![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
The stress versus strain graphs for wires of two materials A and B are as shown in the figure.
If $ {Y_A} $ and $ {Y_B} $ are the Young’s moduli of the materials, then
A) $ {Y_B} = 2{Y_A} $
B) $ {Y_A} = {Y_B} $
C) $ {Y_B} = 3{Y_A} $
D) $ {Y_A} = 3{Y_B} $
![seo images](https://www.vedantu.com/question-sets/12175eb2-605e-4a09-855d-38b5d21780503609788722035395329.png)
Answer
455.4k+ views
Hint : The young’s modulus of wire is the slope of the line of the wire in the graph of stress versus strain. The slope of a line can be determined using the angle the line makes with the $ x $ -axis.
Complete step by step answer
We know that the slope of the line in a stress-strain curve represents the young’s modulus for a wire. We also know that the slope of a line can be calculated as the tangent of the angle the line makes with the positive x-axis of the graph.
So, for wire A, the stress-strain line of the material is at an angle of $ 60^\circ $ from the positive $ x $ -axis. So the slope of the line $ ({m_A}) $ will be
$ {m_A} = \tan 60^\circ $
$ \Rightarrow {m_A} = \sqrt 3 $
Hence the young’s modulus of wire A will also be $ \sqrt 3 $ .
Similarly, for wire B, the stress-strain line of the material is at an angle of $ 30^\circ $ from the positive $ x $ -axis. So, the slope of the line $ ({m_B}) $ will be
$ {m_B} = \tan 30^\circ $
$ \Rightarrow {m_B} = \dfrac{1}{{\sqrt 3 }} $
Hence the young’s modulus of wire B will also be $ 1/\sqrt 3 $ .
Then taking the ratio of the young’s modulus for wire A and B, we get
$ \dfrac{{{Y_A}}}{{{Y_B}}} = \dfrac{{\sqrt 3 }}{{1/\sqrt 3 }} $
$ \therefore \dfrac{{{Y_A}}}{{{Y_B}}} = 3 $
Hence, we can write
$ {Y_A} = 3{Y_B} $ which corresponds to option (D) which is the correct choice.
Note
We can only calculate the slope of the line in such a way if the stress-strain curve for a wire is a straight line. For practical wires, the stress-strain curve is linear only for a range of values of stress applied on the wire. While calculating the slope of the wire, we must calculate the tangent of the line made with the $ x $ -axis and not the $ y $ -axis if the strain is represented on the $ x $ -axis of the graph.
Complete step by step answer
We know that the slope of the line in a stress-strain curve represents the young’s modulus for a wire. We also know that the slope of a line can be calculated as the tangent of the angle the line makes with the positive x-axis of the graph.
So, for wire A, the stress-strain line of the material is at an angle of $ 60^\circ $ from the positive $ x $ -axis. So the slope of the line $ ({m_A}) $ will be
$ {m_A} = \tan 60^\circ $
$ \Rightarrow {m_A} = \sqrt 3 $
Hence the young’s modulus of wire A will also be $ \sqrt 3 $ .
Similarly, for wire B, the stress-strain line of the material is at an angle of $ 30^\circ $ from the positive $ x $ -axis. So, the slope of the line $ ({m_B}) $ will be
$ {m_B} = \tan 30^\circ $
$ \Rightarrow {m_B} = \dfrac{1}{{\sqrt 3 }} $
Hence the young’s modulus of wire B will also be $ 1/\sqrt 3 $ .
Then taking the ratio of the young’s modulus for wire A and B, we get
$ \dfrac{{{Y_A}}}{{{Y_B}}} = \dfrac{{\sqrt 3 }}{{1/\sqrt 3 }} $
$ \therefore \dfrac{{{Y_A}}}{{{Y_B}}} = 3 $
Hence, we can write
$ {Y_A} = 3{Y_B} $ which corresponds to option (D) which is the correct choice.
Note
We can only calculate the slope of the line in such a way if the stress-strain curve for a wire is a straight line. For practical wires, the stress-strain curve is linear only for a range of values of stress applied on the wire. While calculating the slope of the wire, we must calculate the tangent of the line made with the $ x $ -axis and not the $ y $ -axis if the strain is represented on the $ x $ -axis of the graph.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the value of x if the mode of the following data class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
10 examples of friction in our daily life
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Prokaryotic Cells and Eukaryotic Cells
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
State and prove Bernoullis theorem class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What organs are located on the left side of your body class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How many valence electrons does nitrogen have class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)