Heat and Thermodynamics
811 Questions
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
An engine operates by taking a monoatomic ideal gas through the cycle shown in the figure. The
percentage efficiency of the engine is close to __________.
Correct Answer: 19
Explanation:
% efficiency of carnot engine $\eta $ = ${W \over {{Q_{in}}}} \times 100$
Work = Area of ABCD = (2P0)(V0)
Qin = QAB + QBC
QAB = isochoric process
= nCV(TB - TA)
= 1 $ \times $ ${3 \over 2}R\left( {{T_B} - {T_A}} \right)$
= ${3 \over 2}R\left( {{{3{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over R}} \right)$
= 3P0V0
QBC = isobaric process
= nCP$\Delta $T
= $1 \times {5 \over 2}R$(TC - TB)
= ${5 \over 2}R\left( {{{6{P_0}{V_0}} \over R} - {{3{P_0}{V_0}} \over R}} \right)$
= 7.5 P0V0
$ \therefore $ $\eta $ = ${{{2{P_0}{V_0}} \over {3{P_0}{V_0} + 7.5{P_0}{V_0}}}}$ $ \times $ 100
$ \simeq $ 19%
Work = Area of ABCD = (2P0)(V0)
Qin = QAB + QBC
QAB = isochoric process
= nCV(TB - TA)
= 1 $ \times $ ${3 \over 2}R\left( {{T_B} - {T_A}} \right)$
= ${3 \over 2}R\left( {{{3{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over R}} \right)$
= 3P0V0
QBC = isobaric process
= nCP$\Delta $T
= $1 \times {5 \over 2}R$(TC - TB)
= ${5 \over 2}R\left( {{{6{P_0}{V_0}} \over R} - {{3{P_0}{V_0}} \over R}} \right)$
= 7.5 P0V0
$ \therefore $ $\eta $ = ${{{2{P_0}{V_0}} \over {3{P_0}{V_0} + 7.5{P_0}{V_0}}}}$ $ \times $ 100
$ \simeq $ 19%
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Morning Slot
Initially a gas of diatomic molecules is contained in a cylinder of volume V1
at a pressure P1
and
temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in
number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a
volume 2V1
is given by P2
. The ratio ${{{P_2}} \over {{P_1}}}$
is ________.
Correct Answer: 5
Explanation:
We know, PV = nRT
$ \therefore $ P1V1 = nR (250)
and P2(2V1) = ${{5n} \over 4}R \times \left( {2000} \right)$
By Dividing
${{{P_1}} \over {2{P_2}}}$ = ${{4 \times 250} \over {5 \times 2000}}$
$ \Rightarrow $ ${{{P_1}} \over {{P_2}}} = {1 \over 5}$
$ \Rightarrow $ ${{{P_2}} \over {{P_1}}} = 5$
$ \therefore $ P1V1 = nR (250)
and P2(2V1) = ${{5n} \over 4}R \times \left( {2000} \right)$
By Dividing
${{{P_1}} \over {2{P_2}}}$ = ${{4 \times 250} \over {5 \times 2000}}$
$ \Rightarrow $ ${{{P_1}} \over {{P_2}}} = {1 \over 5}$
$ \Rightarrow $ ${{{P_2}} \over {{P_1}}} = 5$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Evening Slot
Nitrogen gas is at 300oC temperature. The
temperature (in K) at which the rms speed of a
H2 molecule would be equal to the rms speed
of a nitrogen molecule, is _______.
(Molar mass of N2 gas 28 g).
(Molar mass of N2 gas 28 g).
Correct Answer: 40TO41
Explanation:
Vrms = $\sqrt {{{3RT} \over M}} $
VN2 = $\sqrt {{{3R(573)} \over 28}} $
VH2 = $\sqrt {{{3RT} \over 2}} $
Given, VN2 = VH2
$\sqrt {{{3RT} \over 2}} $ = $\sqrt {{{3R(573)} \over 28}} $
$ \Rightarrow $ ${T \over 2} = {{573} \over {28}}$
$ \Rightarrow $ T = 41 K
VN2 = $\sqrt {{{3R(573)} \over 28}} $
VH2 = $\sqrt {{{3RT} \over 2}} $
Given, VN2 = VH2
$\sqrt {{{3RT} \over 2}} $ = $\sqrt {{{3R(573)} \over 28}} $
$ \Rightarrow $ ${T \over 2} = {{573} \over {28}}$
$ \Rightarrow $ T = 41 K
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta $P is
applied at a constant temperature, is the same as the change when the temperature is reduced by
a small quantity $\Delta $T at constant pressure. The initial temperature and pressure of the gas were 300
K and 2 atm. respectively.
If |$\Delta $T| = C|$\Delta $P| then value of C in (K/atm.) is _________.
If |$\Delta $T| = C|$\Delta $P| then value of C in (K/atm.) is _________.
