141031
The modulus of elasticity is dimensionally equivalent to:
1 strain
2 force
3 stress
4 coefficient of viscosity
5 surface tension
Explanation:
C Modulus of elasticity is defined as the ratio of stress to strain of material $\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{N}}{\mathrm{m}^{2}}$ Since, strain is unitless/dimension less quantity. So, unit of $\mathrm{Y}=$ unit of stress $\mathrm{Y}=\mathrm{N} / \mathrm{m}^{2} \quad \text { [Strain is unit less] }$ Hence, modulus of elasticity is dimensionally equivalent to stress.
EAMCET 1996
Mechanical Properties of Solids
141032
Compressibility of water is $5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}$. The change in volume of $100 \mathrm{~mL}$ water subjected to $15 \times 10^{6} \mathrm{~Pa}$ pressure will :
141033
If the longitudinal strain in a cubical body is three times the lateral strain then the bulk modulus K, Young's modulus $Y$ and rigidity $\eta$ are related by
1 $\mathrm{K}=\mathrm{Y}$
2 $\eta=\frac{3 Y}{8}$
3 $Y=\frac{3 \eta}{8}$
4 $Y=\eta$
Explanation:
A Relation between Y, K, $\eta$ and $\mu$ which are young modulus compressibility, modulus of rigidity and Poisson ratio respectively. Poisson's ratio $=\frac{\text { Lateral strain }}{\text { Longitudnal strain }}$ $\left.\mu=\frac{1}{3} \text { [longitudnal strain }=3 \text { laternal strain }\right]$ We know, $\mathrm{Y}=3 \mathrm{~K}(1-2 \mu)$ $\mathrm{Y}=3 \mathrm{~K}\left(1-2 \times \frac{1}{3}\right)$ $\mathrm{Y}=\mathrm{K}$ And $Y=2 \eta(1+\mu)$ $Y=2 \eta\left(1+\frac{1}{3}\right)$ $\eta=\frac{3 Y}{8}$
UPSEE - 2011
Mechanical Properties of Solids
141034
A copper rod of length $L$ and radius $r$ is suspended from the ceiling by one of its ends. What will be elongation of the rod due to its own weight where $\rho$ and $Y$ are the density and Young's modulus of the copper respectively?
1 $\frac{\rho^{2} g L^{2}}{2 \mathrm{Y}}$
2 $\frac{\rho g L^{2}}{2 \mathrm{Y}}$
3 $\frac{\rho^{2} g^{2} L^{2}}{2 Y}$
4 $\frac{\rho g L}{2 Y}$
Explanation:
B $\rho$ is density of the copper $\mathrm{r}=$ radius of copper rod $\mathrm{L}=$ Length of copper rod $\mathrm{Y}=$ Young modulus Weigh $(w)=m g$ Weight $\mathrm{w}=\rho \times \mathrm{v} \times \mathrm{g}$ $\mathrm{w}=\rho \times \mathrm{A} \times \mathrm{L} \times \mathrm{g}$ $\mathrm{w}=\pi \rho \mathrm{r}^{2} \mathrm{Lg}$ $\left[\mathrm{A}=\pi \mathrm{r}^{2}\right]$ The above weight act at $\mathrm{L} / 2$ of the rod. Then, $\text { Elongation } =\frac{\mathrm{w} \times \mathrm{L}}{\mathrm{AY}}=\frac{\pi \rho r^{2} \mathrm{Lg} \times(\mathrm{L} / 2)}{\pi \mathrm{r}^{2} \mathrm{Y}}$ $\Delta \mathrm{L} =\frac{\rho \mathrm{L}^{2} \mathrm{~g}}{2 \mathrm{Y}}$
141031
The modulus of elasticity is dimensionally equivalent to:
1 strain
2 force
3 stress
4 coefficient of viscosity
5 surface tension
Explanation:
C Modulus of elasticity is defined as the ratio of stress to strain of material $\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{N}}{\mathrm{m}^{2}}$ Since, strain is unitless/dimension less quantity. So, unit of $\mathrm{Y}=$ unit of stress $\mathrm{Y}=\mathrm{N} / \mathrm{m}^{2} \quad \text { [Strain is unit less] }$ Hence, modulus of elasticity is dimensionally equivalent to stress.
