A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_{0}$ and a resistance $r_{1}$ . An unknown e.m.f. E is balanced at a length $l$ of the potentiometer wire. The e.m.f. E will be given by

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$\frac{L E_{0} r}{l r_{1}}$

$\frac{E_{0} r}{(r + r_{1})} \cdot \frac{l}{L}$

$\frac{L E_{0} r}{(r + r_{1}) l}$

$\frac{E_{0} l}{L}$

answer is D.

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Detailed Solution

Given, length of potentiometer wire = L
Resistance of the wire = r
Resistance per unit length $(ρ) = \frac{r}{L}$
Potentiometer battery emf = E₀
Internal resistance = r₁
$∴$ Current length of the potentiometer wire i is $i = \frac{E_{0}}{(r + r_{1})}$ [From Ohm's Law]
When the unknown emf E is balanced, there will be no flow of current through it.
The balancing length is given as l .
The emf E is equal to the potential drop across length l .
$E = i \times R_{ℓ},$
where i is the current and R is the resistance of the length l .
$\begin{array}{l} \Rightarrow R_{ℓ} = ℓ \times ρ \\ = ℓ \times (\frac{r}{L}) \\ \Rightarrow E = \frac{E_{0}}{(r + r_{1})} \times (\frac{r}{L}) \times ℓ \\ \frac{E_{0} r}{(r + r_{1})} \times \frac{ℓ}{L} \end{array}$