datavorous/entrance-exam-dataset huggingface dataset:
Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel (s = relative density of mercury and ρ= density of water) by

Options:
A n2h(n+1)2s
B h(n2+1) s
C h(n+1)2s
D hn2s

The correct answer is:
h(n2+1) s

Solution:
If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
Now, πr2h1=π(nr)2h2
⇒h1=n2h2
Now, pressure at point A = pressure at point B
hρg=(h1+h2)ρ′g
⇒h =n2h2+h2sgAs s=ρ′ρ
⇒h2=h(n2+1)s

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two complexes [CrH2O)6Cl3A and [CrNH3)6Cl3B are violet and yellow coloured, respectively. The incorrect statement regarding them is:

Options:
A Δ0 value of A is less than that of B
B Δ0 value of A and B are calculated from the enrgies of violet and yellow light, respectively
C Both absorb energies corresponding to their complementary colours
D Both are paramagnetic with three unpaired electrons

The correct answer is:
Δ0 value of A and B are calculated from the enrgies of violet and yellow light, respectively

Solution:
H2O is weaker ligand than NH3
So, (Δ0)B>(Δ0)A
Since, oxidation number of Cr in the complex
Cr(H2O)6x0-1Cl3 is calculated as, x+60+3(-1)=0=x=+3
So, Cr3+ has 3d3 configuration
Similarly, oxidation no. of Cr in
Cr(NH3)6x0-1 Cl3 is calculated as, x+6(0)+3(-1)=0=x=+3
So Cr3+ has 3d3 configuration
Thus, both A and B are paramagnetic with 3 unpaired electrons.
We know, Δ=hcλabsorbed

Since (Δ0)B>(Δ0)A
λA>λB
A is violet coloured so its energy is calculated from yellow colour absorbed.
(Δ0)A or Δviolet=hcλyellow
Similarly, (Δ0)B or Δyellow=hcλyellow
Hence, Δ0 values of A and B are calculated from the energies of yellow and violet light respectively.

Topic: Chemistry, Coordination Compounds, JEE Main
Two complexes CrH2O6 Cl3 A and CrNH36Cl3B are violet and yellow coloured, respectively, The incorrect statement regarding them is:

Options:
A Both are paramagnetic with three unpaired electrons
B Δ0 value of A is less than that of B
C Δ0 values of A and B are calculated from the energies violet and yellow light, respectively.
D Both absorb energies corresponding to their complementary colors.

The correct answer is:
Δ0 values of A and B are calculated from the energies violet and yellow light, respectively.

Solution:
∆0A<∆0B , both complex have CN=6, B has strong field ligand and A has weak field ligand
so ∆0A< ∆0B
In both complexes, Cr has 3 unpaired electrons and hence, will be paramagnetic CrH2OCl3
Cr+3=1s2  2s2  2p6 3s2 3p6  3d3 in strong field ligand or weak field ligand is t2g1,1,1,ego
Δ0 order will be compared by spectrochemical series not by energies of violet and yellow light. Hence incorrect

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Two compounds A and B with same molecular formula C3H6O undergo Grignard reaction with methylmagnesium bromide to give products C and D. Products C and D show following chemical tests.

Test	C	D

Ceric ammonium nitrate Test	Positive	Positive

Lucas Test	Turbidity obtained after five minutes	Turbidity obtained Immediately

Iodoform Test	Positive	Negative

C and D respectively are

Options:
A
B
C
D

The correct answer is:

Solution:
B will be ketone as it gives tertiary alcohol after reaction with grignard while A will be aldehyde which produces secondary alcohol after reaction with grignard reagent.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Two compounds I and II are eluted by column chromato-graphy(adsorption of I $>$ II). Which one of the following is a correct statement?

