Answer
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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.
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