datavorous/entrance-exam-dataset huggingface dataset:
A cyclic process $A B C D$ is shown below in the given $P-V$ diagram. In the following answers the one that represents the same process as in $P-V$ diagram :

Options:
A

B

C

D

The correct answer is:

Solution:
In $P$ - $V$ diagrams process $A B$ is isobaric process in which pressure remains constant i.e, $P=$ constant at all temperatures.
Process $B C$ is isothermal process in which, temperature remains constant i.e., $T=$ constant.
Process $C D$ is isochoric process in which volume remains constant i.e., $P-T$ diagram $C D$ is a straight line passing through origin.
Process $A D$ is adiabatic process which corresponds to process $A D$ in $P$ - $T$ diagram
Hence, the correct $P-T$ diagram is shown in option (a).

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2006
A cyclic process $A B C D$ is shown below in the given $P-V$ diagram. In the following answers the one that represents the same process as in $P-V$ diagram :

Options:
A

B

C

D

The correct answer is:

Solution:
In $P$ - $V$ diagrams process $A B$ is isobaric process in which pressure remains constant i.e, $P=$ constant at all temperatures.
Process $B C$ is isothermal process in which, temperature remains constant i.e., $T=$ constant.
Process $C D$ is isochoric process in which volume remains constant i.e., $P-T$ diagram $C D$ is a straight line passing through origin.
Process $A D$ is adiabatic process which corresponds to process $A D$ in $P$ - $T$ diagram
Hence, the correct $P-T$ diagram is shown in option (a).

Topic: Physics, Thermodynamics, JEE Main
A cyclic process $A B C D$ is shown below in the given $P-V$ diagram. In the following answers the one that represents the same process as in $P-V$ diagram :

Options:
A

B

C

D

The correct answer is:

Solution:
In $P$ - $V$ diagrams process $A B$ is isobaric process in which pressure remains constant i.e, $P=$ constant at all temperatures.
Process $B C$ is isothermal process in which, temperature remains constant i.e., $T=$ constant.
Process $C D$ is isochoric process in which volume remains constant i.e., $P-T$ diagram $C D$ is a straight line passing through origin.
Process $A D$ is adiabatic process which corresponds to process $A D$ in $P$ - $T$ diagram
Hence, the correct $P-T$ diagram is shown in option (a).

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2006
A cyclic process ABCA is shown in the \( V-T \) diagram. Process on the \( P-V \) diagram is

Options:
A
B
C
D

The correct answer is:

Solution:
ABCA is a cyclic process in which

AB Process is isobaric as in AB Process volume is directly proportional to temperature

BC is isochoric process as volume is constant

CA is isothermal process as temperature is constant in it .

For isothermal process P-V graph is rectangular hyperbola

Topic: Physics, Thermodynamics, JEE Main
A cyclic process for 1 mol of an ideal gas is shown in the V-T diagram. The work done in AB, BC and CA respectively are

Options:
A 0,RT1lnV⁡1V⁡2,R⁡T⁡1-T⁡2
B R⁡, T⁡1-T⁡2R⁡,RT1lnV⁡1V⁡2
C 0, RT 2 ln Vâ¡ 2 Vâ¡ 1 ,â¡ RT 1 Vâ¡ 1 Vâ¡ 1 - Vâ¡ 2
D 0, RT 2 ln V&ApplyFunction; 1 V&ApplyFunction; 2 ,&ApplyFunction; R&ApplyFunction; T&ApplyFunction; 1 - T&ApplyFunction; 2

The correct answer is:
0, RT 2 ln V⁡ 2 V⁡ 1 ,⁡ RT 1 V⁡ 1 V⁡ 1 - V⁡ 2

Solution:
During AB process is isochoric.
∴ ΔV = 0 ∴ W = 0
During BC, the process is isothermal.
∴ ΔT=0
∴W⁡=RT2lnV⁡2V⁡1
During CA, the process is isobaric. So, the pressure is constant.
∴ W = P(V1–V2)
But PV1= RT1
∴ P⁡ = RT 1 V⁡ 1 = RT 2 V⁡ 2 ∴ W⁡ = RT 1 V⁡ 1 V⁡ 1 - V⁡ 2 = RT 2 V⁡ 2 V⁡ 1 - V⁡ 2

Topic: Physics, Thermodynamics, JEE Main
A cyclic process is shown in the figure. Work done by the syatem during the cyclic process ABCDA is

Options:
A 160 J

B 150 J

C 600 J

D 900 J

The correct answer is:
150 J

Solution:
Work done during the cyclic process ABCDA
= area enclosed by ABCDA
= AB + CD 2 Ã AD
= 2 + 1 2 Ã 1 0 2
=1.5Ã102 J
= 150 J

Topic: Physics, Thermodynamics, JEE Main
A cyclic silicate having structural formula Si6O18n- . What is value of n?

