}} = \frac{3}{1}$