Question

What is ratio between resistance of wire whose length is l and when it's length increased n times of the original length l.

 1/n square

 n square

 n times

 Same​

Answer 1

Let l be the original length of A.

Original area of cross-section, then

Original resistance R=

A

ρl

Now, length of the wire =l

′

=nl

If A

′

be the new cross-sectional area, then l

′

A

′

=lA

(∵ volume of metal is a constant).

A

′

=

l

′

lA

=

nl

lA

=

n

A

New resistance of the wire is

R

′

=

Al

ρl

′

=

n

A

ρ(nl)

=

l

ρnl

×

A

n

=n

2

(

A

ρl

)=n

2

R

Answer 2

$</p><p></p><p> \rho = R \frac{l}{A} \\ \\</p><p></p><p>R = \rho \frac{A}{l} \\ \\ </p><p></p><p>\sf According \: to \: the \: question, \\ </p><p> \rightarrow \quad R_{1} = \rho \frac{A}{l} \\ </p><p> \rightarrow \quad R_{2} = \rho \frac{A}{ n \cdot l} \\ \\ </p><p></p><p>\sf \quad Ratio = \frac{R_{1} }{ R_{2} } </p><p></p><p>\\ \\</p><p></p><p>$

$</p><p> \rightarrow \quad \frac{ \cancel{\rho} \frac{A}{l} }{ \cancel{\rho } \frac{A}{n \cdot l} } \\ \\ \rightarrow \quad \frac{ \frac{A }{l} }{ \frac{A}{n \cdot l} } \\ \\ \rightarrow \quad \frac{ \cancel{A}}{ \cancel{l}} \times \frac{n \cdot \cancel{l}}{ \cancel{A}} \\ \\ \rightarrow \quad n \\ \\ </p><p>$

Hence, The ratio is n