Options:
A
II moves slower and has higher $\mathrm{R}_{\mathrm{f}}$ value than I

B
II moves faster and has higher $\mathrm{R}_{\mathrm{f}}$ value than I

C
I moves faster and has higher $\mathrm{R}_{\mathrm{f}}$ value than II

D
I moves slower and has higher $\mathrm{R}_{\mathrm{f}}$ value than I

The correct answer is:
II moves faster and has higher $\mathrm{R}_{\mathrm{f}}$ value than I

Solution:
Since, adsorption of I $>$ II, I is firmly attached to column (stationary phase). Hence, it will move slowly and will move little distance. A1so II is loosely attached to column (stationary phase). Hence, it will move faster and will move large distance.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Two concentric circular coils, C1 and C2, are placed in the XY plane. C1 has 500turns, and a radius of 1 cm. C2 has 200 turns and radius of 20 cm. C2 carries a time dependent current It=5t2-2t+3A where t is in s. The emf induced in C1in mV at the instant t=1s is 4x. The value of x is ..........

Options:

The correct answer is:
5

Solution:
I=5t2+2t+C

didt=(10t+2)

ϕsmall =BA=μ0IN22Rπr2

induced emf in small coil

e=dϕdt=μ0N22rπr2N1didt=μ0N1N2πr22R(10t-2)

at t=1

e=μ0N1N2πr22R8=4μ0N1N2πr2R=4(4π)10-7×20020×500×10-410-2π

=80×π2×10-7×10×102×10-2=8×10-4 volt=0.8 mV=4x

x=5

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2020 (05 Sep Shift 1)
Two concentric circular coils with radii 1 cm and 1000 cm and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be _____ ×10–8 H. (Take, π2=10)

Options:

The correct answer is:
4

Solution:
Let the magnetic field of the 200 turns be B1. The formula of magnetic field is given by B=μ0i2r

Let r=1000 cm. Thus,

B1=nμ0×I2×10=200μ0I2×10

The flux through the coil of radius r1=1 cm is ϕ=10×B→·A→=10×B1πr21

⇒ϕ=10×200μ0I2×10×π0.012

L=10×200μ02×10×π0.012=10×2004π×10-72×10×π0.012=4×10-8 H

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Two concentric coils each of radius equal to $2 \pi \mathrm{cm}$ are placed at right angles to each other. 3 Ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber $/ \mathrm{m}^2$ at the centre of the coils will be $\left(\mu_0=4 \pi \times 10^{-7}\right.$ Wb/A-m)

Options:
A
$12 \times 10^{-5}$

B
$10^{-5}$

C
$5 \times 10^{-5}$

D
$7 \times 10^{-5}$

The correct answer is:
$5 \times 10^{-5}$

Solution:
$B=\frac{\mu_o}{2 r} \sqrt{I_1^2+I_2^2}$
$B=\frac{4 \pi \times 10^{-7}}{2 \times 2 \pi \times 10^{-2}} \times 5$
$B=5 \times 10^{-5}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2005
Two concentric coils of 10 turns each are placed in the same plane. Their radii are 20 cm and 40 cm and carry 0.2 A and 0.3 A current respectively in opposite directions. The magnetic induction (in tesla) at the center is

Options:
A 34μ0
B 54μ0
C 74μ0
D 94μ0

The correct answer is:
54μ0

Solution:
Two coils carry currents in opposite directions, hence net magnetic field at centre will be difference of the two fields.
ie, Bnet=μ04π . 2πNi1r1-i2r2
=10μ020.20.2-0.30.4
=54μ0

Topic: Physics, Magnetic Effects of Current, JEE Main
Two concentric coils of 10 turns each are placed in the same plane. Their radii are $20 \mathrm{~cm}$ and $40 \mathrm{~cm}$ and carry 0.2 and $0.3 \mathrm{~A}$. current respectively in opposite directions. The magnetic induction (in T) at the centre is

Options:
A $\frac{3}{4} \mu_0$
B $\frac{5}{4} \mu_0$
C $\frac{7}{4} \mu_0$
D $\frac{9}{4} \mu_0$