Options:

The correct answer is:
12

Solution:
Ring structure of the formula is SiO3n2n-
So, n=12

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
A cyclist is moving with a constant acceleration of 1.2 m s-2 on a straight track. A racer is moving on a circular path of the radius 150 m at a constant speed of 15 m s-1. Find the magnitude of the velocity of racer which is measured by the cyclist has reached a speed of 20 m s-1 for the position represented in the figure.

Options:
A 18.03 m s-1
B 25 m s-1
C 20 m s-1
D 15 m s-1

The correct answer is:
18.03 m s-1

Solution:
Let v→c = actual velocity of cyclist along x-axis
v→r = the velocity of racer along tangent of circular path (shown in figure)
∴ v→rc= v→r- v→c
Hence v→c=20 i^
v→rc=Vrcos⁡30oj^+Vrsin⁡30oi^
Or v→r=15cos⁡30j^+15sin⁡30i^
=1522 j^+152 i^
∴ v→rc= v→r- v→c
=1532 j^+1532 j^
= -12.5 i^+15 32 j^
∴ v→rc= -12.52+15322
=18.03 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A cyclist riding the bicycle at a speed of 143 m s-1 takes a turn around a circular road of radius 203 m without skidding. Given g=9.8 m s-2 What is his inclination to the vertical ?

Options:
A 30°
B 90°
C 45o
D 60°

The correct answer is:
60°

Solution:
θ=tan-1v2rg=tan-1(143)2203×9.8=tan-1[3]=60°

Topic: Physics, Motion In Two Dimensions, JEE Main
A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown Fig. If he maintains constant speed of 10ms–1, what is his acceleration at point R in magnitude

Options:
A 2v2/R
B v2/2R
C v2/R
D v2/4R

The correct answer is:
v2/R

Solution:
Correct Option is : (C)
v2/R

Topic: Chemistry, Motion In Two Dimensions, NEET
A cyclist starts from centre $O$ of a circular park of radius 1 $\mathrm{km}$ and moves along the path $O P R Q O$ as shown in figure. If he maintains constant speed of $10 \mathrm{~ms}^{-1}$, what is his acceleration at point $R$ in magnitude and direction?

Options:

None

Solution:
According to given figure, the cyclist covers the path $O P R Q O$. When an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.
As given that
$$
\begin{aligned}
&R=1 \mathrm{~km}=1000 \mathrm{~m} \\
&v=10 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$
So, acceleration at $R$ is $a=\frac{v^2}{r}$
$$
a=\frac{(10)^2}{1 \mathrm{~km}}=\frac{100}{10^3}=0.1 \mathrm{~m} / \mathrm{s}^2 \text { along } R O
$$

Topic: Physics, Motion In Two Dimensions
A cyclist starts from the point $P$ of a circular ground of radius $2 \mathrm{~km}$ and travels along its circumference to the point $\mathrm{S}$. The displacement of a cyclist is:

Options:
A $\sqrt{8} \mathrm{~km}$
B $8 \mathrm{~km}$
C $6 \mathrm{~km}$
D $4 \mathrm{~km}$

The correct answer is:
$\sqrt{8} \mathrm{~km}$

Solution:
$\therefore$ Displacement $=\mathrm{R} \sqrt{2}=2 \sqrt{2}=\sqrt{8} \mathrm{~km}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A cyclist traversed half the distance of a linear track with a velocity 10 m s-1. The remaining part of the track was covered with a velocity v1 for half the time and a velocity v2 for the other half of the time. If v1+v2=20 m s-1, then the average velocity of the cyclist during the completion of the journey through the track is

