369741
A uniform bar of length ' \(L\) ' and cross-sectional area ' \(A\) ' is subjected to a tensile load ' \(F\) '. ' \(Y\) ' be the Young's modulus and ' \(\sigma\) ' be the poisson's ratio then volumetric strain is
369742
There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is:
1 \(-\dfrac{1}{2}\)
2 \(+\dfrac{1}{2}\)
3 \(-\dfrac{1}{4}\)
4 \(+\dfrac{1}{4}\)
Explanation:
Volume of cylindrical wire, \(V=\dfrac{\pi x^{2} L}{4}\) where \(x\) is diameter of wire Differentiating both sides \(\dfrac{d V}{d x}=\dfrac{\pi}{4}\left[2 x L+x^{2} \cdot \dfrac{d L}{d x}\right]\) Also volume remains constant \(\therefore \dfrac{d V}{d x}=0\) \(\begin{gathered}\therefore \quad 2 x L+x^{2} \dfrac{d L}{d x}=0 \Rightarrow 2 x L=-x^{2} \dfrac{d L}{d x}, \\\Rightarrow \dfrac{\dfrac{d x}{x}}{\dfrac{d L}{L}}=-\dfrac{1}{2}\end{gathered}\) \(\Rightarrow \text { Poisson's ratio }=-\dfrac{1}{2}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369743
Relation between \(Y,\,K\& \,\sigma \) is
1 \(Y=K(1-\sigma)\)
2 \(Y=2 K(1+\sigma)\)
3 \(Y=3 K(1-2 \sigma)\)
4 \(Y=2 K(1-\sigma)\)
Explanation:
This formula relates young's modulus \(\left( Y \right)\), Bulk modulus \(\left( K \right)\), and Poisson's ratio \(\left( \sigma \right)\) for isotropic materials, showing how they influence each other in the material's response to mechanical stress.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369744
For homogenous isotropic material, which one of the following cannot be the value of Poisson's ratio?
1 0.1
2 -1
3 0.5
4 0.8
Explanation:
The Poisson's ratio for homogenous isotropic material must be greater than -1 and less than 0.5 for the elastic moduli to have finite positive values. Hence, 0.8 cannot be value of Poisson's ratio for homogenous isotropic material
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PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369741
A uniform bar of length ' \(L\) ' and cross-sectional area ' \(A\) ' is subjected to a tensile load ' \(F\) '. ' \(Y\) ' be the Young's modulus and ' \(\sigma\) ' be the poisson's ratio then volumetric strain is
369742
There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is:
1 \(-\dfrac{1}{2}\)
2 \(+\dfrac{1}{2}\)
3 \(-\dfrac{1}{4}\)
4 \(+\dfrac{1}{4}\)
Explanation:
Volume of cylindrical wire, \(V=\dfrac{\pi x^{2} L}{4}\) where \(x\) is diameter of wire Differentiating both sides \(\dfrac{d V}{d x}=\dfrac{\pi}{4}\left[2 x L+x^{2} \cdot \dfrac{d L}{d x}\right]\) Also volume remains constant \(\therefore \dfrac{d V}{d x}=0\) \(\begin{gathered}\therefore \quad 2 x L+x^{2} \dfrac{d L}{d x}=0 \Rightarrow 2 x L=-x^{2} \dfrac{d L}{d x}, \\\Rightarrow \dfrac{\dfrac{d x}{x}}{\dfrac{d L}{L}}=-\dfrac{1}{2}\end{gathered}\) \(\Rightarrow \text { Poisson's ratio }=-\dfrac{1}{2}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369743
Relation between \(Y,\,K\& \,\sigma \) is
1 \(Y=K(1-\sigma)\)
2 \(Y=2 K(1+\sigma)\)
3 \(Y=3 K(1-2 \sigma)\)
4 \(Y=2 K(1-\sigma)\)
Explanation:
This formula relates young's modulus \(\left( Y \right)\), Bulk modulus \(\left( K \right)\), and Poisson's ratio \(\left( \sigma \right)\) for isotropic materials, showing how they influence each other in the material's response to mechanical stress.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369744
For homogenous isotropic material, which one of the following cannot be the value of Poisson's ratio?
