datavorous/entrance-exam-dataset huggingface dataset:
A car of $800 \mathrm{~kg}$ is taking turn on a banked road of radius $300 \mathrm{~m}$ and angle of banking $30^{\circ}$. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sqrt{3}=1.73\right)$

Options:
A $264 \mathrm{~m} / \mathrm{s}$
B $51.4 \mathrm{~m} / \mathrm{s}$
C $70.4 \mathrm{~m} / \mathrm{s}$
D $102.8 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$51.4 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{aligned} & \mathrm{m}=800 \mathrm{~kg} \\ & \mathrm{r}=300 \mathrm{~m} \\ & \theta=30^{\circ} \\ & \mu_{\mathrm{s}}=0.2 \\ & \mathrm{~V}_{\max }=\sqrt{\operatorname{Rg}\left[\frac{\tan \theta+\mu}{1-\mu \tan \theta}\right]} \\ & =\sqrt{300 \times \mathrm{g} \times\left[\frac{\tan 30^{\circ}+0.2}{1-0.2 \times \tan 30^{\circ}}\right]} \\ & =\sqrt{300 \times 10 \times\left[\frac{0.57+0.2}{1-0.2 \times 0.57}\right]} \\ & \mathrm{V}_{\max }=51.4 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (06 Apr Shift 2)
A car of mass $1000 \mathrm{~kg}$ accelerates uniformly from rest to a velocity of $54 \mathrm{~km} /$ hour in $5 \mathrm{~s}$. The average power of the engine during this period in watts is (neglect friction)

Options:
A 2000 W
B 22500 W
C 5000 W
D 2250 W

The correct answer is:
22500 W

Solution:
Power $=\frac{\text { Work done }}{\text { time }}=\frac{\text { Increasein K.E. }}{\text { time }}$
$P=\frac{\frac{1}{2} m v^2}{t}=\frac{\frac{1}{2} \times 10^3 \times(15)^2}{5}=22500 \mathrm{~W}$

Topic: Physics, Work Power Energy, JEE Main
A car of mass $1000 \mathrm{~kg}$ is moving at a speed of 30 $\mathrm{m} / \mathrm{s}$. Brakes are applied to bring the car to rest. If the net retarding force is $5000 \mathrm{~N}$, the car comes to stop after travelling $d \mathrm{~m}$ in $t \mathrm{~s}$. Then

Options:
A
$d=150, t=5$

B
$d=120, t=8$

C
$d=180, t=6$

D
$d=90, t=6$

The correct answer is:
$d=90, t=6$

Solution:
Given: mass of car $m=1000 \mathrm{~kg}$
$$
\begin{aligned}
& u=30 \mathrm{~m} / \mathrm{s} \\
& v=0 \mathrm{~m} / \mathrm{s} \\
& \text { retarding force } f=5000 \mathrm{~N} \\
& \therefore \text { retardation, }-a=\frac{5000}{1000}=5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
By equation, $v^2-u^2=2$ as
$$
\begin{aligned}
& 0-(30)^2=-2 \times 5 \times d \\
& \therefore d=\frac{900}{10}=90 \mathrm{~m}
\end{aligned}
$$
and $a=\frac{v-u}{t} \therefore t=\frac{v-u}{a}=\frac{0-30}{-5}=6 \mathrm{~s}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (07 May Online)
A car of mass $1200 \mathrm{~kg}$ (together with the driver) is moving with a constant acceleration of $2 \mathrm{~m} / \mathrm{s}^2$. How much power does the engine generate at the instance, when the speed reaches $20 \mathrm{~m} / \mathrm{s}$ ? (Assume that the coefficient of friction between the car and the road is 0.5 ).