Options:
A 30 m s-1
B 20 m s-1
C 10 m s-1
D 15 m s-1

The correct answer is:
10 m s-1

Solution:
Let the complete distance of the track is, 4x
Time of completion of half the distance is given by, t1=2x10
Time of completion of remaining by v1 is, T2
Time of completion of remaining by v2 is, T2
Average velocity for last half distance is, vavg=V1T2+V2T2T
⇒vavg=V1+V22=202=10 m s-1
Magnitude of average velocity for whole journey is given by, vavg.=total displacementtotal time
⇒vavg.=4xt1+T
⇒vavg.=4x2x10+2x10
⇒vavg.=4x2×2x10
⇒vavg.=10 m s-1

Topic: Physics, Motion In Two Dimensions, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be :
[use mp=1.6×10-27 kg, e=1.6×10-19 C]

Options:
A 12
B 18
C 16
D 32

The correct answer is:
18

Solution:
When the proton is rotated in a cyclotron its maximum velocity will be corresponding to the maximum radius of the circular path. We know that the radius of the circle in a magnetic field is given by,

r=mvBq⇒v=Bqrm.

The kinetic energy will be,

K=12mv2=B2q2r22m⇒K=12×1.6×10-192×0.622×1.6×10-27×1.6×10-13 MeV⇒K=18 MeV

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A cyclotron's oscillator frequency is 10 MHz and the radius of its dees is 60 cm, then the kinetic energy of the proton beam produced by the accelerator is [mproton=1.67Ã10-27 kg]

Options:
A 9 MeV
B 10 MeV
C 7 MeV
D 11 MeV

The correct answer is:
7 MeV

Solution:
KE=q2B2R22m
Given, f=10 MHz=107 Hz
R=60 cm=0.6 m
f=qB2πm
∴ KE=2 π2 m f2 R2
=2×3.142×1.67×10-27×1014×0.36 J
≈7 MeV

Topic: Physics, Magnetic Effects of Current, JEE Main
A cylinder A rolls without slipping on a plank $B$. The velocities of center of the cylinder and that of the plank are $4 \mathrm{~m} / \mathrm{s}$ and $2 \mathrm{~m} / \mathrm{s}$ respectively in same direction, with respect to the ground. Find the angular velocity of the cylinder (in rad/s) if its radius is $1 \mathrm{~m}$.

Options:
A $2 \mathrm{rad} / \mathrm{sec}$
B $4 \mathrm{rad} / \mathrm{sec}$
C $6 \mathrm{rad} / \mathrm{sec}$
D $10 \mathrm{rad} / \mathrm{sec}$

The correct answer is:
$2 \mathrm{rad} / \mathrm{sec}$

Solution:
For rolling, $v_A=2 m / s$
or $4-1 . \omega=2$ or $\omega=2 \mathrm{rad} / \mathrm{sec}$ (clockwise)

Topic: Physics, Rotational Motion, JEE Main
A cylinder closed at both ends is separated into two equal parts (45 cm each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 K and a pressure of 1 atm. If now the gas in one of the parts is heated such that the piston shifts by 5 cm, then the temperature and the pressure of the gas in this part after heating is

Options:
A T=365 K and P=1.125 atm
B T=350 K and P=1.125 atm
C T=375 K and P=2.125 atm
D T=375 K and P=1.125 atm

The correct answer is:
T=375 K and P=1.125 atm

Solution:
Let the crossectional area of the cylinder is A . Initially
Representing a closed cylinder
separated by a piston
The piston is shifted by 5 cm in the cylinder
For 1, p1V1T1=p2V2T2 ⇒ pV300=p2V+AxT2 .....(i)
For 2, p1V1=p2V2 ⇒ pV=p2V-Ax
p2=pVV-Ax ....(ii)
From Equations (i) and (ii),
pV300=pVV-Ax⋅V+AxT2
⇒ T2=V+AxV-Ax300=A45+A5A45-A5×300=5040300
T2=375 K
Putting T2 in Equation (i),
P2=PVV+Ax×375300=1×A×45A450+5×375300
=1.125 atm

Topic: Physics, Thermodynamics, JEE Main
A cylinder consists of a gas mixture of helium and oxygen. If the mass of helium is $4 \mathrm{~g}$ and the mass of oxygen is $32 \mathrm{~g}$, then the ratio of specific heat of the mixture $\left(C_p / C_V\right)$

