369865
Two wire of same material having radius in ratio \(2: 1\) and lengths in ratio \(1: 2\). If same force is applied on them, then ratio of their change in length will be
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 4\)
4 \(1: 8\)
Explanation:
Given, ratio of radius of two wires, \(\dfrac{r_{1}}{r_{2}}=\dfrac{2}{1}\) and ratio in their lengths, \(\dfrac{l_{1}}{l_{2}}=\dfrac{1}{2}\) When same force is applied on them, then ratio change in their lengths, \(\dfrac{\Delta l_{1}}{\Delta l_{2}}=?\) We know that, Young's modulus, \(Y = \frac{{Fl}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{Fl}}{{AY}}\) \(\Delta l \propto \frac{l}{A}\) \(\therefore \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{A_1}}}/\frac{{{l_2}}}{{{A_2}}} = \frac{{{l_1}{A_2}}}{{{l_2}{A_1}}}\) \( \Rightarrow \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{l_2}}} \cdot \frac{{r_2^2}}{{r_1^2}} = \frac{1}{2} \cdot {\left( {\frac{1}{2}} \right)^2} = \frac{1}{8}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369866
A thick rope of density \(\rho\) and length \(L\) is hung from a rigid support. The increase in length of the rope due to its own weight is (Y is the Young's modulus)
1 \(\dfrac{1}{2 Y} \rho L^{2} g\)
2 \(\dfrac{1}{4 Y} \rho L^{2} g\)
3 \(\dfrac{\rho L g}{Y}\)
4 \(\dfrac{\rho L^{2} g}{Y}\)
Explanation:
As the weight of wire acts at centre of gravity. \(\therefore \) Only half the length of wire gets extended. Now \(Y = \frac{F}{A}.\frac{{(L/2)}}{{\Delta l}} = \frac{{Mg(L/2)}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{MgL}}{{2AY}} \Rightarrow \Delta l = \frac{{AL\rho gL}}{{2AY}}\) \(\therefore \;\quad \Delta l = \frac{{\rho {L^2}g}}{{2Y}}\) So the correct choice is (1)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369867
In experiment to find young's modulus, if length of wire and radius both are doubled then the value of \(Y\) will become
1 4 times
2 2 times
3 Half
4 Remains same
Explanation:
Conceptual Question
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369868
Longitudinal stress of \(1\;kg/m{m^2}\) is applied on a wire. The percentage increase in length is \(\left( {y = {{10}^{11}}\;N/{m^2}} \right)\)
369869
Young's modules of material of a wire of length ' \(L\) ' and cross-sectional area \(A\) is \(Y\). If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be
1 \(2 Y\)
2 \(Y\)
3 \(4 Y\)
4 \(\dfrac{Y}{4}\)
Explanation:
Young's modulus is the property of the material and it does not depend upon the physical dimensions of the wire. Therefore, Young's modulus will be \(Y\). So, correct option is \((2)\).
369865
Two wire of same material having radius in ratio \(2: 1\) and lengths in ratio \(1: 2\). If same force is applied on them, then ratio of their change in length will be
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 4\)
4 \(1: 8\)
Explanation:
Given, ratio of radius of two wires, \(\dfrac{r_{1}}{r_{2}}=\dfrac{2}{1}\) and ratio in their lengths, \(\dfrac{l_{1}}{l_{2}}=\dfrac{1}{2}\) When same force is applied on them, then ratio change in their lengths, \(\dfrac{\Delta l_{1}}{\Delta l_{2}}=?\) We know that, Young's modulus, \(Y = \frac{{Fl}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{Fl}}{{AY}}\) \(\Delta l \propto \frac{l}{A}\) \(\therefore \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{A_1}}}/\frac{{{l_2}}}{{{A_2}}} = \frac{{{l_1}{A_2}}}{{{l_2}{A_1}}}\) \( \Rightarrow \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{l_2}}} \cdot \frac{{r_2^2}}{{r_1^2}} = \frac{1}{2} \cdot {\left( {\frac{1}{2}} \right)^2} = \frac{1}{8}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369866
A thick rope of density \(\rho\) and length \(L\) is hung from a rigid support. The increase in length of the rope due to its own weight is (Y is the Young's modulus)
1 \(\dfrac{1}{2 Y} \rho L^{2} g\)
2 \(\dfrac{1}{4 Y} \rho L^{2} g\)
3 \(\dfrac{\rho L g}{Y}\)
4 \(\dfrac{\rho L^{2} g}{Y}\)
Explanation:
As the weight of wire acts at centre of gravity. \(\therefore \) Only half the length of wire gets extended. Now \(Y = \frac{F}{A}.\frac{{(L/2)}}{{\Delta l}} = \frac{{Mg(L/2)}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{MgL}}{{2AY}} \Rightarrow \Delta l = \frac{{AL\rho gL}}{{2AY}}\) \(\therefore \;\quad \Delta l = \frac{{\rho {L^2}g}}{{2Y}}\) So the correct choice is (1)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369867
In experiment to find young's modulus, if length of wire and radius both are doubled then the value of \(Y\) will become
1 4 times
2 2 times
3 Half
4 Remains same
Explanation:
Conceptual Question
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369868
Longitudinal stress of \(1\;kg/m{m^2}\) is applied on a wire. The percentage increase in length is \(\left( {y = {{10}^{11}}\;N/{m^2}} \right)\)
369869
Young's modules of material of a wire of length ' \(L\) ' and cross-sectional area \(A\) is \(Y\). If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be
1 \(2 Y\)
2 \(Y\)
3 \(4 Y\)
4 \(\dfrac{Y}{4}\)
Explanation:
Young's modulus is the property of the material and it does not depend upon the physical dimensions of the wire. Therefore, Young's modulus will be \(Y\). So, correct option is \((2)\).
