140881
The value of $\tan \left(90^{\circ}-\theta\right)$ in the graph gives
1 Young's modulus of elasticity
2 compressibility
3 shear strain
4 tensile strength
Explanation:
A Young's modulus of elasticity is define as the ratio of stress upon strain. Basically young's modulus denote slope of the course in stress $\mathrm{v} / \mathrm{s}$ strain graph. But in question, the graph as shown interchanges the symbol along $\mathrm{x}$ axis and $\mathrm{y}$ axis. $\tan (90-\theta)=\frac{\Delta \varepsilon}{\Delta \sigma} .$ $\cot (\theta)=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\frac{1}{\tan \theta}=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\tan \theta=\frac{\Delta \sigma}{\Delta \varepsilon}$ $\therefore \tan (90-\theta)$ denote young's modulus of elasticity.
VITEEE-2017
Mechanical Properties of Solids
140882
A $2 \mathbf{m m}^{2}$ cross-sectional area wire is stretched by $4 \mathrm{~mm}$ by a certain weight. If the same material wire of cross-sectional area $8 \mathrm{~mm}^{2}$ is stretched by the same weight, the stretch length is,
1 $2 \mathrm{~mm}$
2 $0.5 \mathrm{~mm}$
3 $1 \mathrm{~mm}$
4 $1.5 \mathrm{~mm}$
Explanation:
C Given that, $\Delta \mathrm{L}=4 \mathrm{~mm}, \Delta \mathrm{L}^{\prime}=$ ?, $\mathrm{A}=2 \mathrm{~mm}^{2}$ and $\mathrm{A}^{\prime}=8 \mathrm{~mm}^{2}$ From Hooke's Law, Y $=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}$ $\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}} \text { and } \Delta \mathrm{L}^{\prime}=\frac{\mathrm{FL}}{\mathrm{A}^{\prime} \mathrm{Y}}$ For same F, $\mathrm{L}$ and $\mathrm{Y}$, we get, $\frac{\Delta \mathrm{L}^{\prime}}{\Delta \mathrm{L}} =\frac{\mathrm{A}}{\mathrm{A}^{\prime}}$ $\frac{\Delta \mathrm{L}}{4} =\frac{2}{8}$ $\Delta \mathrm{L}^{\prime} =1 \mathrm{~mm}$
TSEAMCET(Medical)-2017
Mechanical Properties of Solids
140883
A material has a Poisson's 0.5. If a uniform rod of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in its volume is
1 0.6
2 0.4
3 0.2
4 zero
Explanation:
D Given that, Poissons ratio $=0.5$, strain $\left(\frac{\mathrm{dL}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, $\frac{\mathrm{dv}}{\mathrm{v}} =(1-2 \sigma) \frac{\mathrm{dL}}{\mathrm{L}}$ $\frac{\mathrm{dv}}{\mathrm{v}} =\{1-(2 \times 0.5)\} \times\left(2 \times 10^{-3}\right)$ $\frac{\mathrm{dv}}{\mathrm{v}} =0 \times 2 \times 10^{-3}$ $\frac{\mathrm{dv}}{\mathrm{v}} =0$
Shift-II]
Mechanical Properties of Solids
140884
The maximum length of a copper wire which has density $8.94 \mathrm{gcm}^{-3}$ and breaking stress $3 \times 10^{8} \mathrm{Nm}^{-2}$ so that it does not break when it is vertically suspended is
1 $7.70 \mathrm{~km}$
2 $3.35 \mathrm{~km}$
3 $33.5 \mathrm{~km}$
4 $335 \mathrm{~km}$
Explanation:
B Given, Density $(\rho)=8.94 \mathrm{~g} / \mathrm{cm}^{3}=8940 \mathrm{~kg} / \mathrm{m}^{3}$ Breaking stress $(\sigma)=3 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}$ Maximum Length $(\mathrm{L})=$ ? Cross section area $=\mathrm{A}$ Weight of copper wire $(\mathrm{W})=\rho . \mathrm{AL} \cdot \mathrm{g}\{\mathrm{m}=\mathrm{P} \cdot \mathrm{V}=\rho . \mathrm{AL}\}$ Stress $=\frac{W}{A}=\frac{\rho A L g}{A}$ $\sigma=\rho \mathrm{Lg}$ $3 \times 10^{8}=8940 \times \mathrm{L} \times 10$ $\mathrm{L}=\frac{3 \times 10^{8}}{8940 \times 10}=3355.7 \mathrm{~m}$ $=3.35 \mathrm{~km}$
140881
The value of $\tan \left(90^{\circ}-\theta\right)$ in the graph gives
1 Young's modulus of elasticity
2 compressibility
3 shear strain
4 tensile strength
Explanation:
A Young's modulus of elasticity is define as the ratio of stress upon strain. Basically young's modulus denote slope of the course in stress $\mathrm{v} / \mathrm{s}$ strain graph. But in question, the graph as shown interchanges the symbol along $\mathrm{x}$ axis and $\mathrm{y}$ axis. $\tan (90-\theta)=\frac{\Delta \varepsilon}{\Delta \sigma} .$ $\cot (\theta)=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\frac{1}{\tan \theta}=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\tan \theta=\frac{\Delta \sigma}{\Delta \varepsilon}$ $\therefore \tan (90-\theta)$ denote young's modulus of elasticity.
