NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148540
An electrical refrigerator with $\beta=5$ extracts $5000 \mathrm{~J}$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.
1 $1 \mathrm{~kJ}$
2 $0.5 \mathrm{~kJ}$
3 $0.8 \mathrm{~kJ}$
4 $1.2 \mathrm{~kJ}$
Explanation:
A Given, Coefficient of performance of refrigerator- $\beta=5$ Amount of heat removed $(\mathrm{Q})=5000 \mathrm{~J}$ Electrical energy utilized by the motor $=$ Work done by the motor $(\mathrm{W})$ $\beta =\frac{\mathrm{Q}}{\mathrm{W}}$ $\therefore \quad \mathrm{W} =\frac{\mathrm{Q}}{\beta}=\frac{5000}{5}=1000 \mathrm{~J}$ $=1 \mathrm{~kJ}$
AP EAMCET (22.09.2020) Shift-I
Thermodynamics
148541
The efficiency of a Carnot's engine is $100 \%$ only when
1 ideal gas is used as a working substance
2 temperature of the sink is equal to absolute zero
3 source temperature is equal to the temperature of the sink
4 source temperature is equal to absolute zero
Explanation:
B The efficiency of Carnot's engine is given as $\eta=1-\frac{T_{2}}{T_{1}}$ Where, $T_{1}=$ temperature of source $\mathrm{T}_{2}=$ temperature of sink For efficiency $100 \%, \eta=1$ $1=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0$ $\therefore \quad \mathrm{T}_{2}=0$
AP EAMCET (21.09.2020) Shift-II
Thermodynamics
148542
An ideal heat engine has an efficiency $\eta$. The coefficient of performance of the engine when driven backward will be
1 $1-\left(\frac{1}{\eta}\right)$
2 $\eta-\left(\frac{1}{\eta}\right)$
3 $\left(\frac{1}{\eta}\right)-1$
4 $\frac{1}{1-\eta}$
Explanation:
C Ideal heat engine, efficiency is given as- $\eta=\frac{W}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}$ $\eta=1-\frac{Q_{2}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\eta$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{1}{1-\eta}$ When heat engine operated in backward direction, then coefficient of performance given as- $\alpha=\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}=\frac{1}{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}-1}$ From equation (i) and (ii), $\alpha=\frac{1}{\frac{1}{1-\eta}-1}=\frac{1}{\frac{1}{1-\eta}-1}$ $\frac{1-\eta}{1-1+\eta}=\frac{1-\eta}{\eta}=\left(\frac{1}{\eta}\right)-1$
AP EAMCET (21.09.2020) Shift-I
Thermodynamics
148543
When the temperature difference between the source and sink increases, the efficiency of the heat engine
1 Decreases
2 Increases
3 Is not affected
4 May increases or decrease
Explanation:
B The efficiency ( $\eta$ ) of the Carnot engine is given by- $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}=\text { Temperature of the sink }$ $T_{H}=\text { Temperature of source }$ - When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases. - If the temperature of the source $\left(\mathrm{T}_{\mathrm{H}}\right)$ is increases then the efficiency of the Carnot engine increases.
148540
An electrical refrigerator with $\beta=5$ extracts $5000 \mathrm{~J}$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.