Correct Answer: 150
Explanation:
We know, $PV = nRT$
$ \therefore $ $P\Delta V + V\Delta P = 0$ (for constant temp.)
and $P\Delta V$ = $nR\Delta T$ (for constant pressure)
$\Delta T = {{P\Delta V} \over {nR}}$
$\Delta P = - {{P\Delta V} \over V}$ ($\Delta V$ is same in both cases)
${{\Delta T} \over {\Delta P}} = {{P\Delta V} \over {nR}}{V \over { - P\Delta V}} = {{ - V} \over {nR}} = - {T \over P}$
[As PV = nRT
$ \Rightarrow $ $ {{V \over {nR}} = {T \over P}} $]
$ \therefore $ $\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$
$ \therefore $ $P\Delta V + V\Delta P = 0$ (for constant temp.)
and $P\Delta V$ = $nR\Delta T$ (for constant pressure)
$\Delta T = {{P\Delta V} \over {nR}}$
$\Delta P = - {{P\Delta V} \over V}$ ($\Delta V$ is same in both cases)
${{\Delta T} \over {\Delta P}} = {{P\Delta V} \over {nR}}{V \over { - P\Delta V}} = {{ - V} \over {nR}} = - {T \over P}$
[As PV = nRT
$ \Rightarrow $ $ {{V \over {nR}} = {T \over P}} $]
$ \therefore $ $\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas
at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close
to _______.
Correct Answer: 266
Explanation:
As work done on gas and heat supplied to the gas are zero,
$ \therefore $ Total internal energy of gases remain same.
u1 + u2 = u1' + u2'
We know, $\Delta $U = nCv$\Delta $T
$ \therefore $ $\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$ = $\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$
$ \Rightarrow $ (20 + 20) = 0.15 Tf
$ \Rightarrow $ Tf = 266.67
$ \therefore $ Total internal energy of gases remain same.
u1 + u2 = u1' + u2'
We know, $\Delta $U = nCv$\Delta $T
$ \therefore $ $\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$ = $\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$
$ \Rightarrow $ (20 + 20) = 0.15 Tf
$ \Rightarrow $ Tf = 266.67
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
If minimum possible work is done by a refrigerator in converting 100 grams of water at 0oC to ice,
how much heat (in calories) is released to the surroundings at temperature 27oC (Latent heat of
ice = 80 Cal/gram) to the nearest integer ?
Correct Answer: 8791
Explanation:
Q1 = mL = 8000cal
Q2 = W + Q1
Coefficient of performance (C.O.P)
$ = {{{Q_1}} \over W} = {{{Q_1}} \over {{Q_2} - {Q_1}}} = {{{T_1}} \over {{T_2} - {T_1}}}$
${{{Q_1}} \over W} = {{273} \over {300 - 273}}$
${{{Q_1}} \over W} = {{273} \over {27}}$
$W = {{27} \over {273}}{Q_1}$
$W = {{27} \over {273}}mL$
$W = {{27} \over {273}} \times 80 \times 100$
${Q_2} = {{27} \over {273}} \times 80 \times 100 + 80 \times 100$
$ = 8791.2$ cal
Q2 = W + Q1
Coefficient of performance (C.O.P)
$ = {{{Q_1}} \over W} = {{{Q_1}} \over {{Q_2} - {Q_1}}} = {{{T_1}} \over {{T_2} - {T_1}}}$
${{{Q_1}} \over W} = {{273} \over {300 - 273}}$
${{{Q_1}} \over W} = {{273} \over {27}}$
$W = {{27} \over {273}}{Q_1}$
$W = {{27} \over {273}}mL$
$W = {{27} \over {273}} \times 80 \times 100$
${Q_2} = {{27} \over {273}} \times 80 \times 100 + 80 \times 100$
$ = 8791.2$ cal
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
A bakelite beaker has volume capacity of 500 cc at 30oC. When it is partially filled with Vm
volume
(at 30oC) of mercury, it is found that the unfilled volume of the beaker remains constant as
temperature is varied. If $\gamma $(beaker) = 6 × 10–6 oC–1 and $\gamma $(mercury) = 1.5 × 10–4 oC–1, where $\gamma $ is the
coefficient of volume expansion, then Vm
(in cc) is close to ____.
Correct Answer: 20
Explanation:
$\Delta $V = V$\gamma $$\Delta $T
$ \therefore $ V1$\gamma $1 = V2$\gamma $2
$ \Rightarrow $ 500 $ \times $ 6 $ \times $ 10-6 = Vm $ \times $ 1.5 $ \times $ 10-4
$ \Rightarrow $ Vm = ${{500 \times 6 \times {{10}^{ - 6}}} \over {1.5 \times {{10}^{ - 4}}}}$
$ \Rightarrow $ Vm = 20 cc
$ \therefore $ V1$\gamma $1 = V2$\gamma $2
$ \Rightarrow $ 500 $ \times $ 6 $ \times $ 10-6 = Vm $ \times $ 1.5 $ \times $ 10-4
$ \Rightarrow $ Vm = ${{500 \times 6 \times {{10}^{ - 6}}} \over {1.5 \times {{10}^{ - 4}}}}$
$ \Rightarrow $ Vm = 20 cc
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Morning Slot
An engine takes in 5 moles of air at 20oC and
1 atm, and compresses it adiabatically to
1/10th of the original volume. Assuming air to
be a diatomic ideal gas made up of rigid
molecules, the change in its internal energy
during this process comes out to be X kJ. The
value of X to the nearest integer is________.
