datavorous/entrance-exam-dataset huggingface dataset:
A vertical pole fixed to the horizontal ground is divided in the ratio 3: 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :

Options:
A 810
B 610
C 1210
D 1215

The correct answer is:
1210

Solution:
Given that

A vertical pole fixed to the horizontal ground is divided in the ratio 3: 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole.

tanθ=x6
tan2θ=10x18=5x9

We know that tan 2θ = 2 tan θ1 - tan2θ

5x9=x31-x236

⇒ 5=31-x236

1-x236=35

x236=25

⇒ x2=725 ⇒ x=625

From the figure ;

Length of pole =10x=6025×55

=60105=1210m

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A vertical spring of force constant 100 Nm-1 is attached to the ceiling, with a block of mass 10 kg suspended from it. Now an external force F is applied on the block so that the spring is stretched by an additional length of 2 m. The work done by the force F is [Take g=10 ms-2]

Options:
A 200 J
B 400 J
C 100 J
D 600 J

The correct answer is:
200 J

Solution:
x = elongation in spring due to mass 10 kg
= 1 0 Ã 1 0 1 0 0 = 1 m
W F = 1 2 Ã 1 0 0 Ã 3 2 - 1 2 - 1 0 Ã 1 0 Ã 2
= 200 J

Topic: Physics, Work Power Energy, JEE Main
A vertical tower subtends an angle of 60o at a point on the same level as the foot of the tower. On moving 100 m further from the first point in line with the tower, it subtends an angle of 30o at the point. If the height of the tower is H m, then the value of H253 (in meters) is

Options:

The correct answer is:
2

Solution:
tan⁡60o=Hx, tan⁡30o=Hx+100
⇒ Htan60∘=Htan30∘-100
⇒ H3=H13-100
⇒ 3H-H3=100⇒3H-H=1003
⇒H=503
H253=2

Topic: Mathematics, Heights and Distances, JEE Main
A vertical tower subtends an angle of 60 o at a point on the same level as the foot of the tower. On moving 100 m further in line with the tower, it subtends an angle of 30 o at the point. Then, the height of the tower is

Options:
A 203 m
B 303 m
C 503 m
D 103 m

The correct answer is:
503 m

Solution:
From ∆ABD and ∆ACD,

tan60o=Hx, tan30o=Hx+100
⇒Htan⁡60°=Htan⁡30°-100
⇒H3=H13-100
⇒3H-H3=100
⇒3H-H=1003
⇒H=503 m

Topic: Mathematics, Heights and Distances, JEE Main
A vertical U-tube of uniform inner cross section contains mercury in both sides of its arms. $\mathrm{A}$ glycerin (density = $1.3 \mathrm{~g} / \mathrm{cm}^3$) column of length $10 \mathrm{~cm}$ is introduced into one of its arms. Oil of density $0.8 \mathrm{gm} / \mathrm{cm}^3$ is poured into the other arm until the upper surfaces of the oil and glycerin are in the same horizontal level. Find the length of the oil column, Density of mercury $=13.6 \mathrm{~g} / \mathrm{cm}^3$

Options:
A $10.4 \mathrm{~cm}$
B $8.2 \mathrm{~cm}$
C $7.2 \mathrm{~cm}$
D $9.6 \mathrm{~cm}$

The correct answer is:
$9.6 \mathrm{~cm}$

Solution:
At the condition of equilibrium Pressure at point $A=$ Pressure at point $B$
$P_A=P_B \Rightarrow 10 \times 1.3 \times g=h \times 0.8 \times g+(10-h) \times 13.6 \times g$
By solving we get $h=9.7 \mathrm{~cm}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A very light, rectangular wire-frame of dimensions 7 cmÃ5 cm hangs just above the free surface of a liquid of surface tension T, with its plane parallel to the free surface. The wire-frame is just brought in contact with the liquid surface and then, lifted up. If the force required to lift the wire-frame is 3.36 N, then what is the value of T (in N m-1)?

