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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369947 A mild steel wire of length $2 L$ and crosssectional area A is stretched well within elastic limit, horizontally between two pillars as shown in figure. A mass $m$ is suspended from the mid point of the wire. Strain in the wire is
supporting img

1

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

2

\frac{x}{L}

3

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

4

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

Explanation:

According to the figure,

supporting img

Increases in length $= B O + O C - B C$
$Δ L = 2 B O - 2 L = 2 {(L^{2} + x^{2})}^{1 / 2} - 2 L$
or $Δ L = 2 L [1 + \frac{x^{2}}{2 L^{2}}] - 2 L = \frac{x^{2}}{L}$
$∴ Strain = \frac{Δ L}{2 L} = \frac{x^{2} / L}{2 L} = \frac{x^{2}}{2 L^{2}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369948 If the angle of shear is $30^{\circ}$ for a cubical body and the change in length is $250 c m,$ then the original length of cube is

1

4.9 m

2

5.5 m

3

5.3 m

4

4.3 m

Explanation:

Given, angle of shear. $θ = 30^{\circ}$
Change in length, $Δ L = 250 c m = 2.5 m$
As,shear strain, $\tan θ = \frac{Δ L}{L}$
$\Rightarrow \tan 30^{\circ} = \frac{2.5}{L}$
$L = 2.5 \sqrt{3} = 4.33 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369949 A man grows into a giant such that his linear dimensions increase by a factor of 9 . Assuming that his density remains same, the stress in the leg will change by a factor of :

1

\frac{1}{9}

2 81

3 9

4

\frac{1}{81}

Explanation:

$Stress = \frac{Force}{area} = \frac{m g}{A} = \frac{volume \times density \times g}{Area}$
Stress $= \frac{L^{3} ρ g}{L^{2}}$
Stress $\propto L$
Stress will become 9 times the initial value

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369950 Match the Column I (stress) with Column II (characteristic) and select the correct answer from the codes given below.
Column I
Column II
A
Longitudinal stress
P
Independent of area of cross-section of wire
B
Shear stress
Q
Change in pressure
C
Breaking stress
R
Length of wire increases or decreases
D
Bulk stress
S
Shape changes

1

A - R, B - S, C - Q D - P

2

A - S, B - R, C - P D - Q

3

A - R, B - S, C - P D - Q

4

A - P, B - Q, C - R D - S

Explanation:

A. During longitudinal stress, length of wire increases or decreases.
$B$ . When force is applied tangentially to a surface of a body, then tangential or shear stress is produced and shape of the body changes.
C. Breaking stress is independent of area of crosssection of wire
D. Bulk stress is equal to change in pressure.
Hence, option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369947 A mild steel wire of length $2 L$ and crosssectional area A is stretched well within elastic limit, horizontally between two pillars as shown in figure. A mass $m$ is suspended from the mid point of the wire. Strain in the wire is
supporting img

1

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

2

\frac{x}{L}

3

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

4

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

Explanation:

According to the figure,

supporting img

Increases in length $= B O + O C - B C$
$Δ L = 2 B O - 2 L = 2 {(L^{2} + x^{2})}^{1 / 2} - 2 L$
or $Δ L = 2 L [1 + \frac{x^{2}}{2 L^{2}}] - 2 L = \frac{x^{2}}{L}$
$∴ Strain = \frac{Δ L}{2 L} = \frac{x^{2} / L}{2 L} = \frac{x^{2}}{2 L^{2}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369948 If the angle of shear is $30^{\circ}$ for a cubical body and the change in length is $250 c m,$ then the original length of cube is

1

4.9 m

2

5.5 m

3

5.3 m

4

4.3 m

Explanation:

Given, angle of shear. $θ = 30^{\circ}$
Change in length, $Δ L = 250 c m = 2.5 m$
As,shear strain, $\tan θ = \frac{Δ L}{L}$
$\Rightarrow \tan 30^{\circ} = \frac{2.5}{L}$
$L = 2.5 \sqrt{3} = 4.33 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369949 A man grows into a giant such that his linear dimensions increase by a factor of 9 . Assuming that his density remains same, the stress in the leg will change by a factor of :

1

\frac{1}{9}

2 81

3 9

4

\frac{1}{81}

Explanation:

$Stress = \frac{Force}{area} = \frac{m g}{A} = \frac{volume \times density \times g}{Area}$
Stress $= \frac{L^{3} ρ g}{L^{2}}$
Stress $\propto L$
Stress will become 9 times the initial value

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369950 Match the Column I (stress) with Column II (characteristic) and select the correct answer from the codes given below.
Column I
Column II
A
Longitudinal stress
P
Independent of area of cross-section of wire
B
Shear stress
Q
Change in pressure
C
Breaking stress
R
Length of wire increases or decreases
D
Bulk stress
S
Shape changes

1

A - R, B - S, C - Q D - P

2

A - S, B - R, C - P D - Q

3

A - R, B - S, C - P D - Q

4

A - P, B - Q, C - R D - S

Explanation:

A. During longitudinal stress, length of wire increases or decreases.
$B$ . When force is applied tangentially to a surface of a body, then tangential or shear stress is produced and shape of the body changes.
C. Breaking stress is independent of area of crosssection of wire
D. Bulk stress is equal to change in pressure.
Hence, option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369947 A mild steel wire of length $2 L$ and crosssectional area A is stretched well within elastic limit, horizontally between two pillars as shown in figure. A mass $m$ is suspended from the mid point of the wire. Strain in the wire is
supporting img

