369848
If the ratio of radii of two wires of same material is \(2: 1\) and ratio of their lengths is \(4: 1\), then the ratio of the normal forces that will produce the same extension in the length of two wires is
369849
The Young's modulus of a rubber string \(8\;cm\) long and density \(1.5\;kg{\rm{/}}{m^3}\) is \(5 \times {10^8}\;N{\rm{/}}{m^2}\), is suspended on the ceiling in a room. The increase in length due to its own weight will be :
369850
For steel \(Y = 2 \times {10^{11}}\;N/{m^2}\). The force required to double the length of a steel wire of area \(1\;c{m^2}\) is
1 \(2 \times {10^7}\;N\)
2 \(2 \times {10^6}\;N\)
3 \(2 \times {10^8}\;N\)
4 \(2 \times {10^5}\;N\)
Explanation:
Stress required to double the length is called Young's modulus \(\begin{aligned}& \therefore \mathrm{Y}=\dfrac{\mathrm{F}}{\mathrm{A}} \text { or } \mathrm{F}=\mathrm{Y} . \mathrm{A} \\& \mathrm{A}=2 \times 10^{11} \times 10^{-4}=2 \times 10^{7} \mathrm{~N}\end{aligned}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369851
The dimensions of two wires \(A\) and \(B\) are the same. But their materials are different. Their load-extension graphs are shown. If \(Y_{A}\) and \(Y_{B}\) are the values of Young's modulus of elasticity of \(A\) and \(B\) respectively the
1 \(Y_{A}>Y_{B}\)
2 \(Y_{A} < Y_{B}\)
3 \(Y_{A}=Y_{B}\)
4 \(Y_{B}=2 Y_{A}\)
Explanation:
\(Y=\dfrac{F \cdot L}{A . \Delta \ell}\) for same \(\Delta l\). wire A required more load. \(\text { so } \quad Y_{A}>Y_{B}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369852
A constant force \(F_{0}\) is applied on a uniform elastic rod placed over a smooth horizontal surface as shown in figure. Young's modulus of rod is \(Y\) and area of cross section is \(S\). The strain produced in the rod in the direction of force is
1 \(\dfrac{F_{0}}{S Y}\)
2 \(\dfrac{F_{0} Y}{S}\)
3 \(\dfrac{F_{0} Y}{2 S}\)
4 \(\dfrac{F_{0}}{2 S Y}\)
Explanation:
Acceleartion \(=\dfrac{F_{0}}{\lambda L}\), where \(\lambda=\) mass per unit length, tension at a distance \(x\) from front end \(T=\lambda(L-x) a=\dfrac{\lambda(L-x) F_{0}}{\lambda L}=F_{0}\left(1-\dfrac{x}{L}\right)\) Consider a small element of length \(\mathrm{d} x\) at a distance \(x\) from the front end.
The change in the length of the element is \(d l\). \(d l=\dfrac{T d x}{S Y}\) \(\therefore\) Total change in length \(=\int_{0}^{L} \dfrac{F_{0}}{S Y}\left(1-\dfrac{x}{L}\right) \cdot d x\) \(\begin{aligned}& \Delta l=\dfrac{F_{0}}{S Y}\left|x-\dfrac{x^{2}}{2 L}\right|_{0}^{L}=\dfrac{F_{0}}{S Y}\left(L-\dfrac{L}{2}\right)=\dfrac{F_{0} L}{2 S Y} \\& \therefore \text { Total strain }=\dfrac{F_{0}}{2 S Y}\end{aligned}\)
369848
If the ratio of radii of two wires of same material is \(2: 1\) and ratio of their lengths is \(4: 1\), then the ratio of the normal forces that will produce the same extension in the length of two wires is
369849
The Young's modulus of a rubber string \(8\;cm\) long and density \(1.5\;kg{\rm{/}}{m^3}\) is \(5 \times {10^8}\;N{\rm{/}}{m^2}\), is suspended on the ceiling in a room. The increase in length due to its own weight will be :
369850
For steel \(Y = 2 \times {10^{11}}\;N/{m^2}\). The force required to double the length of a steel wire of area \(1\;c{m^2}\) is
1 \(2 \times {10^7}\;N\)
2 \(2 \times {10^6}\;N\)
3 \(2 \times {10^8}\;N\)
4 \(2 \times {10^5}\;N\)
Explanation:
Stress required to double the length is called Young's modulus \(\begin{aligned}& \therefore \mathrm{Y}=\dfrac{\mathrm{F}}{\mathrm{A}} \text { or } \mathrm{F}=\mathrm{Y} . \mathrm{A} \\& \mathrm{A}=2 \times 10^{11} \times 10^{-4}=2 \times 10^{7} \mathrm{~N}\end{aligned}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369851
The dimensions of two wires \(A\) and \(B\) are the same. But their materials are different. Their load-extension graphs are shown. If \(Y_{A}\) and \(Y_{B}\) are the values of Young's modulus of elasticity of \(A\) and \(B\) respectively the