The correct answer is:
$\frac{5}{4} \mu_0$

Solution:
Two coils carry currents in opposite directions, hence net magnetic field at centre will be difference of the two fields. ie,
$$
\begin{aligned}
B_{\text {net }} & =\frac{\mu_0}{4 \pi} \cdot 2 \pi N\left[\frac{i_1}{r_1}-\frac{i_2}{r_2}\right] \\
& =\frac{10 \mu_0}{2}\left[\frac{0.2}{0.2}-\frac{0.3}{0.4}\right] \\
& =\frac{5}{4} \mu_0
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
Two concentric hollow spherical shells have radii $r$ and $R(R \gg r)$. A charge $Q$ is distributed on them such that the surface charge densities are equal. The electric potential at the centre is

Options:
A $\frac{Q(R+r)}{4 \pi \varepsilon_0\left(R^2+r^2\right)}$
B $\frac{Q\left(R^2+r^2\right)}{4 \pi \varepsilon_0(R+r)}$
C $\frac{Q}{(R+r)}$
D zero

The correct answer is:
$\frac{Q(R+r)}{4 \pi \varepsilon_0\left(R^2+r^2\right)}$

Solution:
We know that, the surface charge densities is
$$
\sigma=\frac{Q}{4 \pi\left(r^2+R^2\right)}
$$
The electric potential at the centre
$$
\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{\sigma \times 4 \pi r^2}{r}+\frac{\sigma \times 4 \pi R^2}{R}\right] \\
& =\frac{1}{4 \pi \varepsilon_0} \times \sigma \times 4 \pi\left[\frac{r^2}{r}+\frac{R^2}{R}\right] \\
& =\frac{\sigma}{\varepsilon_0}(r+R)
\end{aligned}
$$
and $\quad V=\frac{Q}{\varepsilon_0 \cdot 4 \pi\left(r^2+R^2\right)} \cdot(r+R)$
$$
\Rightarrow \quad V=\frac{Q}{4 \pi \varepsilon_0} \cdot \frac{(r+R)}{\left(r^2+R^2\right)}
$$

Topic: Physics, Electrostatics, JEE Main
Two concentric spherical shells of masses and radii m1, r1 and m2,r2 respectively are as shown. The gravitational field intensity at point P is

Options:
A Gm1r22-r12
B Gm1r12
C Gm1r12+Gm2r22
D Gm1r2

The correct answer is:
Gm1r2

Solution:
Since field due to the outer sphere is zero at point P Hence,

intensity will be Gm1r2

Topic: Physics, Gravitation, JEE Main
Two conducting circular loops A and B are placed in the same plane with their centers coinciding as shown in figure. The mutual inductance between them is :

Options:
A $\frac{\mu_0 \pi b^2}{2 a}$
B $\frac{\mu_o}{2 \pi} \cdot \frac{b^2}{a}$
C $\frac{\mu_0}{2 \pi} \cdot \frac{a^2}{b}$
D $\frac{\mu_0 \pi a^2}{2 b}$

The correct answer is:
$\frac{\mu_0 \pi a^2}{2 b}$

Solution:
$\begin{aligned} & \phi=\mathrm{Mi}=\mathrm{BA} \\ & \Rightarrow \mathrm{Mi}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{~b}} \pi \mathrm{a}^2 \\ & \therefore \mathrm{M}=\frac{\mu_0 \pi \mathrm{a}^2}{2 \mathrm{~b}}\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1≫R2, the mutual inductance M between them will be directly proportional to:

Options:
A R1R2
B R2R1
C R22R1
D R12R2

The correct answer is:
R22R1

Solution:
M=ϕ12I1=B1A2I1=μ0I12R1πR22I1

M=μ0πR222R1

∴ M∝R22R1

Topic: Physics, Electromagnetic Induction, JEE Main
Two conducting spheres A and B of radius a and b respectively are at the same electric potential. The ratio of the surface charge densities of A and B is

