148571
A Cannot engine whose efficiency is $50 \%$ has an exhaust temperature of $500 \mathrm{~K}$. If the efficiency is to be $60 \%$ with the same intake temperature, the exhaust temperature must be (in K
1 800
2 200
3 400
4 600
Explanation:
C $\text { Carnot efficiency }(\eta)=1-\frac{T_{2}}{T_{1}}$ $\frac{50}{100}=1-\frac{500}{T_{1}}$ $\mathrm{~T}_{1}=1000 \mathrm{~K}$ Similar, According to condition intake temp. is same $\frac{60}{100}=1-\frac{\mathrm{T}_{2}}{1000}$ $\mathrm{~T}_{2}=400 \mathrm{~K}$
AIIMS-27.05.2018(M)
Thermodynamics
148572
In a heat engine, the temperature of the source and sink are $500 \mathrm{~K}$ and $375 \mathrm{~K}$. If the engine consumes $25 \times 10^{5} \mathrm{~J}$ per cycle, the work done per cycle is
1 $6.25 \times 10^{5} \mathrm{~J}$
2 $3 \times 10^{5} \mathrm{~J}$
3 $2.19 \times 10^{5} \mathrm{~J}$
4 $4 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that Temperature of source $T_{1}=500 \mathrm{~K}$ Temperature of sink $\mathrm{T}_{2}=375 \mathrm{~K}$ $\mathrm{Q}_{1}=25 \times 10^{5} \mathrm{~J}$ Efficiency of a Carnot's heat engine $\eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\mathrm{W}=\mathrm{Q}_{1}\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)$ $\mathrm{W}=25 \times 10^{5}\left(1-\frac{375}{500}\right)$ $\mathrm{W}=6.25 \times 10^{5} \mathrm{~J}$
AIIMS-2017
Thermodynamics
148573
N moles of a monoatomic gas is carried round the reversible rectangular cycle $A B C D A$ as shown in the diagram. The temperature at $A$ is $T_{0}$. The thermodynamic efficiency of the cycle is:
1 $15 \%$
2 $50 \%$
3 $20 \%$
4 $25 \%$
Explanation:
B Given, At A temperature $\left(\mathrm{T}_{0}\right)$ volume $\left(\mathrm{V}_{0}\right)$ and pressure $\left(\mathrm{P}_{0}\right)$ Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ At point $\mathrm{A}$ $\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}_{0}$ At B $2 \mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}$ $\mathrm{T}=2 \mathrm{~T}_{0}$ Maximum thermal efficiency of the reversible engine of Carnot's cycle $\eta=1-\frac{T_{0}}{T}=1-\frac{T_{0}}{2 T_{0}}=\frac{1}{2}=50 \%$ $\eta=50 \%$
AIIMS-2004
Thermodynamics
148575
Determine efficiency of Carnot cycle if in adiabatic expansion volume becomes 3 times of initial value and $\gamma=1.5$
1 $1-\frac{1}{\sqrt{2}}$
2 $1-\frac{1}{\sqrt{3}}$
3 $1+\frac{1}{\sqrt{2}}$
4 $1+\frac{1}{\sqrt{3}}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=3 \mathrm{~V}, \gamma=1.5$ For a adiabatic process, $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\text { or } \quad \mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\mathrm{T}_{1}(\mathrm{~V})_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ $\mathrm{~T}_{1}(\mathrm{~V})^{\gamma-1}=\mathrm{T}_{2} \times(3 \mathrm{~V})^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{3 \mathrm{~V}}{\mathrm{~V}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=(3)^{\gamma-1}=(3)^{1.5-1}=\sqrt{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{\sqrt{3}}$ $\eta=1-\frac{1}{\sqrt{3}}$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148569
Assertion: The efficiency of a reversible engine is maximum. Reason: In such a device no dissipation of energy takes place.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Its efficiency is maximum as no dissipation of energy takes place against friction in such a heat engine.
148571
A Cannot engine whose efficiency is $50 \%$ has an exhaust temperature of $500 \mathrm{~K}$. If the efficiency is to be $60 \%$ with the same intake temperature, the exhaust temperature must be (in K
1 800
2 200
3 400
4 600
Explanation:
C $\text { Carnot efficiency }(\eta)=1-\frac{T_{2}}{T_{1}}$ $\frac{50}{100}=1-\frac{500}{T_{1}}$ $\mathrm{~T}_{1}=1000 \mathrm{~K}$ Similar, According to condition intake temp. is same $\frac{60}{100}=1-\frac{\mathrm{T}_{2}}{1000}$ $\mathrm{~T}_{2}=400 \mathrm{~K}$
AIIMS-27.05.2018(M)
Thermodynamics
148572
In a heat engine, the temperature of the source and sink are $500 \mathrm{~K}$ and $375 \mathrm{~K}$. If the engine consumes $25 \times 10^{5} \mathrm{~J}$ per cycle, the work done per cycle is
1 $6.25 \times 10^{5} \mathrm{~J}$
2 $3 \times 10^{5} \mathrm{~J}$
3 $2.19 \times 10^{5} \mathrm{~J}$
4 $4 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that Temperature of source $T_{1}=500 \mathrm{~K}$ Temperature of sink $\mathrm{T}_{2}=375 \mathrm{~K}$ $\mathrm{Q}_{1}=25 \times 10^{5} \mathrm{~J}$ Efficiency of a Carnot's heat engine $\eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\mathrm{W}=\mathrm{Q}_{1}\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)$ $\mathrm{W}=25 \times 10^{5}\left(1-\frac{375}{500}\right)$ $\mathrm{W}=6.25 \times 10^{5} \mathrm{~J}$
AIIMS-2017
Thermodynamics
148573
N moles of a monoatomic gas is carried round the reversible rectangular cycle $A B C D A$ as shown in the diagram. The temperature at $A$ is $T_{0}$. The thermodynamic efficiency of the cycle is:
1 $15 \%$
2 $50 \%$
3 $20 \%$
4 $25 \%$
Explanation:
B Given, At A temperature $\left(\mathrm{T}_{0}\right)$ volume $\left(\mathrm{V}_{0}\right)$ and pressure $\left(\mathrm{P}_{0}\right)$ Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ At point $\mathrm{A}$ $\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}_{0}$ At B $2 \mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}$ $\mathrm{T}=2 \mathrm{~T}_{0}$ Maximum thermal efficiency of the reversible engine of Carnot's cycle $\eta=1-\frac{T_{0}}{T}=1-\frac{T_{0}}{2 T_{0}}=\frac{1}{2}=50 \%$ $\eta=50 \%$
AIIMS-2004
Thermodynamics
148575
Determine efficiency of Carnot cycle if in adiabatic expansion volume becomes 3 times of initial value and $\gamma=1.5$
1 $1-\frac{1}{\sqrt{2}}$
2 $1-\frac{1}{\sqrt{3}}$
3 $1+\frac{1}{\sqrt{2}}$
4 $1+\frac{1}{\sqrt{3}}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=3 \mathrm{~V}, \gamma=1.5$ For a adiabatic process, $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\text { or } \quad \mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\mathrm{T}_{1}(\mathrm{~V})_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ $\mathrm{~T}_{1}(\mathrm{~V})^{\gamma-1}=\mathrm{T}_{2} \times(3 \mathrm{~V})^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{3 \mathrm{~V}}{\mathrm{~V}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=(3)^{\gamma-1}=(3)^{1.5-1}=\sqrt{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{\sqrt{3}}$ $\eta=1-\frac{1}{\sqrt{3}}$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148569
Assertion: The efficiency of a reversible engine is maximum. Reason: In such a device no dissipation of energy takes place.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Its efficiency is maximum as no dissipation of energy takes place against friction in such a heat engine.
148571
A Cannot engine whose efficiency is $50 \%$ has an exhaust temperature of $500 \mathrm{~K}$. If the efficiency is to be $60 \%$ with the same intake temperature, the exhaust temperature must be (in K
1 800
2 200
3 400
4 600
Explanation:
C $\text { Carnot efficiency }(\eta)=1-\frac{T_{2}}{T_{1}}$ $\frac{50}{100}=1-\frac{500}{T_{1}}$ $\mathrm{~T}_{1}=1000 \mathrm{~K}$ Similar, According to condition intake temp. is same $\frac{60}{100}=1-\frac{\mathrm{T}_{2}}{1000}$ $\mathrm{~T}_{2}=400 \mathrm{~K}$
AIIMS-27.05.2018(M)
Thermodynamics
148572
In a heat engine, the temperature of the source and sink are $500 \mathrm{~K}$ and $375 \mathrm{~K}$. If the engine consumes $25 \times 10^{5} \mathrm{~J}$ per cycle, the work done per cycle is
1 $6.25 \times 10^{5} \mathrm{~J}$
2 $3 \times 10^{5} \mathrm{~J}$
3 $2.19 \times 10^{5} \mathrm{~J}$
4 $4 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that Temperature of source $T_{1}=500 \mathrm{~K}$ Temperature of sink $\mathrm{T}_{2}=375 \mathrm{~K}$ $\mathrm{Q}_{1}=25 \times 10^{5} \mathrm{~J}$ Efficiency of a Carnot's heat engine $\eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\mathrm{W}=\mathrm{Q}_{1}\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)$ $\mathrm{W}=25 \times 10^{5}\left(1-\frac{375}{500}\right)$ $\mathrm{W}=6.25 \times 10^{5} \mathrm{~J}$
AIIMS-2017
Thermodynamics
148573
N moles of a monoatomic gas is carried round the reversible rectangular cycle $A B C D A$ as shown in the diagram. The temperature at $A$ is $T_{0}$. The thermodynamic efficiency of the cycle is:
1 $15 \%$
2 $50 \%$
3 $20 \%$
4 $25 \%$
Explanation:
B Given, At A temperature $\left(\mathrm{T}_{0}\right)$ volume $\left(\mathrm{V}_{0}\right)$ and pressure $\left(\mathrm{P}_{0}\right)$ Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ At point $\mathrm{A}$ $\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}_{0}$ At B $2 \mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}$ $\mathrm{T}=2 \mathrm{~T}_{0}$ Maximum thermal efficiency of the reversible engine of Carnot's cycle $\eta=1-\frac{T_{0}}{T}=1-\frac{T_{0}}{2 T_{0}}=\frac{1}{2}=50 \%$ $\eta=50 \%$
AIIMS-2004
Thermodynamics
148575
Determine efficiency of Carnot cycle if in adiabatic expansion volume becomes 3 times of initial value and $\gamma=1.5$
1 $1-\frac{1}{\sqrt{2}}$
2 $1-\frac{1}{\sqrt{3}}$
3 $1+\frac{1}{\sqrt{2}}$
4 $1+\frac{1}{\sqrt{3}}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=3 \mathrm{~V}, \gamma=1.5$ For a adiabatic process, $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\text { or } \quad \mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\mathrm{T}_{1}(\mathrm{~V})_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ $\mathrm{~T}_{1}(\mathrm{~V})^{\gamma-1}=\mathrm{T}_{2} \times(3 \mathrm{~V})^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{3 \mathrm{~V}}{\mathrm{~V}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=(3)^{\gamma-1}=(3)^{1.5-1}=\sqrt{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{\sqrt{3}}$ $\eta=1-\frac{1}{\sqrt{3}}$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148569
Assertion: The efficiency of a reversible engine is maximum. Reason: In such a device no dissipation of energy takes place.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Its efficiency is maximum as no dissipation of energy takes place against friction in such a heat engine.
148571
A Cannot engine whose efficiency is $50 \%$ has an exhaust temperature of $500 \mathrm{~K}$. If the efficiency is to be $60 \%$ with the same intake temperature, the exhaust temperature must be (in K
1 800
2 200
3 400
4 600
Explanation:
C $\text { Carnot efficiency }(\eta)=1-\frac{T_{2}}{T_{1}}$ $\frac{50}{100}=1-\frac{500}{T_{1}}$ $\mathrm{~T}_{1}=1000 \mathrm{~K}$ Similar, According to condition intake temp. is same $\frac{60}{100}=1-\frac{\mathrm{T}_{2}}{1000}$ $\mathrm{~T}_{2}=400 \mathrm{~K}$
AIIMS-27.05.2018(M)
Thermodynamics
148572
In a heat engine, the temperature of the source and sink are $500 \mathrm{~K}$ and $375 \mathrm{~K}$. If the engine consumes $25 \times 10^{5} \mathrm{~J}$ per cycle, the work done per cycle is
1 $6.25 \times 10^{5} \mathrm{~J}$
2 $3 \times 10^{5} \mathrm{~J}$
3 $2.19 \times 10^{5} \mathrm{~J}$
4 $4 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that Temperature of source $T_{1}=500 \mathrm{~K}$ Temperature of sink $\mathrm{T}_{2}=375 \mathrm{~K}$ $\mathrm{Q}_{1}=25 \times 10^{5} \mathrm{~J}$ Efficiency of a Carnot's heat engine $\eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\mathrm{W}=\mathrm{Q}_{1}\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)$ $\mathrm{W}=25 \times 10^{5}\left(1-\frac{375}{500}\right)$ $\mathrm{W}=6.25 \times 10^{5} \mathrm{~J}$
AIIMS-2017
Thermodynamics
148573
N moles of a monoatomic gas is carried round the reversible rectangular cycle $A B C D A$ as shown in the diagram. The temperature at $A$ is $T_{0}$. The thermodynamic efficiency of the cycle is:
1 $15 \%$
2 $50 \%$
3 $20 \%$
4 $25 \%$
Explanation:
B Given, At A temperature $\left(\mathrm{T}_{0}\right)$ volume $\left(\mathrm{V}_{0}\right)$ and pressure $\left(\mathrm{P}_{0}\right)$ Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ At point $\mathrm{A}$ $\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}_{0}$ At B $2 \mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}$ $\mathrm{T}=2 \mathrm{~T}_{0}$ Maximum thermal efficiency of the reversible engine of Carnot's cycle $\eta=1-\frac{T_{0}}{T}=1-\frac{T_{0}}{2 T_{0}}=\frac{1}{2}=50 \%$ $\eta=50 \%$
AIIMS-2004
Thermodynamics
148575
Determine efficiency of Carnot cycle if in adiabatic expansion volume becomes 3 times of initial value and $\gamma=1.5$
1 $1-\frac{1}{\sqrt{2}}$
2 $1-\frac{1}{\sqrt{3}}$
3 $1+\frac{1}{\sqrt{2}}$
4 $1+\frac{1}{\sqrt{3}}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=3 \mathrm{~V}, \gamma=1.5$ For a adiabatic process, $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\text { or } \quad \mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\mathrm{T}_{1}(\mathrm{~V})_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ $\mathrm{~T}_{1}(\mathrm{~V})^{\gamma-1}=\mathrm{T}_{2} \times(3 \mathrm{~V})^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{3 \mathrm{~V}}{\mathrm{~V}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=(3)^{\gamma-1}=(3)^{1.5-1}=\sqrt{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{\sqrt{3}}$ $\eta=1-\frac{1}{\sqrt{3}}$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148569
Assertion: The efficiency of a reversible engine is maximum. Reason: In such a device no dissipation of energy takes place.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Its efficiency is maximum as no dissipation of energy takes place against friction in such a heat engine.
148571
A Cannot engine whose efficiency is $50 \%$ has an exhaust temperature of $500 \mathrm{~K}$. If the efficiency is to be $60 \%$ with the same intake temperature, the exhaust temperature must be (in K
1 800
2 200
3 400
4 600
Explanation:
C $\text { Carnot efficiency }(\eta)=1-\frac{T_{2}}{T_{1}}$ $\frac{50}{100}=1-\frac{500}{T_{1}}$ $\mathrm{~T}_{1}=1000 \mathrm{~K}$ Similar, According to condition intake temp. is same $\frac{60}{100}=1-\frac{\mathrm{T}_{2}}{1000}$ $\mathrm{~T}_{2}=400 \mathrm{~K}$
AIIMS-27.05.2018(M)
Thermodynamics
148572
In a heat engine, the temperature of the source and sink are $500 \mathrm{~K}$ and $375 \mathrm{~K}$. If the engine consumes $25 \times 10^{5} \mathrm{~J}$ per cycle, the work done per cycle is
1 $6.25 \times 10^{5} \mathrm{~J}$
2 $3 \times 10^{5} \mathrm{~J}$
3 $2.19 \times 10^{5} \mathrm{~J}$
4 $4 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that Temperature of source $T_{1}=500 \mathrm{~K}$ Temperature of sink $\mathrm{T}_{2}=375 \mathrm{~K}$ $\mathrm{Q}_{1}=25 \times 10^{5} \mathrm{~J}$ Efficiency of a Carnot's heat engine $\eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\mathrm{W}=\mathrm{Q}_{1}\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)$ $\mathrm{W}=25 \times 10^{5}\left(1-\frac{375}{500}\right)$ $\mathrm{W}=6.25 \times 10^{5} \mathrm{~J}$
AIIMS-2017
Thermodynamics
148573
N moles of a monoatomic gas is carried round the reversible rectangular cycle $A B C D A$ as shown in the diagram. The temperature at $A$ is $T_{0}$. The thermodynamic efficiency of the cycle is:
1 $15 \%$
2 $50 \%$
3 $20 \%$
4 $25 \%$
Explanation:
B Given, At A temperature $\left(\mathrm{T}_{0}\right)$ volume $\left(\mathrm{V}_{0}\right)$ and pressure $\left(\mathrm{P}_{0}\right)$ Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ At point $\mathrm{A}$ $\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}_{0}$ At B $2 \mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{nRT}$ $\mathrm{T}=2 \mathrm{~T}_{0}$ Maximum thermal efficiency of the reversible engine of Carnot's cycle $\eta=1-\frac{T_{0}}{T}=1-\frac{T_{0}}{2 T_{0}}=\frac{1}{2}=50 \%$ $\eta=50 \%$
AIIMS-2004
Thermodynamics
148575
Determine efficiency of Carnot cycle if in adiabatic expansion volume becomes 3 times of initial value and $\gamma=1.5$
1 $1-\frac{1}{\sqrt{2}}$
2 $1-\frac{1}{\sqrt{3}}$
3 $1+\frac{1}{\sqrt{2}}$
4 $1+\frac{1}{\sqrt{3}}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=3 \mathrm{~V}, \gamma=1.5$ For a adiabatic process, $\mathrm{PV}^{\gamma}=\mathrm{C}$ $\text { or } \quad \mathrm{TV}^{\gamma-1}=\mathrm{C}$ $\mathrm{T}_{1}(\mathrm{~V})_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ $\mathrm{~T}_{1}(\mathrm{~V})^{\gamma-1}=\mathrm{T}_{2} \times(3 \mathrm{~V})^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{3 \mathrm{~V}}{\mathrm{~V}}\right)^{\gamma-1}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=(3)^{\gamma-1}=(3)^{1.5-1}=\sqrt{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{\sqrt{3}}$ $\eta=1-\frac{1}{\sqrt{3}}$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148569
Assertion: The efficiency of a reversible engine is maximum. Reason: In such a device no dissipation of energy takes place.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Its efficiency is maximum as no dissipation of energy takes place against friction in such a heat engine.