Options:
A 2/3
B 3/2
C 1/2
D 1/3

The correct answer is:
3/2

Solution:
Mass of helium, $m_{\mathrm{He}}=4 \mathrm{~g}$
Mass of oxygen, $m_{\mathrm{O}}=32 \mathrm{~g}$
Helium is a monoatomic gas.
$\therefore$ Degrees of freedom, $f_1=3$
Oxygen is a diatomic gas.
$\therefore$ Degrees of freedom, $f_2=5$
$4 \mathrm{~g}$ helium contains $N_1=N_A=6.023 \times 10^{23}$ number of molecules
$32 \mathrm{~g}$ oxygen contains $N_2=N_A$ number of molecules.
(As atomic mass of oxygen is $16 \mathrm{~g}$ )
$$
\begin{aligned}
\therefore \quad \frac{C_p}{C_V} & =\gamma \\
& =\frac{N_1\left(2+f_1\right)+N_2\left(2+f_2\right)}{N_1 f_1+N_2 f_2} \\
& =\frac{N_A(2+3)+N_A(2+5)}{3 N_A+5 N_A} \\
& =\frac{5+7}{3+5}=\frac{12}{8}=\frac{3}{2}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A cylinder containing an ideal gas is in vertical position and has a piston of mass $M$ that is able to move up or down without friction (figure). If the temperature is increased

Options:
A
both $\mathrm{p}$ and $\mathrm{V}$ of the gas will change

B
only $p$ will increase according to Charles' law

C
$V$ will change but not $p$

D
$p$ will change but not $V$

The correct answer is:
$V$ will change but not $p$

Solution:
Let us consider the given diagram where an ideal gas is contained in a cylinder, having a piston of mass $M$. The pressure on gas does not change.
The pressure inside the gas will be $p=p_a+M g / A$
where, $p_a=$ atmospheric pressure
$A=$ area of cross-section of the piston.
$M g=$ weight of piston
Hence, $p=$ constant.
As the piston and cylinder is frictionless so the equation for ideal gas
$p V=n R T$, volume $(V)$ increases at constant pressure. as $p, R, n$ are constant so,
$V \propto T$
so on increasing temperature of system its volume increased but $p$ will remain constant.

Topic: Physics, Kinetic Theory of Gases
A cylinder of capacity 20 litres is filled with $\mathrm{H}_2$ gas. The total average kinetic energy of translatory motion of its molecules is $1.5 \times 10^5 \mathrm{~J}$. The pressure of hydrogen in the cylinder is

Options:
A $2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
B $3 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
C $4 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
D $5 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

The correct answer is:
$5 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

Solution:
$E=\frac{3}{2} P V \Rightarrow P=\frac{2 E}{3 V}=\frac{2}{3} \times \frac{1.52 \times 10^5}{20 \times 10^{-3}}$ $=\frac{10^5}{20 \times 10^{-2}}=5 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A cylinder of compressed gas that bears no label is supposed to contain either ethane or ethene. Combustion of the sample shows that 16 cm3 of the gas require 48 cm3 of oxygen for complete combustion. This shows that the gas is

Options:
A only ethane
B only ethene
C 1:1 mixtures of two gases
D some unknown mixture of the two gases

The correct answer is:
only ethene

Solution:
CxHy1vol16cm3+x+y4O2x+y4vol16x+y4cm3→xCO2+y2H2O
Now we have 16x+y4=48
or y = 4
Hence, the hydrocarbon is C2H4 .

Topic: Chemistry, Thermodynamics (C), JEE Main
A cylinder of height $20 \mathrm{~m}$ is completely filled with water. The velocity of efflux of water (in $\mathrm{m} / \mathrm{s}$ ) through a small hole on the side wall of the cylinder near its bottom is

Options:
A 10
B 20
C 25.5
D 5

The correct answer is:
20

Solution:
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20}=20 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A cylinder of height $20 \mathrm{~m}$ is completely filled with water. The velocity of efflux of water (in $\mathrm{ms}^{-1}$ ) through a small hole on the side wall of the cylinder near its bottom is

Options:
A
10

B
20

C
25.5

D
5

The correct answer is:
20

Solution:
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20}=20 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2002
A cylinder of mass M=2 kg and radius R=12 cm lies on a plank of the same mass as shown in the figure. The surface between plank and ground is smooth but there is friction between cylinder and plank. If the coefficient of friction between the cylinder and the plank is Î¼=0.4, then what maximum initial compression (in cm) can be given to the spring such that the cylinder moves without slipping with respect to the plank? [Given, k=200 N m-1]

Options:

The correct answer is:
16

Solution:
Using no Slipping condition:
ac+αR=ap
fM+2fRMR2R=kx-fM
⇒4f=kx
xmax=4μMgk