Options:

The correct answer is:
7

Solution:
The downward force of surface tension acting on the wire frame is

Fsurface tension=T×l, where l is the total length of the wire which is in contact with the liquid.

Now since it is a wire frame, the free surface of the liquid will be in contact with the inner as well as the outer perimeter of the wire.

Hence, l=2×2×7+5=48 cm

T×48100=3.36 N

⇒T=7 N m-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A very long cylindrical wire is carrying a current Io distributed uniformly over its cross-section area. O is the centre of the cross-section of the wire and the direction of current in into the plane of the figure. The value of â«ABBâ.dIâ along the path AB (from A to B) is

Options:
A μ0I0
B -μ0I06
C μ0I06
D μ0I03

The correct answer is:
-μ0I06

Solution:
According to Ampere's circuital law

∮B→.dl→=μ0Inet

For the closed path OAB
∮B→.dl→=∫OAB→.dl→+∫ABB→.dl→+ ∫BOB→.dl→=μ0-I06

Since the magnetic field is perpendicular to the paths OA and BO

∫ABB→.dl→=-μ0I06

Topic: Physics, Magnetic Effects of Current, JEE Main
A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius R (R<<L). A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of the star is T and its distance from the galaxy's axis is r, then

Options:
A TÂ âÂ r
B T âr
C TÂ âr2
D T2âr3

The correct answer is:
T âr

Solution:
Due to a long solid cylinder gravitational field strength can be given as,
g= 2 G λr
Where,
λ= linear mass density of galaxy.
For the orbital motion around the galaxy,
fg=fcentripetal
⇒ mg=m ω2r
⇒ 2Gλr=ω2r
⇒ ω2 ∝1r2
⇒ ω ∝1r
⇒ 2πT ∝1r ⇒ T∝r
So option B is correct.

Topic: Physics, Gravitation, JEE Main, JEE Main 2015 (10 Apr Online)
A very long solenoid is made out of a wire with n turns per unit length. The radius of the cylinder is a and is negligible compared to its length l. The interior of the cylinder is filled with materials such that the linear magnetic permeability varies with the distance r from axis according to
Î¼r=Î¼1=constant;for 0<r<bÎ¼2=constant;for b<r<a
The self-inductance of the solenoid is

Options:
A πn2lμ1b2+μ2b2
B πn2lμ1+μ2a2
C πn2lμ1b2+μ2a2-b2
D πn2lμ1b2+μ1+μ2a2

The correct answer is:
πn2lμ1b2+μ2a2-b2

Solution:
Self-inductance of the solenoid is given by
L=Î¼rn2Ïa2l
Here, Î¼ varies with distance r.
So, the self-inductance can be written as
L=Î¼1n2lÏb2 0<r<b+Î¼2n2lÏa2-Ïb2b<r<a
=Ïn2lÎ¼1b2+Î¼2a2-b2

Topic: Physics, Electromagnetic Induction, JEE Main
A very long solenoid of radius R is carrying current It=kte-αtk>0, as a function of time t≥0. Counterclockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equitorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:

Options:
A
B
C
D

The correct answer is:

Solution:
ϕouter=μ0nKte-αt4πR2
ε=-dϕdt=-μ0nK4πR2(e-αt1-αt)
iinduced=-ce-αt1-αtResistance
At t=0, Iinduced=-ve
From the above equation and initial value of current we can say the correct graph is B.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A very long straight conducting wire, lying along the z-axis, carries a current of 2 A. The integral â« B â . d â â is computed along the straight line PQ, where P has the coordinates (2 cm, 0, 0) and Q has the coordinates (2 cm, 2 cm, 0). The integral has the magnitude (in T m)

Options:
A Zero
B 8 π × 1 0 - 7
C 2 π × 1 0 - 7
D Ï Ã 1 0 - 7

The correct answer is:
π × 1 0 - 7

Solution:
â« PAQ B â . d â â = â« RBQ B â . d â â â« PCR B â . d â â = 0

=Î¼0I2ÏRÏ4R
=4ÏÃ10-728=ÏÃ10-7 T m

Topic: Physics, Magnetic Effects of Current, JEE Main
A very long straight wire of radius \( r \) carries current \( I \). Intensity of magnetic field \( B \) at a point lying at a perpendicular distance \( a \) from the axis is \( \propto \) (where \( a < r \) ).

