369919
A steel wire of length \(5\;m\) is pulled to have an extension of \(1\;mm\). Its \(Y\) is \(1.9 \times {10^4}\;N{\rm{/}}{m^2}\). The energy per unit volume stored in it is
369920
Young's modulus of the material of a wire is \(Y\). On pulling the wire by a force \(F\), the increase in its length is \(x\). The potential energy of the stretched wire is
1 \(\dfrac{1}{2} Y x\)
2 \(\dfrac{1}{2} F x\)
3 \(\dfrac{1}{2} F x^{2}\)
4 None of these
Explanation:
When a wire is stretched through a length, then work has to be done, this work is stored in the wire in as elastic potential energy. \(U=\dfrac{1}{2} K x^{2}=\dfrac{1}{2}(K x) x=\dfrac{1}{2} F x\) where \(K\) is the effective spring constant of the rod.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369921
When a metal wire elongates by hanging a load \(M g\) on it, the gravitational potential energy of mass \(M\) decreases by \(M g l\). This energy appears
1 as elastic potential energy completely
2 as thermal energy completely
3 half as elastic potential energy and half as thermal energy
4 as kinetic energy of the load completely
Explanation:
When a metal wire elongates by hanging a load \(M g\) on it. Decrease in potential energy of the load \(=M g l\), where \(l=\) elongation in metal wire. Elastic potential energy stored in stretched \(\text { wire }=\dfrac{1}{2} \times M g l\) Difference of \(M g l\) and \(\dfrac{1}{2} M g l\) appears as thermal energy in the stretched wire. Energy appearing as thermal energy \( = Mgl - \frac{1}{2}Mgl\) \( = \frac{1}{2}Mgl\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369922
If in a wire of cross-sectional area \(A\) and length \(l\) and Young's modulus \(Y\), longitudinal strain \(x\) is produced then the potential energy stored in it will be
1 \(0.5 Y^{2} x l\)
2 \(Y{x^2}Al/2\)
3 \(Y x^{2} A\)
4 \(2Y{x^2}Al\)
Explanation:
Elastic potential energy per unit volume is \(u=\dfrac{1}{2} \times\) stress \(\times\) strain \(Y = \frac{{{\text{ stress }}}}{{{\text{ strain }}}} \Rightarrow {\text{ stress }} = Y \times {\text{ strain}}\) Given, strain \(=x\) Therefore, \(u=\dfrac{1}{2} Y x^{2} \Rightarrow u=0.5 Y x^{2}\) \(U=u V=u A l=\dfrac{Y x^{2} A l}{2}\)
369919
A steel wire of length \(5\;m\) is pulled to have an extension of \(1\;mm\). Its \(Y\) is \(1.9 \times {10^4}\;N{\rm{/}}{m^2}\). The energy per unit volume stored in it is
369920
Young's modulus of the material of a wire is \(Y\). On pulling the wire by a force \(F\), the increase in its length is \(x\). The potential energy of the stretched wire is
1 \(\dfrac{1}{2} Y x\)
2 \(\dfrac{1}{2} F x\)
3 \(\dfrac{1}{2} F x^{2}\)
4 None of these
Explanation:
When a wire is stretched through a length, then work has to be done, this work is stored in the wire in as elastic potential energy. \(U=\dfrac{1}{2} K x^{2}=\dfrac{1}{2}(K x) x=\dfrac{1}{2} F x\) where \(K\) is the effective spring constant of the rod.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369921
When a metal wire elongates by hanging a load \(M g\) on it, the gravitational potential energy of mass \(M\) decreases by \(M g l\). This energy appears
1 as elastic potential energy completely
2 as thermal energy completely
3 half as elastic potential energy and half as thermal energy
4 as kinetic energy of the load completely
Explanation:
When a metal wire elongates by hanging a load \(M g\) on it. Decrease in potential energy of the load \(=M g l\), where \(l=\) elongation in metal wire. Elastic potential energy stored in stretched \(\text { wire }=\dfrac{1}{2} \times M g l\) Difference of \(M g l\) and \(\dfrac{1}{2} M g l\) appears as thermal energy in the stretched wire. Energy appearing as thermal energy \( = Mgl - \frac{1}{2}Mgl\) \( = \frac{1}{2}Mgl\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369922
If in a wire of cross-sectional area \(A\) and length \(l\) and Young's modulus \(Y\), longitudinal strain \(x\) is produced then the potential energy stored in it will be
1 \(0.5 Y^{2} x l\)
2 \(Y{x^2}Al/2\)
3 \(Y x^{2} A\)
4 \(2Y{x^2}Al\)
Explanation:
Elastic potential energy per unit volume is \(u=\dfrac{1}{2} \times\) stress \(\times\) strain \(Y = \frac{{{\text{ stress }}}}{{{\text{ strain }}}} \Rightarrow {\text{ stress }} = Y \times {\text{ strain}}\) Given, strain \(=x\) Therefore, \(u=\dfrac{1}{2} Y x^{2} \Rightarrow u=0.5 Y x^{2}\) \(U=u V=u A l=\dfrac{Y x^{2} A l}{2}\)