Options:
A $48000 \mathrm{~W}$
B $120000 \mathrm{~W}$
C $168000 \mathrm{~W}$
D $288000 \mathrm{~W}$

The correct answer is:
$168000 \mathrm{~W}$

Solution:
Power of engine overcomes friction and provides necessary acceleration to the car.
Now friction, $f=\mu m g=0.5 \times 1200 \times 10=6000 \mathrm{~N}$ and accelerating force, $F=m a=1200 \times 2=2400 \mathrm{~N}$ So, total force produced by engine
$$
=F_T=8400 \mathrm{~N}
$$
We know that, the formula of power is
$$
P=F_T v
$$
Then, $P=8400 \times 20=168000 \mathrm{~W}$

Topic: Physics, Work Power Energy, JEE Main
A car of mass $m$ is moving on a level circular track of radius $R$. If $\mu_s$ represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by

Options:
A $\sqrt{\mu_s m R g}$
B $\sqrt{R g / \mu_s}$
C $\sqrt{m R g / \mu_s}$
D $\sqrt{\mu_s R g}$

The correct answer is:
$\sqrt{\mu_s R g}$

Solution:
In this condition, centripetal force is equal to static frictional force between road and tyres, so
$$
\begin{aligned}
\mu_s m g & =\frac{m v^2}{R} \\
v_{\max } & =\sqrt{\mu_s R g}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main
A car of mass $m$ is moving on a level circular track of radius $R$. If $\mu_s$ represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by

Options:
A $\sqrt{\mu_s m R g}$
B $\sqrt{R g / \mu_s}$
C $\sqrt{m R g / \mu_s}$
D $\sqrt{\mu_s R g}$

The correct answer is:
$\sqrt{\mu_s R g}$

Solution:
In this condition, centripetal force is equal to static frictional force between road and tyres, so
$$
\begin{aligned}
\mu_s m g & =\frac{m v^2}{R} \\
v_{\max } & =\sqrt{\mu_s R g}
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2012 (Mains)
A car of mass $m$ starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude $P_0$. The instantaneous velocity of this car is proportional to

Options:
A $t^2 P_0$
B $t^{1 / 2}$
C $t^{-1 / 2}$
D $t / \sqrt{m}$

The correct answer is:
$t^{1 / 2}$

Solution:
Power $P_0=F v$
$$
\begin{aligned}
& =\left(m \frac{d v}{d t}\right) v \\
& =m v \frac{d v}{d t}
\end{aligned}
$$
Integrating both sides
$$
\begin{aligned}
\int P_0 d t & =\int m v d v \\
P_0 t & =\frac{m v^2}{2} \\
v^2 & =\frac{2 P_0 t}{m} \\
v & \propto t^{1 / 2}
\end{aligned}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2012 (Mains)
A car of weight W is on an inclined road that rises by 100 m over a distance of 1km and applies a constant frictional force W20 on the car. While moving uphill on the road at a speed of 10 ms-1 , the car needs power P. If it needs power P2 while moving downhill at speed υ then value of υ is:

Options:
A 20Â ms-1
B 5Â ms-1
C 15 ms-1
D 10Â ms-1

The correct answer is:
15 ms-1

Solution:
P=wsin⁡θ+w2010
P=w10+w2010=3w2
P2=3w4=w10-w20v
34=v20
15m/s=v

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2016 (09 Apr Online)
A car P travelling at 20 m s-1 sounds its horn at a frequency of 400 Hz. Another car Q is travelling behind the first car in the same direction with a velocity 40 m s-1. The frequency heard by the passenger of the car Q is approximately [Take, velocity of sound =360 m s-1]

Options:
A 421 Hz
B 471 Hz
C 485 Hz
D 514 Hz

The correct answer is:
471 Hz

Solution:
Let velocity of sound in air be v m s-1. Using Doppler formula,

f=f0v+vov-vs=f0v+40v-20
=400360+40340
≈471 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A car runs at a constant speed on a circular track of radius $100 \mathrm{~m}$, taking 62.8 seconds for every circular lap. The average velocity and average speed for each circular lap respectively is:

Options:
A $10 \mathrm{~m} / \mathrm{s}, 0$
B 0,0
C $0,10 \mathrm{~m} / \mathrm{s}$
D $10 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$10 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}$

