datavorous/entrance-exam-dataset huggingface dataset:
A Carnot engine, having an efficiency of Î·=1/10 as a heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is

Options:
A 99 J
B 90 J
C 1 J
D 100 J

The correct answer is:
90 J

Solution:
Relation between coefficient of performance and efficiency of refrigerator is
β=1-ηη
∴ β=1-110110=9
Coefficient of performance, β=Heat absorbed(Q2)Work done (W)
⇒9=Q210
Q2=90 J

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine having an efficiency of 110 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at a lower temperature is

Options:
A 99 J
B 100 J
C 1 J
D 90 J

The correct answer is:
90 J

Solution:
For Carnot engine using as refrigerator
W=Q2T1T2-1
It is given η=110⇒η=1-T2T1⇒T2T1=910
So ,Q2=90 J as W=10 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A Carnot engine, having an efficiency of $\eta=1 / 10$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is

Options:
A
$99 \mathrm{~J}$

B
$90 \mathrm{~J}$

C
$1 \mathrm{~J}$

D
$100 \mathrm{~J}$

The correct answer is:
$90 \mathrm{~J}$

Solution:
$\begin{aligned} \mathrm{W} & =\mathrm{Q}_2\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}-1\right) \quad \eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \\ 10 & =\mathrm{Q}_2\left(\frac{10}{9}-1\right) \quad \frac{1}{10}=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{1}{10}=\frac{9}{10} \\ 10 & =\mathrm{Q}_2\left(\frac{1}{9}\right) \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{10}{9} \\ \mathrm{Q}_2 & =90 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine, having an efficiency of $\eta=1 / 10$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is

Options:
A
$99 \mathrm{~J}$

B
$90 \mathrm{~J}$

C
$1 \mathrm{~J}$

D
$100 \mathrm{~J}$

The correct answer is:
$90 \mathrm{~J}$

Solution:
$\begin{aligned} \mathrm{W} & =\mathrm{Q}_2\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}-1\right) \quad \eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \\ 10 & =\mathrm{Q}_2\left(\frac{10}{9}-1\right) \quad \frac{1}{10}=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{1}{10}=\frac{9}{10} \\ 10 & =\mathrm{Q}_2\left(\frac{1}{9}\right) \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{10}{9} \\ \mathrm{Q}_2 & =90 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2007
A Carnot engine is made to work between 200 Â°C and 0 Â°C first and then between 0 Â°C to -200 Â°C. The ratio of efficiencies of the engine in the two cases is

Options:
A 1 : 2
B 1 : 1
C 1.73 : 1
D 1 : 1.73

The correct answer is:
1 : 1.73

Solution:
T1=200 â=200+273=473 K
T2=0 â=0+273=273 K
Î·1=1-T2T1=1-273473=200473
Again, T1â²=0â=0+273K=273 K
T2â²=-200â=-200+273K=73 K
Î·2=1-T2T1â²=1-73273=200273
Î·1Î·2=200473Ã273200=273473=11.732

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine of efficiency 40 %, takes heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60 %. Then, the source temperature for the same sink temperature must be

Options:
A 650 K
B 750 K
C 550 K
D 850 K

The correct answer is:
750 K

Solution:
The formula for the efficiency of Carnot engine Working between temperature limits T1 and T2.T1>T2

η=1-T2T1×100

For η=40 %, T1=500 K

40=1-T2500×100

T2=300 K

For η=60 %,T2=300 K (the source temperature of Carnot
engine for the same sink temp),

⇒ 60=1-300T1×100

⇒ T1=750 K

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
A Carnot engine operates between $227^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. Efficiency of the engine will be

Options:
A $\frac{1}{3}$
B $\frac{2}{5}$
C $\frac{3}{4}$
D $\frac{3}{5}$

The correct answer is:
$\frac{2}{5}$

Solution:
$\eta=1-\frac{T_2}{T_1}=1-\frac{300}{500}=\frac{2}{5}$

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine operates between two reservoirs of temperatures 900K and 300K . The engine performs 1200J of work per cycle. The heat energy (in J ) delivered by the engine to the low temperature reservoir, in a cycle, is _____________

Options:

The correct answer is:
600

Solution:
η=WQh=1-300900=23
Qh=32W=1800J
QL=Qh-W=600J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by 62 $\mathrm{K}$, its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are, respectively :

