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Q.

In given potentiometer, e.m.f of primary cell is 40 V and its internal resistance is $2 Ω$ , resistance of wire AB is $4 Ω$ . In primary circuit, there is also a rheostat whose resistance varies as $R_{h} = (2 x + 2) Ω$ . If balance length obtained for x = 0 is $l_{1}$ and x = 2 is $l_{2}$ respectively, then find $\frac{l_{1}}{l_{2}}$ .

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a

$\frac{3}{4}$

b

$\frac{2}{3}$

c

$\frac{4}{3}$

d

$\frac{3}{2}$

answer is C.

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Detailed Solution

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Let resistance per unit length of wire $' A B'$ be k
Case (i) :
x = 0 balance length : $l_{1}, R_{h} = 2 Ω$
$i = \frac{40}{4 + 2 + 2} \Rightarrow i = 5 A$ $e = (10 / 3) k l_{2}$
At balance length
$e = 5 k l_{1}$ ………(1)
Case (ii) :
$x = 2 \to$ Balance length $l_{2}, R_{h} = 6 Ω$ $e = (10 / 3) k l_{2}$

$i = \frac{40}{4 + 2 + 6} \Rightarrow i = 10 / 3 A$

At balance length,
$e = (10 / 3) k l_{2}$ ……..(2)
From (1) and (2), $5 x l_{1} = (10 / 3) k l_{2}$

$\Rightarrow \frac{l_{1}}{l_{2}} = \frac{2}{3}$

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