Options:
A \( 1 / a \)
B \( 1 / a^{2} \)
C \( a^{2} \)
D \( a \)

The correct answer is:
\( a \)

Solution:
Magnetic field inside a wire is given by,

B=μ0ia2πr2

⇒B∝a

Topic: Physics, Magnetic Properties of Matter, JEE Main
A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic filed at the center of circle is:

Options:
A μ0I2πRπ+12
B μ0I2πRπ-12
C μ0I2πRπ+1
D μ0I2R

The correct answer is:
μ0I2πRπ+12

Solution:
B0→=B0→1+B0→2+B0→3+B0→4
μ0i4πRsin⁡90°-sin45°⨂+μ0i2R⨀+ μ0i4πR(sin45°+sin90°)⨀
-μ0i4πR1-12+μ0i2R+μ0i4πR12+1⨀
=μ0i4πR-1+12+2π+12+1⨀=μ0i4πR2+2π⨀=μ0i2πR12+π⨀

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A very narrow capillary tube records a rise of 20 cm when dipped in water. When the area of cross-section is reduced to one-fourth of former value, water will rise to a height of

Options:
A 10Â cm
B 20Â cm
C 40 cm
D 80Â cm

The correct answer is:
40 cm

Solution:
A∝r2
∴ radius becomes half.
As Hr=constant.
∴H becomes double.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A very small sphere having a charge q, uniformly distributed throughout its volume, is placed at the vertex of a cube of side a. The electric flux through the cube is

Options:
A qε0
B q3ε0
C q6ε0
D q8ε0

The correct answer is:
q8ε0

Solution:
If charge q is placed at a corner of cube, it will be divided into 8 such cubes. Therefore, electric flux through the cube is
ϕ′=18qε0

Topic: Physics, Electrostatics, JEE Main
A vessel ABCD of 10 cm width has two small slits S 1 and S 2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm . POQ is the line perpendicular to the plane AB and passing through O , the

middle point of S 1 and S 2, A monochromatic light source is kept at S, 40 cm cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure alongside. After a liquid is poured into the vessel and filled upto OQ, it was found that the central bright fringe is now located at Q. Calculate the refractive index of that liquid.

Options:
A 1.0016
B 2.0032
C 1.0010
D 1.0026

The correct answer is:
1.0016

Solution:
Given Y 1 =40 cm, D 1 =2âm=200âcm, D 2 =10âcm
tan Î± = y 1 D 1 = 4 0 2 0 0 = 1 5
â´ Î± = tan -1 1 / 5
sin Î± = 1 2 6 â 1 5 = tan Î±
Initial path difference Îx 1 ,â= SS 1 â SS 2 = y 1 d D 1 = y 1 d D 1 & geometrical path difference Î x 2 S 1 Râ S 2 R= - y 2 d D 2
For central maxima, total path difference has to be zero.
i.e., y 1 d D 1 - y 2 d D 2 = 0
â y 2 = y 1 D 2 D 1 = 4 0 cm Ã 1 0 cm 2 0 0 cm = 2 cm
Now as central bright fringe is at Q. them the optical path difference is Î¼ - 1 t & for central maxima Î¼ - 1 t = Î x 1
Î¼ = 1 + Î x 1 t = 1 + y 1 d t D 1
= 1 + 0 . 4 m Ã 0 . 8 mm 0 . 1 m Ã 2 m
= 1 + 2 Ã 0 . 8 Ã 1 0 - 3
= 1 + 1 . 6 Ã 1 0 - 3
= 1 . 0 0 1 6