369865
Two wire of same material having radius in ratio \(2: 1\) and lengths in ratio \(1: 2\). If same force is applied on them, then ratio of their change in length will be
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 4\)
4 \(1: 8\)
Explanation:
Given, ratio of radius of two wires, \(\dfrac{r_{1}}{r_{2}}=\dfrac{2}{1}\) and ratio in their lengths, \(\dfrac{l_{1}}{l_{2}}=\dfrac{1}{2}\) When same force is applied on them, then ratio change in their lengths, \(\dfrac{\Delta l_{1}}{\Delta l_{2}}=?\) We know that, Young's modulus, \(Y = \frac{{Fl}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{Fl}}{{AY}}\) \(\Delta l \propto \frac{l}{A}\) \(\therefore \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{A_1}}}/\frac{{{l_2}}}{{{A_2}}} = \frac{{{l_1}{A_2}}}{{{l_2}{A_1}}}\) \( \Rightarrow \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{l_2}}} \cdot \frac{{r_2^2}}{{r_1^2}} = \frac{1}{2} \cdot {\left( {\frac{1}{2}} \right)^2} = \frac{1}{8}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369866
A thick rope of density \(\rho\) and length \(L\) is hung from a rigid support. The increase in length of the rope due to its own weight is (Y is the Young's modulus)
1 \(\dfrac{1}{2 Y} \rho L^{2} g\)
2 \(\dfrac{1}{4 Y} \rho L^{2} g\)
3 \(\dfrac{\rho L g}{Y}\)
4 \(\dfrac{\rho L^{2} g}{Y}\)
Explanation:
As the weight of wire acts at centre of gravity. \(\therefore \) Only half the length of wire gets extended. Now \(Y = \frac{F}{A}.\frac{{(L/2)}}{{\Delta l}} = \frac{{Mg(L/2)}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{MgL}}{{2AY}} \Rightarrow \Delta l = \frac{{AL\rho gL}}{{2AY}}\) \(\therefore \;\quad \Delta l = \frac{{\rho {L^2}g}}{{2Y}}\) So the correct choice is (1)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369867
In experiment to find young's modulus, if length of wire and radius both are doubled then the value of \(Y\) will become
1 4 times
2 2 times
3 Half
4 Remains same
Explanation:
Conceptual Question
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369868
Longitudinal stress of \(1\;kg/m{m^2}\) is applied on a wire. The percentage increase in length is \(\left( {y = {{10}^{11}}\;N/{m^2}} \right)\)
369869
Young's modules of material of a wire of length ' \(L\) ' and cross-sectional area \(A\) is \(Y\). If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be
1 \(2 Y\)
2 \(Y\)
3 \(4 Y\)
4 \(\dfrac{Y}{4}\)
Explanation:
Young's modulus is the property of the material and it does not depend upon the physical dimensions of the wire. Therefore, Young's modulus will be \(Y\). So, correct option is \((2)\).