VITEEE-2017
Mechanical Properties of Solids
140882
A $2 \mathbf{m m}^{2}$ cross-sectional area wire is stretched by $4 \mathrm{~mm}$ by a certain weight. If the same material wire of cross-sectional area $8 \mathrm{~mm}^{2}$ is stretched by the same weight, the stretch length is,
1 $2 \mathrm{~mm}$
2 $0.5 \mathrm{~mm}$
3 $1 \mathrm{~mm}$
4 $1.5 \mathrm{~mm}$
Explanation:
C Given that, $\Delta \mathrm{L}=4 \mathrm{~mm}, \Delta \mathrm{L}^{\prime}=$ ?, $\mathrm{A}=2 \mathrm{~mm}^{2}$ and $\mathrm{A}^{\prime}=8 \mathrm{~mm}^{2}$ From Hooke's Law, Y $=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}$ $\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}} \text { and } \Delta \mathrm{L}^{\prime}=\frac{\mathrm{FL}}{\mathrm{A}^{\prime} \mathrm{Y}}$ For same F, $\mathrm{L}$ and $\mathrm{Y}$, we get, $\frac{\Delta \mathrm{L}^{\prime}}{\Delta \mathrm{L}} =\frac{\mathrm{A}}{\mathrm{A}^{\prime}}$ $\frac{\Delta \mathrm{L}}{4} =\frac{2}{8}$ $\Delta \mathrm{L}^{\prime} =1 \mathrm{~mm}$
TSEAMCET(Medical)-2017
Mechanical Properties of Solids
140883
A material has a Poisson's 0.5. If a uniform rod of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in its volume is
1 0.6
2 0.4
3 0.2
4 zero
Explanation:
D Given that, Poissons ratio $=0.5$, strain $\left(\frac{\mathrm{dL}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, $\frac{\mathrm{dv}}{\mathrm{v}} =(1-2 \sigma) \frac{\mathrm{dL}}{\mathrm{L}}$ $\frac{\mathrm{dv}}{\mathrm{v}} =\{1-(2 \times 0.5)\} \times\left(2 \times 10^{-3}\right)$ $\frac{\mathrm{dv}}{\mathrm{v}} =0 \times 2 \times 10^{-3}$ $\frac{\mathrm{dv}}{\mathrm{v}} =0$
Shift-II]
Mechanical Properties of Solids
140884
The maximum length of a copper wire which has density $8.94 \mathrm{gcm}^{-3}$ and breaking stress $3 \times 10^{8} \mathrm{Nm}^{-2}$ so that it does not break when it is vertically suspended is
1 $7.70 \mathrm{~km}$
2 $3.35 \mathrm{~km}$
3 $33.5 \mathrm{~km}$
4 $335 \mathrm{~km}$
Explanation:
B Given, Density $(\rho)=8.94 \mathrm{~g} / \mathrm{cm}^{3}=8940 \mathrm{~kg} / \mathrm{m}^{3}$ Breaking stress $(\sigma)=3 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}$ Maximum Length $(\mathrm{L})=$ ? Cross section area $=\mathrm{A}$ Weight of copper wire $(\mathrm{W})=\rho . \mathrm{AL} \cdot \mathrm{g}\{\mathrm{m}=\mathrm{P} \cdot \mathrm{V}=\rho . \mathrm{AL}\}$ Stress $=\frac{W}{A}=\frac{\rho A L g}{A}$ $\sigma=\rho \mathrm{Lg}$ $3 \times 10^{8}=8940 \times \mathrm{L} \times 10$ $\mathrm{L}=\frac{3 \times 10^{8}}{8940 \times 10}=3355.7 \mathrm{~m}$ $=3.35 \mathrm{~km}$
140881
The value of $\tan \left(90^{\circ}-\theta\right)$ in the graph gives
1 Young's modulus of elasticity
2 compressibility
3 shear strain
4 tensile strength
Explanation:
A Young's modulus of elasticity is define as the ratio of stress upon strain. Basically young's modulus denote slope of the course in stress $\mathrm{v} / \mathrm{s}$ strain graph. But in question, the graph as shown interchanges the symbol along $\mathrm{x}$ axis and $\mathrm{y}$ axis. $\tan (90-\theta)=\frac{\Delta \varepsilon}{\Delta \sigma} .$ $\cot (\theta)=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\frac{1}{\tan \theta}=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\tan \theta=\frac{\Delta \sigma}{\Delta \varepsilon}$ $\therefore \tan (90-\theta)$ denote young's modulus of elasticity.
