
A wire of certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively.
(A) Both remain the same
(B) $ 1.1 $ times, $ 1.1 $ times
(C) $ 1.2 $ times, $ 1.1 $ times
(D) $ 1.21 $ times, same
Answer
520.5k+ views
Hint :Resistance is defined as the measure of opposition to current flow in an electrical circuit and the resistance is directly proportional to the resistivity of the material and the length of the material and it is inversely proportional to the area of the component of which resistance is being calculated.
Complete Step By Step Answer:
In the given problem we have been asked that if the wire is stretched ten percent along the length $ l $ , let area be $ A $ and the resistivity or specific resistance be $ \rho $ .
According to the definition of the resistance, it is given by $ R $ :
$ R = \dfrac{{\rho l}}{A} $ ( resistance of wire at length $ l $ ) $ (1) $
Now, the length of the wire is stretched by $ 10\% $
Therefore,
$ l' = l + \dfrac{{10l}}{{100}} = \dfrac{{110l}}{{100}} = \dfrac{{11l}}{{10}} $ ….. $ (2) $
Also, the area of the wire after stretching is
$ A' = \dfrac{{10A}}{{11}} $ ….. $ (3) $
So, put these values from $ (2) $ and $ (3) $ in the new formula for resistance as below
$ R' = \dfrac{{\rho l'}}{{A'}} $ (specific resistance will remain same since the material is same)
$ \Rightarrow R' = \dfrac{{\rho \left( {\dfrac{{11l}}{{10}}} \right)}}{{\dfrac{{10A}}{{11}}}} = \dfrac{{\rho l}}{A} \times {\left( {\dfrac{{11}}{{10}}} \right)^2} = 1.21R $
Thus the new resistance of the stretched wire is $ 1.21R $
Hence, we have observed that the new resistance of the stretched wire becomes $ 1.21 $ times to that of $ R $ (original resistance) and specific resistance is the intrinsic property thus it remains the same.
The correct answer is option D.
Note :
Here, we are asked to calculate new resistance using the old resistance and we have obtained the new resistance as above. But we have obtained the specific resistance of new length of wire as same as it was of the old length because the reason is that the specific resistance is the intrinsic property of the material and it cannot be affected by the length or shape of the material.
Complete Step By Step Answer:
In the given problem we have been asked that if the wire is stretched ten percent along the length $ l $ , let area be $ A $ and the resistivity or specific resistance be $ \rho $ .
According to the definition of the resistance, it is given by $ R $ :
$ R = \dfrac{{\rho l}}{A} $ ( resistance of wire at length $ l $ ) $ (1) $
Now, the length of the wire is stretched by $ 10\% $
Therefore,
$ l' = l + \dfrac{{10l}}{{100}} = \dfrac{{110l}}{{100}} = \dfrac{{11l}}{{10}} $ ….. $ (2) $
Also, the area of the wire after stretching is
$ A' = \dfrac{{10A}}{{11}} $ ….. $ (3) $
So, put these values from $ (2) $ and $ (3) $ in the new formula for resistance as below
$ R' = \dfrac{{\rho l'}}{{A'}} $ (specific resistance will remain same since the material is same)
$ \Rightarrow R' = \dfrac{{\rho \left( {\dfrac{{11l}}{{10}}} \right)}}{{\dfrac{{10A}}{{11}}}} = \dfrac{{\rho l}}{A} \times {\left( {\dfrac{{11}}{{10}}} \right)^2} = 1.21R $
Thus the new resistance of the stretched wire is $ 1.21R $
Hence, we have observed that the new resistance of the stretched wire becomes $ 1.21 $ times to that of $ R $ (original resistance) and specific resistance is the intrinsic property thus it remains the same.
The correct answer is option D.
Note :
Here, we are asked to calculate new resistance using the old resistance and we have obtained the new resistance as above. But we have obtained the specific resistance of new length of wire as same as it was of the old length because the reason is that the specific resistance is the intrinsic property of the material and it cannot be affected by the length or shape of the material.
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