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A potentiometer having a wire `10 m` long stretched on it is connected to a battery having a steady voltage. A length of potentiometer wire is increase by `100 cm`, find the new position of null point. |
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Answer» Correct Answer - ` 825 cm` Let `x` be the emf of battery used in potentiometer wire `= 10 m = 1000 cm ` emf of Lenlache cell `= (epsilon)/(1000) xx 750 V` New length of potentimeter wire `= 1000 + 100 = 1100 cm ` Let `l_(1)` be the balancing length of potentiometer wire be the emf of lachacles cell .Then `(epsilon)/(1000) xx 750 = (epsilon)/(1100) l_(1)` or `l_(1) = (1100 xx 750)/(1000) = 825 cm` |
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