NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148683
A cyclic process $A B C D$ is shown below in the given P-V diagram. In the following answers the one that represents the same process as in P-T diagram
1
2
3
4
Explanation:
A :Process $(\mathrm{AB})=$ Isobaric process it means pressure is constant. Process $(\mathrm{BC})=$ Isothermal expansion process, it means temperature is constant Process $(\mathrm{CD})=$ Isochoric process, it means volume constant and temperature decrease. Process $(\mathrm{DA})=$ Isothermal compression, it means temperature constant and volume reduced.
BITSAT- 2008
Thermodynamics
148684
A Carnot engine takes 300 calories of heat at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is
1 $1000 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $250 \mathrm{~K}$
Explanation:
D Given, Heat reject $\left(\mathrm{Q}_{\mathrm{L}}\right)=150 \mathrm{cal}$ Heat takes $\left(\mathrm{Q}_{\mathrm{H}}\right)=300 \mathrm{cal}$ Temperature of source $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Temperature of sink $\left(\mathrm{T}_{\mathrm{L}}\right)=$ ? Efficiency of engine $=\frac{\text { Output }}{\text { Input }}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{H}}-\mathrm{Q}_{\mathrm{L}}}=\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{300-150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $2 \times\left(500-\mathrm{T}_{\mathrm{L}}\right)=500$ $1000-2 \mathrm{~T}_{\mathrm{L}}=500$ $-2 \mathrm{~T}_{\mathrm{L}}=500-1000$ $-2 \mathrm{~T}_{\mathrm{L}}=-500$ $\mathrm{T}_{\mathrm{L}}=250 \mathrm{~K}$
UPSEE 2019
Thermodynamics
148685
A Carnot engine works between $727^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. The efficiency of the engine is
1 $30 \%$
2 $70 \%$
3 $96 \%$
4 $100 \%$
Explanation:
B Given, $\mathrm{T}_{\mathrm{L}}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{H}}=727^{\circ} \mathrm{C}+273=1000 \mathrm{~K}$ The efficiency of Carnot engine- $\eta=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{1000}$ $=\frac{700}{1000} \times 100$ $\eta=70 \%$
UPSEE 2020
Thermodynamics
148686
When the temperature of the source of a Carnot engine is at $400 \mathrm{~K}$, its efficiency is $25 \%$. The required increase in temperature of the source to increase the efficiency to $50 \%$ is
1 $800 \mathrm{~K}$
2 $600 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $400 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\eta=\frac{1}{4}=25 \%$ $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\eta^{\prime}=50 \%=\frac{1}{2}$ $\mathrm{~T}_{\mathrm{H}}=?$ The efficiency of Carnot engine is- $\eta =1-\frac{T_{L}}{T_{H}}$ $\frac{1}{4} =1-\frac{T_{L}}{400}$ $\frac{3}{4} =\frac{T_{L}}{400}$ $T_{L} =\frac{400 \times 3}{4}=300 \mathrm{~K}$ For 50\% efficiency, $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $\frac{1}{2}=1-\frac{300}{T_{H}}$ $\frac{1}{2}=\frac{300}{T_{H}}$ $T_{H}=600 \mathrm{~K}$
148683
A cyclic process $A B C D$ is shown below in the given P-V diagram. In the following answers the one that represents the same process as in P-T diagram
1
2
3
4
Explanation:
A :Process $(\mathrm{AB})=$ Isobaric process it means pressure is constant. Process $(\mathrm{BC})=$ Isothermal expansion process, it means temperature is constant Process $(\mathrm{CD})=$ Isochoric process, it means volume constant and temperature decrease. Process $(\mathrm{DA})=$ Isothermal compression, it means temperature constant and volume reduced.
