
When the load on a wire is slowly increased from 3 to 5kgwt, the elongation increases from 0.61 to 1.02mm. The work done during the extension of wire is:
A) 0.16J
B) 0.016J
C) 1.6J
D) 16J
Answer
232.8k+ views
Hint: Here, we need to form a relation between work, stress and strain. The general formula for work which is W =F.d, will not work as the work is being applied to the wire and not on a particular object. The wire will get stress and strain.
Complete step by step solution:
Find the work done during extension of wire:
Here, the wire is experiencing stress and strain, so the work done for the initial elongation would be:
${W_i} = \dfrac{1}{2}\left( {\dfrac{F}{A} \times \dfrac{{\Delta l}}{l}} \right) \times A \times l$;
Here:
$\dfrac{F}{A}$= Stress;
F = Force;
$\dfrac{{\Delta l}}{l}$= Strain;
$\Delta l$= Change in length;
$A$= Area;
$l$= Length;
Cancel out the common and put F = mg;
$ \Rightarrow {W_i} = \dfrac{1}{2}\left( {mg \times \Delta l} \right)$;
Now, put the given value in the above equation:
$ \Rightarrow {W_i} = \dfrac{1}{2}\left( {3 \times 9.8 \times 0.61 \times {{10}^{ - 3}}J} \right)$;
$ \Rightarrow {W_i} = \dfrac{1}{2}\left( {3 \times 0.61 \times 9.8 \times {{10}^{ - 3}}J} \right)$;
Do the needed calculation:
$ \Rightarrow {W_i} = 8.965 \times {10^{ - 3}}J$;
Now, we can do this for the final elongation by applying the formula for work $W = \dfrac{1}{2}\left( {mg \times \Delta l} \right):$
$\Rightarrow {W_f} = \dfrac{1}{2}\left( {mg \times \Delta l} \right)$;
$\Rightarrow {W_f} = \dfrac{1}{2}\left( {5 \times 9.8 \times 1.02 \times {{10}^{ - 3}}J} \right)$;
Now, do the needed calculation:
$\Rightarrow {W_f} = 24.99 \times {10^{ - 3}}J;$
The Net work done on the wire is: ${W_f} - {W_i}$
$\Rightarrow {W_f} - {W_i} = 24.99 \times {10^{ - 3}} - 8.965 \times {10^{ - 3}}$;
Do, the needed calculation:
$ \Rightarrow {W_{net}} = 16.025 \times {10^{ - 3}}J$;
$ \Rightarrow {W_{net}} = 0.016J$;
Final answer is option B. The work done during the extension of wire is 0.016J.
Note: Here, the work done on the wire will include the (1/2) times the stress and strain on the wire times the area and length. Here the stress on the wire would be forced upon the area and the strain on the wire will be changed of length upon original length. After calculating the initial work done and the final work done, subtract the final work done from the initial work done and that would be the net or the required work done. Be careful when converting the units, here the units are given in mm, kg-wt.
Complete step by step solution:
Find the work done during extension of wire:
Here, the wire is experiencing stress and strain, so the work done for the initial elongation would be:
${W_i} = \dfrac{1}{2}\left( {\dfrac{F}{A} \times \dfrac{{\Delta l}}{l}} \right) \times A \times l$;
Here:
$\dfrac{F}{A}$= Stress;
F = Force;
$\dfrac{{\Delta l}}{l}$= Strain;
$\Delta l$= Change in length;
$A$= Area;
$l$= Length;
Cancel out the common and put F = mg;
$ \Rightarrow {W_i} = \dfrac{1}{2}\left( {mg \times \Delta l} \right)$;
Now, put the given value in the above equation:
$ \Rightarrow {W_i} = \dfrac{1}{2}\left( {3 \times 9.8 \times 0.61 \times {{10}^{ - 3}}J} \right)$;
$ \Rightarrow {W_i} = \dfrac{1}{2}\left( {3 \times 0.61 \times 9.8 \times {{10}^{ - 3}}J} \right)$;
Do the needed calculation:
$ \Rightarrow {W_i} = 8.965 \times {10^{ - 3}}J$;
Now, we can do this for the final elongation by applying the formula for work $W = \dfrac{1}{2}\left( {mg \times \Delta l} \right):$
$\Rightarrow {W_f} = \dfrac{1}{2}\left( {mg \times \Delta l} \right)$;
$\Rightarrow {W_f} = \dfrac{1}{2}\left( {5 \times 9.8 \times 1.02 \times {{10}^{ - 3}}J} \right)$;
Now, do the needed calculation:
$\Rightarrow {W_f} = 24.99 \times {10^{ - 3}}J;$
The Net work done on the wire is: ${W_f} - {W_i}$
$\Rightarrow {W_f} - {W_i} = 24.99 \times {10^{ - 3}} - 8.965 \times {10^{ - 3}}$;
Do, the needed calculation:
$ \Rightarrow {W_{net}} = 16.025 \times {10^{ - 3}}J$;
$ \Rightarrow {W_{net}} = 0.016J$;
Final answer is option B. The work done during the extension of wire is 0.016J.
Note: Here, the work done on the wire will include the (1/2) times the stress and strain on the wire times the area and length. Here the stress on the wire would be forced upon the area and the strain on the wire will be changed of length upon original length. After calculating the initial work done and the final work done, subtract the final work done from the initial work done and that would be the net or the required work done. Be careful when converting the units, here the units are given in mm, kg-wt.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