EAMCET 1996
Mechanical Properties of Solids
141032
Compressibility of water is $5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}$. The change in volume of $100 \mathrm{~mL}$ water subjected to $15 \times 10^{6} \mathrm{~Pa}$ pressure will :
141033
If the longitudinal strain in a cubical body is three times the lateral strain then the bulk modulus K, Young's modulus $Y$ and rigidity $\eta$ are related by
1 $\mathrm{K}=\mathrm{Y}$
2 $\eta=\frac{3 Y}{8}$
3 $Y=\frac{3 \eta}{8}$
4 $Y=\eta$
Explanation:
A Relation between Y, K, $\eta$ and $\mu$ which are young modulus compressibility, modulus of rigidity and Poisson ratio respectively. Poisson's ratio $=\frac{\text { Lateral strain }}{\text { Longitudnal strain }}$ $\left.\mu=\frac{1}{3} \text { [longitudnal strain }=3 \text { laternal strain }\right]$ We know, $\mathrm{Y}=3 \mathrm{~K}(1-2 \mu)$ $\mathrm{Y}=3 \mathrm{~K}\left(1-2 \times \frac{1}{3}\right)$ $\mathrm{Y}=\mathrm{K}$ And $Y=2 \eta(1+\mu)$ $Y=2 \eta\left(1+\frac{1}{3}\right)$ $\eta=\frac{3 Y}{8}$
UPSEE - 2011
Mechanical Properties of Solids
141034
A copper rod of length $L$ and radius $r$ is suspended from the ceiling by one of its ends. What will be elongation of the rod due to its own weight where $\rho$ and $Y$ are the density and Young's modulus of the copper respectively?
1 $\frac{\rho^{2} g L^{2}}{2 \mathrm{Y}}$
2 $\frac{\rho g L^{2}}{2 \mathrm{Y}}$
3 $\frac{\rho^{2} g^{2} L^{2}}{2 Y}$
4 $\frac{\rho g L}{2 Y}$
Explanation:
B $\rho$ is density of the copper $\mathrm{r}=$ radius of copper rod $\mathrm{L}=$ Length of copper rod $\mathrm{Y}=$ Young modulus Weigh $(w)=m g$ Weight $\mathrm{w}=\rho \times \mathrm{v} \times \mathrm{g}$ $\mathrm{w}=\rho \times \mathrm{A} \times \mathrm{L} \times \mathrm{g}$ $\mathrm{w}=\pi \rho \mathrm{r}^{2} \mathrm{Lg}$ $\left[\mathrm{A}=\pi \mathrm{r}^{2}\right]$ The above weight act at $\mathrm{L} / 2$ of the rod. Then, $\text { Elongation } =\frac{\mathrm{w} \times \mathrm{L}}{\mathrm{AY}}=\frac{\pi \rho r^{2} \mathrm{Lg} \times(\mathrm{L} / 2)}{\pi \mathrm{r}^{2} \mathrm{Y}}$ $\Delta \mathrm{L} =\frac{\rho \mathrm{L}^{2} \mathrm{~g}}{2 \mathrm{Y}}$
141031
The modulus of elasticity is dimensionally equivalent to:
1 strain
2 force
3 stress
4 coefficient of viscosity
5 surface tension
Explanation:
C Modulus of elasticity is defined as the ratio of stress to strain of material $\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{N}}{\mathrm{m}^{2}}$ Since, strain is unitless/dimension less quantity. So, unit of $\mathrm{Y}=$ unit of stress $\mathrm{Y}=\mathrm{N} / \mathrm{m}^{2} \quad \text { [Strain is unit less] }$ Hence, modulus of elasticity is dimensionally equivalent to stress.
EAMCET 1996
Mechanical Properties of Solids
141032
Compressibility of water is $5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}$. The change in volume of $100 \mathrm{~mL}$ water subjected to $15 \times 10^{6} \mathrm{~Pa}$ pressure will :
141033
If the longitudinal strain in a cubical body is three times the lateral strain then the bulk modulus K, Young's modulus $Y$ and rigidity $\eta$ are related by
1 $\mathrm{K}=\mathrm{Y}$
2 $\eta=\frac{3 Y}{8}$
3 $Y=\frac{3 \eta}{8}$
4 $Y=\eta$
Explanation:
A Relation between Y, K, $\eta$ and $\mu$ which are young modulus compressibility, modulus of rigidity and Poisson ratio respectively. Poisson's ratio $=\frac{\text { Lateral strain }}{\text { Longitudnal strain }}$ $\left.\mu=\frac{1}{3} \text { [longitudnal strain }=3 \text { laternal strain }\right]$ We know, $\mathrm{Y}=3 \mathrm{~K}(1-2 \mu)$ $\mathrm{Y}=3 \mathrm{~K}\left(1-2 \times \frac{1}{3}\right)$ $\mathrm{Y}=\mathrm{K}$ And $Y=2 \eta(1+\mu)$ $Y=2 \eta\left(1+\frac{1}{3}\right)$ $\eta=\frac{3 Y}{8}$
UPSEE - 2011
Mechanical Properties of Solids
141034
A copper rod of length $L$ and radius $r$ is suspended from the ceiling by one of its ends. What will be elongation of the rod due to its own weight where $\rho$ and $Y$ are the density and Young's modulus of the copper respectively?
