Question

The resistivity of a material of a wire is $ 2.7 \times {10^{ - 8}} $ , if the length of the wire is triple of initial what will be the resistivity of that wire. Give reason for your answer.

Answer 1

Hint :We are here provided with the initial resistivity and the factor by which the length of the wire is increased, we need to find the final resistivity which can be found by finding the area to length ratio from the initial resistivity and the factor of increase of length. This gives the final resistivity.
The resistivity of any material of length $ L $ and area $ A $ is given by
 $ \rho = \dfrac{{RA}}{L} $
Where $ R $ is the resistance of that material.

Complete Step By Step Answer:
Let us consider the initial length of the wire to be $ L $ and the area of cross section of the wire be $ A $
Let $ R $ be the resistance of the material
Thus, the initial resistivity will be equal to
 $ \rho = \dfrac{{RA}}{L} $
Given $ \rho = 2.7 \times {10^{ - 8}} $
Now when the length of the wire becomes triple of the initial value, area and resistance remains the same, thus the final resistivity will be calculated as
Here, $ L' = 3L $
 $ \rho ' = RA \times \dfrac{1}{{3L}} $
Simplifying this,
 $ \rho ' = \dfrac{1}{3} \times \dfrac{{RA}}{L} = \dfrac{1}{3} \times \rho \\
   \Rightarrow \rho ' = \dfrac{\rho }{3} \\ $
Putting the value of the initial resistance in the above equation, we get
 $ \Rightarrow \rho ' = \dfrac{\rho }{3} = \dfrac{{2.7 \times {{10}^{ - 8}}}}{3} = 0.9 \times {10^{ - 8}}\Omega m $
Hence the final resistivity decreases by a factor of three because it is inversely proportional to the length of the wire.

Note :
It may be thought of that stretching the wire by length might shrink it horizontally, such that the area of cross section of the wire decreases, however, this doesn’t happen, the final area of the cross section remains the same irrespective of the change in length, so area and resistance remains constant, only resistivity changes.