Question

A wire of resistance 32 ohm is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire and the percentage change in resistance?

Answer 1

I hope you got your answer

Answer 2

Answer:

The new resistance of the wire is 16 ohm and the percentage change is 50%.

Explanation:

Given that,

Resistance of wire = 32 ohm

Let be the length of the wire l.

The resistance of the wire is defined as,

$R = \dfrac{\rho l}{A}$

Where,

R = resistance

$\rho$ = resistivity of the wire

l = length

A = area of cross section

The resistance of the wire is

$R = \dfrac{\rho l_{1}}{A_{1}}$....(I)

When the wire is melt and reacts into a wire of half length, its volume will remain unchanged.

Volume of cylindrical wire = length x area of cross section

$l_{1}A_{1}=l_{2}A_{2}$

$l_{1}A_{1}=\dfrac{l_{1}}{2}A_{2}$

$A_{2}=2A_{1}$

The new resistance of the wire is

$R'=\dfrac{\rho\times l_{2}}{A_{2}}$....(II)

From equation (I) and (II)

$\dfrac{R}{R'}=\dfrac{\dfrac{\rho l_{1}}{A_{1}}}{\dfrac{\rho l_{2}}{A_{2}}}$

$\dfrac{R}{R'}=\dfrac{2l_{1}\times2A_{2}}{l_{1}A_{2}}$

$\dfrac{32}{R'}=4$

$R'=\dfrac{32}{4}$

$R'=16\Omega$

Therefore, 50% change in resistance.

Hence, The new resistance of the wire is 16 ohm and the percentage change is 50%.