
A wire is loaded by 6kg at its one end, the increase in length is 12mm. If the radius of the wire is doubled and all other magnitudes are unchanged, then increase in length will be
A. 6mm
B. 3mm
C. 24mm
D. 48mm
Answer
161.7k+ views
Hint: To solve this question, we need to use the formula for the young’s modulus of a string in terms of its geometrical parameters. Then, putting the values given in the question, we can get the required value of the increase in length of the wire.
Formula used:
The formula which has been used to solve this question is given by
$Y = \dfrac{{FL}}{{Al}}$
Here Y is the young’s modulus of a string of length L and area of cross section A , F is the force applied on it due to which its length gets changed by l .
Complete answer:
We know that the Young’s modulus for a wire can be written as
$Y = \dfrac{{FL}}{{Al}}$
$ \Rightarrow l = \dfrac{{FL}}{{\pi {R^2}Y}}$
Since all other magnitudes are unchanged,
$ \Rightarrow l \propto \dfrac{1}{{{R^2}}}$
Then we can say that
$ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{R_2}^2}}{{{R_1}^2}}$
Given that ${R_2} = 2{R_1}$and ${l_1} = 12mm$
Then
$ \Rightarrow \dfrac{{12}}{{{l_2}}} = \dfrac{{4{R_1}^2}}{{{R_1}^2}}$
Now we can say that ${l_2} = 3mm$. So, the correct option is B.
Note: We should not forget to convert the values of the quantities given in the question into their respective SI units. In this question the length of the wire was given in millimetres, which is not an SI unit. So, it was supposed to be converted into meters.
Formula used:
The formula which has been used to solve this question is given by
$Y = \dfrac{{FL}}{{Al}}$
Here Y is the young’s modulus of a string of length L and area of cross section A , F is the force applied on it due to which its length gets changed by l .
Complete answer:
We know that the Young’s modulus for a wire can be written as
$Y = \dfrac{{FL}}{{Al}}$
$ \Rightarrow l = \dfrac{{FL}}{{\pi {R^2}Y}}$
Since all other magnitudes are unchanged,
$ \Rightarrow l \propto \dfrac{1}{{{R^2}}}$
Then we can say that
$ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{R_2}^2}}{{{R_1}^2}}$
Given that ${R_2} = 2{R_1}$and ${l_1} = 12mm$
Then
$ \Rightarrow \dfrac{{12}}{{{l_2}}} = \dfrac{{4{R_1}^2}}{{{R_1}^2}}$
Now we can say that ${l_2} = 3mm$. So, the correct option is B.
Note: We should not forget to convert the values of the quantities given in the question into their respective SI units. In this question the length of the wire was given in millimetres, which is not an SI unit. So, it was supposed to be converted into meters.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

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

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
