NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Solids
140946
If the coefficient of linear expansion of a metal is $0.00002 \mathrm{~K}^{-1}$, then the necessary increase in temperature of the metal rod in order to increase its length by $2 \%$ is
1 $100 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
D Given that, $\frac{\Delta \mathrm{L}}{\mathrm{L}}=2 \%=\frac{2}{100}$ Coefficient of linear expansion $(\alpha)=0.00002 \mathrm{~K}^{-1}$ We know that, linear expansion $(\alpha)=\frac{\Delta \mathrm{L}}{\mathrm{L}} \frac{1}{\Delta \mathrm{T}}$ $\Delta \mathrm{T} =\frac{\Delta \mathrm{L}}{\alpha \cdot \mathrm{L}}$ $\Delta \mathrm{T} =\frac{2}{100 \times 0.00002}$ $\Delta \mathrm{T} =\frac{1}{0.001}$ $\Delta \mathrm{T}=10^{3}=1000 \mathrm{~K}$
J and K-CET-2016
Mechanical Properties of Solids
140947
A $3 \mathbf{m}$ long steel wire is stretched to increase its length by $0.3 \mathrm{~cm}$. Poisson's ratio for steel is 0.26. The lateral strain produces in the wires is
140856
According to the Hooke's law, the force required to change the length of a wire by $l$ is proportional to
1 $l^{-2}$
2 $\Gamma^{-1}$
3 $l$
4 $l^{2}$
Explanation:
C Given, change in length $\Delta \mathrm{L}=l$ $\mathrm{L}=$ original length According to Hook's law - Stress $\propto$ Strain $\frac{\text { Stress }}{\text { Strain }}=$ Elastic constant $\frac{\mathrm{F} / \mathrm{A}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}=\mathrm{Y}$ $\mathrm{F}=\frac{\mathrm{AY \Delta L}}{\mathrm{L}}$ Here, A, Y, L are constant $\mathrm{F} \propto l$
AP EAMCET (22.09.2020) Shift-II
Mechanical Properties of Solids
140858
The energy stored in a strained wire is given by
1 $\frac{1}{2} \times$ load $\times$ extension
2 $\frac{1}{2} \times$ extension $\times$ stress
3 $\frac{1}{2} \times$ stress $\times$ strain
4 $\frac{1}{2} \times$ strain $\times$ load
Explanation:
A Energy stored in a strained wire is equal to work done by the load to increase the length of wire. $\therefore$ Energy $(\mathrm{U})=$ Average force $\times$ Extension in the wire $U=\left(\frac{0+F}{2}\right) \times \text { Extension }$ $U=\frac{F}{2} \times \text { Extension }$ $U=\frac{1}{2} \times \text { Load } \times \text { Extension }$
140946
If the coefficient of linear expansion of a metal is $0.00002 \mathrm{~K}^{-1}$, then the necessary increase in temperature of the metal rod in order to increase its length by $2 \%$ is
1 $100 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
D Given that, $\frac{\Delta \mathrm{L}}{\mathrm{L}}=2 \%=\frac{2}{100}$ Coefficient of linear expansion $(\alpha)=0.00002 \mathrm{~K}^{-1}$ We know that, linear expansion $(\alpha)=\frac{\Delta \mathrm{L}}{\mathrm{L}} \frac{1}{\Delta \mathrm{T}}$ $\Delta \mathrm{T} =\frac{\Delta \mathrm{L}}{\alpha \cdot \mathrm{L}}$ $\Delta \mathrm{T} =\frac{2}{100 \times 0.00002}$ $\Delta \mathrm{T} =\frac{1}{0.001}$ $\Delta \mathrm{T}=10^{3}=1000 \mathrm{~K}$
J and K-CET-2016
Mechanical Properties of Solids
140947
A $3 \mathbf{m}$ long steel wire is stretched to increase its length by $0.3 \mathrm{~cm}$. Poisson's ratio for steel is 0.26. The lateral strain produces in the wires is
140856
According to the Hooke's law, the force required to change the length of a wire by $l$ is proportional to
1 $l^{-2}$
2 $\Gamma^{-1}$
3 $l$
4 $l^{2}$
Explanation:
C Given, change in length $\Delta \mathrm{L}=l$ $\mathrm{L}=$ original length According to Hook's law - Stress $\propto$ Strain $\frac{\text { Stress }}{\text { Strain }}=$ Elastic constant $\frac{\mathrm{F} / \mathrm{A}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}=\mathrm{Y}$ $\mathrm{F}=\frac{\mathrm{AY \Delta L}}{\mathrm{L}}$ Here, A, Y, L are constant $\mathrm{F} \propto l$
AP EAMCET (22.09.2020) Shift-II
Mechanical Properties of Solids
140858
The energy stored in a strained wire is given by
1 $\frac{1}{2} \times$ load $\times$ extension
2 $\frac{1}{2} \times$ extension $\times$ stress
3 $\frac{1}{2} \times$ stress $\times$ strain
4 $\frac{1}{2} \times$ strain $\times$ load
Explanation:
A Energy stored in a strained wire is equal to work done by the load to increase the length of wire. $\therefore$ Energy $(\mathrm{U})=$ Average force $\times$ Extension in the wire $U=\left(\frac{0+F}{2}\right) \times \text { Extension }$ $U=\frac{F}{2} \times \text { Extension }$ $U=\frac{1}{2} \times \text { Load } \times \text { Extension }$
140946