Options:
A
$372 \mathrm{~K}$ and $330 \mathrm{~K}$

B
$330 \mathrm{~K}$ and $268 \mathrm{~K}$

C
$310 \mathrm{~K}$ and $248 \mathrm{~K}$

D
$372 \mathrm{~K}$ and $310 \mathrm{~K}$

The correct answer is:
$372 \mathrm{~K}$ and $310 \mathrm{~K}$

Solution:
$\eta_1=\frac{T_1-T_2}{T_1}=\frac{1}{6}$
$\eta_2=\frac{T_1-\left(T_2-62\right)}{T_1}=\frac{1}{3}$
$\Rightarrow \frac{T_1-T_2}{T_1}+\frac{62}{T_1}=\frac{1}{3}$
$\frac{1}{6}+\frac{62}{T_1}=\frac{1}{3}$
$\frac{62}{T_1}=\frac{1}{6}$
$\therefore T_1=62 \times 6=372 \mathrm{~K}$
$\frac{T_1-T_2}{T_1}=\frac{1}{6}$
$1-\frac{T_2}{T_1}=\frac{1}{6}$
$\frac{T_2}{372}=\frac{5}{6}$
$\Rightarrow T_2=310 \mathrm{~K}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2011
A Carnot engine operating between two reservoirs has efficiency 13. When the temperature of cold reservoir raised by x, its efficiency decreases to 16. The value of x, if the temperature of hot reservoir is 99° C, will be

Options:
A 16.5 K
B 33 K
C 66 K
D 62 K

The correct answer is:
62 K

Solution:
A carnot engine works between two reservoirs, one hot reservoir and one cold reservoir.

The absolute temperature TH of the hot reservoir can be calculated as follows-

TH=99°C=99+273 K= 372 K

The formula to calculate the efficiency η of the engine is given by

η= 1-TCTH.....................(1)

where, TC is the temperature of the cold reservoir.

Substituting the values of the known parameters in equation (1), we get

13= 1-TCTH⇒TCTH=23⇒TC=23×372 K= 248 K

The new efficiency of the engine η' is given by

η'= 1-TC+XTH.........................(2)

Substituting 16 for η' and the known values of the temperatures into equation (2), we get

16= 1-TC+XTH⇒56=TC+XTH⇒248+X=5×62⇒X=310-248=62 K

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (01 Feb Shift 2)
A Carnot engine takes $3 \times 10^6 \mathrm{cal}$. of heat from a reservoir at $627^{\circ} \mathrm{C}$, and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is

Options:
A
$4.2 \times 10^6 \mathrm{~J}$

B
$8.4 \times 10^6 \mathrm{~J}$

C
$16.8 \times 10^6 \mathrm{~J}$

D
Zero

The correct answer is:
$8.4 \times 10^6 \mathrm{~J}$

Solution:
$\eta=\frac{(627+273)-(273+27)}{627+273}$
$=\frac{900-300}{900}=\frac{600}{900}=\frac{2}{3}$
work $=(\eta) \times$ Heat
$=\frac{2}{3} \times 3 \times 10^6 \times 4.2 \mathrm{~J}=8.4 \times 10^6 \mathrm{~J}$

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine takes $3 \times 10^6 \mathrm{cal}$. of heat from a reservoir at $627^{\circ} \mathrm{C}$, and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is

Options:
A
$4.2 \times 10^6 \mathrm{~J}$

B
$8.4 \times 10^6 \mathrm{~J}$

C
$16.8 \times 10^6 \mathrm{~J}$

D
Zero

The correct answer is:
$8.4 \times 10^6 \mathrm{~J}$

Solution:
$\eta=\frac{(627+273)-(273+27)}{627+273}$
$=\frac{900-300}{900}=\frac{600}{900}=\frac{2}{3}$
work $=(\eta) \times$ Heat
$=\frac{2}{3} \times 3 \times 10^6 \times 4.2 \mathrm{~J}=8.4 \times 10^6 \mathrm{~J}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2003
A Carnot engine takes 5000 kcal of heat from a reservoir at 727°C and gives heat to a sink at 127°C. The work done by the engine is

