NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148653
In $1^{\text {st }}$ case, Carnot engine operates between temperatures $300 \mathrm{~K}$ and $100 \mathrm{~K}$. In $2^{\text {nd }}$ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in $2^{\text {nd }}$ case) will be:
1 same as the $1^{\text {st }}$ case
2 always greater than the $1^{\text {st }}$ case
3 always less than the $1^{\text {st }}$ case
4 may increase or decrease with respect to the $1^{\text {st }}$ case
Explanation:
A Efficiency of Carnot engine, $\eta=1-\frac{T_{\text {Lower }}}{T_{\text {Higher }}}$ First case, $\eta_{\mathrm{I}}=1-\frac{100}{300}=\frac{2}{3}$ Second case, $\eta_{\mathrm{II}}=\eta_{\mathrm{I}}+\eta_{\mathrm{II}}-\eta_{\mathrm{I}} \cdot \eta_{\mathrm{II}}$ $=\left(1-\frac{200}{300}\right)+\left(1-\frac{100}{200}\right)-\left(1-\frac{200}{300}\right)\left(1-\frac{100}{200}\right)$ $=\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)-\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)$ $=\frac{1}{3}+\frac{1}{2}-\frac{1}{6}$ $=\frac{2+3-1}{6}=\frac{4}{6}=\frac{2}{3}$ Therefore, $\eta_{\text {(first case) }}=\eta_{\text {(second case) }}$
JEE Main-27.07.2022
Thermodynamics
148654
The efficiency of a Carnot's engine, working between steam point and ice point, will be:
1 $26.81 \%$
2 $37.81 \%$
3 $47.81 \%$
4 $57.81 \%$
Explanation:
A According to question- $\text { Steam point }\left(\mathrm{T}_{1}\right)=100^{\circ} \mathrm{C}=373 \mathrm{~K}$ $\text { Ice point }\left(\mathrm{T}_{2}\right)=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ We know that, steam point and ice point for water are $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{273}{373}\right) \times 100$ $\eta =\frac{100}{373} \times 100$ $\eta =26.81 \%$ Hence, the efficiency of the Carnot engine working between the ice point and the steam point would be $26.81 \%$.
JEE Main-26.06.2022
Thermodynamics
148655
A Carnot engine whose heat sinks at $27^{\circ} \mathrm{C}$, has an efficiency of $25 \%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency?
1 Increases by $18^{\circ} \mathrm{C}$
2 Increases by $200^{\circ} \mathrm{C}$
3 Increases by $120^{\circ} \mathrm{C}$
4 Increases by $73^{\circ} \mathrm{C}$
Explanation:
B Given, $\eta=25 \%$ $\mathrm{T}_{\text {sink }}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Efficiency is given by, $\eta=1-\frac{T_{\text {sink }}}{T_{\text {source }}}$ $25 \%=1-\frac{300}{T_{\text {source }}}$ $\frac{25}{100}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{1}{4}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{1}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{3}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=400 \mathrm{~K}=127^{\circ} \mathrm{C}$ According to question, if efficiency increased by $100 \%$ then new efficiency, $\eta^{\prime}=50 \%$ $1-\frac{300}{\mathrm{~T}_{\text {source }}}=0.5$ $1-0.5=\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=600 \mathrm{~K}=327^{\circ} \mathrm{C}$ Increase in temperature $=327^{\circ} \mathrm{C}-127^{\circ} \mathrm{C}$ $=200^{\circ} \mathrm{C}$
JEE Main-24.06.2022
Thermodynamics
148656
A Carnot engine takes $5000 \mathrm{kcal}$ of heat from a reservoir at $727^{\circ} \mathrm{C}$ and gives heat to a sink at $127^{\circ} \mathrm{C}$. The work done by the engine is
148653
In $1^{\text {st }}$ case, Carnot engine operates between temperatures $300 \mathrm{~K}$ and $100 \mathrm{~K}$. In $2^{\text {nd }}$ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in $2^{\text {nd }}$ case) will be:
1 same as the $1^{\text {st }}$ case
2 always greater than the $1^{\text {st }}$ case
3 always less than the $1^{\text {st }}$ case
4 may increase or decrease with respect to the $1^{\text {st }}$ case
Explanation:
A Efficiency of Carnot engine, $\eta=1-\frac{T_{\text {Lower }}}{T_{\text {Higher }}}$ First case, $\eta_{\mathrm{I}}=1-\frac{100}{300}=\frac{2}{3}$ Second case, $\eta_{\mathrm{II}}=\eta_{\mathrm{I}}+\eta_{\mathrm{II}}-\eta_{\mathrm{I}} \cdot \eta_{\mathrm{II}}$ $=\left(1-\frac{200}{300}\right)+\left(1-\frac{100}{200}\right)-\left(1-\frac{200}{300}\right)\left(1-\frac{100}{200}\right)$ $=\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)-\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)$ $=\frac{1}{3}+\frac{1}{2}-\frac{1}{6}$ $=\frac{2+3-1}{6}=\frac{4}{6}=\frac{2}{3}$ Therefore, $\eta_{\text {(first case) }}=\eta_{\text {(second case) }}$
JEE Main-27.07.2022
Thermodynamics
148654
The efficiency of a Carnot's engine, working between steam point and ice point, will be:
1 $26.81 \%$
2 $37.81 \%$
3 $47.81 \%$
4 $57.81 \%$
Explanation:
A According to question- $\text { Steam point }\left(\mathrm{T}_{1}\right)=100^{\circ} \mathrm{C}=373 \mathrm{~K}$ $\text { Ice point }\left(\mathrm{T}_{2}\right)=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ We know that, steam point and ice point for water are $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{273}{373}\right) \times 100$ $\eta =\frac{100}{373} \times 100$ $\eta =26.81 \%$ Hence, the efficiency of the Carnot engine working between the ice point and the steam point would be $26.81 \%$.
JEE Main-26.06.2022
Thermodynamics
148655
A Carnot engine whose heat sinks at $27^{\circ} \mathrm{C}$, has an efficiency of $25 \%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency?
1 Increases by $18^{\circ} \mathrm{C}$
2 Increases by $200^{\circ} \mathrm{C}$
3 Increases by $120^{\circ} \mathrm{C}$
4 Increases by $73^{\circ} \mathrm{C}$
Explanation:
B Given, $\eta=25 \%$ $\mathrm{T}_{\text {sink }}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Efficiency is given by, $\eta=1-\frac{T_{\text {sink }}}{T_{\text {source }}}$ $25 \%=1-\frac{300}{T_{\text {source }}}$ $\frac{25}{100}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{1}{4}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{1}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{3}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=400 \mathrm{~K}=127^{\circ} \mathrm{C}$ According to question, if efficiency increased by $100 \%$ then new efficiency, $\eta^{\prime}=50 \%$ $1-\frac{300}{\mathrm{~T}_{\text {source }}}=0.5$ $1-0.5=\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=600 \mathrm{~K}=327^{\circ} \mathrm{C}$ Increase in temperature $=327^{\circ} \mathrm{C}-127^{\circ} \mathrm{C}$ $=200^{\circ} \mathrm{C}$
JEE Main-24.06.2022
Thermodynamics
148656
A Carnot engine takes $5000 \mathrm{kcal}$ of heat from a reservoir at $727^{\circ} \mathrm{C}$ and gives heat to a sink at $127^{\circ} \mathrm{C}$. The work done by the engine is
148653
In $1^{\text {st }}$ case, Carnot engine operates between temperatures $300 \mathrm{~K}$ and $100 \mathrm{~K}$. In $2^{\text {nd }}$ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in $2^{\text {nd }}$ case) will be:
1 same as the $1^{\text {st }}$ case
2 always greater than the $1^{\text {st }}$ case
3 always less than the $1^{\text {st }}$ case
4 may increase or decrease with respect to the $1^{\text {st }}$ case
Explanation:
A Efficiency of Carnot engine, $\eta=1-\frac{T_{\text {Lower }}}{T_{\text {Higher }}}$ First case, $\eta_{\mathrm{I}}=1-\frac{100}{300}=\frac{2}{3}$ Second case, $\eta_{\mathrm{II}}=\eta_{\mathrm{I}}+\eta_{\mathrm{II}}-\eta_{\mathrm{I}} \cdot \eta_{\mathrm{II}}$ $=\left(1-\frac{200}{300}\right)+\left(1-\frac{100}{200}\right)-\left(1-\frac{200}{300}\right)\left(1-\frac{100}{200}\right)$ $=\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)-\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)$ $=\frac{1}{3}+\frac{1}{2}-\frac{1}{6}$ $=\frac{2+3-1}{6}=\frac{4}{6}=\frac{2}{3}$ Therefore, $\eta_{\text {(first case) }}=\eta_{\text {(second case) }}$
JEE Main-27.07.2022
Thermodynamics
148654
The efficiency of a Carnot's engine, working between steam point and ice point, will be:
1 $26.81 \%$
2 $37.81 \%$
3 $47.81 \%$
4 $57.81 \%$
Explanation:
A According to question- $\text { Steam point }\left(\mathrm{T}_{1}\right)=100^{\circ} \mathrm{C}=373 \mathrm{~K}$ $\text { Ice point }\left(\mathrm{T}_{2}\right)=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ We know that, steam point and ice point for water are $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{273}{373}\right) \times 100$ $\eta =\frac{100}{373} \times 100$ $\eta =26.81 \%$ Hence, the efficiency of the Carnot engine working between the ice point and the steam point would be $26.81 \%$.
JEE Main-26.06.2022
Thermodynamics
148655
A Carnot engine whose heat sinks at $27^{\circ} \mathrm{C}$, has an efficiency of $25 \%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency?
1 Increases by $18^{\circ} \mathrm{C}$
2 Increases by $200^{\circ} \mathrm{C}$
3 Increases by $120^{\circ} \mathrm{C}$
4 Increases by $73^{\circ} \mathrm{C}$
Explanation:
B Given, $\eta=25 \%$ $\mathrm{T}_{\text {sink }}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Efficiency is given by, $\eta=1-\frac{T_{\text {sink }}}{T_{\text {source }}}$ $25 \%=1-\frac{300}{T_{\text {source }}}$ $\frac{25}{100}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{1}{4}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{1}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{3}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=400 \mathrm{~K}=127^{\circ} \mathrm{C}$ According to question, if efficiency increased by $100 \%$ then new efficiency, $\eta^{\prime}=50 \%$ $1-\frac{300}{\mathrm{~T}_{\text {source }}}=0.5$ $1-0.5=\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=600 \mathrm{~K}=327^{\circ} \mathrm{C}$ Increase in temperature $=327^{\circ} \mathrm{C}-127^{\circ} \mathrm{C}$ $=200^{\circ} \mathrm{C}$
JEE Main-24.06.2022
Thermodynamics
148656
A Carnot engine takes $5000 \mathrm{kcal}$ of heat from a reservoir at $727^{\circ} \mathrm{C}$ and gives heat to a sink at $127^{\circ} \mathrm{C}$. The work done by the engine is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148653
In $1^{\text {st }}$ case, Carnot engine operates between temperatures $300 \mathrm{~K}$ and $100 \mathrm{~K}$. In $2^{\text {nd }}$ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in $2^{\text {nd }}$ case) will be:
1 same as the $1^{\text {st }}$ case
2 always greater than the $1^{\text {st }}$ case
3 always less than the $1^{\text {st }}$ case
4 may increase or decrease with respect to the $1^{\text {st }}$ case
Explanation:
A Efficiency of Carnot engine, $\eta=1-\frac{T_{\text {Lower }}}{T_{\text {Higher }}}$ First case, $\eta_{\mathrm{I}}=1-\frac{100}{300}=\frac{2}{3}$ Second case, $\eta_{\mathrm{II}}=\eta_{\mathrm{I}}+\eta_{\mathrm{II}}-\eta_{\mathrm{I}} \cdot \eta_{\mathrm{II}}$ $=\left(1-\frac{200}{300}\right)+\left(1-\frac{100}{200}\right)-\left(1-\frac{200}{300}\right)\left(1-\frac{100}{200}\right)$ $=\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)-\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)$ $=\frac{1}{3}+\frac{1}{2}-\frac{1}{6}$ $=\frac{2+3-1}{6}=\frac{4}{6}=\frac{2}{3}$ Therefore, $\eta_{\text {(first case) }}=\eta_{\text {(second case) }}$
JEE Main-27.07.2022
Thermodynamics
148654
The efficiency of a Carnot's engine, working between steam point and ice point, will be:
1 $26.81 \%$
2 $37.81 \%$
3 $47.81 \%$
4 $57.81 \%$
Explanation:
A According to question- $\text { Steam point }\left(\mathrm{T}_{1}\right)=100^{\circ} \mathrm{C}=373 \mathrm{~K}$ $\text { Ice point }\left(\mathrm{T}_{2}\right)=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ We know that, steam point and ice point for water are $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{273}{373}\right) \times 100$ $\eta =\frac{100}{373} \times 100$ $\eta =26.81 \%$ Hence, the efficiency of the Carnot engine working between the ice point and the steam point would be $26.81 \%$.
JEE Main-26.06.2022
Thermodynamics
148655
A Carnot engine whose heat sinks at $27^{\circ} \mathrm{C}$, has an efficiency of $25 \%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency?
1 Increases by $18^{\circ} \mathrm{C}$
2 Increases by $200^{\circ} \mathrm{C}$
3 Increases by $120^{\circ} \mathrm{C}$
4 Increases by $73^{\circ} \mathrm{C}$
Explanation:
B Given, $\eta=25 \%$ $\mathrm{T}_{\text {sink }}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Efficiency is given by, $\eta=1-\frac{T_{\text {sink }}}{T_{\text {source }}}$ $25 \%=1-\frac{300}{T_{\text {source }}}$ $\frac{25}{100}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{1}{4}=1-\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{1}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{3}{4}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=400 \mathrm{~K}=127^{\circ} \mathrm{C}$ According to question, if efficiency increased by $100 \%$ then new efficiency, $\eta^{\prime}=50 \%$ $1-\frac{300}{\mathrm{~T}_{\text {source }}}=0.5$ $1-0.5=\frac{300}{\mathrm{~T}_{\text {source }}}$ $1-\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\frac{5}{10}=\frac{300}{\mathrm{~T}_{\text {source }}}$ $\mathrm{T}_{\text {source }}=600 \mathrm{~K}=327^{\circ} \mathrm{C}$ Increase in temperature $=327^{\circ} \mathrm{C}-127^{\circ} \mathrm{C}$ $=200^{\circ} \mathrm{C}$
JEE Main-24.06.2022
Thermodynamics
148656
A Carnot engine takes $5000 \mathrm{kcal}$ of heat from a reservoir at $727^{\circ} \mathrm{C}$ and gives heat to a sink at $127^{\circ} \mathrm{C}$. The work done by the engine is
