The potentiometer consists of a wire of length 4 m and resistance 10 Ω. It is connected to a cell of emf 2V the potential difference per unit length of the wire will be :

Question

The potentiometer consists of a wire of length 4 m and resistance 10 Ω. It is connected to a cell of emf 2V the potential difference per unit length of the wire will be : (A) 10 V/m (B) 5 V/m (C) 2 V/m (D) 0.5 V/m

Tardigrade Admin · Accepted Answer

Potential difference across any portion of the wire.is proportional to the length of that portion.
We know that
voltage

=

current

\times

resistance

∴ i = \frac{E}{R}

= \frac{2}{10} = 0.2 A

From the principle of potentiometer when a constant current is passed through a wire of uniform cross-section the potential difference across any portion of the wire is directly proportional to the length of that portion i.e.,

V = k L

So, potential difference per unit length of the wire, i.e., potential gradient

k = \frac{V}{l}

∴ = \frac{i R}{l} = \frac{0.2 \times 10}{4}

= 0.5 V / m .

Q. The potentiometer consists of a wire of length 4m and resistance 10Ω. It is connected to a cell of emf 2V the potential difference per unit length of the wire will be :

10 V/m

5 V/m

2 V/m

0.5 V/m

Q. The potentiometer consists of a wire of length $4 m$ and resistance $10 Ω.$ It is connected to a cell of emf $2 V$ the potential difference per unit length of the wire will be :