141026
The average depth of Indian Ocean is about $3000 \mathrm{~m}$. The fractional compression, $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ of water at the bottom of the water $=2.2 \times 10^{9}$ $\mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ) is
141027
The compressibility of water is $6 \times 10^{-10} \mathrm{~N}^{-1} \mathrm{~m}^{2}$. If one litre is subjected to a pressure of $4 \times 10^{\dagger}$ $\mathrm{Nm}^{-2}$, the decrease in its volume is
141028
The Young's modulus of the material of a wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$. If the elongation strain is $1 \%$ then the energy stored in the wire per unit volume in $\mathbf{J m}^{-3}$ is
1 $10^{6}$
2 $10^{8}$
3 $2 \times 10^{6}$
4 $2 \times 10^{8}$
5 $0.5 \times 10^{6}$
Explanation:
A Given, Young's modulus $(\mathrm{Y})=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$ And $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{1}{100}$ We know, Energy stored per unit volume, $=\frac{1}{2} \times(\mathrm{Y}) \times(\text { Strain })^{2}$ $=\frac{1}{2} \times(\mathrm{Y}) \times\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)^{2}$ $=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{1}{100}\right)^{2}$ $=1 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3}$
Kerala CEE - 2009
Mechanical Properties of Solids
141029
If the volume of a block of aluminium is decreased by $1 \%$, the pressure (stress) on its surface is increased by (Bulk modulus of $\mathrm{A} l=$ $7.5 \times 10^{10} \mathrm{Nm}^{-2}$ )
141030
A work of $2 \times 10^{-2} \mathrm{~J}$ is done on a wire of length $50 \mathrm{~cm}$ and area of cross-section $0.5 \mathrm{~mm}^{2}$. If the Young's modulus of the material of the wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$, then the wire must be :
1 elongated to $50.1414 \mathrm{~cm}$
2 contracted by $2.0 \mathrm{~mm}$
3 stretched by $0.707 \mathrm{~mm}$
4 of length changed to $49.293 \mathrm{~cm}$
5 of length changed to $50.2 \mathrm{~cm}$
Explanation:
C Given, $\mathrm{Y}=2 \times 10^{10} \mathrm{Nm}^{-2}$, Strain $=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}, \mathrm{U}=2 \times 10^{-2} \mathrm{~J}$ The work done by wire is stored as potential energy in the wire $\mathrm{U}=\frac{1}{2} \times \text { Young's modulus } \times(\text { strain })^{2}$ $\therefore \quad 2 \times 10^{-2}=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}\right)^{2}$ $\Delta \mathrm{L} \approx 0.707 \mathrm{~mm}$ (stretched)
141026
The average depth of Indian Ocean is about $3000 \mathrm{~m}$. The fractional compression, $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ of water at the bottom of the water $=2.2 \times 10^{9}$ $\mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ) is
141027
The compressibility of water is $6 \times 10^{-10} \mathrm{~N}^{-1} \mathrm{~m}^{2}$. If one litre is subjected to a pressure of $4 \times 10^{\dagger}$ $\mathrm{Nm}^{-2}$, the decrease in its volume is
141028
The Young's modulus of the material of a wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$. If the elongation strain is $1 \%$ then the energy stored in the wire per unit volume in $\mathbf{J m}^{-3}$ is
1 $10^{6}$
2 $10^{8}$
3 $2 \times 10^{6}$
4 $2 \times 10^{8}$
5 $0.5 \times 10^{6}$
Explanation:
A Given, Young's modulus $(\mathrm{Y})=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$ And $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{1}{100}$ We know, Energy stored per unit volume, $=\frac{1}{2} \times(\mathrm{Y}) \times(\text { Strain })^{2}$ $=\frac{1}{2} \times(\mathrm{Y}) \times\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)^{2}$ $=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{1}{100}\right)^{2}$ $=1 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3}$
Kerala CEE - 2009
Mechanical Properties of Solids
141029
If the volume of a block of aluminium is decreased by $1 \%$, the pressure (stress) on its surface is increased by (Bulk modulus of $\mathrm{A} l=$ $7.5 \times 10^{10} \mathrm{Nm}^{-2}$ )
141030