Topic: Physics, Wave Optics, JEE Main
A vessel at $1000 \mathrm{~K}$ contains $\mathrm{CO}_2$ with a pressure of $0.5 \mathrm{~atm}$. Some of the $\mathrm{CO}_2$ is converted into $\mathrm{CO}$ on the addition of graphite. If the total pressure at equilibrium is $0.8 \mathrm{~atm}$, the value of $\mathrm{K}$ is

Options:
A
$3 \mathrm{~atm}$

B
$0.3 \mathrm{~atm}$

C
$0.18 \mathrm{~atm}$

D
$1.8 \mathrm{~atm}$

The correct answer is:
$1.8 \mathrm{~atm}$

Solution:
Total pressure at equilibrium $=0.8 \mathrm{~atm}$; Total no.of moles $=p+x$.
Therefore $\mathrm{p} \propto \mathrm{n} ; \frac{0.5}{0.8}=\frac{\mathrm{p}}{\mathrm{p}+\mathrm{x}} \Rightarrow \mathrm{x}=0.3$
$$
\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{CO}}^2}{\mathrm{P}_{\mathrm{CO}_2}}=\frac{0.6 \times 0.6}{0.2}=1.8 \mathrm{~atm}
$$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2011
A vessel completely filled with water has holes A and B at depths h and 3h from the top respectively. Hole A is a square of side L and B is circle of radius r. The water flowing out per second from both the holes is same. Then L is equal to

Options:
A r12π12312
B rπ14314
C rπ12314
D r12π13312

The correct answer is:
rπ12314

Solution:
A 1 V 1 = A 1 V 2
L 2 2 gh = π r 2 6 gh
L 4 gh = π 2 r 4 3 gh
L = π 1 2 r 3 1 4
L = r π 1 2 3 1 4

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A vessel completely filled with water has holes A and B at depths h and 3h from the top respectively. The hole A is a square of side L and B is a circle of radius r. The water flowing out per second from both the holes is same. Then L is equal to

Options:
A r12π12312
B rπ14314
C rπ12314
D r12π13312

The correct answer is:
rπ12314

Solution:
A 1 V 1 = A 1 V 2
L 2 2 gh = π r 2 6 gh
L 4 gh = π 2 r 4 3 gh
L = π 1 2 r 3 1 4
L = r π 1 2 3 1 4

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A vessel containing 1 more of $\mathrm{O}_{2}$ gas (molar mass 32) at temperature $\mathrm{T}$. The pressure of the gas is p. An identical vessel containing one mole of he gas (molar mass 4 ) at temperature $2 \mathrm{~T}$ has a pressure of

Options:
A $\mathrm{p} / 8$
B $\mathrm{p}$
C $2 \mathrm{p}$
D $8 \mathrm{p}$

The correct answer is:
$2 \mathrm{p}$

Solution:
Applying gas equation, $\mathrm{pV}=\mathrm{nRT}$ We can write, $\mathrm{p}_{1} \mathrm{~V}=\mathrm{n}_{1} \mathrm{RT}_{1}$ and $\mathrm{p}_{2} \mathrm{~V}=\mathrm{n}_{2} \mathrm{RT}_{2}$

$\Rightarrow \quad \frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1}{1} \times \frac{2 \mathrm{~T}}{\mathrm{~T}}=2$

$\Rightarrow \mathrm{p}_{2}=2 \mathrm{p}$

Topic: Physics, Thermodynamics, JEE Main
A vessel containing water is given a constant acceleration a towards the right, along a straight horizontal path, which of the following diagram represents the surface of the liquid?

Options:
A
B
C
D

The correct answer is:

Solution:
Due to acceleration in forward direction, vessel is an accelerated frame therefore a Pseudo force will be exerted in backward direction. Therefore water will be displaced in backward direction.Hence, C is the right option

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A vessel contains 1 g of oxygen at a pressure of 10 atm and a temperature of 47 Â°C. It is found that because of a leak, the pressure drops to 58th of its original value and the temperature falls to 27 Â°C. Find the mass of oxygen that is leaked out.

