datavorous/entrance-exam-dataset huggingface dataset:
A fighter jet is flying horizontally at a certain altitude with a speed of 200 m s-1. When it passes directly overhead an anti-aircraft gun, a bullet is fired from the gun, at an angle θ with the horizontal, to hit the jet. If the bullet speed is 400 m s-1, the value of θ will be _____ °.

Options:

The correct answer is:
60

Solution:
In order to hit, horizontal velocity of both moving object should be same,

V1=V2cosθ ⇒200=400cosθ

⇒cosθ=12 ⇒θ=60°

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A fighter plane enters inside the enemy territory, at time t=0 with velocity v0=250 m s-1 and moves horizontally with constant acceleration a=20 m s-2 (see figure). An enemy tank, at the border, spots the plane and fire shots at an angle Î¸=60Â°with the horizontal and with velocityu=600 m s-1.The altitude H, where the plane can be hit by the shot, is

Options:
A 15003m
B 125 m
C 1400 m
D 2473 m

The correct answer is:
2473 m

Solution:
If being hit, then horizontal distance remains same.
vx=600 cos 60ο=300 m s-1
250 m s-1t+1220t2=300 t
50 t = 1 2 20 t 2
t = 0 or t = 5
H = 600 sin 60 ο t - 1 2 g t 2
=60032×5-12·1025
H=30035-125 m
H=2473 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A fighter plane enters inside the enemy territory, at time t=0 with velocity v0=250 m s-1 and moves horizontally with constant acceleration a=20 m s-2 (see figure). An enemy tank, at the border, spots the plane and fire shots at an angle Î¸=60Â°with the horizontal and with velocityu=600 m s-1.The altitude H, where the plane can be hit by the shot, is

Options:
A 15003m
B 125 m
C 1400 m
D 2473 m

The correct answer is:
2473 m

Solution:
If being hit, then horizontal distance remains same.
vx=600 cos 60ο=300 m s-1
250 m s-1t+1220t2=300 t
50 t = 1 2 20 t 2
t = 0 or t = 5
H = 600 sin 60 ο t - 1 2 g t 2
=60032×5-12·1025
H=30035-125 m
H=2473 m

Topic: Physics, Motion In Two Dimensions, NEET
A fighter plane, flying horizontally with a speed $360 \mathrm{kmph}$ at an altitude of $500 \mathrm{~m}$ drops a bomb for a target straight ahead of it on the ground. The bomb .should be dropped at what approximate distance ahead of the target ? Assume that acceleration due to gravity $(\mathrm{g})$ is $10 \mathrm{~ms}^{-2}$. Also neglect air drag

Options:
A $1000 \mathrm{~m}$
B $50 \sqrt{5} \mathrm{~m}$
C $500 \sqrt{5} \mathrm{~m}$
D $866 \mathrm{~m}$

The correct answer is:
$1000 \mathrm{~m}$

Solution:
Hint:
$v=360 \mathrm{~km} / \mathrm{h}=100 \mathrm{~m} / \mathrm{s} .$
$\mathrm{h}=500 \mathrm{~m}$
$\mathrm{R}=\mathrm{u} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=100 \times \sqrt{\frac{2 \times 500}{10}}=1000 \mathrm{~m}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A fighter plane, flying horizontally with a speed \( 360 \mathrm{~km} / \mathrm{h} \) at an altitude of \( 500 \mathrm{~m} \) drops a bomb for a target straight ahead of it on the ground. The bomb should be dropped at what approximate distance ahead of the target? Assume that acceleration due to gravity \( (g) \) is \( 10 \mathrm{~ms}^{-2} \). Also, neglect air drag.

Options:
A \( 1000 \mathrm{~m} \)
B \( 50 \sqrt{5} \mathrm{~m} \)
C \( 500 \sqrt{5} \mathrm{~m} \)
D \( 866 \mathrm{~m} \)

The correct answer is:
\( 1000 \mathrm{~m} \)

Solution:
Given, speed of fighter plane, v=360 km/h

=360×518 m/s=100 m/s

Altitude, h=500 m

Now, from equation of motion,
h=ut+12gt2

⇒ 500=0t+12×10×t2 ∵u=0 and g=10 m/s2

⇒ t2=100⇒t=10 s
∴ The bomb should be dropped at the distance,
x=vt=100×10=1000 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A fighter plane is moving in a vertical circle of radius ' $r$ '. Its minimum velocity at the highest point of the circle will be