1 \(Y_{A}>Y_{B}\)
2 \(Y_{A} < Y_{B}\)
3 \(Y_{A}=Y_{B}\)
4 \(Y_{B}=2 Y_{A}\)
Explanation:
\(Y=\dfrac{F \cdot L}{A . \Delta \ell}\) for same \(\Delta l\). wire A required more load. \(\text { so } \quad Y_{A}>Y_{B}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369852
A constant force \(F_{0}\) is applied on a uniform elastic rod placed over a smooth horizontal surface as shown in figure. Young's modulus of rod is \(Y\) and area of cross section is \(S\). The strain produced in the rod in the direction of force is
1 \(\dfrac{F_{0}}{S Y}\)
2 \(\dfrac{F_{0} Y}{S}\)
3 \(\dfrac{F_{0} Y}{2 S}\)
4 \(\dfrac{F_{0}}{2 S Y}\)
Explanation:
Acceleartion \(=\dfrac{F_{0}}{\lambda L}\), where \(\lambda=\) mass per unit length, tension at a distance \(x\) from front end \(T=\lambda(L-x) a=\dfrac{\lambda(L-x) F_{0}}{\lambda L}=F_{0}\left(1-\dfrac{x}{L}\right)\) Consider a small element of length \(\mathrm{d} x\) at a distance \(x\) from the front end.
The change in the length of the element is \(d l\). \(d l=\dfrac{T d x}{S Y}\) \(\therefore\) Total change in length \(=\int_{0}^{L} \dfrac{F_{0}}{S Y}\left(1-\dfrac{x}{L}\right) \cdot d x\) \(\begin{aligned}& \Delta l=\dfrac{F_{0}}{S Y}\left|x-\dfrac{x^{2}}{2 L}\right|_{0}^{L}=\dfrac{F_{0}}{S Y}\left(L-\dfrac{L}{2}\right)=\dfrac{F_{0} L}{2 S Y} \\& \therefore \text { Total strain }=\dfrac{F_{0}}{2 S Y}\end{aligned}\)
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PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369848
If the ratio of radii of two wires of same material is \(2: 1\) and ratio of their lengths is \(4: 1\), then the ratio of the normal forces that will produce the same extension in the length of two wires is
369849
The Young's modulus of a rubber string \(8\;cm\) long and density \(1.5\;kg{\rm{/}}{m^3}\) is \(5 \times {10^8}\;N{\rm{/}}{m^2}\), is suspended on the ceiling in a room. The increase in length due to its own weight will be :
369850
For steel \(Y = 2 \times {10^{11}}\;N/{m^2}\). The force required to double the length of a steel wire of area \(1\;c{m^2}\) is
1 \(2 \times {10^7}\;N\)
2 \(2 \times {10^6}\;N\)
3 \(2 \times {10^8}\;N\)
4 \(2 \times {10^5}\;N\)
Explanation:
Stress required to double the length is called Young's modulus \(\begin{aligned}& \therefore \mathrm{Y}=\dfrac{\mathrm{F}}{\mathrm{A}} \text { or } \mathrm{F}=\mathrm{Y} . \mathrm{A} \\& \mathrm{A}=2 \times 10^{11} \times 10^{-4}=2 \times 10^{7} \mathrm{~N}\end{aligned}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369851
The dimensions of two wires \(A\) and \(B\) are the same. But their materials are different. Their load-extension graphs are shown. If \(Y_{A}\) and \(Y_{B}\) are the values of Young's modulus of elasticity of \(A\) and \(B\) respectively the
1 \(Y_{A}>Y_{B}\)
2 \(Y_{A} < Y_{B}\)
3 \(Y_{A}=Y_{B}\)
4 \(Y_{B}=2 Y_{A}\)
Explanation:
\(Y=\dfrac{F \cdot L}{A . \Delta \ell}\) for same \(\Delta l\). wire A required more load. \(\text { so } \quad Y_{A}>Y_{B}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369852
A constant force \(F_{0}\) is applied on a uniform elastic rod placed over a smooth horizontal surface as shown in figure. Young's modulus of rod is \(Y\) and area of cross section is \(S\). The strain produced in the rod in the direction of force is
1 \(\dfrac{F_{0}}{S Y}\)
2 \(\dfrac{F_{0} Y}{S}\)
3 \(\dfrac{F_{0} Y}{2 S}\)
4 \(\dfrac{F_{0}}{2 S Y}\)
Explanation:
Acceleartion \(=\dfrac{F_{0}}{\lambda L}\), where \(\lambda=\) mass per unit length, tension at a distance \(x\) from front end \(T=\lambda(L-x) a=\dfrac{\lambda(L-x) F_{0}}{\lambda L}=F_{0}\left(1-\dfrac{x}{L}\right)\) Consider a small element of length \(\mathrm{d} x\) at a distance \(x\) from the front end.