Options:
A 13 g
B 148 g
C 1 g
D 23 g

The correct answer is:
13 g

Solution:
The pressure, temperature and the number of moles of oxygen in the vessel change due to leak, while the volume remains fixed. Hence using PV=nRT, we have
P1n1T1=P2n2T2ân2=320300Ã58Ã132=148 moles
Mass leaking out=1-n232=1-2/3=13 g

Topic: Physics, Thermodynamics, NEET
A vessel contains 1 g of oxygen at a pressure of 10 atm and a temperature of 47 Â°C. It is found that because of a leak, the pressure drops to 58th of its original value and the temperature falls to 27 Â°C. Find the mass of oxygen that is leaked out.

Options:
A 1 3 g
B 1 48 g
C 1 g
D 2 3 g

The correct answer is:
1 3 g

Solution:
The pressure, temperature and the number of moles of oxygen in the vessel change due to leak, while the volume remains fixed. Hence using PV=nRT, we have
P1n1T1=P2n2T2⇒n2=320300×58×132=148 moles
Mass leaking out=1-n232=1-2/3=13 g

Topic: Physics, Thermodynamics, NEET
A vessel contains 1 mole of O2 at 270C and 1 atm pressure.A certain amount of the gas was withdrawn and the vessel was heated to 3270C to maintain the pressure of 1 atm. The amount of gas removed was

Options:
A 0.2 mole
B 0.5 mole
C 0.25 mole
D 0.1 mole

The correct answer is:
0.5 mole

Solution:
P1V1=n1RT1
P2V2=n2RT2
P1=P2, V1=V2
âµn1T1=n2T2
n2=300600=12=0.5 moles

Topic: Chemistry, States of Matter, NEET
A vessel contains 14 g of nitrogen gas at a temperature of 27°C. The amount of heat to be transferred to the gas to double the r.m.s. speed of its molecules will be : (Take R=8.32 J mol-1k-1)

Options:
A 2229 J
B 5616 J
C 9360 J
D 13,104 J

The correct answer is:
9360 J

Solution:
The RMS speed of a gas at a given temperature is given by, vrms=3RTM.

Therefore, to double the RMS speed, the temperature should be increased to four times the initial temperature. Therefore,

Tf=1200 K, Ti=300 K, and n=1428=12

The volume of the vessel will remain the same therefore, the process is an isochoric process. Therefore,

Q=nCvΔT=12×5R2×900

⇒Q=9360 J

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A vessel contains 16 g of hydrogen and 128 g of oxygen at standard temperature and pressure. The volume of the vessel in cm3 is :

Options:
A 72×105
B 32×105
C 27×104
D 54×104

The correct answer is:
27×104

Solution:
Number of moles of Hydrogen H2=mM=162=8

Number of moles of oxygen O2=12832=4

Using ideal gas equation, PV=nRT,

Where pressure P=1 atm=1.01×105 Pa STP

Putting the values,

1.01×105 V=8+4×8.314×273

V=12×8.314×2731.01×105 m3

=0.27×106 cm3=27×104 cm3

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (29 Jun Shift 2)
A vessel contains $110 \mathrm{~g}$ of water. The heat capacity of the vessel is equal to $10 \mathrm{~g}$ of water. The initial temperature of water in vessel is $10^{\circ} \mathrm{C}$. If $220 \mathrm{~g}$ of hot water at $70^{\circ} \mathrm{C}$ is poured in the vessel, the final temperature neglecting radiation loss, will be

Options:
A $70^{\circ} \mathrm{C}$
B $80^{\circ} \mathrm{C}$
C $60^{\circ} \mathrm{C}$
D $50^{\circ} \mathrm{C}$

