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The length of a wire of a potentiometer is 100 cm and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is If the balance point is obtained at $l=30$ cm from the positive end, the e.m.f. of the battery is :

Option 1)
$\frac{30E}{100.5}$

Option 2)
$\frac{30E}{100-0.5}$

Option 3)
$\frac{30E}{100}-0.5 i$ where i is the current in the potentiometer wire

Option 4)
$\frac{30E}{100}$

Answers (1)

best_answer

As we learnt in

Potential gradient -

$V=iR= \left (\frac{e}{R+R_{n}+r} \right )R$

$x=\frac{V}{L}=\frac{e}{(R+R_{n}+r)}\frac{R}{L}$

- wherein

$r-$ internal resistance

Equation of cell -

$E=V+iR$

- wherein

$(E>V)$

Potential gradient along wire $= \frac{E}{100}$

$K= \frac{E}{100}\frac{v}{cm}$

For battery $V=E-ir$

$E$ = E.m.f. of battery

$K\times 30= E-ir$ Where current $i$ is drawn from battery.

$\frac{E}{100}\times30=E-ir$

$\frac{E}{100}\times30=E+ir$

$\Rightarrow \frac{E}{100}\times300=E+0.5i$

$E=\frac{30E}{100}-0.5i$

Option 1)

$\frac{30E}{100.5}$

This option is incorrect.

Option 2)

$\frac{30E}{100-0.5}$

This option is incorrect.

Option 3)

$\frac{30E}{100}-0.5 i$ where i is the current in the potentiometer wire

This option is correct.

Option 4)

$\frac{30E}{100}$

This option is incorrect.

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