Options:
A $\sqrt{3 g r}$
B $\sqrt{2 g r}$
C $\sqrt{g r}$
D $\sqrt{g r / 2}$

The correct answer is:
$\sqrt{g r}$

Solution:
Minimum velocity at the highest point of vertical circle $\mathrm{VA}=\sqrt{\mathrm{gr}}$

Topic: Physics, Work Power Energy, JEE Main
A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is 240 ms-1. The earth's magnetic field over Delhi is 5×10-5T with the declination angle ~0o and dip of θ such that sin⁡θ=23. If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to :

Options:
A VB=40 mV;VW=135 mV with left side of pilot at higher voltage
B VB=45Â mV;VW=120Â mV with right side of pilot at higher voltage
C VB=40Â mV;VW=135Â mV with right side of pilot at high voltage
D VB=45 mV;VW=120 mV with left side of pilot at higher voltage

The correct answer is:
VB=45 mV;VW=120 mV with left side of pilot at higher voltage

Solution:
V B = B H 5 2 4 0
B H = B cos θ, ....
Bv=Bsinθ
B H = 5 5 × 1 0 - 5 3
B v = 1 0 3 × 1 0 - 5 T
V B = 5 5 3 × 1 0 - 5 × 5 × 2 4 0
V B = 44.6 mV = 4 5 mV
V w = B v ℓ V = 10 3 × 10 −5 ×15×240
= 1 0 - 4 × 1 2 0 0
V w = 1 2 0 mV
(left side at fighter voltage)

Topic: Physics, Magnetic Properties of Matter, JEE Main
A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is 240 ms-1. The earth's magnetic field over Delhi is 5×10-5T with the declination angle ~0o and dip of θ such that sin⁡θ=23. If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to :

Options:
A VB=40 mV;VW=135 mV with left side of pilot at higher voltage
B VB=45Â mV;VW=120Â mV with right side of pilot at higher voltage
C VB=40Â mV;VW=135Â mV with right side of pilot at high voltage
D VB=45 mV;VW=120 mV with left side of pilot at higher voltage

The correct answer is:
VB=45 mV;VW=120 mV with left side of pilot at higher voltage

Solution:
V B = B H 5 2 4 0
B H = B cos θ, ....
Bv=Bsinθ
B H = 5 5 × 1 0 - 5 3
B v = 1 0 3 × 1 0 - 5 T
V B = 5 5 3 × 1 0 - 5 × 5 × 2 4 0
V B = 44.6 mV = 4 5 mV
V w = B v ℓ V = 10 3 × 10 −5 ×15×240
= 1 0 - 4 × 1 2 0 0
V w = 1 2 0 mV
(left side at fighter voltage)

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2016 (10 Apr Online)
A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance, R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is

Options:
A 230 Ω
B 46 Ω
C 26 Ω
D 13 Ω

The correct answer is:
26 Ω

Solution:
In the series, the current is the same in the bulb and the extra resistance, which is given by,

I=PV=500100=5 A.

But, Vbulb+VR=Vnet.

So, voltage across resistance R will be,

VR=Vnet-Vbulb
⇒VR=230-100=130 V.

From Ohm's law,

R=VRI=1305=26 Ω.

Topic: Physics, Current Electricity, NEET, NEET 2016 (Phase 2)
A film of soap solution is formed between two straight parallel wires of length $10 \mathrm{~cm}$ each separated by $0.5 \mathrm{~cm}$. If their separation is increased by $1 \mathrm{~mm}$ while still maintaining their parallelism. How much work will have to be done?
(surface tension of solution $=65 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ )

Options:
A $7.22 \times 10^{-6} \mathrm{~J}$
B $13.0 \times 10^{-5} \mathrm{~J}$
C $2.88 \times 10^{-5} \mathrm{~J}$
D $5.76 \times 10^{-5} \mathrm{~J}$

The correct answer is:
$13.0 \times 10^{-5} \mathrm{~J}$

Solution:
The increase in surface area of the film is, $\Delta \mathrm{A}=\mathrm{A}_2-\mathrm{A}_1$
$\mathrm{A}_1=2 \times l \times \mathrm{b}=2 \times 10 \times 10^{-2} \times 0.5 \times 10^{-2}$
$\mathrm{A}_2=2 \times l \times(\mathrm{b}+1)=2 \times 10 \times 10^{-2} \times(0.5+0.1)$
$\times 10^{-2}$
$A_2-A_1=\left[2 \times 10 \times 10^{-2} \times(0.5+0.1) \times 10^{-2}\right]$
$-\left[2 \times 10 \times 10^{-2} \times 0.5 \times 10^{-2}\right]$
$=2 \times 10^{-4} \mathrm{~m}^2$
$\therefore \quad$ Work done $=$ Increase in surface energy
$=\mathrm{TdA}$
$=\left(65 \times 10^{-2}\right) \times\left(2 \times 10^{-4}\right)$
$=1.3 \times 10^{-4}$
$=13 \times 10^{-5} \mathrm{~J}$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2023 (09 May Shift 2)
A film projector magnifies a $100 \mathrm{~cm}^2$ film strip on a screen. If the linear magnification is 4 , the area of magnified film on the screen is

