JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Current Electricity

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In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm . If this cell is replaced by another cell of emf

E

. the length-of null point increases by 40 cm . The value of

E

\frac{x}{10} V

. The value of

x

is ____ .

Answer: 25

Solution

On a potentiometer the balancing length is proportional to the emf, so

\frac{E _{1}}{E _{2}} = \frac{l _{1}}{l _{2}}

. The new null point is at

l_{2} = 60 + 40 = 100 cm

, hence

E_{2} = E_{1} \frac{l _{2}}{l _{1}} = 1.5 \times \frac{100}{60} = 2.5 V = \frac{25}{10} V,

x = 25

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer

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