Options:
A 3×106J
B Zero
C 8.4×106 J
D 12.6×106 J

The correct answer is:
12.6×106 J

Solution:
Given,

QHeat=5000 kcal=5×106 cal

T2Sink=127°C=127+273K=400 K

T1Source=727°C=727+273K=1000 K

The efficiency of the Carnot cycle is given by,

η=1-TsinkTsource=WQ

⇒1-4001000=610=W5000⇒W=3000 kcal

=3××4.2×106 J=12.6×106 J.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A Carnot engine, whose efficiency is $40 \%$, takes in heat from a source maintained at a temperature of $500 \mathrm{~K}$ It is desired to have an engine of efficiency $60 \%$. Then, the intake temperature for the same exhaust (sink) temperature must be

Options:
A
efficiency of Carnot engine cannot be made larger than $50 \%$

B
$1200 \mathrm{~K}$

C
$750 \mathrm{~K}$

D
$600 \mathrm{~K}$

The correct answer is:
$750 \mathrm{~K}$

Solution:
$\frac{40}{100}=\frac{500-T_S}{500}, T_S=300 \mathrm{~K}$
$\frac{600}{100}=\frac{T-300}{T} \Rightarrow T=750 \mathrm{~K}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (Offline)
A Carnot engine whose efficiency is 40% takes in heat from source maintained at a temperature of 500 K. It is desired to have an
engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

Options:
A 1200 K
B 750 K
C 600 K
D 800 K

The correct answer is:
750 K

Solution:
Efficiency of Carnot engine, $\eta_1=40 \%=0.4$
Temperature of source $T_1=500 \mathrm{~K}$
Temperature of $\operatorname{sink} T_2=T$
Desired efficiency, $\eta_{\varepsilon}=60 \%=0.6$
Temperature of $\operatorname{sink} T_2=T$
Temperature of source $T_1=$ ?
For case 1 Efficiency, $\eta_1=1-\frac{T_2}{T_1}=1-\frac{T}{500}$
$$
\begin{aligned}
& \Rightarrow & \frac{4}{10} & =\frac{500-T}{500} \\
\Rightarrow & & 200 & =500-T \\
\Rightarrow & & T & =300 \mathrm{~K}
\end{aligned}
$$
For case 2 Efficiency, $\eta_2=1-\frac{T_2}{T_1}=1-\frac{T}{T_1^{\prime}}$
$$
\begin{aligned}
\frac{6}{10} & =1-\frac{300}{T_1^{\prime}} \\
\Rightarrow \quad \frac{300}{T_1^{\prime}} & =1-\frac{6}{10} \Rightarrow \frac{300}{T_1^{\prime}}=\frac{4}{10} \\
\Rightarrow \quad \frac{300}{4} \times 10 & =T_1^{\prime} \Rightarrow T_1^{\prime}=750 \mathrm{~K}
\end{aligned}
$$

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
A Carnot engine whose heat sinks at 27°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 ?

Options:
A Increases by 18°C
B Increases by 200°C
C Increases by 120°C
D Increases by 73°C

The correct answer is:
Increases by 200°C

Solution:
Efficiency of carnot engine is given by, η=1-TsinkTsource.

For initial case: Tsink=27+273 °C=300 K

So we can write, 14=1-300T1⇒T1=400 K

Now for the second case: Tsink=300 K

After increasing the previous efficiency by 100%, value of efficiency will get doubled.

Therefore, 12=1-300T2 ⇒T2=600 K

Increase in temperature required will be, 600 K-400 K=200 K

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A Carnot engine whose low temperature reservoir is at $7^{\circ} \mathrm{C}$ has an efficiency of $50 \%$. It is desired to increase the efficiency to $70 \%$. By how many degrees should the temperature of the high temperature reservoir be increased

Options:
A $840 \mathrm{~K}$
B $280 \mathrm{~K}$
C $560 \mathrm{~K}$
D $380 \mathrm{~K}$

The correct answer is:
$380 \mathrm{~K}$

Solution:
\(\begin{aligned}
& \text {Initially } \eta=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1} \\
& \Rightarrow 0.5=\frac{\mathrm{T}_1-(273+7)}{\mathrm{T}_1} \\
& \Rightarrow \frac{1}{2}=\frac{\mathrm{T}_1-280}{\mathrm{~T}_1} \Rightarrow \mathrm{T}_1=560 \mathrm{~K}
\end{aligned}\)
Finally \(\eta_1^{\prime}=\frac{\mathrm{T}_1^{\prime}-\mathrm{T}_2}{\mathrm{~T}_1^{\prime}} \Rightarrow 0.7=\frac{\mathrm{T}_1^{\prime}-(273+7)}{\mathrm{T}_1^{\prime}}\)
\(\Rightarrow T_1^{\prime}=933 \mathrm{~K}\)
\(\therefore\) increase in temperature
\(=933-560=373 \mathrm{~K} \approx 380 \mathrm{~K}\)

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine whose low-temperature reservoir is at 350 K has an efficiency of 50% . It is desired to increase the efficiency to 60%. If the temperature of the low-temperature reservoir is kept constant, then by how many kelvins the temperature of the high-temperature reservoir must be increased to get the desired efficiency?

