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Mechanical Properties of Solids

141114 A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 %$ . If the Poisson's ratio for the material of the rod be 0.2 , which of the following statements is correct?
"The volume approximately .......".

1 decreases by

1 %

2 decreases by

0.8 %

3 decreases by

0.6 %

4 increases by

0.2 %

Explanation:

C Given, Decrement in the length $= 1 %$
$Possion's ratio (v) = 0.2$
Volume $(V) = π r^{2} l$
$\frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
Possion's ratio $(v) = \frac{- Δ r / r}{Δ l / l} = \frac{Lateralstrain}{Longitudinalstrain}$
From equation (i), we get -
$\frac{Δ V}{V} = \frac{Δ l}{l} (1 - 2 v)$
$\frac{Δ V}{V} \times 100 = \frac{Δ l}{l} \times 100 (1 - 2 v)$
$= - 1 \times (1 - 2 \times 0.2) = - 0.6 %$
Here, negative sign shows the decrement in the volume.

WB JEE 2019

Mechanical Properties of Solids

141115 A wire is stretched such that its volume remains constant. The poission's ratio of the material of the wire is :

1 0.50

2 -0.50

3 0.25

4 -0.25

Explanation:

B Let,
$Length of wire = L$
$Radius = r$
$Volume of wire = π r^{2} L$
$Differentiating both the side,$
$We get, \frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{1}$
$Because volume of wire constant,$
$Δ V = 0$
$0 = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\frac{- Δ r}{r} = \frac{1}{2} \frac{Δ l}{l} \Rightarrow \frac{- Δ r / r}{Δ l / l} = \frac{1}{2}$
We know that,
$Poisson 's ratio = \frac{Lateral Strain}{Longitudinal Strain}$
$= \frac{\frac{- Δ r}{\frac{r}{l}}}{\frac{Δ l}{l}} = \frac{- 1}{2} = - 0.5$

Karnataka CET-2019

Mechanical Properties of Solids

141116 modulus.
possess maximum value for rigidity

1 iron

2 copper

3 steel

4 tungsten
[SRM JEE-2018]

Explanation:

D Tungsten possess maximum value for rigidity modus.
Material $G (10^{9} {NM}^{- 2})$
Iron 82
Steel $79$
Copper 49
Tungsten 160

Mechanical Properties of Solids

141117 The fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm is: [Bulk modulus of glass $= 37 \times 10^{9} N / m^{2}$ ]

1

2.74 \times 10^{- 5}

2

5.12 \times 10^{- 3}

3

8.31 \times 10^{- 4}

4

6.11 \times 10^{- 4}

Explanation:

A Given,
$P = 10 atm = 10 \times 1.013 \times 10^{5} Pa$
$B = 37 \times 10^{9} {Nm}^{- 2}$
We know that,
$\frac{Δ V}{V} = \frac{P}{B}$
$\frac{Δ V}{V} = \frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$
$\frac{Δ V}{V} = 2.73 \times 10^{- 5}$
Hence fractional change in volume of glass slab is 2.73 $\times 10^{- 5}$

AIIMS-26.05.2018(E)

Mechanical Properties of Solids

141114 A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 %$ . If the Poisson's ratio for the material of the rod be 0.2 , which of the following statements is correct?
"The volume approximately .......".

1 decreases by

1 %

2 decreases by

0.8 %

3 decreases by

0.6 %

4 increases by

0.2 %

Explanation:

C Given, Decrement in the length $= 1 %$
$Possion's ratio (v) = 0.2$
Volume $(V) = π r^{2} l$
$\frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
Possion's ratio $(v) = \frac{- Δ r / r}{Δ l / l} = \frac{Lateralstrain}{Longitudinalstrain}$
From equation (i), we get -
$\frac{Δ V}{V} = \frac{Δ l}{l} (1 - 2 v)$
$\frac{Δ V}{V} \times 100 = \frac{Δ l}{l} \times 100 (1 - 2 v)$
$= - 1 \times (1 - 2 \times 0.2) = - 0.6 %$
Here, negative sign shows the decrement in the volume.