Solution:
Distance trevelled in one revolution
$\begin{aligned}
& =2 \pi r \\
& \therefore \text { Average speed }=\frac{\text { Totaldistance }}{\text { Total time }} \\
& \Rightarrow \quad \frac{2 \pi r}{t}=\frac{2 \times 3.14 \times 100}{62.8} \\
& =10 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Displacement in one revolution is zero.
$\begin{aligned}
\therefore \text { Average Velocity }= & \frac{\text { Net displacement }}{\text { True }} \\
& =\frac{0}{t}=0
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A car starting from rest accelerates at the rate f through a distance S, then continues at a constant speed for time t and then decelerates at rate f/2 to come to rest. If the total distance travelled is 15 S, then

Options:
A S=ft
B S=16ft2
C S=12 ft2
D S=172 ft2

The correct answer is:
S=172 ft2

Solution:
The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope of OA=f
And slope of BC=f2
v=f t1=f2t2
t2=2t1
In graph area of âOAD gives
Distance, S=12 f t12 â¦.i
Area of rectangle ABED gives distance travelled in time t.
S2=f t1t
Distance travelled in time t2=
S3=12f22t12
Thus, S1+S2+S3=15 S
S+ft1t+ft12=15 S
S+ft1t+2S=15 S S=12 ft12
ft1t=12 S â¦ii
From Eqs. (i) and (ii), we have
12 SS=ft1t12ft1t1
â´t1=t6
From Eq. (i), we get
â´S=12 ft12
â´S=12 ft62=172 ft2

Topic: Physics, Motion In One Dimension, JEE Main
A car starting from rest accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at the rate $f / 2$ to come to rest. If the total distance traversed is $15 \mathrm{~S}$, then

Options:
A
$S=f t$

B
$\mathrm{S}=1 / 6 \mathrm{ft}^2$

C
$\mathrm{S}=1 / 2 \mathrm{ft}^2$

D
None of these

The correct answer is:
None of these

Solution:
$$
\begin{aligned}
& \mathrm{S}=\frac{\mathrm{ft}_1^2}{2} \\
& \mathrm{v}_0=\sqrt{2 \mathrm{Sf}}
\end{aligned}
$$
During retardation
$$
S_2=2 S
$$
During constant velocity
$$
\begin{aligned}
& 15 \mathrm{~S}-3 \mathrm{~S}=12 \mathrm{~S}=\mathrm{v}_0 \mathrm{t} \\
& \Rightarrow \mathrm{S}=\frac{\mathrm{ft}^2}{72}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2005
A car travels a distance of x with speed v1 and then same distance x with speed v2 in the same direction. The average speed of the car is:

Options:
A v1v22v1+v2
B v1+v22
C 2xv1+v2
D 2v1v2v1+v2

The correct answer is:
2v1v2v1+v2

Solution:
Average velocity of car is:

vavg=Total displacementTotal time

As the car is travelling along a straight line without turning, the magnitude of total displacement is same as the distance covered.

The time taken to cover the first half is xv1 and the time taken to cover the second half is xv2

∴ vavg = 2xxv1+xv2

⇒ vavg=2xv1v2v1x+v2x=2v1v2v1+v2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A car travels from $A$ to $B$ at a speed of $20 \mathrm{~km} / \mathrm{hr}$ and returns at a speed of $30 \mathrm{~km} / \mathrm{hr}$. The average speed of the car for the whole journey is

Options:
A 25 km/hr
B 24 km/hr
C 50 km/hr
D 5 km/hr

The correct answer is:
24 km/hr

Solution:
Distance average speed $=\frac{2 v_1 v_2}{v_1+v_2}=\frac{2 \times 20 \times 30}{20+30}$ $=\frac{120}{5}=24 \mathrm{~km} / \mathrm{hr}$

Topic: Physics, Motion In One Dimension, JEE Main
A car tyre is filled with nitrogen gas at 35 psi at 27 °C. It will burst if pressure exceeds 40 psi. The temperature in C∘ at which the car tyre will burst is (Rounded-off to the nearest integer)

Options:

The correct answer is:
70

Solution:
P∝T

P2P1=T2 T1⇒4035=T2300

T2=342.857 K

=69.70 °C≃70 °C

Hence answer is 70

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A carbohydrate consists of

Options:
A $C$ and $O$
B $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$
C $C, H, N$ and $O$
D $\mathrm{C}$ and $\mathrm{H}$

The correct answer is:
$\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$

Solution:
$C, H$ and $O$

Topic: Chemistry, Biomolecules, JEE Main
A carbohydrate that cannot be hydrolysed to simpler forms is called

Options:
A Disaccharide
B Monosaccharide
C Polysaccharide
D Trisaccharide

The correct answer is:
Monosaccharide

Solution:
Monosaccharide cannot be hydrolysed to simple forms.

