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Q.

A student finds the balancing length to be ‘l’ with a cell of constant emf in the secondary circuit. Another student connects the same cell in the secondary circuit of a potentiometer of double the length but with a cell of half the emf in the primary circuit. The balancing length will be [cell in the primary is ideal and no series resistance is present in the primary circuit.]

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a

$\frac{l}{4}$

b

2l

c

4l

d

$\frac{l}{2}$

answer is A.

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Detailed Solution

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From $\frac{E_{0}}{E} = \frac{L}{ℓ} =$ constant
$\begin{array}{l} \Rightarrow \frac{E_{O_{1}}}{E_{O_{2}}} = \frac{L_{2}}{L_{1}} \times \frac{ℓ_{1}}{ℓ_{2}} \frac{1}{2} \frac{E_{O_{1}}}{E_{O_{2}}} = \frac{2 L_{1}}{L_{1}} \times \frac{ℓ}{ℓ_{2}} \\ \Rightarrow ℓ_{2} = 4 ℓ \end{array}$

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