If the coefficient of linear expansion of a metal is $0.00002 \mathrm{~K}^{-1}$, then the necessary increase in temperature of the metal rod in order to increase its length by $2 \%$ is
1 $100 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
D Given that, $\frac{\Delta \mathrm{L}}{\mathrm{L}}=2 \%=\frac{2}{100}$ Coefficient of linear expansion $(\alpha)=0.00002 \mathrm{~K}^{-1}$ We know that, linear expansion $(\alpha)=\frac{\Delta \mathrm{L}}{\mathrm{L}} \frac{1}{\Delta \mathrm{T}}$ $\Delta \mathrm{T} =\frac{\Delta \mathrm{L}}{\alpha \cdot \mathrm{L}}$ $\Delta \mathrm{T} =\frac{2}{100 \times 0.00002}$ $\Delta \mathrm{T} =\frac{1}{0.001}$ $\Delta \mathrm{T}=10^{3}=1000 \mathrm{~K}$
J and K-CET-2016
Mechanical Properties of Solids
140947
A $3 \mathbf{m}$ long steel wire is stretched to increase its length by $0.3 \mathrm{~cm}$. Poisson's ratio for steel is 0.26. The lateral strain produces in the wires is
140856
According to the Hooke's law, the force required to change the length of a wire by $l$ is proportional to
1 $l^{-2}$
2 $\Gamma^{-1}$
3 $l$
4 $l^{2}$
Explanation:
C Given, change in length $\Delta \mathrm{L}=l$ $\mathrm{L}=$ original length According to Hook's law - Stress $\propto$ Strain $\frac{\text { Stress }}{\text { Strain }}=$ Elastic constant $\frac{\mathrm{F} / \mathrm{A}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}=\mathrm{Y}$ $\mathrm{F}=\frac{\mathrm{AY \Delta L}}{\mathrm{L}}$ Here, A, Y, L are constant $\mathrm{F} \propto l$
AP EAMCET (22.09.2020) Shift-II
Mechanical Properties of Solids
140858
The energy stored in a strained wire is given by
1 $\frac{1}{2} \times$ load $\times$ extension
2 $\frac{1}{2} \times$ extension $\times$ stress
3 $\frac{1}{2} \times$ stress $\times$ strain
4 $\frac{1}{2} \times$ strain $\times$ load
Explanation:
A Energy stored in a strained wire is equal to work done by the load to increase the length of wire. $\therefore$ Energy $(\mathrm{U})=$ Average force $\times$ Extension in the wire $U=\left(\frac{0+F}{2}\right) \times \text { Extension }$ $U=\frac{F}{2} \times \text { Extension }$ $U=\frac{1}{2} \times \text { Load } \times \text { Extension }$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Mechanical Properties of Solids
140946
If the coefficient of linear expansion of a metal is $0.00002 \mathrm{~K}^{-1}$, then the necessary increase in temperature of the metal rod in order to increase its length by $2 \%$ is
1 $100 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
D Given that, $\frac{\Delta \mathrm{L}}{\mathrm{L}}=2 \%=\frac{2}{100}$ Coefficient of linear expansion $(\alpha)=0.00002 \mathrm{~K}^{-1}$ We know that, linear expansion $(\alpha)=\frac{\Delta \mathrm{L}}{\mathrm{L}} \frac{1}{\Delta \mathrm{T}}$ $\Delta \mathrm{T} =\frac{\Delta \mathrm{L}}{\alpha \cdot \mathrm{L}}$ $\Delta \mathrm{T} =\frac{2}{100 \times 0.00002}$ $\Delta \mathrm{T} =\frac{1}{0.001}$ $\Delta \mathrm{T}=10^{3}=1000 \mathrm{~K}$
J and K-CET-2016
Mechanical Properties of Solids
140947
A $3 \mathbf{m}$ long steel wire is stretched to increase its length by $0.3 \mathrm{~cm}$. Poisson's ratio for steel is 0.26. The lateral strain produces in the wires is
140856
According to the Hooke's law, the force required to change the length of a wire by $l$ is proportional to
1 $l^{-2}$
2 $\Gamma^{-1}$
3 $l$
4 $l^{2}$
Explanation:
C Given, change in length $\Delta \mathrm{L}=l$ $\mathrm{L}=$ original length According to Hook's law - Stress $\propto$ Strain $\frac{\text { Stress }}{\text { Strain }}=$ Elastic constant $\frac{\mathrm{F} / \mathrm{A}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}=\mathrm{Y}$ $\mathrm{F}=\frac{\mathrm{AY \Delta L}}{\mathrm{L}}$ Here, A, Y, L are constant $\mathrm{F} \propto l$
AP EAMCET (22.09.2020) Shift-II
Mechanical Properties of Solids
140858
The energy stored in a strained wire is given by
1 $\frac{1}{2} \times$ load $\times$ extension
2 $\frac{1}{2} \times$ extension $\times$ stress
3 $\frac{1}{2} \times$ stress $\times$ strain
4 $\frac{1}{2} \times$ strain $\times$ load
Explanation:
A Energy stored in a strained wire is equal to work done by the load to increase the length of wire. $\therefore$ Energy $(\mathrm{U})=$ Average force $\times$ Extension in the wire $U=\left(\frac{0+F}{2}\right) \times \text { Extension }$ $U=\frac{F}{2} \times \text { Extension }$ $U=\frac{1}{2} \times \text { Load } \times \text { Extension }$