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.