Correct Answer: 46
Explanation:
For diatomic ideal gas :
f = 5
$\gamma $ = ${7 \over 5}$
Ti = T = 273 + 20 = 293 K
Vi = V
Vf = ${V \over {10}}$
For adiabatic process TV$\gamma $ - 1 = constant
${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$
$ \Rightarrow $ $\left( {293} \right){V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {10}}} \right)^{{7 \over 5} - 1}}$
$ \Rightarrow $ ${T_2} = 293 \times {\left( {10} \right)^{{2 \over 5}}}$
$\Delta $U = ${{nfR\left( {{T_2} - {T_1}} \right)} \over 2}$
= ${{5 \times 5 \times {{25} \over 3} \times \left( {{{293.10}^{{2 \over 5}}} - 293} \right)} \over 2}$
= ${{625 \times 293 \times \left( {{{10}^{{2 \over 5}}} - 1} \right)} \over 6}$
= 46.14 $ \times $ 103 J
$ \simeq $ 46 kJ
f = 5
$\gamma $ = ${7 \over 5}$
Ti = T = 273 + 20 = 293 K
Vi = V
Vf = ${V \over {10}}$
For adiabatic process TV$\gamma $ - 1 = constant
${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$
$ \Rightarrow $ $\left( {293} \right){V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {10}}} \right)^{{7 \over 5} - 1}}$
$ \Rightarrow $ ${T_2} = 293 \times {\left( {10} \right)^{{2 \over 5}}}$
$\Delta $U = ${{nfR\left( {{T_2} - {T_1}} \right)} \over 2}$
= ${{5 \times 5 \times {{25} \over 3} \times \left( {{{293.10}^{{2 \over 5}}} - 293} \right)} \over 2}$
= ${{625 \times 293 \times \left( {{{10}^{{2 \over 5}}} - 1} \right)} \over 6}$
= 46.14 $ \times $ 103 J
$ \simeq $ 46 kJ
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
Starting at temperature 300 K, one mole of an
ideal diatomic gas ($\gamma $ = 1.4) is first compressed
adiabatically from volume V1 to V2 = ${{{V_1}} \over {16}}$. It is
then allowed to expand isobarically to volume 2V2. If all the processes are the quasi-static then
the final temperature of the gas (in oK) is (to the nearest integer) _____.
ideal diatomic gas ($\gamma $ = 1.4) is first compressed
adiabatically from volume V1 to V2 = ${{{V_1}} \over {16}}$. It is
then allowed to expand isobarically to volume 2V2. If all the processes are the quasi-static then
the final temperature of the gas (in oK) is (to the nearest integer) _____.
Correct Answer: 1818TO1819
Explanation:
T1V1$\gamma $–1 = T2V2
$\gamma $–1
$300 \times {V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {16}}} \right)^{{7 \over 5} - 1}}$
$ \Rightarrow $ T2 = 300 × (16)0.4
Isobaric process
V = ${{nRT} \over P}$
V2 = kT2... (1)
2V 2 = KTf... (2)
Tf = 2T2 = 300 × 2 × (16)0.4 = 1818 K
$300 \times {V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {16}}} \right)^{{7 \over 5} - 1}}$
$ \Rightarrow $ T2 = 300 × (16)0.4
Isobaric process
V = ${{nRT} \over P}$
V2 = kT2... (1)
2V 2 = KTf... (2)
Tf = 2T2 = 300 × 2 × (16)0.4 = 1818 K
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Evening Slot
Three containers C1, C2 and C3 have water at
different temperatures. The table below shows
the final temperature T when different amounts
of water (given in litres) are taken from each
containers and mixed (assume no loss of heat
during the process)
The value of $\theta $ (in °C to the nearest integer) is ..........
The value of $\theta $ (in °C to the nearest integer) is ..........
Correct Answer: 50
Explanation:
Let containers C1, C2, C3 contain water at T1, T2 and T3 temperatures respectively.