VITEEE-2017
Mechanical Properties of Solids
140882
A $2 \mathbf{m m}^{2}$ cross-sectional area wire is stretched by $4 \mathrm{~mm}$ by a certain weight. If the same material wire of cross-sectional area $8 \mathrm{~mm}^{2}$ is stretched by the same weight, the stretch length is,
1 $2 \mathrm{~mm}$
2 $0.5 \mathrm{~mm}$
3 $1 \mathrm{~mm}$
4 $1.5 \mathrm{~mm}$
Explanation:
C Given that, $\Delta \mathrm{L}=4 \mathrm{~mm}, \Delta \mathrm{L}^{\prime}=$ ?, $\mathrm{A}=2 \mathrm{~mm}^{2}$ and $\mathrm{A}^{\prime}=8 \mathrm{~mm}^{2}$ From Hooke's Law, Y $=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}$ $\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}} \text { and } \Delta \mathrm{L}^{\prime}=\frac{\mathrm{FL}}{\mathrm{A}^{\prime} \mathrm{Y}}$ For same F, $\mathrm{L}$ and $\mathrm{Y}$, we get, $\frac{\Delta \mathrm{L}^{\prime}}{\Delta \mathrm{L}} =\frac{\mathrm{A}}{\mathrm{A}^{\prime}}$ $\frac{\Delta \mathrm{L}}{4} =\frac{2}{8}$ $\Delta \mathrm{L}^{\prime} =1 \mathrm{~mm}$
TSEAMCET(Medical)-2017
Mechanical Properties of Solids
140883
A material has a Poisson's 0.5. If a uniform rod of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in its volume is
1 0.6
2 0.4
3 0.2
4 zero
Explanation:
D Given that, Poissons ratio $=0.5$, strain $\left(\frac{\mathrm{dL}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, $\frac{\mathrm{dv}}{\mathrm{v}} =(1-2 \sigma) \frac{\mathrm{dL}}{\mathrm{L}}$ $\frac{\mathrm{dv}}{\mathrm{v}} =\{1-(2 \times 0.5)\} \times\left(2 \times 10^{-3}\right)$ $\frac{\mathrm{dv}}{\mathrm{v}} =0 \times 2 \times 10^{-3}$ $\frac{\mathrm{dv}}{\mathrm{v}} =0$
Shift-II]
Mechanical Properties of Solids
140884
The maximum length of a copper wire which has density $8.94 \mathrm{gcm}^{-3}$ and breaking stress $3 \times 10^{8} \mathrm{Nm}^{-2}$ so that it does not break when it is vertically suspended is
1 $7.70 \mathrm{~km}$
2 $3.35 \mathrm{~km}$
3 $33.5 \mathrm{~km}$
4 $335 \mathrm{~km}$
Explanation:
B Given, Density $(\rho)=8.94 \mathrm{~g} / \mathrm{cm}^{3}=8940 \mathrm{~kg} / \mathrm{m}^{3}$ Breaking stress $(\sigma)=3 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}$ Maximum Length $(\mathrm{L})=$ ? Cross section area $=\mathrm{A}$ Weight of copper wire $(\mathrm{W})=\rho . \mathrm{AL} \cdot \mathrm{g}\{\mathrm{m}=\mathrm{P} \cdot \mathrm{V}=\rho . \mathrm{AL}\}$ Stress $=\frac{W}{A}=\frac{\rho A L g}{A}$ $\sigma=\rho \mathrm{Lg}$ $3 \times 10^{8}=8940 \times \mathrm{L} \times 10$ $\mathrm{L}=\frac{3 \times 10^{8}}{8940 \times 10}=3355.7 \mathrm{~m}$ $=3.35 \mathrm{~km}$
140881
The value of $\tan \left(90^{\circ}-\theta\right)$ in the graph gives
1 Young's modulus of elasticity
2 compressibility
3 shear strain
4 tensile strength
Explanation:
A Young's modulus of elasticity is define as the ratio of stress upon strain. Basically young's modulus denote slope of the course in stress $\mathrm{v} / \mathrm{s}$ strain graph. But in question, the graph as shown interchanges the symbol along $\mathrm{x}$ axis and $\mathrm{y}$ axis. $\tan (90-\theta)=\frac{\Delta \varepsilon}{\Delta \sigma} .$ $\cot (\theta)=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\frac{1}{\tan \theta}=\frac{\Delta \varepsilon}{\Delta \sigma}$ $\tan \theta=\frac{\Delta \sigma}{\Delta \varepsilon}$ $\therefore \tan (90-\theta)$ denote young's modulus of elasticity.
