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The emf of a battery A is balanced by a length of 80cm on a potentiometer wire. The emf of a standard cell 1V is balanced by 50cm. The emf of A is

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1.4v

1.5v

1.6v

answer is D.

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Detailed Solution

$E \propto l$

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

$\Rightarrow \frac{E_{1}}{1} = \frac{80}{50}$

$\Rightarrow E_{1} = 1.6 V$

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