In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell?

≃ 1.57 V

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≃ 1.67 V

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≃ 1.47 V

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≃ 1.37 V

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Solution

The correct option is B $≃ 1.67 V$
Given : $E_{1} = 1.25$ V $l_{1} = 30$ cm $l_{2} = 40$ cm
Emf of the cell $E \propto l$
$⟹$ $\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}}$
$∴$ $\frac{E_{2}}{1.25} = \frac{40}{30}$
$⟹$ $E_{2} = 1.67$ V

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In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell?

The correct option is B ≃1.67VGiven : E1=1.25 V l1=30 cm l2=40 cmEmf of the cell E∝l⟹ E2E1=l2l1∴ E21.25=4030⟹ E2=1.67 V

The correct option is B $≃ 1.67 V$
Given : $E_{1} = 1.25$ V $l_{1} = 30$ cm $l_{2} = 40$ cm
Emf of the cell $E \propto l$
$⟹$ $\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}}$
$∴$ $\frac{E_{2}}{1.25} = \frac{40}{30}$
$⟹$ $E_{2} = 1.67$ V