Options:
A ba
B ab
C a2b2
D b2a2

The correct answer is:
ba

Solution:
Given electric potential of spheres are same i.e.,
VA=VB
14ÏÎµ0Q1a=14ÏÎµ0.Q2b
Q1Q2=ab
A surface charge density
Ï=Q4Ïr2
⇒Ï1Ï2=Q1Q2Ãb2a2
=abÃb2a2=ba

Topic: Physics, Electrostatics, JEE Main
Two conducting spheres of radii R1 and R2 having charges Q1 and Q2 respectively are placed very far apart from each other. Now they are connected to each other by a connecting wire, then there is

Options:
A no change in the energy of the system
B an increase in the energy of the system
C always a decrease in the energy of the system
D a decrease in energy of the system unless q1R2 = q2R1

The correct answer is:
a decrease in energy of the system unless q1R2 = q2R1

Solution:
ΔU = 12 C1C2C1+C2 V1 – V22
ΔU = 0, If V1 = V2
or kq1R1 = kq2R2
or q1R2=q2R1

Topic: Physics, Electrostatics, JEE Main
Two conductors have the same resistance at $0^{\circ} \mathrm{C}$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients of their series and parallel combinations are nearly

Options:
A
$\frac{\alpha_1+\alpha_2}{2}, \alpha_1+\alpha_2$

B
$\alpha_1+\alpha_2, \frac{\alpha_1+\alpha_2}{2}$

C
$\alpha_1+\alpha_2, \frac{\alpha_1 \alpha_2}{\alpha_1+\alpha_2}$

D
$\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}$

The correct answer is:
$\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}$

Solution:
Let $R_0$ be the initial resistance of both conductors $\therefore \quad$ At temperature $\theta$ their resistance will be, $R_1=R_0\left(1+\alpha_1 \theta\right)$ and $R_2=R_0\left(1+\alpha_2 \theta\right)$
for, series combination, $\mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\mathrm{R}_2$ $R_{s 0}\left(1+\alpha_s \theta\right)=R_0\left(1+\alpha_1 \theta\right)+R_0\left(1+\alpha_2 \theta\right)$
where $\mathrm{R}_{\mathrm{s} 0}=\mathrm{R}_0+\mathrm{R}_0=2 \mathrm{R}_0$
$\therefore \quad 2 \mathrm{R}_0\left(1+\alpha_s \theta\right)=2 \mathrm{R}_0+\mathrm{R}_0 \theta\left(\alpha_1+\alpha_2\right)$
or $\quad \alpha_s=\frac{\alpha_1+\alpha_2}{2}$
for parallel combination,
$$
R_p=\frac{R_1 R_2}{R_1+R_2}
$$
where, $R_{p 0}=\frac{R_0 R_0}{R_0+R_0}=\frac{R_0}{2}$
$\therefore \quad \frac{\mathrm{R}_0}{2}\left(1+\alpha_{\mathrm{p}} \theta\right)=\frac{\mathrm{R}_0^2\left(1+\alpha_1 \theta+\alpha_2 \theta+\alpha_1 \alpha_2 \theta\right)}{\mathrm{R}_0\left(2+\alpha_1 \theta+\alpha_2 \theta\right)}$
as $\alpha_1$ and $\alpha_2$ are small quantities
$\therefore \quad \alpha_1 \alpha_2$ is negligible
or $\quad \alpha_{\mathrm{p}}=\frac{\alpha_1+\alpha_2}{2+\left(\alpha_1+\alpha_2\right) \theta}=\frac{\alpha_1+\alpha_2}{2}\left[1-\left(\alpha_1+\alpha_2\right) \theta\right]$
as $\left(\alpha_1+\alpha_2\right)^2$ is negligible
$\therefore \quad \alpha_p=\frac{\alpha_1+\alpha_2}{2}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2010
Two containers C1 and C2 of volumes V and 4V respectively, hold the same ideal gas and are connected by a thin horizontal tube of negligible volume with a valve which is initially closed. The initial pressures of the gas in C1 and C2 are p and 5p, respectively. Heat baths are employed to maintain the temperatures in the containers at 300 K and 400 K, respectively. The valve is now opened. Select the correct statement.