⇒xmax=4×0.4×2×10200=0.16 m=16 cm

Topic: Physics, Oscillations, JEE Main
A cylinder of mass M and radius R lies on a plank of mass M as shown. The surface between plank and ground is smooth, and between cylinder and plank is rough. Assuming no slipping between cylinder and plank, the time period of oscillations (when displaced from equilibrium) of the system is

Options:
A 2πm3k
B 4π2M3K
C 4πM3k
D π3M2K

The correct answer is:
4πM3k

Solution:
Using no Slipping condition:
ac+αR=ap
fM+2fRMR2R=kx-fM
⇒4f=kx
∴kx-kx4=Map⇒ap=3k4Mx
∴ω=3k4M
T=2π4M3k=4πM3k

Topic: Physics, Oscillations, JEE Main
A cylinder of mass Mc and sphere of mass Ms are placed at points A and B of two inclines, respectively. (See figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio sin θ c sin θ s is :

Options:
A 8 7
B 8 7
C 15 14
D 1 5 1 4

The correct answer is:
1 5 1 4

Solution:
a = g sin θ 1 + I/ mR 2
a S = g sin θ s 1 + 2 / 5
a C = g sin θ C 1 + 1 / 2
Now, aS = aC
g sin θ S 7 5 = g sin θ C 3 2
5 7 sin θ S = 2 3 sin θ C
∴ sin θ C sin θ S = 5 × 3 7 × 2 = 1 5 1 4

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2014 (09 Apr Online)
A cylinder of mass m is kept on an inclined plane having an angle of inclination 30Â°. Axis of the cylinder makes an angle 30Â° with the line along the greatest slope. Assuming that cylinder rolls without slipping, the acceleration of the cylinder is

Options:
A g6
B g4
C g3
D g2

The correct answer is:
g6

Solution:
Acceleration of the cylinder is given by
a= mgsinâ¡Î¸sinâ¡Î±m+m2
= 2gsinâ¡Î¸sinâ¡Î±3=g6

Topic: Physics, Rotational Motion, JEE Main
A cylinder of mass M, length L and radius R has two cords wound around it with their ends attached to the ceiling, as shown in the figure. The cylinder is held horizontally with the two cords vertical. When the system is released from rest, the linear acceleration of the cylinder as it falls will be

Options:
A g3
B 2g3
C 3g2
D g2

The correct answer is:
2g3

Solution:
The cylinder rotating under gravity has both translational and rotational motions. Let v be the linear velocity of its centre of mass and ω its angular velocity about the axis of rotation, then in descending a distance h it will lose PE = Mgh while the gain in KE

= 1 2 I ω 2 + 1 2 M v 2 , so
Mgh = 1 2 I ω 2 + 1 2 Mv 2
Here I = 1 2 MR 2 and v = R ω
so Mgh = 1 2 Mv 2 + 1 2 1 2 MR 2 v 2 R 2
sov2=43gh
Now differentiating both sides with respect to time
2vdvdt=43gdhdt
ora=23gasdvdt=aanddhdt=v

Topic: Physics, Rotational Motion, JEE Main
A cylinder of radius r and thermal conductivity K1 is surrounded by a cylindrical shell of linear radius r and outer radius 2r, whose thermal conductivity is K2. There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temperatures T1 and T2. The effective thermal conductivity of the system, in the steady state is

Options:
A K1K2K1+K2
B K1+K2
C K1+3K24
D 3K1+K24

The correct answer is:
K1+3K24

Solution:
∆Q∆tinner+∆Q∆touter=∆Q∆ttotal
K1πr2(T2-T1)l+ K2π2r2-r2(T2-T1)l=Kπ2r2T2-T1l
or K1+3K2πr2T2-T1l = Kπ 4 r2(T2-T1)l
∴ K=K1+3K24

Topic: Physics, Thermal Properties of Matter, JEE Main
A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

Options:
A 2K1+3K25
B K1+K22
C K1+K2
D K1+3K24

The correct answer is:
K1+3K24

Solution:
Let q1 and q2 be the rate of flow of heat through inner part from outer part respecting net flow of heat
q=q1+q2
=K1πR2l.ΔT+K2.π2R2-R2l.ΔT
=πR2lΔTK1+3K2
If the cylinder is replaced by a single material of thermal conductivity K then
q=K.2R2l.ΔT
On comparison,
K=K1+3K24.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity K2. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady-state. the effective thermal conductivity of the system is

