Question

A 100 ohm thick wire is stretched so that its length becomes three times find the new resistance and
resistivity.
​

Answer 1

Answer:

New resistance = 900 ohm and resistivity will be same.

Explanation:

We know that

R = ρL/A

[Where, R is resistance, ρ is resistivity (specific resistance), L is length of wire and A is the area of cross section of the wire.]

Multiplying L in both numerator and denominator we get

R = ρL/A × L/L

⇒ R = ρL²/(A × L)

We know that Area of cross section × length of wire = volume of wire

So, we can write R = ρL²/V

Now, the wire is stretched 3 times, so new length L' = 3L

So, new resistance R' =

R' = ρL'²/V  (Volume will be constant as the wire is only stretched)

R' = ρ(3L)²/V

R' = ρ9L²/V

Taking ratio of R to R' we get

⇒ R/R' = ρL²/V × V/9L²ρ

⇒ R/R' = 1/9

We are given that the wire had resistance 100 ohm earlier. So R = 100 ohm. Substituting for R we get

100/R' = 1/9

⇒ R' = 900 ohm.

So the new resistance of wire is 900 ohm. The resistivity depends on material and not length or area, so the resistivity will remain constant.

Answer 2

Answer:

New resistance = 900 ohm.

Resistivity will be same.

Explanation:

Given Problem:

A 100 ohm thick wire is stretched so that its length becomes three times find the new resistance and  resistivity.

​

Solution:

To Find:

The new resistance and  resistivity.

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Method:

We know that,

$\bigstar R = \dfrac{pL}{A} \bigstar$

Where,

R is resistance,

p is resistivity

L is length of wire

A is the area of cross section of the wire.

Now,

Multiply L in both numerator and denominator we will get

$R = \dfrac{pL}{A} \times\dfrac{L}{L}$

$\implies\ R = \dfrac{pl^2}{(A\times\L)}$

We also know that,

$\bigstar Area\: of\: cross\: section \times length\: of\: wire = volume\: of\; wire \bigstar$

We can also write it as:

$\bigstar R = \dfrac{pL^2}{V} \bigstar$

Now,

According to your question;

The wire is stretched 3 times.

It implies that,

New length L = 3L

So,

New resistance R =   ?       (To Find)

So,

$R = \dfrac{pL^2}{V}$

We know that,

⭐If wire only stretched Volume will be constant⭐

$R = \dfrac{p(3L^2)}{V}$

$R = \dfrac{p9L^2}{V}$

Now,

Take ratio of R to R we will get,

$\dfrac{R'}{R} = \dfrac{pL^2}{V} \times V \times\dfrac{V}{9L^2p}$$\dfrac{R'}{R}= \dfrac{1}{9}$

Here,

Given that the wire had resistance 100 ohm earlier. So R = 100 ohm.

Now,

Substituting for R we will get,  

$\dfrac{100}{R'}= \dfrac{1}{9}$

So,

⇒ R' = 900 ohm.

Hence,

It implies that new resistance of wire is 900 ohm.The resistivity will remain constant.