Question

A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.​

Answer 1

Answer:

Hey There!

The resistance R of piece of power ia 20 Ω

Formula to find R are as follows:

$\mathcal{p}= \frac{r \times a}{l} \\ \\ r = \frac{p \times l}{a}$

Here R is the resistance of the wire, A is cross-sectional area of wire, l is the length of wire and p is the resistivity of the wire.

Consider the cross-sectional area A of the wire is halved and become ( A/2), the length l of wire is doubled and becomes 2l, the new resistance R' of the wire is

${R}^{'} = \frac{ \mathcal{P} \times (2l)}{(A/2)} \\ \\ {R}^{'} = 4r \\ \\ {R}^{'} = 4 \times 20 \\ \\ {R}^{'} = 80 Ω$

Hence, the new resistance of the wire is 80 Ω

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Answer 2

Answer:

Hey mate, the resistance of the wire is 80 ohm.

Explanation:

Refer to the attachment the solve the answer.

Hope its help u

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