Options:
A $1600 \mathrm{~cm}^2$
B $400 \mathrm{~cm}^2$
C $800 \mathrm{~cm}^2$
D $200 \mathrm{~cm}^2$

The correct answer is:
$1600 \mathrm{~cm}^2$

Solution:
$\begin{aligned} & m_L=4 \\ & m_A=\left(m_1\right)^2 \text { so that } A^{\prime}=A_0 \times 16=1600 \mathrm{~cm}^2\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
A finite carrying-current wire is placed in a plane along with a very long, straight conductor carrying current as shown in the figure. The torque on the finite current wire about point O is

Options:
A 2μ0πI1I2l
B μ0πI2I2l
C μ02πI1I2l
D 3μ02πI1I2l

The correct answer is:
μ02πI1I2l

Solution:
B(x)=μ02π⋅I1x, dF=B(x)I2dx
dF=μ02π⋅I1I2xdx, dτ=dF⋅x=μ02πI1I2dx
τ=μ02π⋅I1I2∫l2ldx=μ02πI1I2l

Topic: Physics, Magnetic Effects of Current, JEE Main
A fire work gave bright crimson light. It probably contain an element of

Options:
A Ca
B Sr
C Ba
D Mg

The correct answer is:
Sr

Solution:
Ca = Brick red
Sr = red/crimson
Ba = Apple green
Mg = no flame test

Topic: Chemistry, s Block Elements, JEE Main
A first order reaction has a rate constant $0.00813 \mathrm{~min}^{-1}$. How long will it take for $60 \%$
completion?

Options:
A $98.7 \mathrm{~min}$
B $56.35 \mathrm{~min}$
C $112.7 \mathrm{~min}$
D $62 \cdot 77 \mathrm{~min}$

The correct answer is:
$112.7 \mathrm{~min}$

Solution:
(A)
Original amount of reactant $=100$
$[\mathrm{A}]_{t}=$ Reactant remaining unreacted $=100-60=40$
For first order reaction,
$\therefore t=\frac{2.303}{0.00813 \mathrm{~min}^{-1}} \log _{10} \frac{100}{40}=112.7 \mathrm{~min}$

Topic: Chemistry, Chemical Kinetics, MHT CET, MHT CET 2020 (16 Oct Shift 2)
A first order reaction is \( 50 \% \) completed in \( 1.26 \times 10^{14} \mathrm{~s} \). How much time would it take for \( 100 \% \) completion?

Options:
A \( 1.26 \times 10^{15} \mathrm{~s} \)
B \( 2.52 \times 10^{14} \mathrm{~s} \)
C \( 2.52 \times 10^{28} \mathrm{~s} \)
D infinite

The correct answer is:
infinite

Solution:
It is impossible to perform 100% of the reaction, whole of the substance never react. Hence, time taken for 100% completion of a reaction is infinite.

t=1kln(N0(N0-x)),N0-x=0,t=1k×Infinity

So, time taken is infinite to complete the reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of \(3.86 \mathrm{BM}\). The atomic number of the metal is

Options:
A 26
B 25
C 23
D 22

The correct answer is:
23

Solution:
$\begin{aligned} & { }_{22} \mathrm{Ti}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^2 \\ & { }_{23} \mathrm{~V}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^3 \\ & { }_{25} \mathrm{Mn}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 \\ & { }_{26} \mathrm{Fe}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^6\end{aligned}$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula \(\mathrm{M}_2 \mathrm{O}_{\mathrm{n}}\) (where \(\mathrm{n}=3,4\), 5). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is ______ \(\mathrm{BM}\) (near integer)
(Given atomic number: \(\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26, \mathrm{Co}: 27\), \(\mathrm{Ni}: 28, \mathrm{Cu}: 29, \mathrm{Zn}: 30\))

Options:

The correct answer is:
0

Solution:
'$\mathrm{V}$' has highest enthalpy of atomisation ( $515 \mathrm{~kJ} / \mathrm{mol}$ ) among first row transition elements.
$\mathrm{V}_2 \mathrm{O}_5$
Here ' $\mathrm{V}$ ' is in +5 oxidation state
$\mathrm{V}^{+5} \Rightarrow 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 \text { (no unpaired electrons) }$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A fish at a depth of $12 \mathrm{~cm}$ in water is viewed by an observer on the bank of a lake. To what height the image of the fish is raised.