Topic: Chemistry, Biomolecules, JEE Main
A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula of the compound. (Atomic wts: H = 1.008, C = 12.0, Br = 79.9)

Options:
A CH3Br
B CH2Br2I
C C2H4Br2
D C2H3Br3

The correct answer is:
C2H4Br2

Solution:
Element	%	At.wt	Relative number	Simple ratio

C	12.8	12	12.812=1.06	1061.06=1

H	2.1	1	2.11=2.1	2.11.06=2

Br	85.1	80	85.180=1.06	1.061.06=1

E.F=CH2Br
Wt. of EF=12+2+80=94
n=M.Wwt of EF=187.994=2
MF=EFn=CH2Br2
=C2H4Br2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A carbon resistance has a following colour code. What is the value of the resistance?

Options:
A 6.4 MΩ±5%
B 64 kΩ±10%
C 530 kΩ±5%
D 5.3 MΩ±5%

The correct answer is:
530 kΩ±5%

Solution:
Colour band A and B indicate the first two significant figures of resistance in "ohm" , while the C band gives the decimal multiplier i.e., the number of zeros that follows the two significant figures A and B.

The last band D indicates the tolerance in percentage about the indicate value or in another word, it represents the percentage accuracy of the indicated value.

According to given color-coding, the values of the color of rings are

Green	5

Orange	3

Yellow	4

Gold	5%

Value of resistance =53×104Ω
=530 kΩ

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A carbon resistor of power \( 2 \mathrm{~W} \) has resistance according to color codes Green, Black, Red and Silver. Then maximum current through resistor is

Options:
A \( 1 \mathrm{~A} \)
B \( 10 \mathrm{~mA} \)
C \( 20 \mathrm{~mA} \)
D \( 40 \mathrm{~mA} \)

The correct answer is:
\( 20 \mathrm{~mA} \)

Solution:
Colour code of given colours is as follows

Green →5, Black →0, Red →102, Silver →10% tolerance
∴ R=50×102 Ω
Electrical power, P=I2R⇒I2=PR
where,I-electric current,R-electric resistance
I2=250×102=4×10-4 A
I=2×10-2A=20 mA

Topic: Physics, Current Electricity, JEE Main
A carbon resistor with colour code is shown in the figure. There is no fourth band in the resistor. The value of the resistance is

Options:
A $24 \mathrm{M} \Omega \pm 20 \%$
B $14 \mathrm{k} \Omega \pm 5 \%$
C $24 \mathrm{k} \Omega \pm 20 \%$
D $34 \mathrm{k} \Omega \pm 10 \%$

The correct answer is:
$24 \mathrm{k} \Omega \pm 20 \%$

Solution:
RYO
By colour coding of resistors.
\(R=24 \times 10^3 \Omega=24 \mathrm{k} \Omega\)

Topic: Physics, Current Electricity, JEE Main
A carbonyl compound of formula C9H10OA , which is a benzene derivative gives orange precipitate with 2,4 ‐D.N.P. and also gives yellow precipitate with I2 in presence of aqueous NaOH. The total no. of isomers possible for ‘A’ are_____

Options:

The correct answer is:
3

Solution:
Since A is reacting with the 2,4-DNP and showing iodoform reaction, it must be an aromatic methyl ketone.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A carbonyl compound reacts with hydrogen cyanide to form cyanohydrine which on hydrolysis forms a racemic mixture of $\alpha$-hydroxy acid. The carbonyl compound is:

Options:
A formaldehyde
B acetaldehyde
C acetone
D diethyl ketone

The correct answer is:
acetaldehyde

Solution:
Caution
In racemic mixture, $\alpha$-C-atoms must have asymmetric carbon atoms.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A card from a pack of $\mathbf{5} 2$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond?