A work of $2 \times 10^{-2} \mathrm{~J}$ is done on a wire of length $50 \mathrm{~cm}$ and area of cross-section $0.5 \mathrm{~mm}^{2}$. If the Young's modulus of the material of the wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$, then the wire must be :
1 elongated to $50.1414 \mathrm{~cm}$
2 contracted by $2.0 \mathrm{~mm}$
3 stretched by $0.707 \mathrm{~mm}$
4 of length changed to $49.293 \mathrm{~cm}$
5 of length changed to $50.2 \mathrm{~cm}$
Explanation:
C Given, $\mathrm{Y}=2 \times 10^{10} \mathrm{Nm}^{-2}$, Strain $=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}, \mathrm{U}=2 \times 10^{-2} \mathrm{~J}$ The work done by wire is stored as potential energy in the wire $\mathrm{U}=\frac{1}{2} \times \text { Young's modulus } \times(\text { strain })^{2}$ $\therefore \quad 2 \times 10^{-2}=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}\right)^{2}$ $\Delta \mathrm{L} \approx 0.707 \mathrm{~mm}$ (stretched)
141026
The average depth of Indian Ocean is about $3000 \mathrm{~m}$. The fractional compression, $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ of water at the bottom of the water $=2.2 \times 10^{9}$ $\mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ) is
141027
The compressibility of water is $6 \times 10^{-10} \mathrm{~N}^{-1} \mathrm{~m}^{2}$. If one litre is subjected to a pressure of $4 \times 10^{\dagger}$ $\mathrm{Nm}^{-2}$, the decrease in its volume is
141028
The Young's modulus of the material of a wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$. If the elongation strain is $1 \%$ then the energy stored in the wire per unit volume in $\mathbf{J m}^{-3}$ is
1 $10^{6}$
2 $10^{8}$
3 $2 \times 10^{6}$
4 $2 \times 10^{8}$
5 $0.5 \times 10^{6}$
Explanation:
A Given, Young's modulus $(\mathrm{Y})=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$ And $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{1}{100}$ We know, Energy stored per unit volume, $=\frac{1}{2} \times(\mathrm{Y}) \times(\text { Strain })^{2}$ $=\frac{1}{2} \times(\mathrm{Y}) \times\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)^{2}$ $=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{1}{100}\right)^{2}$ $=1 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3}$
Kerala CEE - 2009
Mechanical Properties of Solids
141029
If the volume of a block of aluminium is decreased by $1 \%$, the pressure (stress) on its surface is increased by (Bulk modulus of $\mathrm{A} l=$ $7.5 \times 10^{10} \mathrm{Nm}^{-2}$ )
141030
A work of $2 \times 10^{-2} \mathrm{~J}$ is done on a wire of length $50 \mathrm{~cm}$ and area of cross-section $0.5 \mathrm{~mm}^{2}$. If the Young's modulus of the material of the wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$, then the wire must be :
1 elongated to $50.1414 \mathrm{~cm}$
2 contracted by $2.0 \mathrm{~mm}$
3 stretched by $0.707 \mathrm{~mm}$
4 of length changed to $49.293 \mathrm{~cm}$
5 of length changed to $50.2 \mathrm{~cm}$
Explanation:
C Given, $\mathrm{Y}=2 \times 10^{10} \mathrm{Nm}^{-2}$, Strain $=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}, \mathrm{U}=2 \times 10^{-2} \mathrm{~J}$ The work done by wire is stored as potential energy in the wire $\mathrm{U}=\frac{1}{2} \times \text { Young's modulus } \times(\text { strain })^{2}$ $\therefore \quad 2 \times 10^{-2}=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}\right)^{2}$ $\Delta \mathrm{L} \approx 0.707 \mathrm{~mm}$ (stretched)
141026
The average depth of Indian Ocean is about $3000 \mathrm{~m}$. The fractional compression, $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ of water at the bottom of the water $=2.2 \times 10^{9}$ $\mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ) is
141027
The compressibility of water is $6 \times 10^{-10} \mathrm{~N}^{-1} \mathrm{~m}^{2}$. If one litre is subjected to a pressure of $4 \times 10^{\dagger}$ $\mathrm{Nm}^{-2}$, the decrease in its volume is
141028
The Young's modulus of the material of a wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$. If the elongation strain is $1 \%$ then the energy stored in the wire per unit volume in $\mathbf{J m}^{-3}$ is
1 $10^{6}$
2 $10^{8}$
3 $2 \times 10^{6}$
4 $2 \times 10^{8}$
5 $0.5 \times 10^{6}$
Explanation:
A Given, Young's modulus $(\mathrm{Y})=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$ And $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{1}{100}$ We know, Energy stored per unit volume, $=\frac{1}{2} \times(\mathrm{Y}) \times(\text { Strain })^{2}$ $=\frac{1}{2} \times(\mathrm{Y}) \times\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)^{2}$ $=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{1}{100}\right)^{2}$ $=1 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3}$
Kerala CEE - 2009
Mechanical Properties of Solids
141029
If the volume of a block of aluminium is decreased by $1 \%$, the pressure (stress) on its surface is increased by (Bulk modulus of $\mathrm{A} l=$ $7.5 \times 10^{10} \mathrm{Nm}^{-2}$ )
141030
A work of $2 \times 10^{-2} \mathrm{~J}$ is done on a wire of length $50 \mathrm{~cm}$ and area of cross-section $0.5 \mathrm{~mm}^{2}$. If the Young's modulus of the material of the wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$, then the wire must be :
1 elongated to $50.1414 \mathrm{~cm}$
2 contracted by $2.0 \mathrm{~mm}$
3 stretched by $0.707 \mathrm{~mm}$
4 of length changed to $49.293 \mathrm{~cm}$
5 of length changed to $50.2 \mathrm{~cm}$
Explanation:
C Given, $\mathrm{Y}=2 \times 10^{10} \mathrm{Nm}^{-2}$, Strain $=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}, \mathrm{U}=2 \times 10^{-2} \mathrm{~J}$ The work done by wire is stored as potential energy in the wire $\mathrm{U}=\frac{1}{2} \times \text { Young's modulus } \times(\text { strain })^{2}$ $\therefore \quad 2 \times 10^{-2}=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}\right)^{2}$ $\Delta \mathrm{L} \approx 0.707 \mathrm{~mm}$ (stretched)
141026
The average depth of Indian Ocean is about $3000 \mathrm{~m}$. The fractional compression, $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ of water at the bottom of the water $=2.2 \times 10^{9}$ $\mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ) is
141027
The compressibility of water is $6 \times 10^{-10} \mathrm{~N}^{-1} \mathrm{~m}^{2}$. If one litre is subjected to a pressure of $4 \times 10^{\dagger}$ $\mathrm{Nm}^{-2}$, the decrease in its volume is
141028
The Young's modulus of the material of a wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$. If the elongation strain is $1 \%$ then the energy stored in the wire per unit volume in $\mathbf{J m}^{-3}$ is
1 $10^{6}$
2 $10^{8}$
3 $2 \times 10^{6}$
4 $2 \times 10^{8}$
5 $0.5 \times 10^{6}$
Explanation:
A Given, Young's modulus $(\mathrm{Y})=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$ And $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{1}{100}$ We know, Energy stored per unit volume, $=\frac{1}{2} \times(\mathrm{Y}) \times(\text { Strain })^{2}$ $=\frac{1}{2} \times(\mathrm{Y}) \times\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)^{2}$ $=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{1}{100}\right)^{2}$ $=1 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3}$
Kerala CEE - 2009
Mechanical Properties of Solids
141029
If the volume of a block of aluminium is decreased by $1 \%$, the pressure (stress) on its surface is increased by (Bulk modulus of $\mathrm{A} l=$ $7.5 \times 10^{10} \mathrm{Nm}^{-2}$ )
141030
A work of $2 \times 10^{-2} \mathrm{~J}$ is done on a wire of length $50 \mathrm{~cm}$ and area of cross-section $0.5 \mathrm{~mm}^{2}$. If the Young's modulus of the material of the wire is $2 \times 10^{10} \mathrm{Nm}^{-2}$, then the wire must be :
1 elongated to $50.1414 \mathrm{~cm}$
2 contracted by $2.0 \mathrm{~mm}$
3 stretched by $0.707 \mathrm{~mm}$
4 of length changed to $49.293 \mathrm{~cm}$
5 of length changed to $50.2 \mathrm{~cm}$
Explanation:
C Given, $\mathrm{Y}=2 \times 10^{10} \mathrm{Nm}^{-2}$, Strain $=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}, \mathrm{U}=2 \times 10^{-2} \mathrm{~J}$ The work done by wire is stored as potential energy in the wire $\mathrm{U}=\frac{1}{2} \times \text { Young's modulus } \times(\text { strain })^{2}$ $\therefore \quad 2 \times 10^{-2}=\frac{1}{2} \times 2 \times 10^{10} \times\left(\frac{\Delta \mathrm{L}}{50 \times 10^{-2}}\right)^{2}$ $\Delta \mathrm{L} \approx 0.707 \mathrm{~mm}$ (stretched)
