NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermal Properties of Matter
146539
A metal rod of Young's modulus ' $Y$ ' and coefficient of linear expansion ' $\alpha$ ' has its temperature raised by ' $\Delta \theta$ '. The linear stress to prevent the expansion of rod is $(L$ and $\ell$ is original length of rod and expansion respectively)
A Given, Young's modulus of metal $\operatorname{rod}=\mathrm{Y}$ Coefficient of linear expansion $=\alpha$ Change in temperature $=\Delta \theta$ We know that, $\mathrm{Y} =\frac{\mathrm{FL}}{\mathrm{A} l}$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y} l}{\mathrm{~L}} \quad(\because l=\mathrm{L} \alpha \Delta \theta)$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y}(\mathrm{L} \alpha \Delta \theta)}{\mathrm{L}}$ $\frac{\mathrm{F}}{\mathrm{A}} =\mathrm{Y} \alpha \Delta \theta$
MHT-CET 2020
Thermal Properties of Matter
146541
When a number of thermocouple are joined in series, then the thermo emf
1 is decreased
2 is increased
3 becomes zero
4 remains same
Explanation:
B Exp: When thermocouples are joined in series thermo emfs set up in each thermocouple will help to increase the current in the external load. Therefore, the total thermo emf will be increased.
UP CPMT-2012
Thermal Properties of Matter
146543
If a tensile force is suddenly removed from a wire then its temperature will
1 decrease
2 increase
3 become zero
4 remains constant
Explanation:
B Exp: Sudden removal of tensile force on the wire, increases the kinetic energy of its molecules. Hence temperature of the wire increases.
SRMJEEE - 2015]
Thermal Properties of Matter
146546
Suppose there is a hole in a copper plate, Upon heating the plate, diameter of the hole would
1 always increase
2 increase
3 remain the same
4 None of the above
Explanation:
A On heating any metal (i.e. copper). The metal expand because of phenomenon of thermal expansion. So, hole in a copper plate also expand.
146539
A metal rod of Young's modulus ' $Y$ ' and coefficient of linear expansion ' $\alpha$ ' has its temperature raised by ' $\Delta \theta$ '. The linear stress to prevent the expansion of rod is $(L$ and $\ell$ is original length of rod and expansion respectively)
A Given, Young's modulus of metal $\operatorname{rod}=\mathrm{Y}$ Coefficient of linear expansion $=\alpha$ Change in temperature $=\Delta \theta$ We know that, $\mathrm{Y} =\frac{\mathrm{FL}}{\mathrm{A} l}$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y} l}{\mathrm{~L}} \quad(\because l=\mathrm{L} \alpha \Delta \theta)$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y}(\mathrm{L} \alpha \Delta \theta)}{\mathrm{L}}$ $\frac{\mathrm{F}}{\mathrm{A}} =\mathrm{Y} \alpha \Delta \theta$
MHT-CET 2020
Thermal Properties of Matter
146541
When a number of thermocouple are joined in series, then the thermo emf
1 is decreased
2 is increased
3 becomes zero
4 remains same
Explanation:
B Exp: When thermocouples are joined in series thermo emfs set up in each thermocouple will help to increase the current in the external load. Therefore, the total thermo emf will be increased.
UP CPMT-2012
Thermal Properties of Matter
146543
If a tensile force is suddenly removed from a wire then its temperature will
1 decrease
2 increase
3 become zero
4 remains constant
Explanation:
B Exp: Sudden removal of tensile force on the wire, increases the kinetic energy of its molecules. Hence temperature of the wire increases.
SRMJEEE - 2015]
Thermal Properties of Matter
146546
Suppose there is a hole in a copper plate, Upon heating the plate, diameter of the hole would
1 always increase
2 increase
3 remain the same
4 None of the above
Explanation:
A On heating any metal (i.e. copper). The metal expand because of phenomenon of thermal expansion. So, hole in a copper plate also expand.