Applying law of calorimetry
1(T1 – 60) + 2(T2 – 60) = 0
$ \Rightarrow $ T1 + 2T2 = 180 ....(1)
1(T2 – 30) + 2(T3 – 30) = 0
$ \Rightarrow $ T2 + 2T3 = 90 ......(2)
2(T1 – 60) + 1(T3 – 60) = 0
$ \Rightarrow $ 2T1 + T3 = 180 .....(3)
from (1), (2), (3)
T1 = 80o C
T2 = 50o C
T3 = 20o C
1 × (T1 – $\theta $) + 1 ×(T2 – $\theta $) + 1(T3 – $\theta $) = 0
$ \Rightarrow $ T1 + T2 + T3 = 3$\theta $
$ \Rightarrow $ $\theta $ = ${{80 + 50 + 20} \over 3}$ = 50oC
Applying law of calorimetry
1(T1 – 60) + 2(T2 – 60) = 0
$ \Rightarrow $ T1 + 2T2 = 180 ....(1)
1(T2 – 30) + 2(T3 – 30) = 0
$ \Rightarrow $ T2 + 2T3 = 90 ......(2)
2(T1 – 60) + 1(T3 – 60) = 0
$ \Rightarrow $ 2T1 + T3 = 180 .....(3)
from (1), (2), (3)
T1 = 80o C
T2 = 50o C
T3 = 20o C
1 × (T1 – $\theta $) + 1 ×(T2 – $\theta $) + 1(T3 – $\theta $) = 0
$ \Rightarrow $ T1 + T2 + T3 = 3$\theta $
$ \Rightarrow $ $\theta $ = ${{80 + 50 + 20} \over 3}$ = 50oC
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
M grams of steam at 100oC is mixed with 200 g of ice at its melting point in a thermally insulated
container. If it produced liquid water at 40oC [heat of vaporization of water is 540 cal/g and heat
of fusion of ice is 80 cal/g] the value of M is ____
Correct Answer: 40
Explanation:
M × 540 + M + 60 = 200 × 80 + 200 × 1× (40– 0)
$ \Rightarrow $ 600 M = 24000
$ \Rightarrow $ M = 40
$ \Rightarrow $ 600 M = 24000
$ \Rightarrow $ M = 40
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Morning Slot
A non-isotropic solid metal cube has coefficients of linear expansion as :
5 $ \times $ 10-5/oC along the x-axis and 5 $ \times $ 10-6/oC along the y and the z-axis. If the coefficient of volume expansion of the solid is C $ \times $ 10-6/oC then the value of C is
5 $ \times $ 10-5/oC along the x-axis and 5 $ \times $ 10-6/oC along the y and the z-axis. If the coefficient of volume expansion of the solid is C $ \times $ 10-6/oC then the value of C is
Correct Answer: 60
Explanation:
$\gamma $ = $\alpha $x + $\alpha $y + $\alpha $z
$ \Rightarrow $ C $ \times $ 10–6 = 5 × 10–5 + 5 × 10–6 + 5 × 10–6
$ \Rightarrow $ C $ \times $ 10–6 = 50 × 10–6 + 10 × 10–6
$ \Rightarrow $ C = 60
$ \Rightarrow $ C $ \times $ 10–6 = 5 × 10–5 + 5 × 10–6 + 5 × 10–6
$ \Rightarrow $ C $ \times $ 10–6 = 50 × 10–6 + 10 × 10–6
$ \Rightarrow $ C = 60
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Morning Slot
A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine
performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low
temperature reservoir, in a cycle, is _______.
Correct Answer: 600
Explanation:
So ${{Q + 1200} \over Q} = {{900} \over {300}}$
$ \Rightarrow $ Q + 1200 = 3Q
$ \Rightarrow $ Q = 600 J
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 1 Offline
The filament of a light bulb has surface area 64 mm2
. The filament can be considered as a black
body at temperature 2500 K emitting radiation like a point source when viewed from far. At night
the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to
be circular with radius 3 mm. Then
(Take Stefan-Boltzmann constant = 5.67 $ \times $ 10−8 Wm−2K−4 , Wien’s displacement constant = 2.90 $ \times $ 10−3 m-K, Planck’s constant = 6.63 $ \times $ 10−34 Js, speed of light in vacuum = 3.00 $ \times $ 108 ms−1)
(Take Stefan-Boltzmann constant = 5.67 $ \times $ 10−8 Wm−2K−4 , Wien’s displacement constant = 2.90 $ \times $ 10−3 m-K, Planck’s constant = 6.63 $ \times $ 10−34 Js, speed of light in vacuum = 3.00 $ \times $ 108 ms−1)
A.
power radiated by the filament is in the range 642 W to 645 W
B.
radiated power entering into one eye of the observer is
in the range 3.15 $ \times $ 10−8 W to 3.25 $ \times $ 10−8 W
in the range 3.15 $ \times $ 10−8 W to 3.25 $ \times $ 10−8 W
C.
the wavelength corresponding to the maximum intensity of light is 1160 nm
D.
taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons
entering per second into one eye of the observer is in the range 2.75 $ \times $ 1011 to 2.85 $ \times $ 1011
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two
equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus the
partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below.
Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. The
partition is now released and moves without any gas leaking from one part of the vessel to the other.
When equilibrium is reached, the distance of the partition from the top (in m) will be _______.
(take the acceleration due to gravity = 10 ms−2 and the universal gas constant = 8.3 J mol−1K−1).
(take the acceleration due to gravity = 10 ms−2 and the universal gas constant = 8.3 J mol−1K−1).
Correct Answer: 6
Explanation:
To start with, both the compartments have same volume. Under the action of weight, moving down by distance $x$, length of upper compartment will be $(4+x)$ and that of lower compartment $(4-x)$.
