140917
Two wires $A$ and $B$ are made of the same material. The wire $A$ has a length $l$ and radius $r$ while wire $B$ has a length $2 l$ and radius $2 \mathrm{r}$ are stretched by the same force. The ratio of elongation in A to the elongation in $B$
1 $2: 1$
2 $3: 1$
3 $27: 1$
4 $1: 4$
Explanation:
A Given, Wire A has length $=l_{\mathrm{A}}$ radius $=\mathrm{r}_{\mathrm{A}}$ wire $\mathrm{B}$ has length $=2 l_{\mathrm{B}}$ radius $=2 \mathrm{r}_{\mathrm{B}}$ Since $\mathrm{F}$ and $\mathrm{Y}$ are remain the same i.e, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\Rightarrow \frac{\Delta l_{\mathrm{A}}}{\Delta l_{B}}=\left(\frac{l_{\mathrm{A}}}{l_{B}}\right) \times\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}\right)^{2}$ $=\left(\frac{1}{2 l}\right) \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $=\frac{2}{1}$
J and K CET- 1999
Mechanical Properties of Solids
140918
The area under the stress-strain lines gives
1 work
2 power
3 Young's modulus
4 energy density
Explanation:
D We know, $\frac{\text { Energy }}{\text { Volume }}=\frac{1}{2} \times$ Stress $\times$ Strain Area under stress-strain line is equal to area of triangle. $\therefore \quad$ Area $=\frac{1}{2} \times$ Stress $\times$ Strain $\therefore \quad$ Area $=\frac{\text { Energy }}{\text { Volume }}$ $\therefore \quad=$ Energy density
J and K CET- 1998
Mechanical Properties of Solids
140919
An elastic modulus is the constant of proportionality in a relation of the form
1 stress $=$ constant $\times$ strain
2 strain $=$ constant $\times$ stress
3 stress $\times$ strain $=$ constant
4 the form of the relation depends on whether the constant is the bulk, shear, or Young's modulus
Explanation:
A Elastic modulus $=\frac{\text { stress }}{\text { strain }}$ Young's modulus (also referred to as the elastic modulus or tensile modulus), is a measure of mechanical properties of linear elastic solids like rod, wire and such other things. Young's modulus values of different material are given below: Steel - $210 \mathrm{~N} / \mathrm{m}^{2}$ Glass - $65 \mathrm{~N} / \mathrm{m}^{2}$ Wood - $13 \mathrm{~N} / \mathrm{m}^{2}$ Plastic - $3 \mathrm{~N} / \mathrm{m}^{2}$
J and K CET- 1997
Mechanical Properties of Solids
140920
The length of the wire is increase by $2 \%$ by applying a load of $2.5 \mathrm{~kg}$-wt. What is the linear strain produced in the wire?
1 0.1
2 0.01
3 0.2
4 0.02
Explanation:
D Initial length $=l$ Change in length $=\Delta l$ Linear strain $=\frac{\Delta l}{l}$ Given, $\%$ increase $=\frac{\Delta l}{l} \times 100$ $\frac{\Delta l}{l} \times 100=2$ $\frac{\Delta l}{l}=\frac{2}{100}$ Linear strain $=0.02$
140917
Two wires $A$ and $B$ are made of the same material. The wire $A$ has a length $l$ and radius $r$ while wire $B$ has a length $2 l$ and radius $2 \mathrm{r}$ are stretched by the same force. The ratio of elongation in A to the elongation in $B$
1 $2: 1$
2 $3: 1$
3 $27: 1$
4 $1: 4$
Explanation:
A Given, Wire A has length $=l_{\mathrm{A}}$ radius $=\mathrm{r}_{\mathrm{A}}$ wire $\mathrm{B}$ has length $=2 l_{\mathrm{B}}$ radius $=2 \mathrm{r}_{\mathrm{B}}$ Since $\mathrm{F}$ and $\mathrm{Y}$ are remain the same i.e, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\Rightarrow \frac{\Delta l_{\mathrm{A}}}{\Delta l_{B}}=\left(\frac{l_{\mathrm{A}}}{l_{B}}\right) \times\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}\right)^{2}$ $=\left(\frac{1}{2 l}\right) \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $=\frac{2}{1}$
J and K CET- 1999
Mechanical Properties of Solids
140918
The area under the stress-strain lines gives
1 work
2 power
3 Young's modulus
4 energy density
Explanation:
D We know, $\frac{\text { Energy }}{\text { Volume }}=\frac{1}{2} \times$ Stress $\times$ Strain Area under stress-strain line is equal to area of triangle. $\therefore \quad$ Area $=\frac{1}{2} \times$ Stress $\times$ Strain $\therefore \quad$ Area $=\frac{\text { Energy }}{\text { Volume }}$ $\therefore \quad=$ Energy density
