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In the graph, we are given the angle made by the line in the stress vs strain graph in both cases.

We know that slope is equal to $\tan \theta $

In the case of material (A)

The slope is equal to $\tan 60^\circ $

Material (B)

The slope is equal to $\tan 30^\circ $

We know that slope is equal to Young’s modulus hence

${Y_A} = \tan 60^\circ = \sqrt 3 $

${Y_B} = \tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$

Diving Young’s modulus of (A) with Young’s modulus of (B) we get $\dfrac{{{Y_A}}}{{{Y_B}}} = \dfrac{{\sqrt 3 }}{{\dfrac{1}{{\sqrt 3 }}}} = 3$

Hence $3{Y_B} = {Y_A}$

**Hint.**We solve this question by finding the values of Young’s modulus for both the materials. We find Young’s modulus by finding the slope of the line in the graph for each material. We find the slope by using trigonometric properties. The angle made by the line in the stress vs strain graph is already given. By using these angles in trigonometry properties to find Young’s modulus we get a relation between ${Y_A}$and ${Y_B}$**Complete step by step answer.**The slope of stress vs strain graph gives us Young’s modulus.In the graph, we are given the angle made by the line in the stress vs strain graph in both cases.

We know that slope is equal to $\tan \theta $

In the case of material (A)

The slope is equal to $\tan 60^\circ $

Material (B)

The slope is equal to $\tan 30^\circ $

We know that slope is equal to Young’s modulus hence

${Y_A} = \tan 60^\circ = \sqrt 3 $

${Y_B} = \tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$

Diving Young’s modulus of (A) with Young’s modulus of (B) we get $\dfrac{{{Y_A}}}{{{Y_B}}} = \dfrac{{\sqrt 3 }}{{\dfrac{1}{{\sqrt 3 }}}} = 3$

Hence $3{Y_B} = {Y_A}$

**Option (D) $3{Y_B} = {Y_A}$ is the correct answer.****Additional information**Young’s modulus is defined as the measure of ability of a material to withstand changes in length when it is under a lengthwise tension of compression. It is also referred to as modulus of elasticity.**Note**We find the slope in the graph by using tanθ because we are assuming the graph as a right-angled triangle. And $\tan \theta $is equal to the opposite by the adjacent side of the assumed triangle. In the case of stress vs strain graph opposite is equal to stress and adjacent is equal to the strain. Hence $\tan \theta $ is equal to Young’s modulus. Therefore, we use $\tan \theta $ to find the slope.Recently Updated Pages

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