The correct answer is:
$50^{\circ} \mathrm{C}$

Solution:
Let final temperature of water be $\theta$ Heat taken $=$ Heat given
$\begin{aligned} & 110 \times 1(\theta-10)+10(\theta-10)=220 \times 1(70-\theta) \\ & \Rightarrow \theta=48.8^{\circ} \mathrm{C} \approx 50^{\circ} \mathrm{C} .\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A vessel contains oil (density 0.8   gm/cm 3 ) over mercury (density =13.6   gm/cm 3 ). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in g m/cm 3 is

Options:
A 3.3
B 6.4
C 7.2
D 12.8

The correct answer is:
7.2

Solution:
As the sphere floats in the liquid. Therefore its weight will be equal to the upthrust force on it
Weight of sphere = 4 3 π R 3 ρ g ...... i
Upthrust due to oil and mercury = 2 3 π R 3 × σ oil g + 2 3 π R 3 σ Hg g ...... i i
Equating (i) and (ii)
4 3 π R 3 ρ g = 2 3 π R 3 0 . 8 g + 2 3 π R 3 × 1 3 . 6 g
⇒ 2 ρ = 0 . 8 + 1 3 . 6 = 1 4 . 4
⇒ ρ = 7 . 2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A vessel contains two non-reactive gases neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. the ratio of number of molecules is

Options:
A 32
B 23
C 13
D 12

The correct answer is:
32

Solution:
As partial pressure of a gas in a mixture is the pressure it would exert for the same volume and temperature, if it alone occupied the vessel, therefore, for common V and T, P1 and P2 are partial pressures
P1V=Î¼1RT and P2V=Î¼2RT
Here, 1 and 2 refer to neon gas and oxygen gas respectively
Now, P1P2=Î¼1Î¼2âÎ¼1Î¼2=32 ( given P1P2=32)
If N1 and N2 are number of molecules of two gases, then
â´ââN1N2=Î¼1Î¼2=32, where Î¼1=N1NA and Î¼2=N2NA

Topic: Physics, Thermodynamics, JEE Main
A vessel is filled with an ideal gas at a pressure of 10 atm and temperature $27^{\circ} \mathrm{C} .$ Half of the mass is removed from the vessel and temperature of the remaining gas is increased to $87^{\circ} \mathrm{C}$. Then the pressure of the gas in the vessel will be

Options:
A 5 atm
B $6 \mathrm{~atm}$
C $7 \mathrm{~atm}$
D 8 atm

The correct answer is:
$6 \mathrm{~atm}$

Solution:
Ideal gas equation
$$
\begin{array}{c}
p V=n R T \\
\frac{p V}{T}=n R=\frac{m}{M} R \\
\frac{m}{M} R=\frac{10 \times V}{300} \\
\frac{m}{2 M} R=\frac{p V}{360}
\end{array}
$$
$$
\begin{aligned}
\frac{p V}{360} &=\frac{1}{2} \frac{10 \mathrm{~V}}{300} \\
p &=\frac{3600}{2 \times 300}=6 \mathrm{~atm}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
A vessel is filled with an ideal gas at a pressure of 10 atm and temperature $27^{\circ} \mathrm{C} .$ Half of the mass is removed from the vessel and temperature of the remaining gas is increased to $87^{\circ} \mathrm{C}$. Then the pressure of the gas in the vessel will be

Options:
A 5 atm
B $6 \mathrm{~atm}$
C $7 \mathrm{~atm}$
D 8 atm

The correct answer is:
$6 \mathrm{~atm}$

Solution:
Ideal gas equation
$$
\begin{array}{c}
p V=n R T \\
\frac{p V}{T}=n R=\frac{m}{M} R \\
\frac{m}{M} R=\frac{10 \times V}{300} \\
\frac{m}{2 M} R=\frac{p V}{360}
\end{array}
$$
$$
\begin{aligned}
\frac{p V}{360} &=\frac{1}{2} \frac{10 \mathrm{~V}}{300} \\
p &=\frac{3600}{2 \times 300}=6 \mathrm{~atm}
\end{aligned}
$$