1

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

2

\frac{x}{L}

3

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

4

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

Explanation:

According to the figure,

supporting img

Increases in length $= B O + O C - B C$
$Δ L = 2 B O - 2 L = 2 {(L^{2} + x^{2})}^{1 / 2} - 2 L$
or $Δ L = 2 L [1 + \frac{x^{2}}{2 L^{2}}] - 2 L = \frac{x^{2}}{L}$
$∴ Strain = \frac{Δ L}{2 L} = \frac{x^{2} / L}{2 L} = \frac{x^{2}}{2 L^{2}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369948 If the angle of shear is $30^{\circ}$ for a cubical body and the change in length is $250 c m,$ then the original length of cube is

1

4.9 m

2

5.5 m

3

5.3 m

4

4.3 m

Explanation:

Given, angle of shear. $θ = 30^{\circ}$
Change in length, $Δ L = 250 c m = 2.5 m$
As,shear strain, $\tan θ = \frac{Δ L}{L}$
$\Rightarrow \tan 30^{\circ} = \frac{2.5}{L}$
$L = 2.5 \sqrt{3} = 4.33 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369949 A man grows into a giant such that his linear dimensions increase by a factor of 9 . Assuming that his density remains same, the stress in the leg will change by a factor of :

1

\frac{1}{9}

2 81

3 9

4

\frac{1}{81}

Explanation:

$Stress = \frac{Force}{area} = \frac{m g}{A} = \frac{volume \times density \times g}{Area}$
Stress $= \frac{L^{3} ρ g}{L^{2}}$
Stress $\propto L$
Stress will become 9 times the initial value

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369950 Match the Column I (stress) with Column II (characteristic) and select the correct answer from the codes given below.
Column I
Column II
A
Longitudinal stress
P
Independent of area of cross-section of wire
B
Shear stress
Q
Change in pressure
C
Breaking stress
R
Length of wire increases or decreases
D
Bulk stress
S
Shape changes

1

A - R, B - S, C - Q D - P

2

A - S, B - R, C - P D - Q

3

A - R, B - S, C - P D - Q

4

A - P, B - Q, C - R D - S

Explanation:

A. During longitudinal stress, length of wire increases or decreases.
$B$ . When force is applied tangentially to a surface of a body, then tangential or shear stress is produced and shape of the body changes.
C. Breaking stress is independent of area of crosssection of wire
D. Bulk stress is equal to change in pressure.
Hence, option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369947 A mild steel wire of length $2 L$ and crosssectional area A is stretched well within elastic limit, horizontally between two pillars as shown in figure. A mass $m$ is suspended from the mid point of the wire. Strain in the wire is
supporting img

1

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

2

\frac{x}{L}

3

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

4

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

Explanation:

According to the figure,

supporting img

Increases in length $= B O + O C - B C$
$Δ L = 2 B O - 2 L = 2 {(L^{2} + x^{2})}^{1 / 2} - 2 L$
or $Δ L = 2 L [1 + \frac{x^{2}}{2 L^{2}}] - 2 L = \frac{x^{2}}{L}$
$∴ Strain = \frac{Δ L}{2 L} = \frac{x^{2} / L}{2 L} = \frac{x^{2}}{2 L^{2}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369948 If the angle of shear is $30^{\circ}$ for a cubical body and the change in length is $250 c m,$ then the original length of cube is

1

4.9 m

2

5.5 m

3

5.3 m

4

4.3 m

Explanation:

Given, angle of shear. $θ = 30^{\circ}$
Change in length, $Δ L = 250 c m = 2.5 m$
As,shear strain, $\tan θ = \frac{Δ L}{L}$
$\Rightarrow \tan 30^{\circ} = \frac{2.5}{L}$
$L = 2.5 \sqrt{3} = 4.33 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369949 A man grows into a giant such that his linear dimensions increase by a factor of 9 . Assuming that his density remains same, the stress in the leg will change by a factor of :

1

\frac{1}{9}

2 81

3 9

4

\frac{1}{81}

Explanation:

$Stress = \frac{Force}{area} = \frac{m g}{A} = \frac{volume \times density \times g}{Area}$
Stress $= \frac{L^{3} ρ g}{L^{2}}$
Stress $\propto L$
Stress will become 9 times the initial value

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369950 Match the Column I (stress) with Column II (characteristic) and select the correct answer from the codes given below.
Column I
Column II
A
Longitudinal stress
P
Independent of area of cross-section of wire
B
Shear stress
Q
Change in pressure
C
Breaking stress
R
Length of wire increases or decreases
D
Bulk stress
S
Shape changes

1

A - R, B - S, C - Q D - P

2

A - S, B - R, C - P D - Q

3

A - R, B - S, C - P D - Q

4

A - P, B - Q, C - R D - S

Explanation:

A. During longitudinal stress, length of wire increases or decreases.
$B$ . When force is applied tangentially to a surface of a body, then tangential or shear stress is produced and shape of the body changes.
C. Breaking stress is independent of area of crosssection of wire
D. Bulk stress is equal to change in pressure.
Hence, option (3) is correct.

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