J and K CET- 1998
Mechanical Properties of Solids
140919
An elastic modulus is the constant of proportionality in a relation of the form
1 stress $=$ constant $\times$ strain
2 strain $=$ constant $\times$ stress
3 stress $\times$ strain $=$ constant
4 the form of the relation depends on whether the constant is the bulk, shear, or Young's modulus
Explanation:
A Elastic modulus $=\frac{\text { stress }}{\text { strain }}$ Young's modulus (also referred to as the elastic modulus or tensile modulus), is a measure of mechanical properties of linear elastic solids like rod, wire and such other things. Young's modulus values of different material are given below: Steel - $210 \mathrm{~N} / \mathrm{m}^{2}$ Glass - $65 \mathrm{~N} / \mathrm{m}^{2}$ Wood - $13 \mathrm{~N} / \mathrm{m}^{2}$ Plastic - $3 \mathrm{~N} / \mathrm{m}^{2}$
J and K CET- 1997
Mechanical Properties of Solids
140920
The length of the wire is increase by $2 \%$ by applying a load of $2.5 \mathrm{~kg}$-wt. What is the linear strain produced in the wire?
1 0.1
2 0.01
3 0.2
4 0.02
Explanation:
D Initial length $=l$ Change in length $=\Delta l$ Linear strain $=\frac{\Delta l}{l}$ Given, $\%$ increase $=\frac{\Delta l}{l} \times 100$ $\frac{\Delta l}{l} \times 100=2$ $\frac{\Delta l}{l}=\frac{2}{100}$ Linear strain $=0.02$
140917
Two wires $A$ and $B$ are made of the same material. The wire $A$ has a length $l$ and radius $r$ while wire $B$ has a length $2 l$ and radius $2 \mathrm{r}$ are stretched by the same force. The ratio of elongation in A to the elongation in $B$
1 $2: 1$
2 $3: 1$
3 $27: 1$
4 $1: 4$
Explanation:
A Given, Wire A has length $=l_{\mathrm{A}}$ radius $=\mathrm{r}_{\mathrm{A}}$ wire $\mathrm{B}$ has length $=2 l_{\mathrm{B}}$ radius $=2 \mathrm{r}_{\mathrm{B}}$ Since $\mathrm{F}$ and $\mathrm{Y}$ are remain the same i.e, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\Rightarrow \frac{\Delta l_{\mathrm{A}}}{\Delta l_{B}}=\left(\frac{l_{\mathrm{A}}}{l_{B}}\right) \times\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}\right)^{2}$ $=\left(\frac{1}{2 l}\right) \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $=\frac{2}{1}$
J and K CET- 1999
Mechanical Properties of Solids
140918
The area under the stress-strain lines gives
1 work
2 power
3 Young's modulus
4 energy density
Explanation:
D We know, $\frac{\text { Energy }}{\text { Volume }}=\frac{1}{2} \times$ Stress $\times$ Strain Area under stress-strain line is equal to area of triangle. $\therefore \quad$ Area $=\frac{1}{2} \times$ Stress $\times$ Strain $\therefore \quad$ Area $=\frac{\text { Energy }}{\text { Volume }}$ $\therefore \quad=$ Energy density
J and K CET- 1998
Mechanical Properties of Solids
140919
An elastic modulus is the constant of proportionality in a relation of the form
1 stress $=$ constant $\times$ strain
2 strain $=$ constant $\times$ stress
3 stress $\times$ strain $=$ constant
4 the form of the relation depends on whether the constant is the bulk, shear, or Young's modulus
Explanation:
A Elastic modulus $=\frac{\text { stress }}{\text { strain }}$ Young's modulus (also referred to as the elastic modulus or tensile modulus), is a measure of mechanical properties of linear elastic solids like rod, wire and such other things. Young's modulus values of different material are given below: Steel - $210 \mathrm{~N} / \mathrm{m}^{2}$ Glass - $65 \mathrm{~N} / \mathrm{m}^{2}$ Wood - $13 \mathrm{~N} / \mathrm{m}^{2}$ Plastic - $3 \mathrm{~N} / \mathrm{m}^{2}$
J and K CET- 1997
Mechanical Properties of Solids
140920
The length of the wire is increase by $2 \%$ by applying a load of $2.5 \mathrm{~kg}$-wt. What is the linear strain produced in the wire?
