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 $=1 \%$ $\text { Possion's ratio }(\mathrm{v})=0.2$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} l$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ Possion's ratio $(\mathrm{v})=\frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{\text { Lateralstrain }}{\text { Longitudinalstrain }}$ From equation (i), we get - $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\Delta l}{l}(1-2 \mathrm{v})$ $\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=\frac{\Delta l}{l} \times 100(1-2 \mathrm{v})$ $=-1 \times(1-2 \times 0.2)=-0.6 \%$ 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, $\text {Length of wire }=\mathrm{L}$ $\text {Radius }=\mathrm{r}$ $\text {Volume of wire }=\pi \mathrm{r}^{2} \mathrm{~L}$ $\text {Differentiating both the side, }$ $\text {We get, } \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{l}}{1}$ $\text { Because volume of wire constant, }$ $\Delta \mathrm{V}=0$ $0=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ $\frac{-\Delta \mathrm{r}}{\mathrm{r}}=\frac{1}{2} \frac{\Delta l}{l} \Rightarrow \frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{1}{2}$ We know that, $\text { Poisson 's ratio }= \frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}$ $= \frac{\frac{-\Delta \mathrm{r}}{\frac{\mathrm{r}}{l}}}{\frac{\Delta l}{l}}=\frac{-1}{2}=-0.5$
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 $\quad \mathbf{G}\left(10^{9} \mathrm{NM}^{-2}\right)$ Iron 82 Steel $\quad 79$ 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 $=37 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$ ]
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, $\mathrm{P}=10 \mathrm{~atm}=10 \times 1.013 \times 10^{5} \mathrm{~Pa}$ $\mathrm{~B}=37 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{B}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=2.73 \times 10^{-5}$ Hence fractional change in volume of glass slab is 2.73 $\times 10^{-5}$
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 $=1 \%$ $\text { Possion's ratio }(\mathrm{v})=0.2$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} l$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ Possion's ratio $(\mathrm{v})=\frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{\text { Lateralstrain }}{\text { Longitudinalstrain }}$ From equation (i), we get - $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\Delta l}{l}(1-2 \mathrm{v})$ $\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=\frac{\Delta l}{l} \times 100(1-2 \mathrm{v})$ $=-1 \times(1-2 \times 0.2)=-0.6 \%$ 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, $\text {Length of wire }=\mathrm{L}$ $\text {Radius }=\mathrm{r}$ $\text {Volume of wire }=\pi \mathrm{r}^{2} \mathrm{~L}$ $\text {Differentiating both the side, }$ $\text {We get, } \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{l}}{1}$ $\text { Because volume of wire constant, }$ $\Delta \mathrm{V}=0$ $0=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ $\frac{-\Delta \mathrm{r}}{\mathrm{r}}=\frac{1}{2} \frac{\Delta l}{l} \Rightarrow \frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{1}{2}$ We know that, $\text { Poisson 's ratio }= \frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}$ $= \frac{\frac{-\Delta \mathrm{r}}{\frac{\mathrm{r}}{l}}}{\frac{\Delta l}{l}}=\frac{-1}{2}=-0.5$
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 $\quad \mathbf{G}\left(10^{9} \mathrm{NM}^{-2}\right)$ Iron 82 Steel $\quad 79$ 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 $=37 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$ ]
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, $\mathrm{P}=10 \mathrm{~atm}=10 \times 1.013 \times 10^{5} \mathrm{~Pa}$ $\mathrm{~B}=37 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{B}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=2.73 \times 10^{-5}$ Hence fractional change in volume of glass slab is 2.73 $\times 10^{-5}$
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 $=1 \%$ $\text { Possion's ratio }(\mathrm{v})=0.2$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} l$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ Possion's ratio $(\mathrm{v})=\frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{\text { Lateralstrain }}{\text { Longitudinalstrain }}$ From equation (i), we get - $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\Delta l}{l}(1-2 \mathrm{v})$ $\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=\frac{\Delta l}{l} \times 100(1-2 \mathrm{v})$ $=-1 \times(1-2 \times 0.2)=-0.6 \%$ 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, $\text {Length of wire }=\mathrm{L}$ $\text {Radius }=\mathrm{r}$ $\text {Volume of wire }=\pi \mathrm{r}^{2} \mathrm{~L}$ $\text {Differentiating both the side, }$ $\text {We get, } \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{l}}{1}$ $\text { Because volume of wire constant, }$ $\Delta \mathrm{V}=0$ $0=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ $\frac{-\Delta \mathrm{r}}{\mathrm{r}}=\frac{1}{2} \frac{\Delta l}{l} \Rightarrow \frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{1}{2}$ We know that, $\text { Poisson 's ratio }= \frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}$ $= \frac{\frac{-\Delta \mathrm{r}}{\frac{\mathrm{r}}{l}}}{\frac{\Delta l}{l}}=\frac{-1}{2}=-0.5$
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 $\quad \mathbf{G}\left(10^{9} \mathrm{NM}^{-2}\right)$ Iron 82 Steel $\quad 79$ 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 $=37 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$ ]
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, $\mathrm{P}=10 \mathrm{~atm}=10 \times 1.013 \times 10^{5} \mathrm{~Pa}$ $\mathrm{~B}=37 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{B}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=2.73 \times 10^{-5}$ Hence fractional change in volume of glass slab is 2.73 $\times 10^{-5}$
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 $=1 \%$ $\text { Possion's ratio }(\mathrm{v})=0.2$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} l$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ Possion's ratio $(\mathrm{v})=\frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{\text { Lateralstrain }}{\text { Longitudinalstrain }}$ From equation (i), we get - $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\Delta l}{l}(1-2 \mathrm{v})$ $\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=\frac{\Delta l}{l} \times 100(1-2 \mathrm{v})$ $=-1 \times(1-2 \times 0.2)=-0.6 \%$ 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, $\text {Length of wire }=\mathrm{L}$ $\text {Radius }=\mathrm{r}$ $\text {Volume of wire }=\pi \mathrm{r}^{2} \mathrm{~L}$ $\text {Differentiating both the side, }$ $\text {We get, } \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{l}}{1}$ $\text { Because volume of wire constant, }$ $\Delta \mathrm{V}=0$ $0=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}$ $\frac{-\Delta \mathrm{r}}{\mathrm{r}}=\frac{1}{2} \frac{\Delta l}{l} \Rightarrow \frac{-\Delta \mathrm{r} / \mathrm{r}}{\Delta l / l}=\frac{1}{2}$ We know that, $\text { Poisson 's ratio }= \frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}$ $= \frac{\frac{-\Delta \mathrm{r}}{\frac{\mathrm{r}}{l}}}{\frac{\Delta l}{l}}=\frac{-1}{2}=-0.5$
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 $\quad \mathbf{G}\left(10^{9} \mathrm{NM}^{-2}\right)$ Iron 82 Steel $\quad 79$ 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 $=37 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$ ]
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, $\mathrm{P}=10 \mathrm{~atm}=10 \times 1.013 \times 10^{5} \mathrm{~Pa}$ $\mathrm{~B}=37 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{B}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=2.73 \times 10^{-5}$ Hence fractional change in volume of glass slab is 2.73 $\times 10^{-5}$