1 0.1
2 -1
3 0.5
4 0.8
Explanation:
The Poisson's ratio for homogenous isotropic material must be greater than -1 and less than 0.5 for the elastic moduli to have finite positive values. Hence, 0.8 cannot be value of Poisson's ratio for homogenous isotropic material
369741
A uniform bar of length ' \(L\) ' and cross-sectional area ' \(A\) ' is subjected to a tensile load ' \(F\) '. ' \(Y\) ' be the Young's modulus and ' \(\sigma\) ' be the poisson's ratio then volumetric strain is
369742
There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is:
1 \(-\dfrac{1}{2}\)
2 \(+\dfrac{1}{2}\)
3 \(-\dfrac{1}{4}\)
4 \(+\dfrac{1}{4}\)
Explanation:
Volume of cylindrical wire, \(V=\dfrac{\pi x^{2} L}{4}\) where \(x\) is diameter of wire Differentiating both sides \(\dfrac{d V}{d x}=\dfrac{\pi}{4}\left[2 x L+x^{2} \cdot \dfrac{d L}{d x}\right]\) Also volume remains constant \(\therefore \dfrac{d V}{d x}=0\) \(\begin{gathered}\therefore \quad 2 x L+x^{2} \dfrac{d L}{d x}=0 \Rightarrow 2 x L=-x^{2} \dfrac{d L}{d x}, \\\Rightarrow \dfrac{\dfrac{d x}{x}}{\dfrac{d L}{L}}=-\dfrac{1}{2}\end{gathered}\) \(\Rightarrow \text { Poisson's ratio }=-\dfrac{1}{2}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369743
Relation between \(Y,\,K\& \,\sigma \) is
1 \(Y=K(1-\sigma)\)
2 \(Y=2 K(1+\sigma)\)
3 \(Y=3 K(1-2 \sigma)\)
4 \(Y=2 K(1-\sigma)\)
Explanation:
This formula relates young's modulus \(\left( Y \right)\), Bulk modulus \(\left( K \right)\), and Poisson's ratio \(\left( \sigma \right)\) for isotropic materials, showing how they influence each other in the material's response to mechanical stress.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369744
For homogenous isotropic material, which one of the following cannot be the value of Poisson's ratio?
1 0.1
2 -1
3 0.5
4 0.8
Explanation:
The Poisson's ratio for homogenous isotropic material must be greater than -1 and less than 0.5 for the elastic moduli to have finite positive values. Hence, 0.8 cannot be value of Poisson's ratio for homogenous isotropic material
369741
A uniform bar of length ' \(L\) ' and cross-sectional area ' \(A\) ' is subjected to a tensile load ' \(F\) '. ' \(Y\) ' be the Young's modulus and ' \(\sigma\) ' be the poisson's ratio then volumetric strain is
369742
There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is:
1 \(-\dfrac{1}{2}\)
2 \(+\dfrac{1}{2}\)
3 \(-\dfrac{1}{4}\)
4 \(+\dfrac{1}{4}\)
Explanation:
Volume of cylindrical wire, \(V=\dfrac{\pi x^{2} L}{4}\) where \(x\) is diameter of wire Differentiating both sides \(\dfrac{d V}{d x}=\dfrac{\pi}{4}\left[2 x L+x^{2} \cdot \dfrac{d L}{d x}\right]\) Also volume remains constant \(\therefore \dfrac{d V}{d x}=0\) \(\begin{gathered}\therefore \quad 2 x L+x^{2} \dfrac{d L}{d x}=0 \Rightarrow 2 x L=-x^{2} \dfrac{d L}{d x}, \\\Rightarrow \dfrac{\dfrac{d x}{x}}{\dfrac{d L}{L}}=-\dfrac{1}{2}\end{gathered}\) \(\Rightarrow \text { Poisson's ratio }=-\dfrac{1}{2}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369743
Relation between \(Y,\,K\& \,\sigma \) is
1 \(Y=K(1-\sigma)\)
2 \(Y=2 K(1+\sigma)\)
3 \(Y=3 K(1-2 \sigma)\)
4 \(Y=2 K(1-\sigma)\)
Explanation:
This formula relates young's modulus \(\left( Y \right)\), Bulk modulus \(\left( K \right)\), and Poisson's ratio \(\left( \sigma \right)\) for isotropic materials, showing how they influence each other in the material's response to mechanical stress.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369744
For homogenous isotropic material, which one of the following cannot be the value of Poisson's ratio?
1 0.1
2 -1
3 0.5
4 0.8
Explanation:
The Poisson's ratio for homogenous isotropic material must be greater than -1 and less than 0.5 for the elastic moduli to have finite positive values. Hence, 0.8 cannot be value of Poisson's ratio for homogenous isotropic material
