Question

with the cell of EMF 2.0 volt the balance point is obtained at a distance of 100 cm from one end of the Potentiometer wire calculate the potential difference between the two ends of wire if total length is 400 CM.
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Answer 1

E1=2V

L1=100cm

potential drop across 100 cm =2V

potential drop across 400 CM = 2×400/100

=8V

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Answer 2

Answer:

The potential difference between the two end of the wire will be 8V if the total length of the wire is 400cm.

Explanation:

The potential gradient shows how quickly potential and displacement are changing. It represents the slope along which potential is changing, in other words.

Potential gradient is the potential drop per unit of wire length.

i.e., k= $\frac{V}{L}$

Let  V be the potential across balance point and one end of wire and L be the length of the wire.

Therefore, the potential gradient(K) = $\frac{2}{100}$

K = 0.02 unit

Thus, the potential gradient is 0.02 unit

Now, we will calculate the Potential difference between the ends of wire of length 400Cm by using the formula:

V = K × L

V = 0.02 × 400

V = 8V

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