369919
A steel wire of length \(5\;m\) is pulled to have an extension of \(1\;mm\). Its \(Y\) is \(1.9 \times {10^4}\;N{\rm{/}}{m^2}\). The energy per unit volume stored in it is
369920
Young's modulus of the material of a wire is \(Y\). On pulling the wire by a force \(F\), the increase in its length is \(x\). The potential energy of the stretched wire is
1 \(\dfrac{1}{2} Y x\)
2 \(\dfrac{1}{2} F x\)
3 \(\dfrac{1}{2} F x^{2}\)
4 None of these
Explanation:
When a wire is stretched through a length, then work has to be done, this work is stored in the wire in as elastic potential energy. \(U=\dfrac{1}{2} K x^{2}=\dfrac{1}{2}(K x) x=\dfrac{1}{2} F x\) where \(K\) is the effective spring constant of the rod.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369921
When a metal wire elongates by hanging a load \(M g\) on it, the gravitational potential energy of mass \(M\) decreases by \(M g l\). This energy appears
1 as elastic potential energy completely
2 as thermal energy completely
3 half as elastic potential energy and half as thermal energy
4 as kinetic energy of the load completely
Explanation:
When a metal wire elongates by hanging a load \(M g\) on it. Decrease in potential energy of the load \(=M g l\), where \(l=\) elongation in metal wire. Elastic potential energy stored in stretched \(\text { wire }=\dfrac{1}{2} \times M g l\) Difference of \(M g l\) and \(\dfrac{1}{2} M g l\) appears as thermal energy in the stretched wire. Energy appearing as thermal energy \( = Mgl - \frac{1}{2}Mgl\) \( = \frac{1}{2}Mgl\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369922
If in a wire of cross-sectional area \(A\) and length \(l\) and Young's modulus \(Y\), longitudinal strain \(x\) is produced then the potential energy stored in it will be
1 \(0.5 Y^{2} x l\)
2 \(Y{x^2}Al/2\)
3 \(Y x^{2} A\)
4 \(2Y{x^2}Al\)
Explanation:
Elastic potential energy per unit volume is \(u=\dfrac{1}{2} \times\) stress \(\times\) strain \(Y = \frac{{{\text{ stress }}}}{{{\text{ strain }}}} \Rightarrow {\text{ stress }} = Y \times {\text{ strain}}\) Given, strain \(=x\) Therefore, \(u=\dfrac{1}{2} Y x^{2} \Rightarrow u=0.5 Y x^{2}\) \(U=u V=u A l=\dfrac{Y x^{2} A l}{2}\)
369919
A steel wire of length \(5\;m\) is pulled to have an extension of \(1\;mm\). Its \(Y\) is \(1.9 \times {10^4}\;N{\rm{/}}{m^2}\). The energy per unit volume stored in it is
369920
Young's modulus of the material of a wire is \(Y\). On pulling the wire by a force \(F\), the increase in its length is \(x\). The potential energy of the stretched wire is
1 \(\dfrac{1}{2} Y x\)
2 \(\dfrac{1}{2} F x\)
3 \(\dfrac{1}{2} F x^{2}\)
4 None of these
Explanation:
When a wire is stretched through a length, then work has to be done, this work is stored in the wire in as elastic potential energy. \(U=\dfrac{1}{2} K x^{2}=\dfrac{1}{2}(K x) x=\dfrac{1}{2} F x\) where \(K\) is the effective spring constant of the rod.
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369921
When a metal wire elongates by hanging a load \(M g\) on it, the gravitational potential energy of mass \(M\) decreases by \(M g l\). This energy appears
1 as elastic potential energy completely
2 as thermal energy completely
3 half as elastic potential energy and half as thermal energy
4 as kinetic energy of the load completely
Explanation:
When a metal wire elongates by hanging a load \(M g\) on it. Decrease in potential energy of the load \(=M g l\), where \(l=\) elongation in metal wire. Elastic potential energy stored in stretched \(\text { wire }=\dfrac{1}{2} \times M g l\) Difference of \(M g l\) and \(\dfrac{1}{2} M g l\) appears as thermal energy in the stretched wire. Energy appearing as thermal energy \( = Mgl - \frac{1}{2}Mgl\) \( = \frac{1}{2}Mgl\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369922
If in a wire of cross-sectional area \(A\) and length \(l\) and Young's modulus \(Y\), longitudinal strain \(x\) is produced then the potential energy stored in it will be
1 \(0.5 Y^{2} x l\)
2 \(Y{x^2}Al/2\)
3 \(Y x^{2} A\)
4 \(2Y{x^2}Al\)
Explanation:
Elastic potential energy per unit volume is \(u=\dfrac{1}{2} \times\) stress \(\times\) strain \(Y = \frac{{{\text{ stress }}}}{{{\text{ strain }}}} \Rightarrow {\text{ stress }} = Y \times {\text{ strain}}\) Given, strain \(=x\) Therefore, \(u=\dfrac{1}{2} Y x^{2} \Rightarrow u=0.5 Y x^{2}\) \(U=u V=u A l=\dfrac{Y x^{2} A l}{2}\)