1 $\frac{\rho^{2} g L^{2}}{2 \mathrm{Y}}$
2 $\frac{\rho g L^{2}}{2 \mathrm{Y}}$
3 $\frac{\rho^{2} g^{2} L^{2}}{2 Y}$
4 $\frac{\rho g L}{2 Y}$
Explanation:
B $\rho$ is density of the copper $\mathrm{r}=$ radius of copper rod $\mathrm{L}=$ Length of copper rod $\mathrm{Y}=$ Young modulus Weigh $(w)=m g$ Weight $\mathrm{w}=\rho \times \mathrm{v} \times \mathrm{g}$ $\mathrm{w}=\rho \times \mathrm{A} \times \mathrm{L} \times \mathrm{g}$ $\mathrm{w}=\pi \rho \mathrm{r}^{2} \mathrm{Lg}$ $\left[\mathrm{A}=\pi \mathrm{r}^{2}\right]$ The above weight act at $\mathrm{L} / 2$ of the rod. Then, $\text { Elongation } =\frac{\mathrm{w} \times \mathrm{L}}{\mathrm{AY}}=\frac{\pi \rho r^{2} \mathrm{Lg} \times(\mathrm{L} / 2)}{\pi \mathrm{r}^{2} \mathrm{Y}}$ $\Delta \mathrm{L} =\frac{\rho \mathrm{L}^{2} \mathrm{~g}}{2 \mathrm{Y}}$
141031
The modulus of elasticity is dimensionally equivalent to:
1 strain
2 force
3 stress
4 coefficient of viscosity
5 surface tension
Explanation:
C Modulus of elasticity is defined as the ratio of stress to strain of material $\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{N}}{\mathrm{m}^{2}}$ Since, strain is unitless/dimension less quantity. So, unit of $\mathrm{Y}=$ unit of stress $\mathrm{Y}=\mathrm{N} / \mathrm{m}^{2} \quad \text { [Strain is unit less] }$ Hence, modulus of elasticity is dimensionally equivalent to stress.
EAMCET 1996
Mechanical Properties of Solids
141032
Compressibility of water is $5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}$. The change in volume of $100 \mathrm{~mL}$ water subjected to $15 \times 10^{6} \mathrm{~Pa}$ pressure will :
141033
If the longitudinal strain in a cubical body is three times the lateral strain then the bulk modulus K, Young's modulus $Y$ and rigidity $\eta$ are related by
1 $\mathrm{K}=\mathrm{Y}$
2 $\eta=\frac{3 Y}{8}$
3 $Y=\frac{3 \eta}{8}$
4 $Y=\eta$
Explanation:
A Relation between Y, K, $\eta$ and $\mu$ which are young modulus compressibility, modulus of rigidity and Poisson ratio respectively. Poisson's ratio $=\frac{\text { Lateral strain }}{\text { Longitudnal strain }}$ $\left.\mu=\frac{1}{3} \text { [longitudnal strain }=3 \text { laternal strain }\right]$ We know, $\mathrm{Y}=3 \mathrm{~K}(1-2 \mu)$ $\mathrm{Y}=3 \mathrm{~K}\left(1-2 \times \frac{1}{3}\right)$ $\mathrm{Y}=\mathrm{K}$ And $Y=2 \eta(1+\mu)$ $Y=2 \eta\left(1+\frac{1}{3}\right)$ $\eta=\frac{3 Y}{8}$
UPSEE - 2011
Mechanical Properties of Solids
141034
A copper rod of length $L$ and radius $r$ is suspended from the ceiling by one of its ends. What will be elongation of the rod due to its own weight where $\rho$ and $Y$ are the density and Young's modulus of the copper respectively?
1 $\frac{\rho^{2} g L^{2}}{2 \mathrm{Y}}$
2 $\frac{\rho g L^{2}}{2 \mathrm{Y}}$
3 $\frac{\rho^{2} g^{2} L^{2}}{2 Y}$
4 $\frac{\rho g L}{2 Y}$
Explanation:
B $\rho$ is density of the copper $\mathrm{r}=$ radius of copper rod $\mathrm{L}=$ Length of copper rod $\mathrm{Y}=$ Young modulus Weigh $(w)=m g$ Weight $\mathrm{w}=\rho \times \mathrm{v} \times \mathrm{g}$ $\mathrm{w}=\rho \times \mathrm{A} \times \mathrm{L} \times \mathrm{g}$ $\mathrm{w}=\pi \rho \mathrm{r}^{2} \mathrm{Lg}$ $\left[\mathrm{A}=\pi \mathrm{r}^{2}\right]$ The above weight act at $\mathrm{L} / 2$ of the rod. Then, $\text { Elongation } =\frac{\mathrm{w} \times \mathrm{L}}{\mathrm{AY}}=\frac{\pi \rho r^{2} \mathrm{Lg} \times(\mathrm{L} / 2)}{\pi \mathrm{r}^{2} \mathrm{Y}}$ $\Delta \mathrm{L} =\frac{\rho \mathrm{L}^{2} \mathrm{~g}}{2 \mathrm{Y}}$
