Question

If the length of the wire stretched by 20%. the percentage change in new resistance is

Answer 1

Hint: In order to solve this question, we need to understand that resistance is a measure of the opposition of flow of electrons. It is measured in ohms. Resistance of a material is dependent on the length of the material, area of the material and resistivity of the material. When the nature of material will not change then resistivity will remain constant. Then, resistance will depend on the length and area of the material. Firstly, assume the length of wire and apply the condition when it is stretched by 20%. Secondly assume the area of cross-section after stretching and the original. Then use the formula of volume and then equate the equation.

Formula Used:
\[V = A \times l\]
Where, V is the voltage
A is the Area of cross-section
l is the length
\[R = \rho \dfrac{L}{A}\]
Where, R is the resistance of a wire.
\[\rho \] is the resistivity of wire.
L is the length of wire
A is the area of cross-section

Complete step by step solution:
Let original length of wire be \[{l_1}\]
Let the length of wire after stretching be \[{l_2}\]
Then the length of wire after stretching by 20% be \[{l_1} + \dfrac{{20}}{{100}}({l_1})\]
Let original area of cross-section be \[{A_1}\]
Let area of cross-section after stretching be \[{A_2}\]
Now by assuming that volume remains same before and after the stretching,
\[ \Rightarrow {V_2} = {V_1}\]
\[ \Rightarrow {A_2}{l_2} = {A_1}{l_1}\]
\[ \Rightarrow {A_2} = \dfrac{{{A_1}{l_1}}}{{{l_2}}}\]
Substituting the known values we get,
\[ \Rightarrow {A_2} = \dfrac{{{A_1}{l_1}}}{{1.2{l_1}}}\]
\[ \Rightarrow {A_2} = \dfrac{{{A_1}}}{{1.2}}\]
Now, the resistance of original wire is \[{R_1}\]
\[ \Rightarrow {R_1} = \rho \dfrac{{{l_1}}}{{{A_1}}} - - - (1)\]
And the resistance after stretching is \[{R_2}\]
\[ \Rightarrow {R_2} = \rho \dfrac{{{l_2}}}{{{A_2}}}\]
Substituting the value of \[{A_2}\] and \[{l_1}\] we get,
\[ \Rightarrow {R_2} = \dfrac{{\rho \left( {1.2{l_1}} \right)\left( {1.2} \right)}}{{{A_1}}}\]
\[ \Rightarrow {R_2} = \rho \dfrac{{{l_1}}}{{{A_1}}}\left( {1.44} \right) - - - (2)\]
Comparing equation (1) and (2) we get,
\[ \Rightarrow {R_2} = 1.44{R_1}\]
We can write \[1.44{R_1}\] as below,
\[ \Rightarrow {R_2} = {R_1} + 0.44{R_1}\]
\[ \Rightarrow {R_2} = {R_1} + \left( {\dfrac{{44}}{{100}}} \right){R_1}\]
\[ \therefore {R_2} = {R_1} + 44\% {\text{ }}{{\text{R}}_1}\]
Thus, the resistance of wire increases by 44% after stretching by 20%,

Note:
It should be remembered that resistivity is dependent on the nature of the material only. When the nature of material will change then resistivity will no longer be constant and comes into the formula of resistance. When the stretching of wire will take place then length and area will both change to make the volume constant.