BITSAT- 2008
Thermodynamics
148684
A Carnot engine takes 300 calories of heat at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is
1 $1000 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $250 \mathrm{~K}$
Explanation:
D Given, Heat reject $\left(\mathrm{Q}_{\mathrm{L}}\right)=150 \mathrm{cal}$ Heat takes $\left(\mathrm{Q}_{\mathrm{H}}\right)=300 \mathrm{cal}$ Temperature of source $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Temperature of sink $\left(\mathrm{T}_{\mathrm{L}}\right)=$ ? Efficiency of engine $=\frac{\text { Output }}{\text { Input }}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{H}}-\mathrm{Q}_{\mathrm{L}}}=\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{300-150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $2 \times\left(500-\mathrm{T}_{\mathrm{L}}\right)=500$ $1000-2 \mathrm{~T}_{\mathrm{L}}=500$ $-2 \mathrm{~T}_{\mathrm{L}}=500-1000$ $-2 \mathrm{~T}_{\mathrm{L}}=-500$ $\mathrm{T}_{\mathrm{L}}=250 \mathrm{~K}$
UPSEE 2019
Thermodynamics
148685
A Carnot engine works between $727^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. The efficiency of the engine is
1 $30 \%$
2 $70 \%$
3 $96 \%$
4 $100 \%$
Explanation:
B Given, $\mathrm{T}_{\mathrm{L}}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{H}}=727^{\circ} \mathrm{C}+273=1000 \mathrm{~K}$ The efficiency of Carnot engine- $\eta=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{1000}$ $=\frac{700}{1000} \times 100$ $\eta=70 \%$
UPSEE 2020
Thermodynamics
148686
When the temperature of the source of a Carnot engine is at $400 \mathrm{~K}$, its efficiency is $25 \%$. The required increase in temperature of the source to increase the efficiency to $50 \%$ is
1 $800 \mathrm{~K}$
2 $600 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $400 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\eta=\frac{1}{4}=25 \%$ $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\eta^{\prime}=50 \%=\frac{1}{2}$ $\mathrm{~T}_{\mathrm{H}}=?$ The efficiency of Carnot engine is- $\eta =1-\frac{T_{L}}{T_{H}}$ $\frac{1}{4} =1-\frac{T_{L}}{400}$ $\frac{3}{4} =\frac{T_{L}}{400}$ $T_{L} =\frac{400 \times 3}{4}=300 \mathrm{~K}$ For 50\% efficiency, $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $\frac{1}{2}=1-\frac{300}{T_{H}}$ $\frac{1}{2}=\frac{300}{T_{H}}$ $T_{H}=600 \mathrm{~K}$
148683
A cyclic process $A B C D$ is shown below in the given P-V diagram. In the following answers the one that represents the same process as in P-T diagram
1
2
3
4
Explanation:
A :Process $(\mathrm{AB})=$ Isobaric process it means pressure is constant. Process $(\mathrm{BC})=$ Isothermal expansion process, it means temperature is constant Process $(\mathrm{CD})=$ Isochoric process, it means volume constant and temperature decrease. Process $(\mathrm{DA})=$ Isothermal compression, it means temperature constant and volume reduced.
BITSAT- 2008
Thermodynamics
148684
A Carnot engine takes 300 calories of heat at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is
1 $1000 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $250 \mathrm{~K}$
Explanation:
D Given, Heat reject $\left(\mathrm{Q}_{\mathrm{L}}\right)=150 \mathrm{cal}$ Heat takes $\left(\mathrm{Q}_{\mathrm{H}}\right)=300 \mathrm{cal}$ Temperature of source $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Temperature of sink $\left(\mathrm{T}_{\mathrm{L}}\right)=$ ? Efficiency of engine $=\frac{\text { Output }}{\text { Input }}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{H}}-\mathrm{Q}_{\mathrm{L}}}=\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{300-150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $2 \times\left(500-\mathrm{T}_{\mathrm{L}}\right)=500$ $1000-2 \mathrm{~T}_{\mathrm{L}}=500$ $-2 \mathrm{~T}_{\mathrm{L}}=500-1000$ $-2 \mathrm{~T}_{\mathrm{L}}=-500$ $\mathrm{T}_{\mathrm{L}}=250 \mathrm{~K}$
UPSEE 2019
Thermodynamics
148685
A Carnot engine works between $727^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. The efficiency of the engine is
1 $30 \%$
2 $70 \%$
3 $96 \%$
4 $100 \%$
Explanation:
B Given, $\mathrm{T}_{\mathrm{L}}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{H}}=727^{\circ} \mathrm{C}+273=1000 \mathrm{~K}$ The efficiency of Carnot engine- $\eta=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{1000}$ $=\frac{700}{1000} \times 100$ $\eta=70 \%$
UPSEE 2020
Thermodynamics
148686
When the temperature of the source of a Carnot engine is at $400 \mathrm{~K}$, its efficiency is $25 \%$. The required increase in temperature of the source to increase the efficiency to $50 \%$ is
1 $800 \mathrm{~K}$
2 $600 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $400 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\eta=\frac{1}{4}=25 \%$ $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\eta^{\prime}=50 \%=\frac{1}{2}$ $\mathrm{~T}_{\mathrm{H}}=?$ The efficiency of Carnot engine is- $\eta =1-\frac{T_{L}}{T_{H}}$ $\frac{1}{4} =1-\frac{T_{L}}{400}$ $\frac{3}{4} =\frac{T_{L}}{400}$ $T_{L} =\frac{400 \times 3}{4}=300 \mathrm{~K}$ For 50\% efficiency, $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $\frac{1}{2}=1-\frac{300}{T_{H}}$ $\frac{1}{2}=\frac{300}{T_{H}}$ $T_{H}=600 \mathrm{~K}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148683
A cyclic process $A B C D$ is shown below in the given P-V diagram. In the following answers the one that represents the same process as in P-T diagram
1
2
3
4
Explanation:
A :Process $(\mathrm{AB})=$ Isobaric process it means pressure is constant. Process $(\mathrm{BC})=$ Isothermal expansion process, it means temperature is constant Process $(\mathrm{CD})=$ Isochoric process, it means volume constant and temperature decrease. Process $(\mathrm{DA})=$ Isothermal compression, it means temperature constant and volume reduced.
BITSAT- 2008
Thermodynamics
148684
A Carnot engine takes 300 calories of heat at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is
1 $1000 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $250 \mathrm{~K}$
Explanation:
D Given, Heat reject $\left(\mathrm{Q}_{\mathrm{L}}\right)=150 \mathrm{cal}$ Heat takes $\left(\mathrm{Q}_{\mathrm{H}}\right)=300 \mathrm{cal}$ Temperature of source $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Temperature of sink $\left(\mathrm{T}_{\mathrm{L}}\right)=$ ? Efficiency of engine $=\frac{\text { Output }}{\text { Input }}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{H}}-\mathrm{Q}_{\mathrm{L}}}=\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{300-150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $\frac{300}{150}=\frac{500}{500-\mathrm{T}_{\mathrm{L}}}$ $2 \times\left(500-\mathrm{T}_{\mathrm{L}}\right)=500$ $1000-2 \mathrm{~T}_{\mathrm{L}}=500$ $-2 \mathrm{~T}_{\mathrm{L}}=500-1000$ $-2 \mathrm{~T}_{\mathrm{L}}=-500$ $\mathrm{T}_{\mathrm{L}}=250 \mathrm{~K}$
UPSEE 2019
Thermodynamics
148685
A Carnot engine works between $727^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. The efficiency of the engine is
1 $30 \%$
2 $70 \%$
3 $96 \%$
4 $100 \%$
Explanation:
B Given, $\mathrm{T}_{\mathrm{L}}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{T}_{\mathrm{H}}=727^{\circ} \mathrm{C}+273=1000 \mathrm{~K}$ The efficiency of Carnot engine- $\eta=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{1000}$ $=\frac{700}{1000} \times 100$ $\eta=70 \%$
UPSEE 2020
Thermodynamics
148686
When the temperature of the source of a Carnot engine is at $400 \mathrm{~K}$, its efficiency is $25 \%$. The required increase in temperature of the source to increase the efficiency to $50 \%$ is
1 $800 \mathrm{~K}$
2 $600 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $400 \mathrm{~K}$
5 $200 \mathrm{~K}$
Explanation:
B Given, $\eta=\frac{1}{4}=25 \%$ $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\eta^{\prime}=50 \%=\frac{1}{2}$ $\mathrm{~T}_{\mathrm{H}}=?$ The efficiency of Carnot engine is- $\eta =1-\frac{T_{L}}{T_{H}}$ $\frac{1}{4} =1-\frac{T_{L}}{400}$ $\frac{3}{4} =\frac{T_{L}}{400}$ $T_{L} =\frac{400 \times 3}{4}=300 \mathrm{~K}$ For 50\% efficiency, $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $\frac{1}{2}=1-\frac{300}{T_{H}}$ $\frac{1}{2}=\frac{300}{T_{H}}$ $T_{H}=600 \mathrm{~K}$