Volumes of two compartments are
$ V_1=(4+x) A \text { and } V_2=(4-x) A $
At equilibrium,
$ F_2=F_1+m g $
$ \begin{aligned} & P_2 A=P_1 A+m g \\\\ & P_2=P_1+\frac{m g}{A} \end{aligned} $
From $P V=n R T$, we get $P=\frac{n R T}{V}$
$ \begin{aligned} & \frac{n R T}{V_2}=\frac{n R T}{V_1}+\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{V_2}-\frac{1}{V_1}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{A(4-x)}-\frac{1}{4(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & \frac{n R T}{A}\left[\frac{(4+x)-(4-x)}{(4-x)(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{2 x}{\left(16-x^2\right)}\right]=m g \\\\ \Rightarrow & 0.1 \times 8.3 \times 300\left[\frac{2 x}{\left(16-x^2\right)}\right]=8.3 \times 10 \\\\ \Rightarrow & \frac{6 x}{16-x^2}=1 \Rightarrow 16-x^2=6 x \\\\ \Rightarrow & x^2+6 x-16=0 \\\\ \Rightarrow & x=\frac{-6 \pm \sqrt{36+64}}{2} \\\\ & x=\frac{-6 \pm \sqrt{100}}{2}=\frac{-6 \pm 10}{2} \\\\ & x=\frac{10-6}{2} \text { or } \frac{-10-6}{2}=-8 \text { or } 2 \end{aligned} $
Neglecting negative $\operatorname{sign} x=2$
Partition from top $=4+2=6 \mathrm{~m}$
Volumes of two compartments are
$ V_1=(4+x) A \text { and } V_2=(4-x) A $
At equilibrium,
$ F_2=F_1+m g $
$ \begin{aligned} & P_2 A=P_1 A+m g \\\\ & P_2=P_1+\frac{m g}{A} \end{aligned} $
From $P V=n R T$, we get $P=\frac{n R T}{V}$
$ \begin{aligned} & \frac{n R T}{V_2}=\frac{n R T}{V_1}+\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{V_2}-\frac{1}{V_1}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{A(4-x)}-\frac{1}{4(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & \frac{n R T}{A}\left[\frac{(4+x)-(4-x)}{(4-x)(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{2 x}{\left(16-x^2\right)}\right]=m g \\\\ \Rightarrow & 0.1 \times 8.3 \times 300\left[\frac{2 x}{\left(16-x^2\right)}\right]=8.3 \times 10 \\\\ \Rightarrow & \frac{6 x}{16-x^2}=1 \Rightarrow 16-x^2=6 x \\\\ \Rightarrow & x^2+6 x-16=0 \\\\ \Rightarrow & x=\frac{-6 \pm \sqrt{36+64}}{2} \\\\ & x=\frac{-6 \pm \sqrt{100}}{2}=\frac{-6 \pm 10}{2} \\\\ & x=\frac{10-6}{2} \text { or } \frac{-10-6}{2}=-8 \text { or } 2 \end{aligned} $
Neglecting negative $\operatorname{sign} x=2$
Partition from top $=4+2=6 \mathrm{~m}$
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be p0. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (R $-$ a). For a << R the magnitude of the work done in the process in given by (4$\pi$p0Ra2)X, where X is a constant and $\gamma$ = Cp/Cv = 41/30. The value of X is ________.
Correct Answer: 2.05
Explanation:
As change in the kinetic energy of the surface of the bubble is zero.
$ \therefore $ Wexternal + Wwater + Wgas = 0 ....(i)
where Wexternal = work done by external agent
Wwater = work done by water
Wgas = work done by gas
For adiabatic process, ${P_1}V_1^\gamma = {P_2}V_2^\gamma $
Here, ${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$
$ \Rightarrow $ P = ${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$
Now, Wgas = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$
= ${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$
= -4P0$\pi $R2$a$ - ${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$
= -4P0$\pi $R2$a$ - $4.2{P_0}\pi R{a^2}$
Wwater = P0dV
= P0${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$
= ${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$ [ignore ${{a^3}}$ term as no ${{a^3}}$ term in the question]
$ \therefore $ Wgas + Wwater = -$4{P_0}\pi R{a^2}$ - $4.2{P_0}\pi R{a^2}$
= $ - 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$
= $ - 4{P_0}\pi R{a^2}\left[ {2.05} \right]$
From (i), Wexternal = - (Wgas + Wwater)
= $4{P_0}\pi R{a^2}\left[ {2.05} \right]$
$ \therefore $ X = 2.05
$ \therefore $ Wexternal + Wwater + Wgas = 0 ....(i)
where Wexternal = work done by external agent
Wwater = work done by water
Wgas = work done by gas
For adiabatic process, ${P_1}V_1^\gamma = {P_2}V_2^\gamma $
Here, ${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$
$ \Rightarrow $ P = ${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$
Now, Wgas = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$
= ${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$
= -4P0$\pi $R2$a$ - ${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$
= -4P0$\pi $R2$a$ - $4.2{P_0}\pi R{a^2}$
Wwater = P0dV
= P0${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$
= ${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$ [ignore ${{a^3}}$ term as no ${{a^3}}$ term in the question]
$ \therefore $ Wgas + Wwater = -$4{P_0}\pi R{a^2}$ - $4.2{P_0}\pi R{a^2}$
= $ - 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$