146539
A metal rod of Young's modulus ' $Y$ ' and coefficient of linear expansion ' $\alpha$ ' has its temperature raised by ' $\Delta \theta$ '. The linear stress to prevent the expansion of rod is $(L$ and $\ell$ is original length of rod and expansion respectively)
A Given, Young's modulus of metal $\operatorname{rod}=\mathrm{Y}$ Coefficient of linear expansion $=\alpha$ Change in temperature $=\Delta \theta$ We know that, $\mathrm{Y} =\frac{\mathrm{FL}}{\mathrm{A} l}$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y} l}{\mathrm{~L}} \quad(\because l=\mathrm{L} \alpha \Delta \theta)$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y}(\mathrm{L} \alpha \Delta \theta)}{\mathrm{L}}$ $\frac{\mathrm{F}}{\mathrm{A}} =\mathrm{Y} \alpha \Delta \theta$
MHT-CET 2020
Thermal Properties of Matter
146541
When a number of thermocouple are joined in series, then the thermo emf
1 is decreased
2 is increased
3 becomes zero
4 remains same
Explanation:
B Exp: When thermocouples are joined in series thermo emfs set up in each thermocouple will help to increase the current in the external load. Therefore, the total thermo emf will be increased.
UP CPMT-2012
Thermal Properties of Matter
146543
If a tensile force is suddenly removed from a wire then its temperature will
1 decrease
2 increase
3 become zero
4 remains constant
Explanation:
B Exp: Sudden removal of tensile force on the wire, increases the kinetic energy of its molecules. Hence temperature of the wire increases.
SRMJEEE - 2015]
Thermal Properties of Matter
146546
Suppose there is a hole in a copper plate, Upon heating the plate, diameter of the hole would
1 always increase
2 increase
3 remain the same
4 None of the above
Explanation:
A On heating any metal (i.e. copper). The metal expand because of phenomenon of thermal expansion. So, hole in a copper plate also expand.
146539
A metal rod of Young's modulus ' $Y$ ' and coefficient of linear expansion ' $\alpha$ ' has its temperature raised by ' $\Delta \theta$ '. The linear stress to prevent the expansion of rod is $(L$ and $\ell$ is original length of rod and expansion respectively)
A Given, Young's modulus of metal $\operatorname{rod}=\mathrm{Y}$ Coefficient of linear expansion $=\alpha$ Change in temperature $=\Delta \theta$ We know that, $\mathrm{Y} =\frac{\mathrm{FL}}{\mathrm{A} l}$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y} l}{\mathrm{~L}} \quad(\because l=\mathrm{L} \alpha \Delta \theta)$ $\frac{\mathrm{F}}{\mathrm{A}} =\frac{\mathrm{Y}(\mathrm{L} \alpha \Delta \theta)}{\mathrm{L}}$ $\frac{\mathrm{F}}{\mathrm{A}} =\mathrm{Y} \alpha \Delta \theta$
MHT-CET 2020
Thermal Properties of Matter
146541
When a number of thermocouple are joined in series, then the thermo emf
1 is decreased
2 is increased
3 becomes zero
4 remains same
Explanation:
B Exp: When thermocouples are joined in series thermo emfs set up in each thermocouple will help to increase the current in the external load. Therefore, the total thermo emf will be increased.
UP CPMT-2012
Thermal Properties of Matter
146543
If a tensile force is suddenly removed from a wire then its temperature will
1 decrease
2 increase
3 become zero
4 remains constant
Explanation:
B Exp: Sudden removal of tensile force on the wire, increases the kinetic energy of its molecules. Hence temperature of the wire increases.
SRMJEEE - 2015]
Thermal Properties of Matter
146546
Suppose there is a hole in a copper plate, Upon heating the plate, diameter of the hole would
1 always increase
2 increase
3 remain the same
4 None of the above
Explanation:
A On heating any metal (i.e. copper). The metal expand because of phenomenon of thermal expansion. So, hole in a copper plate also expand.
