
If a $0.1\% $ increase in length due to stretching, the percentage increase in its resistance will be
A. $0.2\% $
B. $2\% $
C. $1\% $
D. $0.1\% $
Answer
164.4k+ views
Hint: Here we know the percentage increase in the length. When a substance is stretched its length increases but the volume remains the same. Using this concept we will find the percentage increase in resistance.
Formula Used:
\[{A_f}{l_f} = {A_i}{l_i}\]
The formula of resistance is,
$\Rightarrow R = \dfrac{{\rho l}}{A}$
The expression of change in resistance is,
$\Rightarrow \vartriangle R = \dfrac{{{R_f} - {R_i}}}{{{R_i}}} \times 100$
Where \[A\] is the area of cross section, $l$ is the length, $R$ is the resistance and $\rho $ is the resistivity of the substance.
Complete step by step solution:
Given: ${l_f} = \left( {l + 0.1\% \,l} \right)$ where ${l_i} = l$ is the initial length and ${l_f}$ is the final length.
Solving this we get, ${l_f} = \left( {l + \dfrac{{0.1\,l}}{{100}}} \right)$
That is, \[{l_f} = \left( {\dfrac{{100.1\,l}}{{100}}} \right)\] ...(1)
Now, we know that when a substance is stretched, its length increases but the volume is conserved.
This implies that, \[{A_f}{l_f} = {A_i}{l_i}\]
Let ${A_i} = A$ be the initial area of the cross section.
We get, ${A_f} = \left( {\dfrac{l}{{{l_f}}}A} \right)$...(2)
Substituting equation (1) in equation (2), we get
${A_f} = \dfrac{{100\,l}}{{100.1\,l}}A$
Solving this we get,
${A_f} = \dfrac{{100}}{{100.1}}A$...(3)
Now we know the relationship between resistance, resistivity, area of cross section and length of the substance.
It is $R = \dfrac{{\rho l}}{A}$ ...(4)
This is the initial resistance. Also, the value of resistivity does not change on stretching. It remains the same for a given substance. Final resistance is,
${R_f} = \dfrac{{\rho {l_f}}}{{{A_f}}}$...(5)
Substituting equations (1) and (3) in equation (5) we get, ${R_f} = \dfrac{{\rho \left( {\dfrac{{100.1\,l}}{{100}}} \right)}}{{\left( {\dfrac{{100}}{{100.1}}A} \right)}}$ ...(6)
Solving this we get, ${R_f} = {\left( {\dfrac{{100.1}}{{100}}} \right)^2}\dfrac{{\rho l}}{A}$
Substitute equation (4) in equation (6), we get
${R_f} = {\left( {\dfrac{{100.1}}{{100}}} \right)^2}R$
We know that to find percentage we use the following formula:
$\vartriangle R = \dfrac{{{R_f} - {R_i}}}{{{R_i}}} \times 100 \\ $
Where $\vartriangle R$ is the percentage change in the resistance.
We get,
$\vartriangle R = \dfrac{{\left[ {{{\left( {\dfrac{{100.1}}{{100}}} \right)}^2}R} \right] - R}}{R} \times 100 \\ $
Solving further,
$\vartriangle R = \left[ {{{\left( {\dfrac{{100.1}}{{100}}} \right)}^2} - 1} \right] \times 100 \\ $
We can write $100.1 = 100 + 0.1$ , we get;
$\vartriangle R = \left[ {{{\left( {\dfrac{{100 + 0.1}}{{100}}} \right)}^2} - 1} \right] \times 100 \\ $
Use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , we get
\[\vartriangle R = \left[ {\dfrac{{{{\left( {100} \right)}^2} + {{\left( {0.1} \right)}^2} + \left( {2 \times 100 \times 0.1} \right) - {{\left( {100} \right)}^2}}}{{{{\left( {100} \right)}^2}}}} \right] \times 100 \\ \]
Now ${\left( {0.1} \right)^2}$ can be ignored in the numerator because it is very small as compared to the other values in the addition. We get,
\[\vartriangle R = \left[ {\dfrac{{\left( {2 \times 100 \times 0.1} \right)}}{{{{\left( {100} \right)}^2}}}} \right] \times 100 =0.2\%\]
Hence, option A is the answer.