= $ - 4{P_0}\pi R{a^2}\left[ {2.05} \right]$
From (i), Wexternal = - (Wgas + Wwater)
= $4{P_0}\pi R{a^2}\left[ {2.05} \right]$
$ \therefore $ X = 2.05
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700 Wm$-$2 and it is absorbed by the water over an effective area of 0.05m2. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in $^\circ$C) in the temperature of water and the surroundings after a long time will be ___________. (Ignore effect of the container, and take constant for Newton's law of cooling = 0.001 s$-$1, Heat capacity of water = 4200 J kg$-$1 K$-$1)
Correct Answer: 8.33
Explanation:
${{dQ} \over {dt}} = e\sigma A({T^4} - T_0^4)$
For small temperature change,
${{dQ} \over {dt}} = e\sigma A{T^3}\Delta T$ .... (i)
${{mCdT} \over {dt}} = e\sigma A{T^3}\Delta T $
$\Rightarrow {{dT} \over {dt}} = {{e\sigma A{T^3}} \over {mC}}\Delta T$
${{e\sigma A{T^3}} \over {mC}}$ $\to$ constant for Newton law of cooling
$ \therefore $ ${{e\sigma A{T^3}} \over {mC}} = 0.001 $
$\Rightarrow e\sigma A{T^3} = mC \times 0.001 = 1 \times 4200 \times 0.001$
$e\sigma A{T^3} = 4.2$ .... (ii)
${{dQ} \over {dt}} = 700 \times 0.05 = 35$ W ..... (iii)
Putting the value of Eqs. (ii) and (iii) in Eq. (i), we get
$35 = 4.2\Delta T \Rightarrow {{35} \over {4.2}} = \Delta T \Rightarrow \Delta T = 8.33$
For small temperature change,
${{dQ} \over {dt}} = e\sigma A{T^3}\Delta T$ .... (i)
${{mCdT} \over {dt}} = e\sigma A{T^3}\Delta T $
$\Rightarrow {{dT} \over {dt}} = {{e\sigma A{T^3}} \over {mC}}\Delta T$
${{e\sigma A{T^3}} \over {mC}}$ $\to$ constant for Newton law of cooling
$ \therefore $ ${{e\sigma A{T^3}} \over {mC}} = 0.001 $
$\Rightarrow e\sigma A{T^3} = mC \times 0.001 = 1 \times 4200 \times 0.001$
$e\sigma A{T^3} = 4.2$ .... (ii)
${{dQ} \over {dt}} = 700 \times 0.05 = 35$ W ..... (iii)
Putting the value of Eqs. (ii) and (iii) in Eq. (i), we get
$35 = 4.2\Delta T \Rightarrow {{35} \over {4.2}} = \Delta T \Rightarrow \Delta T = 8.33$
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 1 Offline
Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1. It
expands isothermally to volume 4V1. After this, the gas expands adiabatically and its volume becomes
32V1. The work done by the gas during isothermal and adiabatic expansion processes are Wiso and
Wadia, respectively. If the ratio ${{{W_{iso}}} \over {{W_{adia}}}}$ = f ln 2, then f is ______.
Correct Answer: 1.77
Explanation:
${{{p_1}} \over 4}{(4{V_1})^{5/3}} = {p_2}{(32{V_1})^{5/3}}$
${p_2} = {{{p_1}} \over 4}{\left( {{1 \over 8}} \right)^{5/3}} = {{{p_1}} \over {128}}$
${W_{adi}} = {{{p_1}{V_1} - {p_2}{V_2}} \over {\gamma - 1}}$
$ = {{{p_1}{V_1} - {{{p_1}} \over {128}}(32{V_1})} \over {{5 \over 3} - 1}}$
$ = {{{p_1}{V_1}(3/4)} \over {2/3}} = {9 \over 8}{p_1}{V_1}$
${W_{iso}} = {p_1}{V_1}\ln \left( {{{4{V_1}} \over {{V_1}}}} \right) = 2{p_1}{V_1}\ln 2$
$ \therefore $ ${{{W_{iso}}} \over {{W_{adi}}}} = {{16} \over 9}\ln 2$
$ \Rightarrow f = {{16} \over 9} = 1.7778 \approx 1.78$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
A sheet of steel at $20^{\circ} \mathrm{C}$ has size as shown in figure below. If the co-efficient of linear expansion for steel is $10^{-5}{ }^{\circ} \mathrm{C}^{-1}$, then what is the change in the area at $60^{\circ} \mathrm{C}$ ?
A.
$0.84 \mathrm{~cm}^2$
B.
$0.64 \mathrm{~cm}^2$
C.
$0.24 \mathrm{~cm}^2$
D.
$0.14 \mathrm{~cm}^2$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
Different material of two identical long bars $A$ and $B$ are coated with wax and have their one end immersed in a hot oil bath. When the steady state is reached, the lengths for which wax melt are $l_A$ and $l_B$. If $k_A$ and $k_B$ are thermal conductivities of materials, then
A.
$\frac{K_A}{K_B}=\sqrt{\frac{I_A}{I_B}}$
B.
$\frac{K_A}{K_B}=\frac{I_B}{I_A}$
C.
$\frac{K_A}{K_B}=\frac{I_A}{I_B}$
D.
$\frac{K_A}{K_B}=\sqrt{\frac{I_B}{I_A}}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
A gas is at constant pressure $4 \times 10^5 \mathrm{~N} / \mathrm{m}^2$. When a heat energy of 2000 J is supplied to the gas, its volume changes by $3 \times 10^{-3} \mathrm{~m}^3$. What is the increase in its internal energy?
A.
650 J
B.
900 J
C.
800 J
D.
400 J
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
Certain amount of heat supplied to an ideal gas under isothermal condition will result in
A.
an increase in the internal energy of the gas
B.
external work done and a change in temperature
C.
a rise in temperature
D.
external work done by the system
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
If $\alpha_V$ and $T$ are the coefficient of volume expansion and temperature for an ideal gas respectively, then
A.
