
A wire of 1Ω has a length of 1 m. It is stretched till its length increases by 25%. What is the percentage change in resistance to the nearest integer?
A. 25%
B. 12.5%
C. 76%
D. 56%
Answer
233.1k+ views
Hint: Resistance of a conductor is known as the obstruction or hindrance in the path of current. It depends on the length and area of cross-section of the conductor. In this problem we are going to use the formula of resistance to find the required answer.
Formula Used:
\[R \propto \dfrac{l}{A}\]
\[R = \dfrac{{\rho l}}{A}\] …… (1)
Where, R is the resistance, l is the length of the wire and A is the cross-sectional area.
Complete step by step solution:
Given:
\[{R_0} = 1\Omega \]
\[\Rightarrow {l_0} = 1m\]
\[\Rightarrow {l_1} = 1.25{l_0}\]
For stretched or compressed wire, volume remains constant i.e., \[V = A \times l\]
So, \[{A_0}{l_0} = {A_1}{l_1}\]
\[ \Rightarrow {A_1} = \dfrac{{{A_0}{l_0}}}{{{l_1}}}\]
From equation (1)
\[{R_0} = \dfrac{{\rho {l_0}}}{{{A_0}}}\] and \[\]\[{R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}\]
Now,
\[\dfrac{{{R_0}}}{{{R_1}}} = \dfrac{{\dfrac{{\rho {l_0}}}{{{A_0}}}}}{{\dfrac{{\rho {l_1}}}{{{A_1}}}}} \\ \]
\[\Rightarrow \dfrac{1}{{{R_1}}} = \dfrac{{\rho {l_0}{A_1}}}{{\rho {l_1}{A_0}}} \times \dfrac{1}{{{R_0}}} \\ \]
\[\Rightarrow {R_1} = \dfrac{{{l_1}}}{{{l_0}}} \times \dfrac{{{A_0}}}{{{A_1}}}\] Where \[{R_0} = 1\Omega \\ \]
Now substituting the value of A0 in the above equation we get,
\[{R_1} = \dfrac{{{l_1}}}{{{l_0}}} \times \dfrac{{{A_0} \times {l_1}}}{{{A_0} \times {l_0}}} \\ \]
\[\Rightarrow {R_1} = \dfrac{{{l_1}^2}}{{{l_0}^2}} \\ \]
\[\Rightarrow {R_1} = \dfrac{{{{(1.25{l_0})}^2}}}{{{l_0}^2}} \\ \]
Substitute the value of l0 by data we obtain
\[{R_{_1}} = 1.5625\Omega \]
Now to find the percentage of change in the resistance the formula is given by,
\[\text{Change in the resistance} = \dfrac{{{R_1} - {R_0}}}{{{R_0}}} \times 100\% \\ \]
\[ \Rightarrow \text{Change in the resistance} = \dfrac{{1.5626 - 1}}{1} \times 100\% \\ \]
\[ \therefore \text{Change in the resistance} = 56.25\% \]
Therefore, the percentage of change in resistance is 56.25
Hence, Option D is the correct answer
Note: The restriction to the flow of electrons is known as resistance of a material. A wire’s resistance increases when its length is increased and the resistance of a wire increases as its cross-sectional area is reduced. In a wire of the same volume, when the length is increased, its area will reduce. Doubling the length, double’s the resistance and since the area also reduces the resistance increases further. Hence the resistance of a material is a major contributor in the circuit.
Formula Used:
\[R \propto \dfrac{l}{A}\]
\[R = \dfrac{{\rho l}}{A}\] …… (1)
Where, R is the resistance, l is the length of the wire and A is the cross-sectional area.
Complete step by step solution:
Given:
\[{R_0} = 1\Omega \]
\[\Rightarrow {l_0} = 1m\]
\[\Rightarrow {l_1} = 1.25{l_0}\]
For stretched or compressed wire, volume remains constant i.e., \[V = A \times l\]
So, \[{A_0}{l_0} = {A_1}{l_1}\]
\[ \Rightarrow {A_1} = \dfrac{{{A_0}{l_0}}}{{{l_1}}}\]
From equation (1)
\[{R_0} = \dfrac{{\rho {l_0}}}{{{A_0}}}\] and \[\]\[{R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}\]
Now,
\[\dfrac{{{R_0}}}{{{R_1}}} = \dfrac{{\dfrac{{\rho {l_0}}}{{{A_0}}}}}{{\dfrac{{\rho {l_1}}}{{{A_1}}}}} \\ \]
\[\Rightarrow \dfrac{1}{{{R_1}}} = \dfrac{{\rho {l_0}{A_1}}}{{\rho {l_1}{A_0}}} \times \dfrac{1}{{{R_0}}} \\ \]
\[\Rightarrow {R_1} = \dfrac{{{l_1}}}{{{l_0}}} \times \dfrac{{{A_0}}}{{{A_1}}}\] Where \[{R_0} = 1\Omega \\ \]
Now substituting the value of A0 in the above equation we get,
\[{R_1} = \dfrac{{{l_1}}}{{{l_0}}} \times \dfrac{{{A_0} \times {l_1}}}{{{A_0} \times {l_0}}} \\ \]
\[\Rightarrow {R_1} = \dfrac{{{l_1}^2}}{{{l_0}^2}} \\ \]
\[\Rightarrow {R_1} = \dfrac{{{{(1.25{l_0})}^2}}}{{{l_0}^2}} \\ \]
Substitute the value of l0 by data we obtain
\[{R_{_1}} = 1.5625\Omega \]
Now to find the percentage of change in the resistance the formula is given by,
\[\text{Change in the resistance} = \dfrac{{{R_1} - {R_0}}}{{{R_0}}} \times 100\% \\ \]
\[ \Rightarrow \text{Change in the resistance} = \dfrac{{1.5626 - 1}}{1} \times 100\% \\ \]
\[ \therefore \text{Change in the resistance} = 56.25\% \]
Therefore, the percentage of change in resistance is 56.25
Hence, Option D is the correct answer
Note: The restriction to the flow of electrons is known as resistance of a material. A wire’s resistance increases when its length is increased and the resistance of a wire increases as its cross-sectional area is reduced. In a wire of the same volume, when the length is increased, its area will reduce. Doubling the length, double’s the resistance and since the area also reduces the resistance increases further. Hence the resistance of a material is a major contributor in the circuit.
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