Question

The wire having resistance r is drawn througg a dye so that is lenght is incrreased 2/3 times. The new value of tthhe resistance wiilll be

Answer 1

Answer:

New resistance will be $\dfrac{4r}{9}$

Explanation:

Given:

The resistance of wire=r

New length of the wire is 2/3 times that of initial length.

Let initial length and radius of the wire be $l_1$ and $r_1$

and final length and radius of the wire be  $l_2$ and $r_2$

The resistivity of the wire which is constant  be$\rho$

when the wire is changed then the volume of the wire remains constant so there will be change in radius of the wire if the length of the wire is changed.

So according to question we have

$l_2=\dfrac{2l_1}{3}\\r_2=\sqrt{\dfrac{3}{2}}r_1$

The resistivity of the wire is given by

$R=\dfrac{\rho l}{A}\\\\R=\dfrac{\rho l}{\pi r^2}$

Now we have

$R_1=\dfrac{\rho l_1}{\pi r_1^2}$

$R=\dfrac{\rho l_2}{\pi r_2^2}$

Putting he value and taking the ratio we have

$\dfrac{R_2}{R_1}=\dfrac{4}{9}\\\\R_2=\dfrac{4r}{9}$

Hence the final resistance is calculated.