1 $1 \mathrm{~kJ}$
2 $0.5 \mathrm{~kJ}$
3 $0.8 \mathrm{~kJ}$
4 $1.2 \mathrm{~kJ}$
Explanation:
A Given, Coefficient of performance of refrigerator- $\beta=5$ Amount of heat removed $(\mathrm{Q})=5000 \mathrm{~J}$ Electrical energy utilized by the motor $=$ Work done by the motor $(\mathrm{W})$ $\beta =\frac{\mathrm{Q}}{\mathrm{W}}$ $\therefore \quad \mathrm{W} =\frac{\mathrm{Q}}{\beta}=\frac{5000}{5}=1000 \mathrm{~J}$ $=1 \mathrm{~kJ}$
AP EAMCET (22.09.2020) Shift-I
Thermodynamics
148541
The efficiency of a Carnot's engine is $100 \%$ only when
1 ideal gas is used as a working substance
2 temperature of the sink is equal to absolute zero
3 source temperature is equal to the temperature of the sink
4 source temperature is equal to absolute zero
Explanation:
B The efficiency of Carnot's engine is given as $\eta=1-\frac{T_{2}}{T_{1}}$ Where, $T_{1}=$ temperature of source $\mathrm{T}_{2}=$ temperature of sink For efficiency $100 \%, \eta=1$ $1=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0$ $\therefore \quad \mathrm{T}_{2}=0$
AP EAMCET (21.09.2020) Shift-II
Thermodynamics
148542
An ideal heat engine has an efficiency $\eta$. The coefficient of performance of the engine when driven backward will be
1 $1-\left(\frac{1}{\eta}\right)$
2 $\eta-\left(\frac{1}{\eta}\right)$
3 $\left(\frac{1}{\eta}\right)-1$
4 $\frac{1}{1-\eta}$
Explanation:
C Ideal heat engine, efficiency is given as- $\eta=\frac{W}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}$ $\eta=1-\frac{Q_{2}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\eta$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{1}{1-\eta}$ When heat engine operated in backward direction, then coefficient of performance given as- $\alpha=\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}=\frac{1}{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}-1}$ From equation (i) and (ii), $\alpha=\frac{1}{\frac{1}{1-\eta}-1}=\frac{1}{\frac{1}{1-\eta}-1}$ $\frac{1-\eta}{1-1+\eta}=\frac{1-\eta}{\eta}=\left(\frac{1}{\eta}\right)-1$
AP EAMCET (21.09.2020) Shift-I
Thermodynamics
148543
When the temperature difference between the source and sink increases, the efficiency of the heat engine
1 Decreases
2 Increases
3 Is not affected
4 May increases or decrease
Explanation:
B The efficiency ( $\eta$ ) of the Carnot engine is given by- $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}=\text { Temperature of the sink }$ $T_{H}=\text { Temperature of source }$ - When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases. - If the temperature of the source $\left(\mathrm{T}_{\mathrm{H}}\right)$ is increases then the efficiency of the Carnot engine increases.
148540
An electrical refrigerator with $\beta=5$ extracts $5000 \mathrm{~J}$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.
1 $1 \mathrm{~kJ}$
2 $0.5 \mathrm{~kJ}$
3 $0.8 \mathrm{~kJ}$
4 $1.2 \mathrm{~kJ}$
Explanation:
A Given, Coefficient of performance of refrigerator- $\beta=5$ Amount of heat removed $(\mathrm{Q})=5000 \mathrm{~J}$ Electrical energy utilized by the motor $=$ Work done by the motor $(\mathrm{W})$ $\beta =\frac{\mathrm{Q}}{\mathrm{W}}$ $\therefore \quad \mathrm{W} =\frac{\mathrm{Q}}{\beta}=\frac{5000}{5}=1000 \mathrm{~J}$ $=1 \mathrm{~kJ}$
AP EAMCET (22.09.2020) Shift-I
Thermodynamics
148541
The efficiency of a Carnot's engine is $100 \%$ only when
1 ideal gas is used as a working substance
2 temperature of the sink is equal to absolute zero
3 source temperature is equal to the temperature of the sink
4 source temperature is equal to absolute zero
Explanation:
B The efficiency of Carnot's engine is given as $\eta=1-\frac{T_{2}}{T_{1}}$ Where, $T_{1}=$ temperature of source $\mathrm{T}_{2}=$ temperature of sink For efficiency $100 \%, \eta=1$ $1=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0$ $\therefore \quad \mathrm{T}_{2}=0$
AP EAMCET (21.09.2020) Shift-II
Thermodynamics
148542
An ideal heat engine has an efficiency $\eta$. The coefficient of performance of the engine when driven backward will be
1 $1-\left(\frac{1}{\eta}\right)$
2 $\eta-\left(\frac{1}{\eta}\right)$
3 $\left(\frac{1}{\eta}\right)-1$
4 $\frac{1}{1-\eta}$
Explanation:
C Ideal heat engine, efficiency is given as- $\eta=\frac{W}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}$ $\eta=1-\frac{Q_{2}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\eta$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{1}{1-\eta}$ When heat engine operated in backward direction, then coefficient of performance given as- $\alpha=\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}=\frac{1}{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}-1}$ From equation (i) and (ii), $\alpha=\frac{1}{\frac{1}{1-\eta}-1}=\frac{1}{\frac{1}{1-\eta}-1}$ $\frac{1-\eta}{1-1+\eta}=\frac{1-\eta}{\eta}=\left(\frac{1}{\eta}\right)-1$
AP EAMCET (21.09.2020) Shift-I
Thermodynamics
148543
When the temperature difference between the source and sink increases, the efficiency of the heat engine
1 Decreases
2 Increases
3 Is not affected
4 May increases or decrease
Explanation:
B The efficiency ( $\eta$ ) of the Carnot engine is given by- $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}=\text { Temperature of the sink }$ $T_{H}=\text { Temperature of source }$ - When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases. - If the temperature of the source $\left(\mathrm{T}_{\mathrm{H}}\right)$ is increases then the efficiency of the Carnot engine increases.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148540
An electrical refrigerator with $\beta=5$ extracts $5000 \mathrm{~J}$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.
