Question

A wire of resistance 40 ohms is stretched to twice its length. Find the value of new resistance of wire.

Answer 1

Answer:

The value of the new resistance of the wire is 160Ω.

Relation between resistance and resistivity: $R = \frac{rho*l }{A}$ where rho is resistivity, l is the length and A is the area of cross-section of the material.

Explanation:

Given, the resistance of the wire is 40Ω.

Let's assume initial resistance as R₁ and new resistance as R₂. Now we have R₁/R₂ = l₁A₂/l₂A₁.

But we know that after stretching volume cannot change so we have

                             l₁A₁ = l₂A₂ but l₂ = 2l₁ which implies that A₂ = A₁/2.

By simplifying the ratio we obtain that R₁/R₂ = 1/4 which implies that R₂ = 4R₁.

Therefore new resistance is 4*40 = 160Ω.

Answer 2

Answer:

Concept:

This comes under current electricity and electrical resistance and resistivity

Explanation:

Make the wire's length and cross-section area l and A, respectively

$\therefore Resistance \mathrm{R}=\rho \frac{1}{\mathrm{~A}} \\Given: \mathrm{R}=40 \Omega \\For 1^{\prime}=21 and A^{\prime}=\frac{A}{2}$

Resistance of the wire be R'

Where $\rho$ is the resistivity of the material.

$\begin{aligned}&\therefore \mathrm{R}^{\prime}=\rho \frac{1^{\prime}}{\mathrm{A}^{\prime}}=\rho \frac{21}{\mathrm{~A} / 2}=4 \rho \frac{1}{\mathrm{~A}}=4 \mathrm{R} \\&\therefore \mathrm{R}^{\prime}=4 \times 40 \Omega=160 \Omega\end{aligned}$

Hence the value of new resistance of wire is $160 \Omega$

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