Options:
A $9 \mathrm{~cm}$
B $12 \mathrm{~cm}$
C $3.8 \mathrm{~cm}$
D $3 \mathrm{~cm}$

The correct answer is:
$3 \mathrm{~cm}$

Solution:
Apparent rise $=d\left(1-\frac{1}{a \mu_w}\right)$
$\begin{aligned} & =12 \times\left(1-\frac{3}{4}\right) \\ & =12 \times \frac{1}{4}=3 \mathrm{~cm}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is $4 / 3$ and the fish is $12 \mathrm{~cm}$ below the surface, the radius of this circle in $\mathrm{cm}$ is

Options:
A
$36 \sqrt{7}$

B
$36 / \sqrt{7}$

C
$36 \sqrt{5}$

D
$4 \sqrt{5}$

The correct answer is:
$36 / \sqrt{7}$

Solution:
$r=\frac{h}{\sqrt{\mu^2-1}}=\frac{36}{\sqrt{7}}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2005
A fish looking upwards through the water seeing the outside world contained in a circular horizon. If the refractive index of water is 43 and the fish is 12 cm below the surface, the radius of this circle in cm is

Options:
A 36 5
B 4 5
C 36 7
D 36 / 7

The correct answer is:
36 / 7

Solution:
For total internal reflection,
μ = 1 sin θ C ⇒ sin θ C = 1 μ = 3 4
∴ tan θ C = sin θ C 1 - sin 2 θ C
= 3 / 4 1 - 9 16 = 3 4 × 4 7 = 3 7
∴ R 12 = 3 7 ⇒ R = 36 7 cm .

Topic: Physics, Ray Optics, JEE Main
A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100 °C, then the internal energy for vaporization in kJmol-1 is integer]

[Assume steam to be an ideal gas. Given ΔvapH⊖ for water at 373 K and 1 bar is 41.1 kJ mol-1:R=8.31 J K-1 mol-1]

Options:

The correct answer is:
38

Solution:
Weight of water on fish =36 gram

H2Ol→P=1barT=100 °CH2Og;ΔHvap=41.1 kJ mole-1

ΔH=ΔU=ΔngRT

41.1×2=ΔU+2×8.3141000373

ΔU=82.2-6.2=76kJ

so change in internal energy in kJ/mol=762=38kJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (26 Jun Shift 2)
A five digit number is formed by the digits 1 , $2,3,4,5$ with no digit being repeated. The probability that the number is divisible by 4 , is

Options:
A $\frac{1}{5}$
B $\frac{2}{5}$
C $\frac{3}{5}$
D $\frac{4}{5}$

The correct answer is:
$\frac{1}{5}$

Solution:
Number is divisible by 4 , if last two digits are 12, 24, 32 and 52.
Remaining 3 place will be filled by 3 ! ways.
$$
\begin{aligned}
\therefore \quad \text { Favourable outcomes } & =3 ! \times 4 \\
\text { Hence, required probability } & =\frac{3 ! \times 4}{5 !} \\
& =\frac{3 ! \times 4}{5 \times 4 \times 3 !}=\frac{1}{5}
\end{aligned}
$$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2016
A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure.

The succeeding operation that enable this transformation of state are

Options:
A heating, cooling, heating, cooling
B cooling, heating, cooling , heating
C heating, cooling, cooling, heating
D cooling , heating , heating , cooling

The correct answer is:
heating, cooling, cooling, heating

Solution:
K→L : Isochoric process with increasing in pressure so by TLTK=PLPK
PL>PK so TL>TK so it is a case of heating
L→M : Isobaric process with increasing in pressure so by TMTL=VMVL
VM<VL so TM<TL so it is a case of cooling
Similarly for
M→N : Isochoric process with increasing in pressure so by TNTM=PNPM
PN<PM so TN<TM so it is a case of cooling

Topic: Chemistry, Thermodynamics (C), JEE Main
A fixed mass of oxygen gas performs a cyclic process ABCA as shown in the figure. Efficiency of the cycle will be

