Question

(b) A wire of length L and resistance R is stretched so that its length is doubled and the
area of cross-section is halved. How will its.
0 Resistance change (ii) Resistivity change​

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

a) Resistance can be given by:

$\cdot\boxed{ \rm{R = \rho \frac{L}{A} }}$

Where $\rho$ is the specific resistance or resistivity, L is the length of the conductor and A is the cross-sectional area.

It is given that,

• Length is doubled.
• Area of cross-section is halved.

New resistance will be:

(Substituting 2L in place of L and A/2 in place of A)

➝ $\rm{R_1 = \rho \dfrac{2L}{ \frac{A}{2} }}$

This is equals to,

➝ $\rm{R_1 = 4 \: \rho \dfrac{L}{A} }$

So, R1 is equals to 4R. Hence, the Resistance is increased by 300% or becomes four times of the initial.

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b) Specific resistance or Resistivity ($\rho$) is independent of the physical dimensions of the conductor like length and Area. It only depends upon the nature of the material. As long as the material remains same, Resistivity is same.

Answer 2

Explanation:

Given :

• Length = L

• Area of cross section =A

• Resistance = R

To Find :

• How will its.0 Resistance change (ii) Resistivity change

Solution :

Resistance = $\sf \rho \frac{Length}{Area of cross section }$

R =$\sf \rho \frac{L}{A}$

The wire is being stretched to double so,

New Length =2L

Resistance is directly proportional to length so if length increase resistance might also have increased.

$\sf R\propto L$

New area of cross section =$\sf \frac{A</p><p>}{2}$

Resistance is inversely proportional to area of cross section so if area of cross section decreases resistance might have increased.

$\sf R\propto \frac{1}{A}$

Let the new resistance be R1.

Resistance =$\sf \rho \frac{Length}{Area of cross section }$

R1=$\sf\rho \frac{2L}{\frac{A}{2}}$

On dividing the original resistance by New resistance,

$\sf\frac{R}{R1}=\rho \frac{L}{A} \div \rho \frac{2L}{\frac{A}{2}}$

$\sf\frac{R}{R1}=\frac{L}{A}\times \frac{\frac{A}{2}}{2L}$

$\sf \frac{R}{R1}=\frac{L}{A}\times \frac{A}{4L}$

ON SOLVING,

$\sf \frac{R}{R1}=\frac{1}{4}$

R1 = 4R.

So, the new resistance will be four times the original resistance.

RESISTIVITY

• The resistivity won't be altered as it doesn't depends on length of area of cross section.

• It only depends on nature and the temperature of the material.