Options:

None

Solution:
$\mathrm{E}_1=$ Event that lost card is diamond,
$\mathrm{E}_2=$ Event that lost card is not diamond.
Thereare 13 diamond cards, out of a pack or 52 cards
$$
\mathrm{P}\left(\mathrm{E}_1\right)=\frac{{ }^{13} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1}=\frac{13}{52}=\frac{1}{4}
$$
There are 39 cards which are not diamond,
$$
\mathrm{P}\left(\mathrm{E}_2\right)=\frac{39}{52}=\frac{3}{4}
$$
(i) When one diamond card is lost, 12 diamond cards are left and in total 51 cards are left. Out of 12 cards 2 may be drawn in ${ }^{12} \mathrm{C}_2$ way.
$\therefore$ Probability of getting 2 diamond cards when one diamond card is lost
$$
\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\frac{{ }^{12} \mathrm{C}_2}{{ }^{51} \mathrm{C}_2}=\frac{12 \times 11}{51 \times 50}
$$
Where A denotes the lost card
When diamond card is not lost, there are 13 diamond cards. The probability of drawing 2 diamond cards
$$
=\frac{{ }^{13} \mathrm{C}_2}{{ }^{51} \mathrm{C}_2}=\frac{13 \times 12}{51 \times 50}
$$
Probability that the lost card is diamond
$$
\begin{aligned}
&=P\left(E_1 / A\right)=\frac{P\left(E_1\right) P\left(A / E_1\right)}{\left(P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)\right.} \\
&=\frac{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50}}{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50}+\frac{3}{4} \times \frac{13}{51} \times \frac{12}{50}}=\frac{11}{50}
\end{aligned}
$$

Topic: Mathematics, Probability
A card is drawn from a pack of 52 cards and a gambler bets that it is a spade or an ace. Which one of the following are the odds against his winning this bet?

Options:
A 13 to 4
B 4 to 13
C 9 to 4
D 4 to 9

The correct answer is:
9 to 4

Solution:
Probability of getting a spade $=\frac{13}{52}$
Probability of an ace $=\frac{4}{52}$
and probability of getting a spade ace $=\frac{1}{52}$
$\therefore \quad$ Required probability $=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$
Odds against his winning $=\frac{1-\frac{4}{13}}{\frac{4}{13}}=\frac{\frac{9}{13}}{\frac{4}{13}}=\frac{9}{4}$

Topic: Mathematics, Probability, JEE Main
A card is lost from a pack of 52 playing cards. From the remainder of the pack, one card is drawn and is found to be spade. The probability that the missing card is a spade is

Options:
A 5 1 7
B 4 1 7
C 3 1 7
D 2 1 7

The correct answer is:
4 1 7

Solution:
Let A and B are the events that card lost is spade and card drawn is a spade.
PA=14, PA¯=34
P B / A = 1 2 5 1 , P B / A ¯ = 1 3 5 1
P A / B = P A · P B / A P A · P B / A + P A ¯ P B / A ¯
= 1 4 · 1 2 5 1 1 4 · 1 2 5 1 + 3 4 · 1 3 5 1 = 1 2 5 1 = 4 1 7

Topic: Mathematics, Probability, JEE Main
A Carnot engine absorbs $6 \times 10^5 \mathrm{cal}$ at $227^{\circ} \mathrm{C}$. The work done per cycle by the engine, if its sink is maintained at $127^{\circ} \mathrm{C}$ is

Options:
A $15 \times 10^8 \mathrm{~J}$
B $15 \times 10^4 \mathrm{~J}$
C $5 \times 10^5 \mathrm{~J}$
D $2 \times 10^4 \mathrm{~J}$