Options:

The correct answer is:
175

Solution:
The efficiency of the Carnot engine,

η=1-TcoldThot×100

When η=50%

then 50100=1-TcoldThot⇒0.5=1-350Thot

or Thot=700 K

When η=60%

then 60100=1-TcoldThot⇒0.6=1-350Thot

or Thot=875 K

Hence, the temperature of the high reservoir is increased by

T′hot-Thot=875 K-700 K=175 K

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine whose low-temperature reservoir is at 350 K has an efficiency of 50% . It is desired to increase this to 60% . If the temperature of the low-temperature reservoir remains constant, then the temperature of the high-temperature reservoir must be increased by how many degrees?

Options:
A 15
B 175
C 100
D 50

The correct answer is:
175

Solution:
Efficiency of Carnot engine,
η=1-TcoldThot×100
When η=50%
then 50100=1-TcoldThot⇒0.5=1-350Thot
or Thot=700 K
When η=60%
then 60100=1-TcoldThot⇒0.6=1-350Thot
or Thot=875 K
Hence, the temperature of high reservoir is increased by
T′hot-Thot=875 K-700 K=175 K

Topic: Physics, Thermodynamics, NEET
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency?

Options:

The correct answer is:
250

Solution:
T2T1=1-Î·=1-40100=35
â´ T1=53T2=53Ã300=500K
Increase in efficiency =50% of 40%=20%
â´ New efficiency Î·â²=40+20=60%
â´ T2Tâ²1=1-Î·â²=1-60100=25
T1â²=52Ã300=750 K
Increase in temperature of source =T1â²-T1
=750-500=250 K

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be:

Options:
A 360 K
B 1000 K
C 900 K
D 300 K

The correct answer is:
1000 K

Solution:
Efficiency of Carnot engine is given by η=1-Tsink Tsource=1-T2T1

Given: Initial efficiency η=12

∴ 12=1-T2600

⇒T2600=12

⇒T2=300 K

When efficiency is increased to 70% and T2=300 K,

Let T' be new temperature of source

⇒710=1-300T'

⇒300T'=1-710

∴ T'=1000 K.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A Carnot engine working between $300 \mathrm{~K}$ and $600 \mathrm{~K}$ has work output of $800 \mathrm{~J}$ per cycle. What is amount of heat energy supplied to the engine from source per cycle

Options:
A $1800 \mathrm{~J} /$ cycle
B $1000 \mathrm{~J} / \mathrm{cycle}$
C $2000 \mathrm{~J} /$ cycle
D $1600 \mathrm{~J} /$ cycle

The correct answer is:
$1600 \mathrm{~J} /$ cycle

Solution:
$\begin{aligned} \eta= & \frac{T_1-T_2}{T_1}-\frac{W}{Q} \Rightarrow Q=\left(\frac{T_1}{T_1-T_2}\right) W \\ & =\frac{600}{(600-300)} \times 800=1600 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
A Carnot freezer takes heat from water at 0oC inside it and rejects it to the room at a temperature of 27oC. The latent heat of ice is 336×103 Jkg-1. If 5kg of water at 0oC is converted into ice at 0oC by the freezer, then the energy consumed by the freezer is close to :

Options:
A 1.51×105 J
B 1.68×106 J
C 1.71×107 J
D 1.67×105 J

The correct answer is:
1.67×105 J

Solution:
Q=mL
=5×336×103=Qsink
QsinkQsource=TsinkTsource
∴ Qsource=TsourceTsinkQsink
∴ Work done =Qsource-Qsink
=QsinkTsourceTsink-1
=5×336×103300273-1
=1.67×105 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2016 (10 Apr Online)
A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot's cycle becomes 1/3. What is the sum of the temperature (in kelvin) of the source and the sink?