WB JEE 2019

Mechanical Properties of Solids

141115 A wire is stretched such that its volume remains constant. The poission's ratio of the material of the wire is :

1 0.50

2 -0.50

3 0.25

4 -0.25

Explanation:

B Let,
$Length of wire = L$
$Radius = r$
$Volume of wire = π r^{2} L$
$Differentiating both the side,$
$We get, \frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{1}$
$Because volume of wire constant,$
$Δ V = 0$
$0 = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\frac{- Δ r}{r} = \frac{1}{2} \frac{Δ l}{l} \Rightarrow \frac{- Δ r / r}{Δ l / l} = \frac{1}{2}$
We know that,
$Poisson 's ratio = \frac{Lateral Strain}{Longitudinal Strain}$
$= \frac{\frac{- Δ r}{\frac{r}{l}}}{\frac{Δ l}{l}} = \frac{- 1}{2} = - 0.5$

Karnataka CET-2019

Mechanical Properties of Solids

141116 modulus.
possess maximum value for rigidity

1 iron

2 copper

3 steel

4 tungsten
[SRM JEE-2018]

Explanation:

D Tungsten possess maximum value for rigidity modus.
Material $G (10^{9} {NM}^{- 2})$
Iron 82
Steel $79$
Copper 49
Tungsten 160

Mechanical Properties of Solids

141117 The fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm is: [Bulk modulus of glass $= 37 \times 10^{9} N / m^{2}$ ]

1

2.74 \times 10^{- 5}

2

5.12 \times 10^{- 3}

3

8.31 \times 10^{- 4}

4

6.11 \times 10^{- 4}

Explanation:

A Given,
$P = 10 atm = 10 \times 1.013 \times 10^{5} Pa$
$B = 37 \times 10^{9} {Nm}^{- 2}$
We know that,
$\frac{Δ V}{V} = \frac{P}{B}$
$\frac{Δ V}{V} = \frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$
$\frac{Δ V}{V} = 2.73 \times 10^{- 5}$
Hence fractional change in volume of glass slab is 2.73 $\times 10^{- 5}$

AIIMS-26.05.2018(E)

Mechanical Properties of Solids

141114 A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 %$ . If the Poisson's ratio for the material of the rod be 0.2 , which of the following statements is correct?
"The volume approximately .......".

1 decreases by

1 %

2 decreases by

0.8 %

3 decreases by

0.6 %

4 increases by

0.2 %

Explanation:

C Given, Decrement in the length $= 1 %$
$Possion's ratio (v) = 0.2$
Volume $(V) = π r^{2} l$
$\frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
Possion's ratio $(v) = \frac{- Δ r / r}{Δ l / l} = \frac{Lateralstrain}{Longitudinalstrain}$
From equation (i), we get -
$\frac{Δ V}{V} = \frac{Δ l}{l} (1 - 2 v)$
$\frac{Δ V}{V} \times 100 = \frac{Δ l}{l} \times 100 (1 - 2 v)$
$= - 1 \times (1 - 2 \times 0.2) = - 0.6 %$
Here, negative sign shows the decrement in the volume.

WB JEE 2019

Mechanical Properties of Solids

141115 A wire is stretched such that its volume remains constant. The poission's ratio of the material of the wire is :

1 0.50

2 -0.50

3 0.25

4 -0.25

Explanation:

B Let,
$Length of wire = L$
$Radius = r$
$Volume of wire = π r^{2} L$
$Differentiating both the side,$
$We get, \frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{1}$
$Because volume of wire constant,$
$Δ V = 0$
$0 = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\frac{- Δ r}{r} = \frac{1}{2} \frac{Δ l}{l} \Rightarrow \frac{- Δ r / r}{Δ l / l} = \frac{1}{2}$
We know that,
$Poisson 's ratio = \frac{Lateral Strain}{Longitudinal Strain}$
$= \frac{\frac{- Δ r}{\frac{r}{l}}}{\frac{Δ l}{l}} = \frac{- 1}{2} = - 0.5$

Karnataka CET-2019

Mechanical Properties of Solids

141116 modulus.
possess maximum value for rigidity

1 iron

2 copper

3 steel

4 tungsten
[SRM JEE-2018]

Explanation:

D Tungsten possess maximum value for rigidity modus.
Material $G (10^{9} {NM}^{- 2})$
Iron 82
Steel $79$
Copper 49
Tungsten 160

Mechanical Properties of Solids

141117 The fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm is: [Bulk modulus of glass $= 37 \times 10^{9} N / m^{2}$ ]

1

2.74 \times 10^{- 5}

2

5.12 \times 10^{- 3}

3

8.31 \times 10^{- 4}

4

6.11 \times 10^{- 4}

Explanation:

A Given,
$P = 10 atm = 10 \times 1.013 \times 10^{5} Pa$
$B = 37 \times 10^{9} {Nm}^{- 2}$
We know that,
$\frac{Δ V}{V} = \frac{P}{B}$
$\frac{Δ V}{V} = \frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$
$\frac{Δ V}{V} = 2.73 \times 10^{- 5}$
Hence fractional change in volume of glass slab is 2.73 $\times 10^{- 5}$

AIIMS-26.05.2018(E)

Mechanical Properties of Solids

141114 A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 %$ . If the Poisson's ratio for the material of the rod be 0.2 , which of the following statements is correct?
"The volume approximately .......".

1 decreases by

1 %

2 decreases by

0.8 %

3 decreases by

0.6 %

4 increases by

0.2 %

Explanation:

C Given, Decrement in the length $= 1 %$
$Possion's ratio (v) = 0.2$
Volume $(V) = π r^{2} l$
$\frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
Possion's ratio $(v) = \frac{- Δ r / r}{Δ l / l} = \frac{Lateralstrain}{Longitudinalstrain}$
From equation (i), we get -
$\frac{Δ V}{V} = \frac{Δ l}{l} (1 - 2 v)$
$\frac{Δ V}{V} \times 100 = \frac{Δ l}{l} \times 100 (1 - 2 v)$
$= - 1 \times (1 - 2 \times 0.2) = - 0.6 %$
Here, negative sign shows the decrement in the volume.

WB JEE 2019

Mechanical Properties of Solids

141115 A wire is stretched such that its volume remains constant. The poission's ratio of the material of the wire is :

1 0.50

2 -0.50

3 0.25

4 -0.25

Explanation:

B Let,
$Length of wire = L$
$Radius = r$
$Volume of wire = π r^{2} L$
$Differentiating both the side,$
$We get, \frac{Δ V}{V} = \frac{2 Δ r}{r} + \frac{Δ l}{1}$
$Because volume of wire constant,$
$Δ V = 0$
$0 = \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\frac{- Δ r}{r} = \frac{1}{2} \frac{Δ l}{l} \Rightarrow \frac{- Δ r / r}{Δ l / l} = \frac{1}{2}$
We know that,
$Poisson 's ratio = \frac{Lateral Strain}{Longitudinal Strain}$
$= \frac{\frac{- Δ r}{\frac{r}{l}}}{\frac{Δ l}{l}} = \frac{- 1}{2} = - 0.5$

Karnataka CET-2019

Mechanical Properties of Solids

141116 modulus.
possess maximum value for rigidity

1 iron

2 copper

3 steel

4 tungsten
[SRM JEE-2018]

Explanation:

D Tungsten possess maximum value for rigidity modus.
Material $G (10^{9} {NM}^{- 2})$
Iron 82
Steel $79$
Copper 49
Tungsten 160

Mechanical Properties of Solids

141117 The fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm is: [Bulk modulus of glass $= 37 \times 10^{9} N / m^{2}$ ]

1

2.74 \times 10^{- 5}

2

5.12 \times 10^{- 3}

3

8.31 \times 10^{- 4}

4

6.11 \times 10^{- 4}

Explanation:

A Given,
$P = 10 atm = 10 \times 1.013 \times 10^{5} Pa$
$B = 37 \times 10^{9} {Nm}^{- 2}$
We know that,
$\frac{Δ V}{V} = \frac{P}{B}$
$\frac{Δ V}{V} = \frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$
$\frac{Δ V}{V} = 2.73 \times 10^{- 5}$
Hence fractional change in volume of glass slab is 2.73 $\times 10^{- 5}$

AIIMS-26.05.2018(E)

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