1 0.1
2 0.01
3 0.2
4 0.02
Explanation:
D Initial length $=l$ Change in length $=\Delta l$ Linear strain $=\frac{\Delta l}{l}$ Given, $\%$ increase $=\frac{\Delta l}{l} \times 100$ $\frac{\Delta l}{l} \times 100=2$ $\frac{\Delta l}{l}=\frac{2}{100}$ Linear strain $=0.02$
140917
Two wires $A$ and $B$ are made of the same material. The wire $A$ has a length $l$ and radius $r$ while wire $B$ has a length $2 l$ and radius $2 \mathrm{r}$ are stretched by the same force. The ratio of elongation in A to the elongation in $B$
1 $2: 1$
2 $3: 1$
3 $27: 1$
4 $1: 4$
Explanation:
A Given, Wire A has length $=l_{\mathrm{A}}$ radius $=\mathrm{r}_{\mathrm{A}}$ wire $\mathrm{B}$ has length $=2 l_{\mathrm{B}}$ radius $=2 \mathrm{r}_{\mathrm{B}}$ Since $\mathrm{F}$ and $\mathrm{Y}$ are remain the same i.e, $\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta l}{l}$ $\Rightarrow \frac{\Delta l_{\mathrm{A}}}{\Delta l_{B}}=\left(\frac{l_{\mathrm{A}}}{l_{B}}\right) \times\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}\right)^{2}$ $=\left(\frac{1}{2 l}\right) \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $=\frac{2}{1}$
J and K CET- 1999
Mechanical Properties of Solids
140918
The area under the stress-strain lines gives
1 work
2 power
3 Young's modulus
4 energy density
Explanation:
D We know, $\frac{\text { Energy }}{\text { Volume }}=\frac{1}{2} \times$ Stress $\times$ Strain Area under stress-strain line is equal to area of triangle. $\therefore \quad$ Area $=\frac{1}{2} \times$ Stress $\times$ Strain $\therefore \quad$ Area $=\frac{\text { Energy }}{\text { Volume }}$ $\therefore \quad=$ Energy density
J and K CET- 1998
Mechanical Properties of Solids
140919
An elastic modulus is the constant of proportionality in a relation of the form
1 stress $=$ constant $\times$ strain
2 strain $=$ constant $\times$ stress
3 stress $\times$ strain $=$ constant
4 the form of the relation depends on whether the constant is the bulk, shear, or Young's modulus
Explanation:
A Elastic modulus $=\frac{\text { stress }}{\text { strain }}$ Young's modulus (also referred to as the elastic modulus or tensile modulus), is a measure of mechanical properties of linear elastic solids like rod, wire and such other things. Young's modulus values of different material are given below: Steel - $210 \mathrm{~N} / \mathrm{m}^{2}$ Glass - $65 \mathrm{~N} / \mathrm{m}^{2}$ Wood - $13 \mathrm{~N} / \mathrm{m}^{2}$ Plastic - $3 \mathrm{~N} / \mathrm{m}^{2}$
J and K CET- 1997
Mechanical Properties of Solids
140920
The length of the wire is increase by $2 \%$ by applying a load of $2.5 \mathrm{~kg}$-wt. What is the linear strain produced in the wire?
1 0.1
2 0.01
3 0.2
4 0.02
Explanation:
D Initial length $=l$ Change in length $=\Delta l$ Linear strain $=\frac{\Delta l}{l}$ Given, $\%$ increase $=\frac{\Delta l}{l} \times 100$ $\frac{\Delta l}{l} \times 100=2$ $\frac{\Delta l}{l}=\frac{2}{100}$ Linear strain $=0.02$