$\alpha_V=\frac{1}{T}$
B.
$\alpha_V=\sqrt{T}$
C.
$\alpha_V=\frac{1}{\sqrt{T}}$
D.
$\alpha_V=\frac{1}{T^2}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
If $\lambda$ denotes the wavelength at which the radiative emission from a black body at a temperature $T$ is maximum, then
A.
$\lambda \propto T^{-1}$
B.
$\lambda \propto T^4$
C.
$\lambda$ is independent of $T$
D.
$\lambda \propto T$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
A Carnot engine $C_1$ operates between temperature $T_1$ and $T_2\left(T_1>T_2\right)$. A second Carnot engine $C_2$ uses all the heat rejected by the engine $C_1$ and operates between temperature $T_2$ and $T_3$ (where $T_2>T_3$ ). The efficiency of this combined ( $C_1$ and $C_2$ together) engine is
A.
$1-\frac{T_3}{T_1}$
B.
$2-\left(\frac{T_2}{T_1}+\frac{T_3}{T_2}\right)$
C.
$1-\frac{\left(T_2+T_3\right)}{T_1}$
D.
$1-\left(1-\frac{T_2}{T_1}\right)\left(1-\frac{T_3}{T_2}\right)$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
All gases deviate from gas laws at
A.
low pressure and high temperature
B.
high pressure and low temperature
C.
low pressure and low temperature
D.
high pressure and high temperature
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
A solid of 2 kg mass absorbs 50 kJ when its temperature is raised from $20^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$. The specific heat capacity of this solid in unit of $\mathrm{J} / \mathrm{kg}{ }^{\circ} \mathrm{C}$ is
A.
500
B.
1000
C.
1500
D.
750
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
A solid cylinder of radius $r_1=2.5 \mathrm{~cm}$, length $l_1=5.0 \mathrm{~cm}$ and temperature $40^{\circ} \mathrm{C}$ is suspended in an environment of temperature $60^{\circ} \mathrm{C}$. The thermal radiation transfer rate for cylinder is 1.0 W . If the cylinder is stretched until its radius becomes $r_2=0.50 \mathrm{~cm}$, the thermal radiation transfer rate is changed to
A.
3.35 W
B.
4.50 W
C.
0.75 W
D.
1.25 W
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
Five moles of an ideal gas has pressure $p_0$, volume $V_0$ and temperature $T_0$. The gas is expanded to volume $3 V_0$ along a path, so that the pressure $p$ is changed as function of volume $V$ as $p=p_0\left(V / V_0\right)$. The pressure is then reduced to $p_0$ maintaining the volume constant. The gas undergoes an isobaric compression till the volume and temperature become $V_0$ and $T_0$, respectively. The total work done by the gas during the entire process is
A.
$p_0 V_0 / 3$
B.
$3 p_0 V_0$
C.
$5 p_0 V_0 / 3$
D.
$2 p_0 V_0$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
How many rotational degrees of freedom does a rigid diatomic molecule have?
A.
0
B.
1
C.
2
D.
3
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
The specific heat of helium at constant volume is 12.6 J $\mathrm{mol}^{-1} \mathrm{~K}^{-1}$. The specific heat of helium at constant pressure in $\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ is approximately (assume, the universal gas constant, $R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )
A.
12.6
B.
16.8
C.
18.9
D.
20.9
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
A composite slab is prepared with two different materials $A$ and $B$. The relation between their coefficient of thermal conductivity and thickness is given as $K_A=\frac{K_B}{2}$ and $X_A=2 X_B$, respectively. If the temperature of faces of $A$ and $B$ are $75^{\circ} \mathrm{C}$ and $50^{\circ} \mathrm{C}$ respectively, what will be the temperature of common surface?
A.
$75^{\circ} \mathrm{C}$
B.
$50^{\circ} \mathrm{C}$
C.
$55^{\circ} \mathrm{C}$
D.
$125^{\circ} \mathrm{C}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
Work done on heating one mole of monoatomic gas adiabatically through $20^{\circ} \mathrm{C}$ is $W$. Then, the work done on heating 6 moles of rigid diatomic gas through the same change in temperature
A.
9 W
B.
10 W
C.
12 W
D.
8 W
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
If a gas has $n$ degrees of freedom, then the ratio of $\frac{C_p}{C_V}$ is
A.
$\frac{n+2}{n}$
B.
$\frac{2 n+1}{n}$
C.
$\frac{n+2}{2 n}$
D.
$\frac{n+4}{2 n}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be
the heat energy absorbed by the gas, in this process ?
A.
35 J
B.
30 J
C.
25 J
D.
40 J
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
One kg of water, at 20oC, heated in an electric kettle whose heating element has a mean (temperature
averaged) resistance of 20 $\Omega $. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time
taken for water to evaporate fully, is close to :
[Specific heat of water = 4200 J/(kg oC), Latent heat of water = 2260 kJ/kg]
A.
10 minutes
B.
22 minutes
C.
3 minutes
D.
16 minutes
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A Carnot engine has an efficiency of ${1 \over 6}$. When the temperature of the sink is reduced by 62ºC, its efficiency
is doubled. The temperatures of the source and the sink are, respectively
A.
