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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369889 When a $8 k g$ mass is hung vertically on a light spring that obey's Hooke's law, the spring stretched by $2 c m$ . The work required to be done by an external agent in stretching this spring by $10 c m$ will be

1

15 J

2

20 J

3

19.6 J

4

10 J

Explanation:

$F = k x = k = \frac{F}{x} = \frac{8 \times 9.8}{2 \times 10^{- 2}} = 4 \times 9.8 \times 10^{2}$
$= 39.2 \times 10^{2} N / m$
$∴ W = \frac{1}{2} \times k x^{2} = 19.6 J$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369890 A metallic rod of length $l$ and cross-section area $A$ is made of a material of Young modulus $Y$ . If the rod is elongated by an amount $y$ , then the work done is proportional to

1

\frac{1}{y}

2

y

3

\frac{1}{y^{2}}

4

y^{2}

Explanation:

$V = A l$
Young's modulus $Y = \frac{Stress}{Strain}$
Work done, $W = \frac{1}{2} \times$ Stress $\times$ Strain $\times$ volume
$W = \frac{1}{2} \times Y \times (Strain)^{2} \times A l$
$= \frac{1}{2} \times Y \times {(\frac{y}{l})}^{2} \times A l = \frac{1}{2} (\frac{Y A}{l}) y^{2} \Rightarrow W \propto y^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369891 The elastic energy stored per unit volume in a stretched wire is

1

\frac{1}{2} (

Stress

) (Strain)^{2}

2

\frac{1}{2} (

young modulus)

(Strain)^{2}

3

\frac{1}{2} (young modulus) (Stress)

4

\frac{1}{2} \frac{Stress}{Strain}

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369892 The graph shows the behaviour of a wire in the region for which the substance obeys Hooke's Law. $P$ and $Q$ represent
supporting img

1

P =

Extension,

Q =

Applied force

2

P

=Applied force,

Q =

Extension

3

P =

Stored elastic energy,

Q =

Extension

4

P =

Extension,

Q =

Stored elastic energy

Explanation:

Graph between applied force and extension will be straight line because in elastic range,
$F α x$
The graph between extension and stored elastic energy will be parabolic in nature
$As U = 1 / 2 k x^{2} or U \propto x^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369893 A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . It is stretched by an amount $x$ . The work done (or energy stored) is

1

\frac{Y x^{2} A}{L}

2

\frac{Y x A}{2 L}

3

\frac{2 Y x^{2} A}{L}

4

\frac{Y x^{2} A}{2 L}

Explanation:

We can treat the rod as a spring and its spring constant is
$K = \frac{Y A}{L}$
Work done in stretching the spring (rod) is
$W = \frac{1}{2} K x^{2} = \frac{Y A x^{2}}{2 L}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369889 When a $8 k g$ mass is hung vertically on a light spring that obey's Hooke's law, the spring stretched by $2 c m$ . The work required to be done by an external agent in stretching this spring by $10 c m$ will be

1

15 J

2

20 J

3

19.6 J

4

10 J

Explanation:

$F = k x = k = \frac{F}{x} = \frac{8 \times 9.8}{2 \times 10^{- 2}} = 4 \times 9.8 \times 10^{2}$
$= 39.2 \times 10^{2} N / m$
$∴ W = \frac{1}{2} \times k x^{2} = 19.6 J$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369890 A metallic rod of length $l$ and cross-section area $A$ is made of a material of Young modulus $Y$ . If the rod is elongated by an amount $y$ , then the work done is proportional to

1

\frac{1}{y}

2

y

3

\frac{1}{y^{2}}

4

y^{2}

Explanation:

$V = A l$
Young's modulus $Y = \frac{Stress}{Strain}$
Work done, $W = \frac{1}{2} \times$ Stress $\times$ Strain $\times$ volume
$W = \frac{1}{2} \times Y \times (Strain)^{2} \times A l$
$= \frac{1}{2} \times Y \times {(\frac{y}{l})}^{2} \times A l = \frac{1}{2} (\frac{Y A}{l}) y^{2} \Rightarrow W \propto y^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369891 The elastic energy stored per unit volume in a stretched wire is

1

\frac{1}{2} (

Stress

) (Strain)^{2}

2

\frac{1}{2} (

young modulus)

(Strain)^{2}

3

\frac{1}{2} (young modulus) (Stress)

4

\frac{1}{2} \frac{Stress}{Strain}

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369892 The graph shows the behaviour of a wire in the region for which the substance obeys Hooke's Law. $P$ and $Q$ represent
supporting img