The change in the length of the element is \(d l\). \(d l=\dfrac{T d x}{S Y}\) \(\therefore\) Total change in length \(=\int_{0}^{L} \dfrac{F_{0}}{S Y}\left(1-\dfrac{x}{L}\right) \cdot d x\) \(\begin{aligned}& \Delta l=\dfrac{F_{0}}{S Y}\left|x-\dfrac{x^{2}}{2 L}\right|_{0}^{L}=\dfrac{F_{0}}{S Y}\left(L-\dfrac{L}{2}\right)=\dfrac{F_{0} L}{2 S Y} \\& \therefore \text { Total strain }=\dfrac{F_{0}}{2 S Y}\end{aligned}\)
369848
If the ratio of radii of two wires of same material is \(2: 1\) and ratio of their lengths is \(4: 1\), then the ratio of the normal forces that will produce the same extension in the length of two wires is
369849
The Young's modulus of a rubber string \(8\;cm\) long and density \(1.5\;kg{\rm{/}}{m^3}\) is \(5 \times {10^8}\;N{\rm{/}}{m^2}\), is suspended on the ceiling in a room. The increase in length due to its own weight will be :
369850
For steel \(Y = 2 \times {10^{11}}\;N/{m^2}\). The force required to double the length of a steel wire of area \(1\;c{m^2}\) is
1 \(2 \times {10^7}\;N\)
2 \(2 \times {10^6}\;N\)
3 \(2 \times {10^8}\;N\)
4 \(2 \times {10^5}\;N\)
Explanation:
Stress required to double the length is called Young's modulus \(\begin{aligned}& \therefore \mathrm{Y}=\dfrac{\mathrm{F}}{\mathrm{A}} \text { or } \mathrm{F}=\mathrm{Y} . \mathrm{A} \\& \mathrm{A}=2 \times 10^{11} \times 10^{-4}=2 \times 10^{7} \mathrm{~N}\end{aligned}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369851
The dimensions of two wires \(A\) and \(B\) are the same. But their materials are different. Their load-extension graphs are shown. If \(Y_{A}\) and \(Y_{B}\) are the values of Young's modulus of elasticity of \(A\) and \(B\) respectively the
1 \(Y_{A}>Y_{B}\)
2 \(Y_{A} < Y_{B}\)
3 \(Y_{A}=Y_{B}\)
4 \(Y_{B}=2 Y_{A}\)
Explanation:
\(Y=\dfrac{F \cdot L}{A . \Delta \ell}\) for same \(\Delta l\). wire A required more load. \(\text { so } \quad Y_{A}>Y_{B}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369852
A constant force \(F_{0}\) is applied on a uniform elastic rod placed over a smooth horizontal surface as shown in figure. Young's modulus of rod is \(Y\) and area of cross section is \(S\). The strain produced in the rod in the direction of force is
1 \(\dfrac{F_{0}}{S Y}\)
2 \(\dfrac{F_{0} Y}{S}\)
3 \(\dfrac{F_{0} Y}{2 S}\)
4 \(\dfrac{F_{0}}{2 S Y}\)
Explanation:
Acceleartion \(=\dfrac{F_{0}}{\lambda L}\), where \(\lambda=\) mass per unit length, tension at a distance \(x\) from front end \(T=\lambda(L-x) a=\dfrac{\lambda(L-x) F_{0}}{\lambda L}=F_{0}\left(1-\dfrac{x}{L}\right)\) Consider a small element of length \(\mathrm{d} x\) at a distance \(x\) from the front end.