Options:
A The gas will flow from the hot container to the cold one and the process is reversible.
B The gas will flow from one container to the other till the number of moles in two containers are equal.
C A long time after the valve is opened, the pressure in both the containers will be 3p.
D A long time after the valve is opened, number of moles of gas in the hot container will be thrice that of the cold one.

The correct answer is:
A long time after the valve is opened, number of moles of gas in the hot container will be thrice that of the cold one.

Solution:
Let n1 and n2 are number of moles of gas present in container C1 and C2 respectively, before the value is opened.

Then, using pV=nRT.

n1=pVR300

and n2=5p4VR400

When value is opened gas flows from C2 to C1 till pressure in C1 and C2 is equal. Let after equalisation of pressure in both C1 and C2, its value is p0,

Then, using pV=nRT.

p0V=n1'R300 (Container C1)

and p04V=n2'R400 (Container C2)

So, n1'=p0VR300

As no gas is leaked from containers,

n1+n2=n1':n2'

⇒pVR300+20pVR400

=p0VR300+4p0VR400

So, p1300+20400=p01300+4400

⇒4p=p0

Now, n2'n1'=4p0VR400p0VR300=3

Finally number of moles of C2 is thrice of C1.

Topic: Physics, Thermodynamics, KVPY, KVPY 2019 (SB/SX)
Two containers of equal volume contain the same gas at the pressure p1 and p2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio is equal to

Options:
A p1T2+p2T1T1×T2
B p1T2+p2T1T1+T2
C 12 p1T2+p2T1T1T2
D p1T2-p2T1T1×T2

The correct answer is:
12 p1T2+p2T1T1T2

Solution:
For a closed system, the total number of moles remains constant. So
p1V=n1RT1 and p2V=n2RT2
∴ p2V=n1+n2RT
∴ pT = n1+n22R=12P1T1+P2T2
= 12p1T2+p2T1T1T2

Topic: Physics, Thermodynamics, JEE Main
Two containers of the equal volume contain the same gas at pressure P1 and P2 and absolute temperatures T1 and T2, respectively. On joining the vessels, the gas reaches a common pressure P and common temperature T. The ratio PT is equal to

Options:
A P1T1+P2T2
B P1T1+P2T2T1+T2
C P1T2+P2T1T1+T22
D P12T1+P22T2

The correct answer is:
P12T1+P22T2

Solution:
Number of moles in the first vessel
Î¼1=P1VRT1
Number of moles in the second vessel
Î¼2=P2VRT2
If both vessels are joined together, then the quantity of gas remains the same, i.e., Î¼ = Î¼ 1 + Î¼ 2
P2VRT=P1VRT1+P2VRT2
PT=P12T1+P22T2

Topic: Physics, Thermodynamics, NEET
Two convex lenses of focal length 20 cm each are placed coaxially with a separation of 60 cm between them. The image of the distant object formed by the combination is at _____ cm from the first lens.

Options:

The correct answer is:
100

Solution:
Let L1 be the first lens and L2 be the second lens.

The first refraction is in L1.

Using lens formula,

u1=∞, f1=20 cm,1v1-0=1f1 ⇒v1=20 cm

Hence, the distance between the second image and the L2 would be 60-20=40 cm.