Options:

The correct answer is:
682

Solution:
The efficiency of heat engine is given by
η=WQ=1-Q2Q1=1-T2T1
where T1 is temperature of source and T2 is temperature of sink.
Given, η1=16, η2=13
∴ 16=T1-T2T1 ...(i)
and 13=T1-(T2-62)T1 ...(ii)
Solving Eqs. (i) and (ii), we get
T1=372 K
and T2=310 K

T1+T2=310+372=682

Topic: Physics, Thermodynamics, JEE Main
A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot's cycle becomes 1/3. The sum of temperature (in kelvin) of the source and sink will be

Options:

The correct answer is:
682

Solution:
The efficiency of heat engine is given by
Î·=WQ=1-Q2Q1=1-T2T1
where T1 is temperature of source and T2 is temperature of sink.
Given, Î·1=16, Î·2=13
â´ 16=T1-T2T1 ...(i)
and 13=T1-(T2-62)T1 ...(ii)
Solving Eqs. (i) and (ii), we get
T1=372 K
and T2=310 K

Topic: Physics, Thermodynamics, JEE Main
A Carnot's engine operates with source at $127^{\circ} \mathrm{C}$ and sink at $27^{\circ} \mathrm{C}$. If the source supplies $40 \mathrm{~kJ}$ of heat energy, the work done by the engine is

Options:
A $30 \mathrm{~kJ}$
B $10 \mathrm{~kJ}$
C $4 \mathrm{~kJ}$
D $1 \mathrm{~kJ}$

The correct answer is:
$10 \mathrm{~kJ}$

Solution:
Efficiency $\eta=1-\frac{T_{2}}{T_{1}}$
$$
=1-\frac{(273+27)}{(273+127)}=1-\frac{300}{400}=\frac{1}{4}
$$
$$
\begin{aligned}
\eta &=\frac{\text { Work done }}{\text { Heat supplied }} \\
\frac{1}{4} &=\frac{W}{40} \Rightarrow W=10 \mathrm{~kJ}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:

Options:
A 3200 J
B 1800 J
C 1600 J
D 2400 J

The correct answer is:
2400 J

Solution:
η=T2 T1=Q2Q1=Q1-WQ1 ∵W=Q1-Q2

400800=1-WQ1

WQ1=1-12=12

Q1=2 W=2400 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A Carnot's engine works as a refrigerator between $250 \mathrm{~K}$ and $300 \mathrm{~K}$. It receives $500 \mathrm{cal}$ heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

Options:
A
$420 \mathrm{~J}$

B
$2100 \mathrm{~J}$

C
$772 \mathrm{~J}$

D
$2520 \mathrm{~J}$

The correct answer is:
$420 \mathrm{~J}$

Solution:
Given: Temperature of cold body, $\mathrm{T}_2=250$ $\mathrm{K}$ temperature of hot body; $\mathrm{T}_1=300 \mathrm{~K}$ Heat received, $Q_2=500 \mathrm{cal}$ work done, $\mathrm{W}=$ ?
Efficiency $=1-\frac{T_2}{T_1}=\frac{W}{Q_2+W}$
$\Rightarrow 1-\frac{250}{300}=\frac{W}{Q_2+W}$
$\mathrm{~W}=\frac{Q_2}{5}=\frac{500 \times 4.2}{5} J=420 \mathrm{~J}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A Carnot's engine works like a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at a lower temperature. The amount of work done in each cycle to operate the refrigerator is,

Options:
A 772 J
B 420 J
C 2100 J
D 2520 J

The correct answer is:
420 J

Solution:
η=1− T 2 T 1 , efficiency =1-T2T1=WQ2+W⇒1-250300=WQ2+W =420 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2018 (15 Apr)
A carrier signal C(t)=25sin2.512×1010t is amplitude modulated by a message signal m(t)=5sin1.57×108t and transmitted through an antenna. What will be the bandwidth of the modulated signal ?