Topic: Physics, Thermodynamics, MHT CET, MHT CET 2011
A vessel is filled with an ideal gas at a pressure of 10 atm and temperature 27 Â°C. Half of the mass is removed from the vessel and temperature of the remaining gas is increased to 87 Â°C. Then the pressure of the gas in the vessel will be

Options:
A 5 atm
B 6 atm
C 7 atm
D 8 atm

The correct answer is:
6 atm

Solution:
Ideal gas equation,
pV=nRT
pVT=nR=mMR
mMR=10ÃV300
m2MR=pV360
pV360=1210V300
p=36002Ã300=6 atm

Topic: Physics, Thermodynamics, NEET
A vessel of \(120 \mathrm{~mL}\) capacity contains a certain amount of gas at \(35^{\circ} \mathrm{C}\) and \(1.2\) bar pressure. The gas is transferred to another vessel of volume \(180 \mathrm{~mL}\) at \(35^{\circ} \mathrm{C}\).
What would be its pressure?

Options:

None

Solution:
\(\mathrm{V}_1=120 \mathrm{~mL}, \mathrm{P}_1=1.2 \mathrm{bar}, \mathrm{T}_1=35^{\circ} \mathrm{C}\)
\(\mathrm{V}_2=180 \mathrm{~mL}, \mathrm{P}_2=?, \mathrm{~T}_2=35^{\circ} \mathrm{C}\)
As température remains constant, \(\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2\)
(1.2 bar \()(120 \mathrm{~mL})=\mathrm{P}_2(180 \mathrm{~mL})\)
or \(P_2=0 \cdot 8 \mathrm{bar}\)

Topic: Chemistry, States of Matter
A vessel of depth d is half filled with oil of refractive index n1 and the other half is filled with water of refractive index n2. The apparent depth of this vessel when viewed from above will be-

Options:
A 2dn1+n2n1+n2
B dn1+n22n1n2
C dn1n22n1+n2
D dn1n2n1+n2

The correct answer is:
dn1+n22n1n2

Solution:
The apparent depth formula is given by n=Real depthApparent depth.

The apparent depth in this case is given by

d'=d1n1+d2n2⇒d'=d2(1n1+1n2)⇒d'=d2(n1+n2n1n2)

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A vessel of volume $V$ contains a mixture of 1 mole of hydrogen and 1 mole oxygen (both considered as ideal). Let $f_1(v) d v$, denote the fraction of molecules with speed between $v$ and $(v+d v)$ with $f_2(v) d v$, similarly for oxygen. Then,

Options:
A
$f_1(v)+f_2(v)=f(v)$ obeys the Maxwell's distribution law

B
$f_1(v), f_2(v)$ will obey the Maxwell's distribution law separately

C
neither $f_1(v)$, nor $f_2(v)$ will obey the Maxwell's distribution law

D
$f_2(v)$ and $f_1(v)$ will be the same

The correct answer is:
$f_1(v), f_2(v)$ will obey the Maxwell's distribution law separately

Solution:
For a function $f_1(v)$ the number of molecules $(n)$ which will have their speeds between $v$ and $v+d v$.
For each function $f_1(v)$ and $f_2(v)$ number of molecules remain same 1 mole each but due to mass difference their speed will be different.
Hence both gases of each function $f_1(v)$ and $f_2(v)$ will obey the Maxwell's distribution law separately.

Topic: Physics, Kinetic Theory of Gases
A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of the combination is 4 s. If one of the magnets is removed, then the time period of other one will be

Options:
A 5.78 s
B 3.36 s
C 4.36 s
D 5.36 s

The correct answer is:
3.36 s

Solution:
As magnets are perpendicular to each other, the resultant magnetic moment
M'=M2+M2=2 M
â´ T1=2Ï2I2MH
In the second case, T2=2ÏIMH
T2T1=121/4
â´ T2=421/4=3.36 s

Topic: Physics, Magnetic Properties of Matter, JEE Main