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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 ' $σ$ ' be the poisson's ratio then volumetric strain is

1

\frac{F}{A Y} (2 - σ)

2

\frac{F}{A Y} (1 - σ)

3

\frac{F}{A Y} \cdot σ

4

\frac{F}{A Y} (1 - 2 σ)

Explanation:

$\frac{Δ V}{V} = \frac{Δ L}{L} + 2 \frac{d r}{r} = \frac{F}{A Y} + 2 (- \frac{σ F}{A Y})$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

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

- \frac{1}{2}

2

+ \frac{1}{2}

3

- \frac{1}{4}

4

+ \frac{1}{4}

Explanation:

Volume of cylindrical wire, $V = \frac{π x^{2} L}{4}$ where $x$ is diameter of wire
Differentiating both sides
$\frac{d V}{d x} = \frac{π}{4} [2 x L + x^{2} \cdot \frac{d L}{d x}]$
Also volume remains constant $∴ \frac{d V}{d x} = 0$
$\begin{matrix} ∴ 2 x L + x^{2} \frac{d L}{d x} = 0 \Rightarrow 2 x L = - x^{2} \frac{d L}{d x}, \\ \Rightarrow \frac{\frac{d x}{x}}{\frac{d L}{L}} = - \frac{1}{2} \end{matrix}$
$\Rightarrow Poisson's ratio = - \frac{1}{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369743 Relation between $Y, K & σ$ is

1

Y = K (1 - σ)

2

Y = 2 K (1 + σ)

3

Y = 3 K (1 - 2 σ)

4

Y = 2 K (1 - σ)

Explanation:

This formula relates young's modulus $(Y)$ , Bulk modulus $(K)$ , and Poisson's ratio $(σ)$ 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

MHTCET - 2019

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 ' $σ$ ' be the poisson's ratio then volumetric strain is

1

\frac{F}{A Y} (2 - σ)

2

\frac{F}{A Y} (1 - σ)

3

\frac{F}{A Y} \cdot σ

4

\frac{F}{A Y} (1 - 2 σ)

Explanation:

$\frac{Δ V}{V} = \frac{Δ L}{L} + 2 \frac{d r}{r} = \frac{F}{A Y} + 2 (- \frac{σ F}{A Y})$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

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

- \frac{1}{2}

2

+ \frac{1}{2}

3

- \frac{1}{4}

4

+ \frac{1}{4}

Explanation:

Volume of cylindrical wire, $V = \frac{π x^{2} L}{4}$ where $x$ is diameter of wire
Differentiating both sides
$\frac{d V}{d x} = \frac{π}{4} [2 x L + x^{2} \cdot \frac{d L}{d x}]$
Also volume remains constant $∴ \frac{d V}{d x} = 0$
$\begin{matrix} ∴ 2 x L + x^{2} \frac{d L}{d x} = 0 \Rightarrow 2 x L = - x^{2} \frac{d L}{d x}, \\ \Rightarrow \frac{\frac{d x}{x}}{\frac{d L}{L}} = - \frac{1}{2} \end{matrix}$
$\Rightarrow Poisson's ratio = - \frac{1}{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369743 Relation between $Y, K & σ$ is

1

Y = K (1 - σ)

2

Y = 2 K (1 + σ)

3

Y = 3 K (1 - 2 σ)

4

Y = 2 K (1 - σ)

Explanation:

This formula relates young's modulus $(Y)$ , Bulk modulus $(K)$ , and Poisson's ratio $(σ)$ 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

MHTCET - 2019

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 ' $σ$ ' be the poisson's ratio then volumetric strain is

1

\frac{F}{A Y} (2 - σ)

2

\frac{F}{A Y} (1 - σ)

3

\frac{F}{A Y} \cdot σ

4

\frac{F}{A Y} (1 - 2 σ)

Explanation:

$\frac{Δ V}{V} = \frac{Δ L}{L} + 2 \frac{d r}{r} = \frac{F}{A Y} + 2 (- \frac{σ F}{A Y})$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

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

- \frac{1}{2}

2

+ \frac{1}{2}

3

- \frac{1}{4}

4

+ \frac{1}{4}

Explanation:

Volume of cylindrical wire, $V = \frac{π x^{2} L}{4}$ where $x$ is diameter of wire
Differentiating both sides
$\frac{d V}{d x} = \frac{π}{4} [2 x L + x^{2} \cdot \frac{d L}{d x}]$
Also volume remains constant $∴ \frac{d V}{d x} = 0$
$\begin{matrix} ∴ 2 x L + x^{2} \frac{d L}{d x} = 0 \Rightarrow 2 x L = - x^{2} \frac{d L}{d x}, \\ \Rightarrow \frac{\frac{d x}{x}}{\frac{d L}{L}} = - \frac{1}{2} \end{matrix}$
$\Rightarrow Poisson's ratio = - \frac{1}{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369743 Relation between $Y, K & σ$ is

1

Y = K (1 - σ)

2

Y = 2 K (1 + σ)

3

Y = 3 K (1 - 2 σ)

4

Y = 2 K (1 - σ)

Explanation:

This formula relates young's modulus $(Y)$ , Bulk modulus $(K)$ , and Poisson's ratio $(σ)$ 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

MHTCET - 2019

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 ' $σ$ ' be the poisson's ratio then volumetric strain is

1

\frac{F}{A Y} (2 - σ)

2

\frac{F}{A Y} (1 - σ)

3

\frac{F}{A Y} \cdot σ

4

\frac{F}{A Y} (1 - 2 σ)

Explanation:

$\frac{Δ V}{V} = \frac{Δ L}{L} + 2 \frac{d r}{r} = \frac{F}{A Y} + 2 (- \frac{σ F}{A Y})$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

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

- \frac{1}{2}

2

+ \frac{1}{2}

3

- \frac{1}{4}

4

+ \frac{1}{4}

Explanation:

Volume of cylindrical wire, $V = \frac{π x^{2} L}{4}$ where $x$ is diameter of wire
Differentiating both sides
$\frac{d V}{d x} = \frac{π}{4} [2 x L + x^{2} \cdot \frac{d L}{d x}]$
Also volume remains constant $∴ \frac{d V}{d x} = 0$
$\begin{matrix} ∴ 2 x L + x^{2} \frac{d L}{d x} = 0 \Rightarrow 2 x L = - x^{2} \frac{d L}{d x}, \\ \Rightarrow \frac{\frac{d x}{x}}{\frac{d L}{L}} = - \frac{1}{2} \end{matrix}$
$\Rightarrow Poisson's ratio = - \frac{1}{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369743 Relation between $Y, K & σ$ is

1

Y = K (1 - σ)

2

Y = 2 K (1 + σ)

3

Y = 3 K (1 - 2 σ)

4

Y = 2 K (1 - σ)

Explanation:

This formula relates young's modulus $(Y)$ , Bulk modulus $(K)$ , and Poisson's ratio $(σ)$ 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

MHTCET - 2019