1

P =

Extension,

Q =

Applied force

2

P

=Applied force,

Q =

Extension

3

P =

Stored elastic energy,

Q =

Extension

4

P =

Extension,

Q =

Stored elastic energy

Explanation:

Graph between applied force and extension will be straight line because in elastic range,
$F α x$
The graph between extension and stored elastic energy will be parabolic in nature
$As U = 1 / 2 k x^{2} or U \propto x^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369893 A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . It is stretched by an amount $x$ . The work done (or energy stored) is

1

\frac{Y x^{2} A}{L}

2

\frac{Y x A}{2 L}

3

\frac{2 Y x^{2} A}{L}

4

\frac{Y x^{2} A}{2 L}

Explanation:

We can treat the rod as a spring and its spring constant is
$K = \frac{Y A}{L}$
Work done in stretching the spring (rod) is
$W = \frac{1}{2} K x^{2} = \frac{Y A x^{2}}{2 L}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369889 When a $8 k g$ mass is hung vertically on a light spring that obey's Hooke's law, the spring stretched by $2 c m$ . The work required to be done by an external agent in stretching this spring by $10 c m$ will be

1

15 J

2

20 J

3

19.6 J

4

10 J

Explanation:

$F = k x = k = \frac{F}{x} = \frac{8 \times 9.8}{2 \times 10^{- 2}} = 4 \times 9.8 \times 10^{2}$
$= 39.2 \times 10^{2} N / m$
$∴ W = \frac{1}{2} \times k x^{2} = 19.6 J$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369890 A metallic rod of length $l$ and cross-section area $A$ is made of a material of Young modulus $Y$ . If the rod is elongated by an amount $y$ , then the work done is proportional to

1

\frac{1}{y}

2

y

3

\frac{1}{y^{2}}

4

y^{2}

Explanation:

$V = A l$
Young's modulus $Y = \frac{Stress}{Strain}$
Work done, $W = \frac{1}{2} \times$ Stress $\times$ Strain $\times$ volume
$W = \frac{1}{2} \times Y \times (Strain)^{2} \times A l$
$= \frac{1}{2} \times Y \times {(\frac{y}{l})}^{2} \times A l = \frac{1}{2} (\frac{Y A}{l}) y^{2} \Rightarrow W \propto y^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369891 The elastic energy stored per unit volume in a stretched wire is

1

\frac{1}{2} (

Stress

) (Strain)^{2}

2

\frac{1}{2} (

young modulus)

(Strain)^{2}

3

\frac{1}{2} (young modulus) (Stress)

4

\frac{1}{2} \frac{Stress}{Strain}

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369892 The graph shows the behaviour of a wire in the region for which the substance obeys Hooke's Law. $P$ and $Q$ represent
supporting img

1

P =

Extension,

Q =

Applied force

2

P

=Applied force,

Q =

Extension

3

P =

Stored elastic energy,

Q =

Extension

4

P =

Extension,

Q =

Stored elastic energy

Explanation:

Graph between applied force and extension will be straight line because in elastic range,
$F α x$
The graph between extension and stored elastic energy will be parabolic in nature
$As U = 1 / 2 k x^{2} or U \propto x^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369893 A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . It is stretched by an amount $x$ . The work done (or energy stored) is

1

\frac{Y x^{2} A}{L}

2

\frac{Y x A}{2 L}

3

\frac{2 Y x^{2} A}{L}

4

\frac{Y x^{2} A}{2 L}

Explanation:

We can treat the rod as a spring and its spring constant is
$K = \frac{Y A}{L}$
Work done in stretching the spring (rod) is
$W = \frac{1}{2} K x^{2} = \frac{Y A x^{2}}{2 L}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369889 When a $8 k g$ mass is hung vertically on a light spring that obey's Hooke's law, the spring stretched by $2 c m$ . The work required to be done by an external agent in stretching this spring by $10 c m$ will be

1

15 J

2

20 J

3

19.6 J

4

10 J

Explanation:

$F = k x = k = \frac{F}{x} = \frac{8 \times 9.8}{2 \times 10^{- 2}} = 4 \times 9.8 \times 10^{2}$
$= 39.2 \times 10^{2} N / m$
$∴ W = \frac{1}{2} \times k x^{2} = 19.6 J$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369890 A metallic rod of length $l$ and cross-section area $A$ is made of a material of Young modulus $Y$ . If the rod is elongated by an amount $y$ , then the work done is proportional to

1

\frac{1}{y}

2

y

3

\frac{1}{y^{2}}

4

y^{2}

Explanation:

