Question

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Answer 1

given, a cell of emf , $E_1$ = 1.25V gives a balance point at $l_1$= 35cm length of the wire.

now cell is replaced by another cell and the balance point shift to , $l_2$ = 63cm.

we have to find emf of another cell.

we know, The potential gradient remains the same , as there is no change in the setting of standard circuit.

so, $\frac{E_1}{E_2}=\frac{l_1}{l_2}$

or, $E_2=\frac{E_1l_2}{l_1}$

= 1.25 × 63/35

= 2.25 V

hence, emf of the second (another ) cell is 2.25 V

Answer 2

Explanation:

$\Large{\orange{\underline{\underline{\sf{\pink{Solution:}}}}}}$

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$\hookrightarrow$ EMF of the cell, $\sf E_1\:=\:1.25V$

$\hookrightarrow$ Balance point of the potentiometer, $\sf l_1\:=\:35cm$

$\hookrightarrow$ The cell is replaced by another cell of EMF $\sf E_2$

$\hookrightarrow$ New balance point of the potentiometer, $\sf l_2\:=\:63cm$

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The balanced condition is given by the relation,

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$\longrightarrow$ $\boxed{\sf{\blue{\dfrac{E_1}{E_2}\:=\: \dfrac{l_1}{l_2}}}}$

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$\leadsto$ $\sf E_2\:=\:E_1\times \dfrac{l_2}{l_1}$

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$\leadsto$ $\sf E_2\:=\:^{0.25}\cancel{1.25}\times \dfrac{^9\cancel{63}}{\cancel{35}_{\cancel{5_1}}}$

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$\leadsto$ $\sf E_2\:=\:2.25\,V$

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$\longrightarrow$ Therefore, emf of second cell is $\sf{\purple{2.25\,V}}$