Question

resistance of a wire is 10 ohm if we increase the area of wire by 3 times than the initial one. Calculate the new resistance of the wire. ​

Answer 1

Given

Original resistance R = 10 ohm

Let be length = l and area of cross section be a.

So ,resistance be

$R= \rho \times \frac{length}{area}$

$10 = \rho \times \frac{l}{a}$

According to question its given ,

The area increases by 3 times,

So , new area will be $a_2 = 3a$

length will be decreased by = $l_2 = \frac{1}{3}$

So new resistance will be $R_2$

$R_2 = \rho \times \frac{lenght }{l area}$

$R_2 = \rho \times { \frac{\frac{l}{3}}{3a}}$

$R_2 = \rho \times \frac{l }{9a}$

On dividing original resistance by new resistance.

$\frac{R}{R_2}= \frac{ \rho \times \frac{l}{a}}{\rho \times \frac{l }{9a}}$

$\frac{10}{R_2}= \frac{ \cancel{ \rho} \times \frac{l}{a}}{ \cancel{ \rho} \times \frac{l }{9a}}$

$\frac{10}{R_2}= \frac{l}{a} \times \frac{9a}{l}$

$\frac{10}{R_2}= \frac{ \cancel{l}}{ \cancel{a}} \times \frac{9 \cancel{a}}{\cancel{l}}$

$\frac{10}{R_2}= \frac{9}{1}$

on cross multipying,

$\frac{10}{9} = R_2$

Hence , the new resistance will be one- ninth of the original one.

Answer 2

Formula for resistance is as follows:

resistance(R) = restivity(r) x Length (L)/ Area (A)

R=rL/A

When the length increases by three, the cross section will reduce by three. Hence;

the Length will be 3L

while area= A/3  

New resistance will be r3L/(A/3)  = r9L/A

If rL/A = 10ohms

Then 9(rL/A)= 10 x9 rL/A x(A/rL)  

                    =90

Therefore the new resistance = 90ohm