99oC, 37oC
B.
124oC, 62oC
C.
37oC, 99oC
D.
62oC, 124oC
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
At 40o C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of
the wire. When the wire is cooled down from 40oC to 20oC it regains its original length of 0.2 m. The value
of M is close to :
(Coefficient of linear expansion and Young’s modulus of brass are 10–5
/oC and 1011 N/m
2
, respectively; g=
10 ms–2
)
A.
1.5 kg
B.
0.5 kg
C.
9 kg
D.
0.9 kg
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the
molar specific heat of mixture at constant volume ? (R = 8.3 J/mol K)
A.
21.6 J/mol K
B.
17.4 J/mol K
C.
15.7 J/mol K
D.
19.7 J/mol K
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
When M1 gram of ice at –10oC (specific heat = 0.5 cal g–1
oC–1
) is added to M2 gram of water at 50C, finally
no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1
is :
A.
${{50{M_2}} \over {{M_1}}} - 5$
B.
${{50{M_2}} \over {{M_1}}}$
C.
${{5{M_2}} \over {{M_1}}} - 5$
D.
${{5{M_1}} \over {{M_2}}} - 50$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the
internal energy of the gas along the path ca is –180 J. The gas absorbs 250 J of heat along the path ab and 60
J along the path bc. The work done by the gas along the path abc is:
A.
120 J
B.
130 J
C.
100 J
D.
140 J
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
One mole of ideal gas passes through a process where pressure and volume obey the relation
$P = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]$.
Here P0 and V0 are constants. Calculate the change in the temperature of the gas if its
volume changes form V0 to 2V0
A.
${3 \over 4}{{{P_0}{V_0}} \over R}$
B.
${1 \over 2}{{{P_0}{V_0}} \over R}$
C.
${5 \over 4}{{{P_0}{V_0}} \over R}$
D.
${1 \over 4}{{{P_0}{V_0}} \over R}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by
$\Delta $T. the heat required to produce the same change in temperature, at a constant pressure is :
A.
${7 \over 5}Q$
B.
${3 \over 2}Q$
C.
${2 \over 3}Q$
D.
${5 \over 3}Q$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
A cylinder with fixed capacity of 67.2 lit
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]
A.
374 J
B.
700 J
C.
748 J
D.
350 J
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
n moles of an ideal gas with constant volume
heat capcity CV undergo an isobaric expansion
by certain volume. The ratio of the work done
in the process, to the heat supplied is :
A.
${{nR} \over {{C_V} - nR}}$
B.
${{4nR} \over {{C_V} - nR}}$
C.
${{4nR} \over {{C_V} + nR}}$
D.
${{nR} \over {{C_V} + nR}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
A 25 × 10–3 m3 volume cylinder is filled with
1 mol of O2 gas at room temperature (300K).
The molecular diameter of O2, and its root
mean square speed, are found to be 0.3 nm, and
200 m/s, respectively. What is the average
collision rate (per second) for an O2 molecule ?
A.
~1013
B.
~1012
C.
~1011
D.
~1010
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
The specific heats, CP and CV of a gas of
diatomic molecules, A, are given (in units of
J mol–1 K–1) by 29 and 22, respectively.
Another gas of diatomic molecules, B, has the
corresponding values 30 and 21. If they are
treated as ideal gases, then :-
A.
A is rigid but B has a vibrational mode
B.
A has a vibrational mode but B has none
C.
A has one vibrational mode and B has two
D.
Both A and B have a vibrational mode each
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
A massless spring (k = 800 N/m), attached with
a mass (500 g) is completely immersed in 1 kg
of water. The spring is stretched by 2 cm and
released so that it starts vibrating. What would
be the order of magnitude of the change in the
temperature of water when the vibrations stop
completely ? (Assume that the water container
and spring receive negligible heat and specific
heat of mass = 400 J/kg K, specific heat of
water = 4184 J/kg K)
A.
10–3 K
B.
10–1 K
C.
10–5K
D.
10–4 K
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Two materials having coefficients of thermal
conductivity '3K' and 'K' and thickness 'd' and
'3d', respectively, are joined to form a slab as
shown in the figure. The temperatures of the
outer surfaces are '$\theta $2' and '$\theta $1' respectively,
($\theta $2 > $\theta $1). The temperature at the interface is :-
A.
${{{\theta _1}} \over {10}} + {{9{\theta _2}} \over {10}}$
B.
${{{\theta _2} + {\theta _1}} \over 2}$
C.
${{{\theta _1}} \over {6}} + {{5{\theta _2}} \over {6}}$
D.
${{{\theta _1}} \over {3}} + {{2{\theta _2}} \over {3}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
An HCl molecule has rotational, translational
and vibrational motions. If the rms velocity of
HCl molecules in its gaseous phase is $\overline v $ , m is
its mass and kB is Boltzmann constant, then its
temperature will be :
A.
${{m{{\overline v }^2}} \over {5{k_B}}}$
B.
${{m{{\overline v }^2}} \over {6{k_B}}}$
C.
${{m{{\overline v }^2}} \over {7{k_B}}}$
D.
${{m{{\overline v }^2}} \over {3{k_B}}}$