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PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369865
Two wire of same material having radius in ratio \(2: 1\) and lengths in ratio \(1: 2\). If same force is applied on them, then ratio of their change in length will be
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 4\)
4 \(1: 8\)
Explanation:
Given, ratio of radius of two wires, \(\dfrac{r_{1}}{r_{2}}=\dfrac{2}{1}\) and ratio in their lengths, \(\dfrac{l_{1}}{l_{2}}=\dfrac{1}{2}\) When same force is applied on them, then ratio change in their lengths, \(\dfrac{\Delta l_{1}}{\Delta l_{2}}=?\) We know that, Young's modulus, \(Y = \frac{{Fl}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{Fl}}{{AY}}\) \(\Delta l \propto \frac{l}{A}\) \(\therefore \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{A_1}}}/\frac{{{l_2}}}{{{A_2}}} = \frac{{{l_1}{A_2}}}{{{l_2}{A_1}}}\) \( \Rightarrow \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{l_2}}} \cdot \frac{{r_2^2}}{{r_1^2}} = \frac{1}{2} \cdot {\left( {\frac{1}{2}} \right)^2} = \frac{1}{8}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369866
A thick rope of density \(\rho\) and length \(L\) is hung from a rigid support. The increase in length of the rope due to its own weight is (Y is the Young's modulus)
1 \(\dfrac{1}{2 Y} \rho L^{2} g\)
2 \(\dfrac{1}{4 Y} \rho L^{2} g\)
3 \(\dfrac{\rho L g}{Y}\)
4 \(\dfrac{\rho L^{2} g}{Y}\)
Explanation:
As the weight of wire acts at centre of gravity. \(\therefore \) Only half the length of wire gets extended. Now \(Y = \frac{F}{A}.\frac{{(L/2)}}{{\Delta l}} = \frac{{Mg(L/2)}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{MgL}}{{2AY}} \Rightarrow \Delta l = \frac{{AL\rho gL}}{{2AY}}\) \(\therefore \;\quad \Delta l = \frac{{\rho {L^2}g}}{{2Y}}\) So the correct choice is (1)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369867
In experiment to find young's modulus, if length of wire and radius both are doubled then the value of \(Y\) will become
1 4 times
2 2 times
3 Half
4 Remains same
Explanation:
Conceptual Question
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369868
Longitudinal stress of \(1\;kg/m{m^2}\) is applied on a wire. The percentage increase in length is \(\left( {y = {{10}^{11}}\;N/{m^2}} \right)\)
369869
Young's modules of material of a wire of length ' \(L\) ' and cross-sectional area \(A\) is \(Y\). If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be
1 \(2 Y\)
2 \(Y\)
3 \(4 Y\)
4 \(\dfrac{Y}{4}\)
Explanation:
Young's modulus is the property of the material and it does not depend upon the physical dimensions of the wire. Therefore, Young's modulus will be \(Y\). So, correct option is \((2)\).
369865
Two wire of same material having radius in ratio \(2: 1\) and lengths in ratio \(1: 2\). If same force is applied on them, then ratio of their change in length will be
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 4\)
4 \(1: 8\)
Explanation:
Given, ratio of radius of two wires, \(\dfrac{r_{1}}{r_{2}}=\dfrac{2}{1}\) and ratio in their lengths, \(\dfrac{l_{1}}{l_{2}}=\dfrac{1}{2}\) When same force is applied on them, then ratio change in their lengths, \(\dfrac{\Delta l_{1}}{\Delta l_{2}}=?\) We know that, Young's modulus, \(Y = \frac{{Fl}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{Fl}}{{AY}}\) \(\Delta l \propto \frac{l}{A}\) \(\therefore \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{A_1}}}/\frac{{{l_2}}}{{{A_2}}} = \frac{{{l_1}{A_2}}}{{{l_2}{A_1}}}\) \( \Rightarrow \quad \frac{{\Delta {l_1}}}{{\Delta {l_2}}} = \frac{{{l_1}}}{{{l_2}}} \cdot \frac{{r_2^2}}{{r_1^2}} = \frac{1}{2} \cdot {\left( {\frac{1}{2}} \right)^2} = \frac{1}{8}\)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369866
A thick rope of density \(\rho\) and length \(L\) is hung from a rigid support. The increase in length of the rope due to its own weight is (Y is the Young's modulus)
1 \(\dfrac{1}{2 Y} \rho L^{2} g\)
2 \(\dfrac{1}{4 Y} \rho L^{2} g\)
3 \(\dfrac{\rho L g}{Y}\)
4 \(\dfrac{\rho L^{2} g}{Y}\)
Explanation:
As the weight of wire acts at centre of gravity. \(\therefore \) Only half the length of wire gets extended. Now \(Y = \frac{F}{A}.\frac{{(L/2)}}{{\Delta l}} = \frac{{Mg(L/2)}}{{A\Delta l}}\) \( \Rightarrow \quad \Delta l = \frac{{MgL}}{{2AY}} \Rightarrow \Delta l = \frac{{AL\rho gL}}{{2AY}}\) \(\therefore \;\quad \Delta l = \frac{{\rho {L^2}g}}{{2Y}}\) So the correct choice is (1)
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369867
In experiment to find young's modulus, if length of wire and radius both are doubled then the value of \(Y\) will become
1 4 times
2 2 times
3 Half
4 Remains same
Explanation:
Conceptual Question
PHXI09:MECHANICAL PROPERTIES OF SOLIDS
369868
Longitudinal stress of \(1\;kg/m{m^2}\) is applied on a wire. The percentage increase in length is \(\left( {y = {{10}^{11}}\;N/{m^2}} \right)\)
369869
Young's modules of material of a wire of length ' \(L\) ' and cross-sectional area \(A\) is \(Y\). If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be
1 \(2 Y\)
2 \(Y\)
3 \(4 Y\)
4 \(\dfrac{Y}{4}\)
Explanation:
Young's modulus is the property of the material and it does not depend upon the physical dimensions of the wire. Therefore, Young's modulus will be \(Y\). So, correct option is \((2)\).