Options:
A 3 ln 3 - 2 5 + 3 ln 3
B 3 ln 4 - 2 4 + 5 ln 3
C 3 ln 4 - 3 4 + 5 ln 3
D 3 ln 3 - 1 6 + 3 ln 4

The correct answer is:
3 ln 3 - 2 5 + 3 ln 3

Solution:
A → B isochoric as P is directly proportional to T
W AB = 0
Q AB = n C v 3 T - T
= 5 2 nR 2 T = 5 nRT
B → C isothermal and thus Q = W as there won't be any change in internal energy.
W BC = nR 3 T ln P B P C = 3 nRT ln 3
= 3 nRT ln 3 = Q BC
C →A isobaric process
W CA = P V A - V C = nR T A - 3 T = - 2 n RT
Efficiency of a cyclic process is given by
η = Work done in cyclic process Heat supplied to the system = W Q in
Efficiency η = 3 nRT A ln 3 - 2 nRT A 5 nRT A + 3 nRT A ln 3
= 3 ln 3 - 2 5 + 3 ln 3

Topic: Physics, Thermodynamics, JEE Main
A fixed mass of oxygen gas performs a cyclic process ABCA as shown in the figure. Efficiency of the cycle will be

Options:
A 3ln3-25+3 ln3
B 3ln4-24+5ln3
C 3ln4-34+5ln3
D 3 ln3-16+3ln4

The correct answer is:
3ln3-25+3 ln3

Solution:
A â B isochoric as P is directly proportional to T
W AB = 0
Q AB = n C v 3 T - T
= 5 2 nR 2 T = 5 nRT
B â C isothermal and thus Q = W as there won't be any change in internal energy.
W BC = nR 3 T ln P B P C = 3 nRT ln 3
= 3 nRT ln 3 = Q BC
C âA isobaric process
W CA = P V A - V C = nR T A - 3 T = - 2 n RT
Efficiency of a cyclic process is given by
Î· = Work done in cyclic process Heat supplied to the system = W Q in
Efficiency Î· = 3 nRT A ln 3 - 2 nRT A 5 nRT A + 3 nRT A ln 3
= 3 ln 3 - 2 5 + 3 ln 3

Topic: Physics, Thermodynamics, JEE Main
A flag - staff of 5 meters high stands on a building of 25 meters height. For an observer at a height of 30 meters, the flag-staff and the building subtend equal angles. The distance of the observer from the top of the flag - staff is

Options:
A 532 m
B 532 m
C 523 m
D None of these

The correct answer is:
532 m

Solution:
We have tanâ¡Î±=5x and tanâ¡2Î±=30x
â´ tanâ¡2Î±=305cotâ¡Î± â tanâ¡2Î±=6tanâ¡Î±

Topic: Mathematics, Heights and Distances, JEE Main
A flagstaff stands vertically on a pillar, the height of the flagstaff being double the height of the pillar. A man on the ground at a distance finds that both the pillar and the flagstaff subtend equal angles at his eyes. The ratio of the height of the pillar and the distance of the man from the pillar is

Options:
A 31
B 13
C 13
D 32

The correct answer is:
13

Solution:
Let hx=a, then tan⁡θ=hx=a
tan⁡2θ=3hx=3a
3a=tan⁡2θ=2tan⁡θ1-tan2⁡θ=2a1-a2⇒a=13

Topic: Mathematics, Heights and Distances, JEE Main
A flagstaff stands vertically on a pillar, the height of the flagstaff being triple the height of the pillar. A man on the ground at a distance finds that both the pillar and the flagstaff subtend equal angles at his eyes. The ratio of the height of the pillar and the distance of the man from the pillar is

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

The correct answer is:
1:2

Solution:
Let, AB is a pillar and BC is the flagstaff on it and man is standing at D
We have, hx=tan⁡α=a (say) (in ∆ABD )
tan⁡2α=4hx=4a (in ∆ACD )
⇒2tan⁡α1-tan2⁡α=4a
⇒2a1-a2=4a⇒2=4-4a2⇒4a2=2⇒a=12
∴h:x=1:2

Topic: Mathematics, Heights and Distances, JEE Main
A flask containing air (open to the atmosphere) is heated from 300K to 500K. Then % of air escaped to the atmosphere is

Options:

The correct answer is:
40

Solution:
P1V1=n1RT1
P1V1=n2RT2
So n1RT1=n2RT2 (P of V are constant)
n1T1=n2T2 ,
n1300=n2500
n2=35n1
No. of moles escaped =n1-35n1=25n1
% escaped =25×100=40%