The correct answer is:
$5 \times 10^5 \mathrm{~J}$

Solution:
Here, $\mathrm{Q}_1=6 \times 10^5 \mathrm{cal}$
$\begin{aligned} & \mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K} \\ & \mathrm{~T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\end{aligned}$
Work done/cycle, $\mathrm{W}=$ ?
As $\frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1}$ or
$\mathrm{Q}_2=\frac{\mathrm{T}_2}{\mathrm{~T}_1} \times \mathrm{Q}_1=\frac{400}{500} \times 6 \times 10^5$
$=4.8 \times 10^5 \mathrm{cal}$
$\begin{aligned} \text { As, } & \mathrm{W}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \\ \mathrm{~W} & =1.2 \times 10^5 \mathrm{cal}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \\ \mathrm{~W} & =5.04 \times 10^5 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JIPMER, JIPMER 2018
A Carnot engine absorbs $1000 \mathrm{~J}$ of heat energy from a reservoir at $127^{\circ} \mathrm{C}$ and rejects $600 \mathrm{~J}$ of heat energy during each cycle. The efficiency of engine and temperature of sink will be:

Options:
A
$20 \%$ and $-43^{\circ} \mathrm{C}$

B
$40 \%$ and $-33^{\circ} \mathrm{C}$

C
$50 \%$ and $-20^{\circ} \mathrm{C}$

D
$70 \%$ and $-10^{\circ} \mathrm{C}$

The correct answer is:
$40 \%$ and $-33^{\circ} \mathrm{C}$

Solution:
Given: $\mathrm{Q}_1=1000 \mathrm{~J}$
$$
\begin{aligned}
&\mathrm{Q}_2=600 \mathrm{~J} \\
&\mathrm{~T}_1=127^{\circ} \mathrm{C}=400 \mathrm{~K} \\
&\mathrm{~T}_2=? \\
&\eta=?
\end{aligned}
$$
Efficiency of carnot engine,
$$
\begin{aligned}
&\eta=\frac{W}{Q_1} \times 100 \% \\
&\text { or, } \eta=\frac{Q_2-Q_1}{Q_1} \times 100 \%
\end{aligned}
$$
or, $\eta=\frac{\mathrm{Q}_2-\mathrm{Q}_1}{\mathrm{Q}_1} \times 100 \%$
or, $\eta=\frac{1000-600}{1000} \times 100 \%$
$$
\eta=40 \%
$$
Now, for carnot cycle $\frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1}$
$$
\begin{aligned}
&\frac{600}{1000}=\frac{\mathrm{T}_2}{400} \\
&\mathrm{~T}_2=\frac{600 \times 400}{1000}=240 \mathrm{~K}=240-273 \\
&\therefore \mathrm{T}_2=-33^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2014 (12 Apr Online)
A Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T / 3$. The amount of heat rejected is

Options:
A $Q / 4$
B $Q / 3$
C $Q / 2$
D $2 Q / 3$

The correct answer is:
$Q / 3$

Solution:
$\boxtimes \eta=1-\frac{T_2}{T_1}=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}$
where $Q_1=$ heat absorbed, $Q_2=$ heat rejected
$\begin{aligned} \Rightarrow & 1-\frac{T / 3}{T}=\frac{W}{Q_1} \Rightarrow \frac{2}{3}=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1} \\ \Rightarrow & \frac{2}{3}=1-\frac{Q_2}{Q_1} \Rightarrow \frac{Q_2}{Q_1}=\frac{1}{3} \Rightarrow Q_2=\frac{Q_1}{3}=\frac{Q}{3}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine absorbs heat from a reservoir maintained at temperature $1000 \mathrm{~K}$. The engine rejects heat to a reservoir whose temperature is $T$. If the magnitude of a absorbed heat is $400 \mathrm{~J}$ and work performed is $300 \mathrm{~J}$, then the value of $T$ is

Options:
A $250 \mathrm{~K}$
B $500 \mathrm{~K}$
C $750 \mathrm{~K}$
D $1750 \mathrm{~K}$