Options:
A K1+K2

B K1K2(K1+K2)

C (K1+3K2)4

D (3K1+K2)4

The correct answer is:
(K1+3K2)4

Solution:
Ler R1 and R2 be the thermal resistances of inner and outer portions. Since, temperature difference at both ends is same, the resistances are in parallel. Hence,

Rnet=R1R2R1+R2
or l⁡Kπ2R2=l⁡K1πR2×l⁡K23πR2l⁡K1πR2+l⁡K23πR2
or 14K=13K1K21K1+13K2=13K2+K1
∴ K=3K2+K14

Topic: Physics, Thermal Properties of Matter, NEET
A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity K2. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system isâ

Options:
A K1+K2
B K1+3K24
C K1K2K1+K2
D 3K1+K24

The correct answer is:
K1+3K24

Solution:
Parallel combination of cross - section area
ÏR2 and Ï2R2-R2=3ÏR2
1Req=1R1+1R2 with R=LKA
Keq.4ÏR2L=K1ÏR2L+K23ÏR2L
â Keq=K1+3K24

Topic: Physics, Thermal Properties of Matter, JEE Main
A cylinder weighing 450 N with a radius of 30 cm is held fixed on an incline that is rotating at 0.5 rad s-1. The cylinder is released when the incline is at position Î¸ equal to 30Â°. If the cylinder is 6 m from the bottom A at the instant of release, what is the initial acceleration of the centre of the cylinder relative to the incline, if there is no slipping? g = 10 m s-2

Options:
A 2.33 m s-2

B 4.66 m s-2

C 1.33 m s-2

D 3.33 m s-2

The correct answer is:
2.33 m s-2

Solution:
Free body diagram of the cylinder w.r.t. the inclined plane. The torque about the point of contact O.
mg sin θ - mr ω 2 R⁡ = I⁡ . α = 3 2 mR 2 × a R⁡ (r= radius of circle, R=6 m)
∴mgsinθ-mrω2=32ma
∴a=23gsinθ-r⁡ω2=2310×12-6×14=2.33m s-2

Topic: Physics, Rotational Motion, JEE Main
A cylinder with fixed capacity of 67.2 litre contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is:
[Given that R=8.31 J mol-1 K-1]

Options:
A 748 J
B 700 J
C 350 J
D 374 J

The correct answer is:
748 J

Solution:
The amount of heat supplied is
Q=nCVΔT
=67.222.43R220
=3×32×8.314×20
=748 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A cylindrical adiabatic container of total volume 2 V 0 divided into two equal parts by a conducting piston which is free to move as shown in figure. Each part containing identical gas at pressure P 0 . Initially temperature of left and right part is 4 T 0 and T 0 respectively. An external force is applied on the piston of area ' A ' to keep the piston at rest. The value of external force required when thermal equilibrium is reached is.

Options:
A 8 5 P 0 A
B 2 5 P 0 A
C 5 6 P 0 A
D 6 5 P 0 A

The correct answer is:
6 5 P 0 A

Solution:
If final temp is T

P0V04RT0Cv4T0-T=P0V0RT0CvT-T0

4T0-T4=T-T0

4T0-T=4T-4T0

8T0=5T

T=8T05

Final pressure

In left Pf=TfT0Pi=85T04T0P0=25P0

In right Pf=85T0T0P0=85P0

Force=8P05-2P05A=6P0A5

Topic: Physics, Thermodynamics, JEE Main
A cylindrical adiabatic container of total volume 2V0 is divided into two equal parts by a conducting piston which is free to move as shown in the figure. Each part contains identical gas at pressure P0. Initially the temperatures of the left and right parts are 4T0 and T0 respectively. An external force is applied on the piston of area A to keep the piston at rest. The value of external force required when thermal equilibrium is reached is

Options:
A 85P0A
B 25P0A
C 56P0A
D 65P0A

The correct answer is:
65P0A

Solution:
If final temp is T

P0V04RT0Cv4T0-T=P0V0RT0CvT-T0

4T0-T4=T-T0

4T0-T=4T-4T0

8T0=5T

T=8T05

Final pressure

In left Pf=TfT0Pi=85T04T0P0=25P0

In right Pf=85T0T0P0=85P0

Force=8P05-2P05A=6P0A5

Topic: Physics, Thermodynamics, JEE Main