Topic: Chemistry, States of Matter, JEE Main
A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is : (Use ln 2=0.693)

Options:
A 180
B 900
C 300
D 120

The correct answer is:
900

Solution:
Ct=C0e-kt

kA=ln2300 k=ln2T1/2

kB=ln2180

CtA=C0Ae-kAt

CtB=C0Be-kBt

Ct,BCt,A=C0,BC0,A×ekA-kBt

14=ekA-kBt

ln14=ln2300-ln2180t

-2ln2=ln2300-ln2180t

2=120180×300t

t=2×180×300120=900 sec

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A flask contains a mixture of isohexane and 3 -methylpentane. One of the liquids boils at 63oC while the other boils at 60oC. What is the best way to separate the two liquids and which one will be distilled out first?

Options:
A fractional distillation, isohexane
B simple distillation, 3 -methylpentane
C simple distillation, isohexane
D fractional distillation, 3 -methylpentane

The correct answer is:
fractional distillation, isohexane

Solution:
Liquid having lower boiling point comes out first in fractional distillation. fractional distillation is generally used if in boiling point is small.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (08 Jan Shift 1)
A flask contains argon and oxygen in the ratio of 3:2 in mass and the mixture is kept at 27°C. The ratio of their average kinetic energy per molecule respectively

Options:
A 3:5
B 9:4
C 2:3
D 1:1

The correct answer is:
3:5

Solution:
Given here, mArmO=32.

As argon is monoatomic gas, so it has three degree of freedom f and oxygen is a diatomic gas, so its degree of freedom is five.

We know that average kinetic energy per molecule is given by KE=f2kT, where, k is Boltzmann's constant and T is temperature.

Here, T=27°C is fixed for both gas.

So, KEArKEO=32kT52kT=35

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A flask contains Hydrogen and Argon in the ratio 2:1 by mass. The temperature of the mixture is 30°C. The ratio of average kinetic energy per molecule of the two gases KargonKhydrogen is: (Given : Atomic Weight of Ar=39.9)

Options:
A 2
B 1
C 39.9
D 39.92

The correct answer is:
1

Solution:
The formula to calculate the average kinetic energy per molecule of a gas at any temperature T is given by

Kavg=32kBT

where, kB is Boltzmann constant.

From above equation, it can be concluded that the average kinetic energy per molecule is independent of the molecular weight of the gas.

Hence, the required ratio is 1:1.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A flask contains hydrogen and oxygen in the ratio of $2: 1$ by mass at temperature $27^{\circ} \mathrm{C}$. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is :

Options:
A
2 : 1
B
1 : 1
C
1 : 4
D
4 : 1

The correct answer is:
1 : 1

Solution:
Average kinetic energy for diatomic gases
$\begin{aligned}
& \mathrm{K}_{\mathrm{av}}=\frac{5}{2} \mathrm{kT} \\
& \therefore \frac{\left(\mathrm{K}_{\mathrm{av}}\right)_{\mathrm{H}}}{\left(\mathrm{K}_{\mathrm{av}}\right)_{\mathrm{O}}}=\frac{(27+273)}{(27+273)}=1
\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A flask contains hydrogen and oxygen in the ratio of 2 : 1 by mass at temperature 27 °C. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is :

Options:
A 2:1
B 1:1
C 1:4
D 4:1

The correct answer is:
1:1

Solution:
Average Kinetic Energy per molecule KEavg=32kBT. Therefore, we can conclude that the average kinetic energy of the molecules is directly proportional to the temperature of the gas and is independent of pressure, volume or the nature of the gas.

Hence, required ratio =1:1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A flask is filled with 13 g of an ideal gas at 27 Â°C and its temperature is raised to 52 Â°C. The mass of the gas that has to be released to maintain the temperature of the gas in the flask at 52 Â°C, the pressure remaining the same is

Options:
A 2.5 g

B 2.0 g
C 1.5 g
D 1.0 g

The correct answer is:
1.0 g

Solution:
PV ∝ Mass of gas × Temperature
In this problem pressure and volume remain constant, so
M1T1 = M2T2 = constant
∴ M 2 M 1 = T 1 T 2 = 2 7 + 2 7 3 5 2 + 2 7 3 = 3 0 0 3 2 5 = 1 2 1 3
⇒ M2=M1×1213=13×1213 g=12g
i.e., the mass of gas released from the flask = 13 g-12 g=1 g.

Topic: Physics, Thermodynamics, NEET