Question

A wire of resistance $4\Omega$ is stretched twice its original length. In the process of stretching its area of cross section gets halved. Now, the resistance of the wire is:
(A) $8\Omega$
(B) $16\Omega$
(C) $1\Omega$
(D) $4\Omega$

Answer 1

Hint : Use the definition of resistance of a material in terms of its length, cross sectional area and resistivity of the material to find the new resistance of the wire when stretched. The resistance of a material or conductor of resistivity $\rho$ wire of length $l$ and cross section $a$ is given by, $R = \dfrac{{\rho l}}{a}$ .

Complete Step By Step Answer:
We know the resistance of a material or conductor of resistivity $\rho$ wire of length $l$ and cross section $a$ is given by, $R = \dfrac{{\rho l}}{a}$ .
Here, at first the length of the wire is stretched to twice its length. Let initially the length of the conductor was $l$ and cross section was $a$ then the resistance of the wire was ${R_1}$ . After stretching its length to twice the initial length, it becomes $l' = 2l$ . Also the cross section is halved , so the cross section becomes $a' = \dfrac{a}{2}$ .Since, the volume of the wire is constant.
Since, resistivity is a property of the material and it only changes if the material is different So, $\rho$ is always constant for the wire.
Hence, the new resistance becomes, ${R_2} = \dfrac{{\rho l'}}{{a'}}$ .
Now, given that the resistance of the wire before stretching, ${R_1} = 4\Omega$ . Therefore, $\dfrac{{\rho l}}{a} = 4$
Now, after stretching the new resistance becomes, ${R_2} = \dfrac{{\rho l'}}{{a'}}$
Putting the values of the new length $l' = 2l$ and cross sectional area $a' = \dfrac{a}{2}$ , we get, the new resistance as,
${R_2} = \dfrac{{\rho 2l}}{{\dfrac{a}{2}}}$
Or, ${R_2} = \dfrac{{4\rho l}}{a}$
In terms of the initial resistance that becomes, ${R_2} = 4{R_1}$
Hence, putting the value of given ${R_1} = 4\Omega$ we get the new resistance ${R_2} = 4 \times 4 = 16\Omega$ .
So, the new resistance of the wire will be $16\Omega$ .
Hence, option ( B ) is correct.

Note :
We can see that if a wire is stretched and the volume is kept constant then the resistance changes as $R' = {n^2}R$ . Where is the ratio of the previous and stretched length. We can remember this formula also for faster calculation.