For the second lens L2,

u2=-40 cm, f2=20 cm,1v2-1u2=1f2⇒1v2+140=120⇒1v2=140⇒v2=40 cm

So, the distance of image from first lens is

x=40 cm+60 cm

=100 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Two cylinders $A$ and $B$ fitted with pistons contain equal amounts of an ideal diatomic gas at $300 \mathrm{~K}$. The piston of $A$ is free to move while that of $B$ is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in $A$ is $30 \mathrm{~K}$, then the rise in temperature of the gas in $B$ is

Options:
A $30 \mathrm{~K}$
B $18 \mathrm{~K}$
C $50 \mathrm{~K}$
D $42 K$

The correct answer is:
$42 K$

Solution:
In both cylinders $A$ and $B$ the gases are diatomic $(\gamma=1.4)$. Piston $A$ is free to move i.e. it is isobaric process. Piston $B$ is fixed i.e. it is isochoric process. If same amount of heat $\Delta Q$ is given to both then
$\begin{aligned} & (\Delta Q)_{\text {isobaric }}=(\Delta Q)_{\text {isochoric }} \Rightarrow \mu C_p(\Delta T)_A=\mu C_v(\Delta T)_B \\ & (\Delta T)_B=\frac{C_p}{C_v}(\Delta T)_A=\gamma(\Delta T)_A=1.4 \times 30=42 \mathrm{~K} . \\ & \end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is.

Options:
A 30 K
B 18 K
C 50 K
D 42K

The correct answer is:
42K

Solution:
A is free to move; therefore, heat will be supplied at constant pressure.
∆QA=nCPA∆TA ...(I)
B is held fixed, therefore, heat will be supplied at constant volume.
∆QB=nCVB∆TB ...(ii)
But ∆QA=∆QB
nCPA∆TA=nCvB∆TB
∆TB=CPCV∆TA
=γ∆TA ( γ = 1.4 (diatomic) (ΔTA = 30 K)
= 1.4 3 0 K
∆TB=42K

Topic: Physics, Thermodynamics, JEE Main
Two cylinders $A$ and $B$ fitted with pistons contain equal number of moles of an ideal monoatomic gas at $400 \mathrm{~K}$. The piston of $A$ is free to move while that of $B$ is held fixed. Same amount of heat energy is given to the gas in each cylinder. If the rise in temperature of the gas in $A$ is $42 \mathrm{~K}$, the rise in temperature of the gas in $B$ is

Options:
A $21 \mathrm{~K}$
B $35 \mathrm{~K}$
C $42 \mathrm{~K}$
D $70 \mathrm{~K}$

The correct answer is:
$42 \mathrm{~K}$

Solution:
From first law of thermodynamics
$Q=\Delta U+W$
For cylinder $A$ pressure remains constant
$\therefore$ Work done by a system
$W=\frac{\mu R}{\gamma-1}\left(T_1-T_2\right)$
For monoatomic gases
$\begin{aligned}
\mu & =1 \\
\gamma & =\frac{5}{3} \\
W & =\frac{1 \times R}{\frac{5}{3}-1}(442-400) \\
& =\frac{3}{2} R \times 42
\end{aligned}$
$or \quad$ $W=63 R$
But $\Delta U=0$, for cylinder $A$
$\begin{aligned}
\therefore \quad Q & =0+63 R \\
Q & =63 R
\end{aligned}$
For cylinder $B$ volume is constant,
$\begin{aligned}
\therefore W =0 \\
\text { and } Q=\mu C_V \Delta T
\end{aligned}$
For monoatomic gas
$\begin{aligned}
C_V & =\frac{3}{2} R \\
Q & =1 \times \frac{3}{2} R \Delta T
\end{aligned}$
As heat given to both cylinder is same
$\begin{aligned}
\therefore \quad 63 R & =\frac{3}{2} R \Delta T \\
\Delta T & =42 \mathrm{~K}
\end{aligned}$

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2009
Two cylinders $A$ and $B$ fitted with pistons contain equal number of moles of an ideal monoatomic gas at $400 \mathrm{~K}$. The piston of $A$ is free to move while that of $B$ is held fixed. Same amount of heat energy is given to the gas in each cylinder. If the rise in temperature of the gas in $A$ is $42 \mathrm{~K}$, the rise in temperature of the gas in $B$ is

Options:
A $21 \mathrm{~K}$
B $35 \mathrm{~K}$
C $42 \mathrm{~K}$
D $70 \mathrm{~K}$