$V = A l$
Young's modulus $Y = \frac{Stress}{Strain}$
Work done, $W = \frac{1}{2} \times$ Stress $\times$ Strain $\times$ volume
$W = \frac{1}{2} \times Y \times (Strain)^{2} \times A l$
$= \frac{1}{2} \times Y \times {(\frac{y}{l})}^{2} \times A l = \frac{1}{2} (\frac{Y A}{l}) y^{2} \Rightarrow W \propto y^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369891 The elastic energy stored per unit volume in a stretched wire is

1

\frac{1}{2} (

Stress

) (Strain)^{2}

2

\frac{1}{2} (

young modulus)

(Strain)^{2}

3

\frac{1}{2} (young modulus) (Stress)

4

\frac{1}{2} \frac{Stress}{Strain}

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369892 The graph shows the behaviour of a wire in the region for which the substance obeys Hooke's Law. $P$ and $Q$ represent
supporting img

1

P =

Extension,

Q =

Applied force

2

P

=Applied force,

Q =

Extension

3

P =

Stored elastic energy,

Q =

Extension

4

P =

Extension,

Q =

Stored elastic energy

Explanation:

Graph between applied force and extension will be straight line because in elastic range,
$F α x$
The graph between extension and stored elastic energy will be parabolic in nature
$As U = 1 / 2 k x^{2} or U \propto x^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369893 A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . It is stretched by an amount $x$ . The work done (or energy stored) is

1

\frac{Y x^{2} A}{L}

2

\frac{Y x A}{2 L}

3

\frac{2 Y x^{2} A}{L}

4

\frac{Y x^{2} A}{2 L}

Explanation:

We can treat the rod as a spring and its spring constant is
$K = \frac{Y A}{L}$
Work done in stretching the spring (rod) is
$W = \frac{1}{2} K x^{2} = \frac{Y A x^{2}}{2 L}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369889 When a $8 k g$ mass is hung vertically on a light spring that obey's Hooke's law, the spring stretched by $2 c m$ . The work required to be done by an external agent in stretching this spring by $10 c m$ will be

1

15 J

2

20 J

3

19.6 J

4

10 J

Explanation:

$F = k x = k = \frac{F}{x} = \frac{8 \times 9.8}{2 \times 10^{- 2}} = 4 \times 9.8 \times 10^{2}$
$= 39.2 \times 10^{2} N / m$
$∴ W = \frac{1}{2} \times k x^{2} = 19.6 J$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369890 A metallic rod of length $l$ and cross-section area $A$ is made of a material of Young modulus $Y$ . If the rod is elongated by an amount $y$ , then the work done is proportional to

1

\frac{1}{y}

2

y

3

\frac{1}{y^{2}}

4

y^{2}

Explanation:

$V = A l$
Young's modulus $Y = \frac{Stress}{Strain}$
Work done, $W = \frac{1}{2} \times$ Stress $\times$ Strain $\times$ volume
$W = \frac{1}{2} \times Y \times (Strain)^{2} \times A l$
$= \frac{1}{2} \times Y \times {(\frac{y}{l})}^{2} \times A l = \frac{1}{2} (\frac{Y A}{l}) y^{2} \Rightarrow W \propto y^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369891 The elastic energy stored per unit volume in a stretched wire is

1

\frac{1}{2} (

Stress

) (Strain)^{2}

2

\frac{1}{2} (

young modulus)

(Strain)^{2}

3

\frac{1}{2} (young modulus) (Stress)

4

\frac{1}{2} \frac{Stress}{Strain}

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369892 The graph shows the behaviour of a wire in the region for which the substance obeys Hooke's Law. $P$ and $Q$ represent
supporting img

1

P =

Extension,

Q =

Applied force

2

P

=Applied force,

Q =

Extension

3

P =

Stored elastic energy,

Q =

Extension

4

P =

Extension,

Q =

Stored elastic energy

Explanation:

Graph between applied force and extension will be straight line because in elastic range,
$F α x$
The graph between extension and stored elastic energy will be parabolic in nature
$As U = 1 / 2 k x^{2} or U \propto x^{2} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369893 A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . It is stretched by an amount $x$ . The work done (or energy stored) is

1

\frac{Y x^{2} A}{L}

2

\frac{Y x A}{2 L}

3

\frac{2 Y x^{2} A}{L}

4

\frac{Y x^{2} A}{2 L}

Explanation:

We can treat the rod as a spring and its spring constant is
$K = \frac{Y A}{L}$
Work done in stretching the spring (rod) is
$W = \frac{1}{2} K x^{2} = \frac{Y A x^{2}}{2 L}$

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