The change in the length of the element is \(d l\). \(d l=\dfrac{T d x}{S Y}\) \(\therefore\) Total change in length \(=\int_{0}^{L} \dfrac{F_{0}}{S Y}\left(1-\dfrac{x}{L}\right) \cdot d x\) \(\begin{aligned}& \Delta l=\dfrac{F_{0}}{S Y}\left|x-\dfrac{x^{2}}{2 L}\right|_{0}^{L}=\dfrac{F_{0}}{S Y}\left(L-\dfrac{L}{2}\right)=\dfrac{F_{0} L}{2 S Y} \\& \therefore \text { Total strain }=\dfrac{F_{0}}{2 S Y}\end{aligned}\)
369848
If the ratio of radii of two wires of same material is \(2: 1\) and ratio of their lengths is \(4: 1\), then the ratio of the normal forces that will produce the same extension in the length of two wires is
369849
The Young's modulus of a rubber string \(8\;cm\) long and density \(1.5\;kg{\rm{/}}{m^3}\) is \(5 \times {10^8}\;N{\rm{/}}{m^2}\), is suspended on the ceiling in a room. The increase in length due to its own weight will be :
369850
For steel \(Y = 2 \times {10^{11}}\;N/{m^2}\). The force required to double the length of a steel wire of area \(1\;c{m^2}\) is
1 \(2 \times {10^7}\;N\)
2 \(2 \times {10^6}\;N\)
3 \(2 \times {10^8}\;N\)
4 \(2 \times {10^5}\;N\)
Explanation:
Stress required to double the length is called Young's modulus \(\begin{aligned}& \therefore \mathrm{Y}=\dfrac{\mathrm{F}}{\mathrm{A}} \text { or } \mathrm{F}=\mathrm{Y} . \mathrm{A} \\& \mathrm{A}=2 \times 10^{11} \times 10^{-4}=2 \times 10^{7} \mathrm{~N}\end{aligned}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369851
The dimensions of two wires \(A\) and \(B\) are the same. But their materials are different. Their load-extension graphs are shown. If \(Y_{A}\) and \(Y_{B}\) are the values of Young's modulus of elasticity of \(A\) and \(B\) respectively the
1 \(Y_{A}>Y_{B}\)
2 \(Y_{A} < Y_{B}\)
3 \(Y_{A}=Y_{B}\)
4 \(Y_{B}=2 Y_{A}\)
Explanation:
\(Y=\dfrac{F \cdot L}{A . \Delta \ell}\) for same \(\Delta l\). wire A required more load. \(\text { so } \quad Y_{A}>Y_{B}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369852
A constant force \(F_{0}\) is applied on a uniform elastic rod placed over a smooth horizontal surface as shown in figure. Young's modulus of rod is \(Y\) and area of cross section is \(S\). The strain produced in the rod in the direction of force is
1 \(\dfrac{F_{0}}{S Y}\)
2 \(\dfrac{F_{0} Y}{S}\)
3 \(\dfrac{F_{0} Y}{2 S}\)
4 \(\dfrac{F_{0}}{2 S Y}\)
Explanation:
Acceleartion \(=\dfrac{F_{0}}{\lambda L}\), where \(\lambda=\) mass per unit length, tension at a distance \(x\) from front end \(T=\lambda(L-x) a=\dfrac{\lambda(L-x) F_{0}}{\lambda L}=F_{0}\left(1-\dfrac{x}{L}\right)\) Consider a small element of length \(\mathrm{d} x\) at a distance \(x\) from the front end.
The change in the length of the element is \(d l\). \(d l=\dfrac{T d x}{S Y}\) \(\therefore\) Total change in length \(=\int_{0}^{L} \dfrac{F_{0}}{S Y}\left(1-\dfrac{x}{L}\right) \cdot d x\) \(\begin{aligned}& \Delta l=\dfrac{F_{0}}{S Y}\left|x-\dfrac{x^{2}}{2 L}\right|_{0}^{L}=\dfrac{F_{0}}{S Y}\left(L-\dfrac{L}{2}\right)=\dfrac{F_{0} L}{2 S Y} \\& \therefore \text { Total strain }=\dfrac{F_{0}}{2 S Y}\end{aligned}\)
 Diagram - 1.png)
