A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3 V and a resistance of 10 Ω. The potential gradient along the wire in V/m is (a) 1.2 (b) 0.10 (c) 0.02 (d) 0.20

Answer: (d). I = (3)/(20 + 10) = 0.1 A. P.d. across wire = 0.1 × 20 = 2 V. k = 2/10 = 0.2 V/m. So the answer is (d) 0.20.

A 10 m long wire of resistance 20 Ω is connected in series with a… | MHT CET 2020 (19 Oct, Shift 2)

Q63MHT CET 2020 · 19 OCT · SHIFT 2Easy

Potentiometer

A 10 m long wire of resistance $20\ \Omega$ is connected in series with a battery of emf 3 V and a resistance of $10\ \Omega$ . The potential gradient along the wire in $\text{V/m}$ is

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Answer: (d)

$I = \frac{3}{20 + 10} = 0.1\ \text{A}$ .
P.d. across wire $= 0.1 \times 20 = 2\ \text{V}$ .
$k = \frac{2}{10} = 0.2\ \text{V/m}$ .
So the answer is (d) 0.20.

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