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Mechanical Properties of Solids

141147 A wire of natural length $l$ , Young's modulus $Y$ and area of cross-section $A$ is extended by $x$ . Then the energy stored in the wire is given by

1

\frac{1}{2} \frac{YA}{l} x^{2}

2

\frac{1}{3} \frac{YA}{l} x^{2}

3

\frac{1}{2} \frac{Yl}{A} x^{2}

4

\frac{1}{2} \frac{YA}{l^{2}} x^{2}

5

\frac{1}{2} \frac{A}{Yl} x^{2}

Explanation:

A Energy stored in the wire,
$U = \frac{1}{2} \times Y \times (Strain)^{2} \times volume$
$U = \frac{1}{2} \times Y \times {(\frac{x}{l})}^{2} \times A \times l$
$U = \frac{1}{2} \frac{Yx^{2} A}{l}$
$U = \frac{1}{2} \frac{YA}{l} x^{2}$

Kerala CEE 2007

Mechanical Properties of Solids

141148 A wire of length $L$ and area of cross-section $A$ is stretched through a distance $x$ meter by applying a force $F$ along length, then the work done in this process is: ( $Y$ is Young's modulus of the material)

1

\frac{1}{2} (A \cdot L) (\frac{Yx}{L}) (\frac{x}{L})

2

(A \cdot L) (YL) (\frac{x}{L})

3

2 (A \cdot L) (YL) (\frac{x}{L})

4

3 (A \cdot L) (YL) (\frac{x}{L})

5

4 (A \cdot L) (YL) (\frac{x}{L})

Explanation:

A work done $= \frac{1}{2} \times$ stress $\times$ strain $\times$ volume
or
$W = \frac{1}{2} Y \frac{X}{L} \times \frac{x}{L} \times AL$
$= \frac{1}{2} (AL) (Y \frac{x}{L}) (\frac{x}{L})$

Kerala CEE 2004

Mechanical Properties of Solids

141141 Proof resilience is related to

1 Potential energy stored in the elastic body

2 Stiffness of a beam

3 Elastic fatigue

4 Elastic relaxation

SRMJEE - 2016

Mechanical Properties of Solids

141147 A wire of natural length $l$ , Young's modulus $Y$ and area of cross-section $A$ is extended by $x$ . Then the energy stored in the wire is given by

1

\frac{1}{2} \frac{YA}{l} x^{2}

2

\frac{1}{3} \frac{YA}{l} x^{2}

3

\frac{1}{2} \frac{Yl}{A} x^{2}

4

\frac{1}{2} \frac{YA}{l^{2}} x^{2}

5

\frac{1}{2} \frac{A}{Yl} x^{2}

Explanation:

A Energy stored in the wire,
$U = \frac{1}{2} \times Y \times (Strain)^{2} \times volume$
$U = \frac{1}{2} \times Y \times {(\frac{x}{l})}^{2} \times A \times l$
$U = \frac{1}{2} \frac{Yx^{2} A}{l}$
$U = \frac{1}{2} \frac{YA}{l} x^{2}$

Kerala CEE 2007

Mechanical Properties of Solids

141148 A wire of length $L$ and area of cross-section $A$ is stretched through a distance $x$ meter by applying a force $F$ along length, then the work done in this process is: ( $Y$ is Young's modulus of the material)

1

\frac{1}{2} (A \cdot L) (\frac{Yx}{L}) (\frac{x}{L})

2

(A \cdot L) (YL) (\frac{x}{L})

3

2 (A \cdot L) (YL) (\frac{x}{L})

4

3 (A \cdot L) (YL) (\frac{x}{L})

5

4 (A \cdot L) (YL) (\frac{x}{L})

Explanation:

A work done $= \frac{1}{2} \times$ stress $\times$ strain $\times$ volume
or
$W = \frac{1}{2} Y \frac{X}{L} \times \frac{x}{L} \times AL$
$= \frac{1}{2} (AL) (Y \frac{x}{L}) (\frac{x}{L})$

Kerala CEE 2004

Mechanical Properties of Solids

141141 Proof resilience is related to

1 Potential energy stored in the elastic body

2 Stiffness of a beam

3 Elastic fatigue

4 Elastic relaxation

SRMJEE - 2016

Mechanical Properties of Solids

141147 A wire of natural length $l$ , Young's modulus $Y$ and area of cross-section $A$ is extended by $x$ . Then the energy stored in the wire is given by

1

\frac{1}{2} \frac{YA}{l} x^{2}

2

\frac{1}{3} \frac{YA}{l} x^{2}

3

\frac{1}{2} \frac{Yl}{A} x^{2}

4

\frac{1}{2} \frac{YA}{l^{2}} x^{2}

5

\frac{1}{2} \frac{A}{Yl} x^{2}

Explanation:

A Energy stored in the wire,
$U = \frac{1}{2} \times Y \times (Strain)^{2} \times volume$
$U = \frac{1}{2} \times Y \times {(\frac{x}{l})}^{2} \times A \times l$
$U = \frac{1}{2} \frac{Yx^{2} A}{l}$
$U = \frac{1}{2} \frac{YA}{l} x^{2}$

Kerala CEE 2007

Mechanical Properties of Solids

141148 A wire of length $L$ and area of cross-section $A$ is stretched through a distance $x$ meter by applying a force $F$ along length, then the work done in this process is: ( $Y$ is Young's modulus of the material)

1

\frac{1}{2} (A \cdot L) (\frac{Yx}{L}) (\frac{x}{L})

2

(A \cdot L) (YL) (\frac{x}{L})

3

2 (A \cdot L) (YL) (\frac{x}{L})

4

3 (A \cdot L) (YL) (\frac{x}{L})

5

4 (A \cdot L) (YL) (\frac{x}{L})

Explanation:

A work done $= \frac{1}{2} \times$ stress $\times$ strain $\times$ volume
or
$W = \frac{1}{2} Y \frac{X}{L} \times \frac{x}{L} \times AL$
$= \frac{1}{2} (AL) (Y \frac{x}{L}) (\frac{x}{L})$

Kerala CEE 2004

Mechanical Properties of Solids

141141 Proof resilience is related to

1 Potential energy stored in the elastic body

2 Stiffness of a beam

3 Elastic fatigue

4 Elastic relaxation

SRMJEE - 2016

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