Options:
A 8GHz
B 2.01GHz
C 1987.5MHz
D 50MHz

The correct answer is:
50MHz

Solution:
Band width =2fm

ωm=1.57×108=2πfm

BW=2fm=1082 Hz=50MHz

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (17 Mar Shift 2)
A carrier wave of amplitude 15 V is modulated by a sinusoidal base band signal of amplitude 3 V. The ratio of maximum amplitude to minimum amplitude in an amplitude modulated wave is

Options:
A 2
B 32
C 1
D 5

The correct answer is:
32

Solution:
The given data is

Ac=15 VAm=3 V

Thus, the maximum amplitude and minimum amplitude is

Amax=Ac+AmAmin=Ac-Am

Hence, the ratio is

AmaxAmin=15+315-3=1812=32

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (10 Apr Shift 1)
A carrier wave of peak voltage 12 V is used to transmit a message signal. The peak voltage of the modulating signal in order to have a modulation index of 75% is

Options:
A 8 V
B 6 V
C 7 V
D 9 V

The correct answer is:
9 V

Solution:
Here Ac=12 V, Am=?

modulation index Î¼c=75%=0.75

Since Î¼c=AmAcâ0.75=Am12

Am=0.75Ã12=9 V

Topic: Physics, Communication System, JEE Main
A carrier wave of peak voltage 14 V is used for transmitting a message signal. The peak voltage of the modulating signal given to achieve a modulation index of 80% will be:

Options:
A 22.4 V
B 7 V
C 11.2 V
D 28 V

The correct answer is:
11.2 V

Solution:
Modulation index, m=AmAc⇒Am=0.8 ×14=11.2 V

Topic: Physics, Communication System, JEE Main, JEE Main 2018 (16 Apr Online)
A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150 V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50: x, then value of x, is _________.

Options:

The correct answer is:
200

Solution:
The value of maximum amplitude, Amax=Ac+Am=250+150 V=400 V
, and the value of minimum amplitude,Amin=Ac-Am=250-150 V=100 V
⇒AminAmax=100 V400 V=14=50200⇒x=200 V

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (01 Sep Shift 2)
A cart loaded with sand having total mass m0=1800 kg moves on a straight horizontal road starting from rest under the action of a force of 120 N. The sand spills through a small hole in the bottom at a rate of K=0.5 kg s-1. What will be the value of the velocity (in m s-1) of the cart after 20 min?

Options:

The correct answer is:
120

Solution:
In this case, sand falls on the ground vertically and comes to rest. Hence, horizontal velocity of ejected sand with respect to ground is zero; so
F⁡ext=ddtmvorF⁡extdt=dmv
Which on integration gives
F⁡ext∫t1t2dt=∫d⁡mv
F⁡extt2-t1=m⁡2v2-m⁡1v1…i
The rate of leakage of sand, K = (–dmdt) is constant which on integration gives
∫m⁡1m⁡2dm⁡=-K⁡∫t1t2dt
(m2 – m1) = –K(t2 – t1) ...(ii)
Here, t1= 0,, t2 = t, v1 = 0, v2 = v
From Eqns. (i) and (ii),

Fext .t = (m0 – Kt)v
v=F⁡ext·tm0-Kt
Substituting for Fext, t, K and m0, we get
v=120×20×601800-0.5×20×60=120 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A catalyst adsorb 100 mL of nitrogen gas at S.T.P. per gram of catalyst surface and forms a mono-molecular layer. The effective surface area occupied by one nitrogen molecule on the surface of catalyst is 0.16Ã10-14 cm2. What is the total surface area occupied by nitrogen molecules per gram of catalyst?
(Given: Volume of gas at STP = 22.4 L)

Options:
A 43.04Ã105 cm2
B 0.18Ã10-15Â cm2
C 42Ã105Â cm2
D 0.19Ã9-15Â cm2

The correct answer is:
43.04Ã105 cm2

Solution:
N=10022400Ã6.023Ã1023=2.69Ã1021
Surface area occupied by one molecule = 0.16Ã10-14 cm2
Total area covered by N2(g) /gram of catalyst =0.16Ã10-14Ã2.69Ã1021=43.04Ã105 cm2

Topic: Chemistry, Surface Chemistry, JEE Main
A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is ρ. An electron (charge e, mass m) is released from rest inside the cavity from point P as shown in the figure. The centre of the sphere and centre of the cavity are separated by a distance a. The time after which the electron again touches the sphere is

Options:
A 62rε0meρa
B 2rε0m6ρa
C 6rε0meρa
D rε0meρa

The correct answer is:
62rε0meρa

Solution:
E = Ïa3Îµ0 F = eE = eÏa3Îµ0
acceleration = Ïea3mÎµ0
&
= 12ÃÏea3mÎµ0Ãt2 = 2r

ât=62rÎµ0meÏa

The electron will move opposite to electronic field with constant acceleration along shown horizontal line of length = 2 r

Topic: Physics, Electrostatics, JEE Main