VITEEE-2017
Mechanical Properties of Solids
140882
A $2 \mathbf{m m}^{2}$ cross-sectional area wire is stretched by $4 \mathrm{~mm}$ by a certain weight. If the same material wire of cross-sectional area $8 \mathrm{~mm}^{2}$ is stretched by the same weight, the stretch length is,
1 $2 \mathrm{~mm}$
2 $0.5 \mathrm{~mm}$
3 $1 \mathrm{~mm}$
4 $1.5 \mathrm{~mm}$
Explanation:
C Given that, $\Delta \mathrm{L}=4 \mathrm{~mm}, \Delta \mathrm{L}^{\prime}=$ ?, $\mathrm{A}=2 \mathrm{~mm}^{2}$ and $\mathrm{A}^{\prime}=8 \mathrm{~mm}^{2}$ From Hooke's Law, Y $=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}$ $\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}} \text { and } \Delta \mathrm{L}^{\prime}=\frac{\mathrm{FL}}{\mathrm{A}^{\prime} \mathrm{Y}}$ For same F, $\mathrm{L}$ and $\mathrm{Y}$, we get, $\frac{\Delta \mathrm{L}^{\prime}}{\Delta \mathrm{L}} =\frac{\mathrm{A}}{\mathrm{A}^{\prime}}$ $\frac{\Delta \mathrm{L}}{4} =\frac{2}{8}$ $\Delta \mathrm{L}^{\prime} =1 \mathrm{~mm}$
TSEAMCET(Medical)-2017
Mechanical Properties of Solids
140883
A material has a Poisson's 0.5. If a uniform rod of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in its volume is
1 0.6
2 0.4
3 0.2
4 zero
Explanation:
D Given that, Poissons ratio $=0.5$, strain $\left(\frac{\mathrm{dL}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, $\frac{\mathrm{dv}}{\mathrm{v}} =(1-2 \sigma) \frac{\mathrm{dL}}{\mathrm{L}}$ $\frac{\mathrm{dv}}{\mathrm{v}} =\{1-(2 \times 0.5)\} \times\left(2 \times 10^{-3}\right)$ $\frac{\mathrm{dv}}{\mathrm{v}} =0 \times 2 \times 10^{-3}$ $\frac{\mathrm{dv}}{\mathrm{v}} =0$
Shift-II]
Mechanical Properties of Solids
140884
The maximum length of a copper wire which has density $8.94 \mathrm{gcm}^{-3}$ and breaking stress $3 \times 10^{8} \mathrm{Nm}^{-2}$ so that it does not break when it is vertically suspended is
1 $7.70 \mathrm{~km}$
2 $3.35 \mathrm{~km}$
3 $33.5 \mathrm{~km}$
4 $335 \mathrm{~km}$
Explanation:
B Given, Density $(\rho)=8.94 \mathrm{~g} / \mathrm{cm}^{3}=8940 \mathrm{~kg} / \mathrm{m}^{3}$ Breaking stress $(\sigma)=3 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}$ Maximum Length $(\mathrm{L})=$ ? Cross section area $=\mathrm{A}$ Weight of copper wire $(\mathrm{W})=\rho . \mathrm{AL} \cdot \mathrm{g}\{\mathrm{m}=\mathrm{P} \cdot \mathrm{V}=\rho . \mathrm{AL}\}$ Stress $=\frac{W}{A}=\frac{\rho A L g}{A}$ $\sigma=\rho \mathrm{Lg}$ $3 \times 10^{8}=8940 \times \mathrm{L} \times 10$ $\mathrm{L}=\frac{3 \times 10^{8}}{8940 \times 10}=3355.7 \mathrm{~m}$ $=3.35 \mathrm{~km}$