The correct answer is:
$250 \mathrm{~K}$

Solution:
Thermal efficiency of Carnot's engine is
$$
\eta=\frac{W}{Q_1}=1-\frac{T_2}{T_1}
$$
Here, $W=300 \mathrm{~J}, Q_1=400 \mathrm{~J}, T_1=1000 \mathrm{~K}, T_2=T$
$$
\begin{array}{ll}
\Rightarrow & \frac{300}{400}=1-\frac{T}{1000} \\
\Rightarrow & \frac{T}{1000}=1-\frac{3}{4}=\frac{1}{4} \Rightarrow T=\frac{1000}{4}=250 \mathrm{~K}
\end{array}
$$

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
A Carnot engine efficiency is equal to 17. If the temperature of the sink is reduced by 65 K, the efficiency becomes 14. The temperature of the source and the sink in the first case are respectively

Options:
A 620 K, 520 K
B 520 K, 606.67 K
C 606.67 K, 520 K
D 520 K, 610 K

The correct answer is:
606.67 K, 520 K

Solution:
âµ1-T2T1=17 (in case 1)
â´ T2T1=67 ... (i)
1-T2-65T1=14 (in case 2)
â´ T2-65T1=34 ...(ii)
From Eqs. (i) and (ii),
T2T1T2-65T1=6734=87
âT2T2-65=87
â7T2=8T2-8Ã65
âT2=520 K
â´T2T1=67
â´T1=T2Ã76=520Ã76
=606.67ââK

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine has an efficiency of 16. When the temperature of the sink is reduced by 62 ℃ , its efficiency is doubled. The temperatures of the source and the sink are, respectively,

Options:
A 124 ℃, 62 ℃
B 37 ℃, 99 ℃
C 99 ℃, 37 ℃
D 62 ℃, 124 ℃

The correct answer is:
37 ℃, 99 ℃

Solution:
η=1-TsinkTsource
16=1-T1T2 ….(1)
13=1-T1-62T2 ….(2)
From equation (1) and (2)
T2=372 K or 99 ℃
And T1=310 K or 37 ℃

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A Carnot engine has efficiency 15. Efficiency becomes 13 when the temperature of sink is decreased by 50 K. Temperature of the sink ( in K ) is

Options:

The correct answer is:
300

Solution:
The efficiency of the Carnot engine
η=Work outputHeat input=WQH
or η=QH-QLQH=1-QLQH
Also we can show that
QLQH=TLTH
∴ η=1-TLTH
where TL is temperature of sink and TH is the temperature of hot reservoir.
According to question
15=1-TLTH ...(i)
13=1-TL-50TH ...(ii)
From Eq. (i)
TLTH=45
⇒TH=54TL
Substituting value of TH in Eq. (ii), we get
13=1-TL-5054TL
or 4(TL-50)5TL=23
or TL-50=23×54TL
or TL-56TL=50
∴TL=50×6=300 K

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine has efficiency 15. Efficiency becomes 13 when the temperature of sink is decreased by 50 K. What is the temperature of the sink (in K)?

Options:

The correct answer is:
300

Solution:
The efficiency of Carnot engine
η=Work outputHeat input=WQH
or η=QH-QLQH=1-QLQH
Also we can show that
QLQH=TLTH
∴ η=1-TLTH
where TL is temperature of sink and TH is temperature of hot reservoir.
According to question
15=1-TLTH ...(i)
and 13=1-TL-50TH ...(ii)
From Eq. (i)
TLTH=45
⇒ TH=54TL
Substituting value of TH in Eq. (ii), we get
13=1-TL-5054TL
or 4(TL-50)5TL=23
or TL-50=23×54TL
or TL-56TL=50
∴ TL=50×6=300 K

Topic: Physics, Thermodynamics, JEE Main
A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be :

Options:
A 166.7 K
B 255.1 K
C 266.7 K
D 367.7 K

The correct answer is:
266.7 K

Solution:
Initial efficiency η0=50%, the increase in efficiency is 30%. Therefore, new efficiency η=501+30100=65%.

Efficiency of the Carnot engin is given by η=1-TLTH

⇒η0=12=1-TLTH

When the temperature of the sink is decreased by 40°C=40 K.

η=121·3=1-TL-40 TH

⇒121·3=12+40TH

⇒TH=266.7 K

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (28 Jul Shift 1)