Question

A wire of resistance x ohm is drawn out, so that its length is increased to twice its original length,and its new resistance becomes 20 Ω then x will be

Answer 1

Step 1: Given that:

Let the initial length of wire is = l,

New length= l'

The initial cross-section of wire = A

New area after stretching = A'.

As given in question l'=2l

That is,

$\frac{l^{\prime }}{l}=2$

$\frac{l}{l^{\prime }}=\frac{1}{2}$

Original resistance = $\text{x}$ ohm

New resistance = R'

$R^{\prime }=20\text{Ω}$

Step 2: Formula used

$R=\frac{(\rho \times l)}{a}$

Where, $\text{ρ}$ = Resistivity of the material of the wire

l = length of the wire

A= Area of cross-section of the wire

Step 3: Finding the relationship with the new area

The volume of the wire remains the same,

Hence, Al must be equal to A'l'

$Al=A^{\prime }{l}^{\prime }$

$\frac{A}{A^{\prime }}=\frac{l^{\prime }}{l}$

$\frac{A^{\prime }}{A}=\frac{l}{l^{\prime }}$

$\frac{A^{\prime }}{A}=\frac{1}{2}$

Step 4: Calculation of the new resistance in terms of original resistance:

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

Thus

$x=\frac{\text{ρ}l}{A}$.............(1)

and after stretching

$R^{\prime }=\frac{(\rho \times l^{\prime })}{A^{\prime }}$ .........(2)

As the resistivity(ρ) of the material will not change with the change in length.

Now, dividing equation (1) from equation (2), we get

$\frac{x}{R^{\prime }}=\frac{\frac{\text{ρ}l}{A}}{\frac{\text{ρ}{l}^{\prime }}{A^{\prime }}}$

$\frac{x}{R^{\prime }}=\frac{l}{l^{\prime }}\times \frac{A^{\prime }}{A}$

$\frac{x}{R^{\prime }}=\frac{1}{2}\times \frac{1}{2}$

$\frac{x}{R^{\prime }}=\frac{1}{4}$

$R^{\prime }=4xohm$

Thus, the new resistance is $4xohm$ .

Step 5: Calculation of the original resistance that is the value of $x$ :

Since, $R^{\prime }=20\text{Ω}$

Then,

$20\text{Ω}=4x$

$4x=20$

$x=\frac{20}{4}$

$x=5\text{Ω}$

Hence, the original resistance of the wire was $\text{5Ω}$ .