Note: While solving this question, to find the percentage change in the area of cross section of the substance because when a substance is stretched its area of cross section is not conserved.
Formula Used:
\[{A_f}{l_f} = {A_i}{l_i}\]
The formula of resistance is,
$\Rightarrow R = \dfrac{{\rho l}}{A}$
The expression of change in resistance is,
$\Rightarrow \vartriangle R = \dfrac{{{R_f} - {R_i}}}{{{R_i}}} \times 100$
Where \[A\] is the area of cross section, $l$ is the length, $R$ is the resistance and $\rho $ is the resistivity of the substance.
Complete step by step solution:
Given: ${l_f} = \left( {l + 0.1\% \,l} \right)$ where ${l_i} = l$ is the initial length and ${l_f}$ is the final length.
Solving this we get, ${l_f} = \left( {l + \dfrac{{0.1\,l}}{{100}}} \right)$
That is, \[{l_f} = \left( {\dfrac{{100.1\,l}}{{100}}} \right)\] ...(1)
Now, we know that when a substance is stretched, its length increases but the volume is conserved.
This implies that, \[{A_f}{l_f} = {A_i}{l_i}\]
Let ${A_i} = A$ be the initial area of the cross section.
We get, ${A_f} = \left( {\dfrac{l}{{{l_f}}}A} \right)$...(2)
Substituting equation (1) in equation (2), we get
${A_f} = \dfrac{{100\,l}}{{100.1\,l}}A$
Solving this we get,
${A_f} = \dfrac{{100}}{{100.1}}A$...(3)
Now we know the relationship between resistance, resistivity, area of cross section and length of the substance.
It is $R = \dfrac{{\rho l}}{A}$ ...(4)
This is the initial resistance. Also, the value of resistivity does not change on stretching. It remains the same for a given substance. Final resistance is,
${R_f} = \dfrac{{\rho {l_f}}}{{{A_f}}}$...(5)
Substituting equations (1) and (3) in equation (5) we get, ${R_f} = \dfrac{{\rho \left( {\dfrac{{100.1\,l}}{{100}}} \right)}}{{\left( {\dfrac{{100}}{{100.1}}A} \right)}}$ ...(6)
Solving this we get, ${R_f} = {\left( {\dfrac{{100.1}}{{100}}} \right)^2}\dfrac{{\rho l}}{A}$
Substitute equation (4) in equation (6), we get
${R_f} = {\left( {\dfrac{{100.1}}{{100}}} \right)^2}R$
We know that to find percentage we use the following formula:
$\vartriangle R = \dfrac{{{R_f} - {R_i}}}{{{R_i}}} \times 100 \\ $
Where $\vartriangle R$ is the percentage change in the resistance.
We get,
$\vartriangle R = \dfrac{{\left[ {{{\left( {\dfrac{{100.1}}{{100}}} \right)}^2}R} \right] - R}}{R} \times 100 \\ $
Solving further,
$\vartriangle R = \left[ {{{\left( {\dfrac{{100.1}}{{100}}} \right)}^2} - 1} \right] \times 100 \\ $
We can write $100.1 = 100 + 0.1$ , we get;
$\vartriangle R = \left[ {{{\left( {\dfrac{{100 + 0.1}}{{100}}} \right)}^2} - 1} \right] \times 100 \\ $
Use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , we get
\[\vartriangle R = \left[ {\dfrac{{{{\left( {100} \right)}^2} + {{\left( {0.1} \right)}^2} + \left( {2 \times 100 \times 0.1} \right) - {{\left( {100} \right)}^2}}}{{{{\left( {100} \right)}^2}}}} \right] \times 100 \\ \]
Now ${\left( {0.1} \right)^2}$ can be ignored in the numerator because it is very small as compared to the other values in the addition. We get,
\[\vartriangle R = \left[ {\dfrac{{\left( {2 \times 100 \times 0.1} \right)}}{{{{\left( {100} \right)}^2}}}} \right] \times 100 =0.2\%\]
Hence, option A is the answer.
Note: While solving this question, to find the percentage change in the area of cross section of the substance because when a substance is stretched its area of cross section is not conserved.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE
