Question

A 10 ohm wire is stretched so that its length becomes 4 times its original length. the new resistance of the wire will be?

Answer 1

The new resistance will be 40 Ohm.

Answer 2

Answer:

The wire's new resistance is 160 ohms.

Explanation:

The wire's resistance is determined as follows:

$\mathrm{R=\frac{\rho l}{A} }$         (1)

Where,

R=resistance of the wire

ρ=specific resistance of the wire (it remains constant for a material)

l=length of the wire

A=area of the cross-section of the wire

From the question we have,

The initial resistance of the wire(R₁)=10 ohm

The initial length of the wire(l₁)=l

The final length of the wire(l₂)=4l

The volume of the wire will remain the same. i.e.

$\mathrm{V_1=V_2}$                   (2)

$\mathrm{A_1l_1=A_2l_2}$             (3)          (Volume=Area×length)

Inserting the value of "l₁" and "l₂" in equation (3) we get;

$\mathrm{A\times l=A_2\times 4l}$

$\mathrm{A_2=\frac{A}{4} }$               (4)

Equation (1) can also be written as,

$\mathrm{\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}}$             (5)

Inserting all the required values in equation (5) we get;

$\mathrm{\frac{10}{R_2}=\frac{l\times \frac{A}{4} }{4l\times A}}$

$\mathrm{\frac{10}{R_2}=\frac{1 }{16}}$

$\mathrm{R_2=10\times 16}$

$\mathrm{R_2=160\ \Omega}$

So, the wire's new resistance is 160 ohms.

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