The correct answer is:
$42 \mathrm{~K}$

Solution:
From first law of thermodynamics
$Q=\Delta U+W$
For cylinder $A$ pressure remains constant
$\therefore$ Work done by a system
$W=\frac{\mu R}{\gamma-1}\left(T_1-T_2\right)$
For monoatomic gases
$\begin{aligned}
\mu & =1 \\
\gamma & =\frac{5}{3} \\
W & =\frac{1 \times R}{\frac{5}{3}-1}(442-400) \\
& =\frac{3}{2} R \times 42
\end{aligned}$
$or \quad$ $W=63 R$
But $\Delta U=0$, for cylinder $A$
$\begin{aligned}
\therefore \quad Q & =0+63 R \\
Q & =63 R
\end{aligned}$
For cylinder $B$ volume is constant,
$\begin{aligned}
\therefore W =0 \\
\text { and } Q=\mu C_V \Delta T
\end{aligned}$
For monoatomic gas
$\begin{aligned}
C_V & =\frac{3}{2} R \\
Q & =1 \times \frac{3}{2} R \Delta T
\end{aligned}$
As heat given to both cylinder is same
$\begin{aligned}
\therefore \quad 63 R & =\frac{3}{2} R \Delta T \\
\Delta T & =42 \mathrm{~K}
\end{aligned}$

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2009
Two cylinders A and B fitted with pistons contain equal number of moles of an ideal monoatomic gas at $400 \mathrm{~K}$. The piston of A is free to move while that of $B$ is held fixed. Same amount of heat energy is given to the gas in each cylinder. If the rise in temperature of the gas in $\mathrm{A}$ is $42 \mathrm{~K}$, the rise in temperature of the gas in $B$ is

Options:
A $21 \mathrm{~K}$
B $42 \mathrm{~K}$
C $35 \mathrm{~K}$
D $70 \mathrm{~K}$

The correct answer is:
$35 \mathrm{~K}$

Solution:
From first law of thermodynamics
$$
Q=\Delta U+W
$$
For cylinder $A$ pressure remains constant.
$\therefore$ Work done by a system
$$
W=\frac{\mu R}{\gamma-1}\left(T_{1}-T_{2}\right)
$$
For monoatomic gases, $\mu=1 ; \gamma=\frac{5}{3}$
$$
\therefore W=\frac{1 \times R}{\frac{5}{3}-1}(442-400)=\frac{3}{2} R \times 42
$$
or $W=63 R$
But $\Delta U=0$, for cylinder $A$
$$
\therefore Q=0+63 R=63 R
$$
For cylinder $B$ volume is constant, $\therefore W=0$ and $Q=\mu C_{V} \Delta T$
For monoatomic gas
$$
C_{V}=\frac{3}{2} R \Rightarrow Q=1 \times \frac{3}{2} R \Delta T
$$
As heat given on both cylinder is same,
$$
\therefore 63 R=\frac{3}{2} R \Delta T \Rightarrow \Delta T=42 \mathrm{~K}
$$

Topic: Physics, Thermodynamics, VITEEE, VITEEE 2009
Two cylinders of \( 10 \) litres and \( 30 \) litres, containing air at STP and vaccum conditions respectively, are connected. What is the final air pressure in the two cylinders if the temperature remains constant?

Options:

The correct answer is:
0.25

Solution:
$\mathbf{P}_{1}=1$ atm (Initial pressure at S.T.P.) $\mathbf{P}_{2}=?, \mathbf{V}_{1}=10$ litres, $\mathbf{V}_{2}=30+10=40$ litres
According to Boyle's law, $P_{1} V_{1}=P_{2} V_{2} ; 1 \times 10=P_{2} \times 40$
$\mathbf{P}_{2}=\frac{1 \times 10}{40}$
$=0.25$ atm

Topic: Chemistry, States of Matter, JEE Main