1 $1 \mathrm{~kJ}$
2 $0.5 \mathrm{~kJ}$
3 $0.8 \mathrm{~kJ}$
4 $1.2 \mathrm{~kJ}$
Explanation:
A Given, Coefficient of performance of refrigerator- $\beta=5$ Amount of heat removed $(\mathrm{Q})=5000 \mathrm{~J}$ Electrical energy utilized by the motor $=$ Work done by the motor $(\mathrm{W})$ $\beta =\frac{\mathrm{Q}}{\mathrm{W}}$ $\therefore \quad \mathrm{W} =\frac{\mathrm{Q}}{\beta}=\frac{5000}{5}=1000 \mathrm{~J}$ $=1 \mathrm{~kJ}$
AP EAMCET (22.09.2020) Shift-I
Thermodynamics
148541
The efficiency of a Carnot's engine is $100 \%$ only when
1 ideal gas is used as a working substance
2 temperature of the sink is equal to absolute zero
3 source temperature is equal to the temperature of the sink
4 source temperature is equal to absolute zero
Explanation:
B The efficiency of Carnot's engine is given as $\eta=1-\frac{T_{2}}{T_{1}}$ Where, $T_{1}=$ temperature of source $\mathrm{T}_{2}=$ temperature of sink For efficiency $100 \%, \eta=1$ $1=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0$ $\therefore \quad \mathrm{T}_{2}=0$
AP EAMCET (21.09.2020) Shift-II
Thermodynamics
148542
An ideal heat engine has an efficiency $\eta$. The coefficient of performance of the engine when driven backward will be
1 $1-\left(\frac{1}{\eta}\right)$
2 $\eta-\left(\frac{1}{\eta}\right)$
3 $\left(\frac{1}{\eta}\right)-1$
4 $\frac{1}{1-\eta}$
Explanation:
C Ideal heat engine, efficiency is given as- $\eta=\frac{W}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}$ $\eta=1-\frac{Q_{2}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\eta$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{1}{1-\eta}$ When heat engine operated in backward direction, then coefficient of performance given as- $\alpha=\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}=\frac{1}{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}-1}$ From equation (i) and (ii), $\alpha=\frac{1}{\frac{1}{1-\eta}-1}=\frac{1}{\frac{1}{1-\eta}-1}$ $\frac{1-\eta}{1-1+\eta}=\frac{1-\eta}{\eta}=\left(\frac{1}{\eta}\right)-1$
AP EAMCET (21.09.2020) Shift-I
Thermodynamics
148543
When the temperature difference between the source and sink increases, the efficiency of the heat engine
1 Decreases
2 Increases
3 Is not affected
4 May increases or decrease
Explanation:
B The efficiency ( $\eta$ ) of the Carnot engine is given by- $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}=\text { Temperature of the sink }$ $T_{H}=\text { Temperature of source }$ - When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases. - If the temperature of the source $\left(\mathrm{T}_{\mathrm{H}}\right)$ is increases then the efficiency of the Carnot engine increases.