datavorous/entrance-exam-dataset huggingface dataset:
The mass of a lift is $500 \mathrm{~kg}$. When it ascends with an acceleration of $2 \mathrm{~m} / \mathrm{s}^2$, the tension in the cable will be $\left[g=10 \mathrm{~m} / \mathrm{s}^2\right]$

Options:
A $6000 \mathrm{~N}$
B $5000 \mathrm{~N}$
C $4000 \mathrm{~N}$
D $50 \mathrm{~N}$

The correct answer is:
$6000 \mathrm{~N}$

Solution:
$T=m(g+a)=500(10+2)=6000 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
The mass of a lift is $2000 \mathrm{~kg}$. When the tension in the supporting cable is $28000 \mathrm{~N}$, then its acceleration is

Options:
A $30 \mathrm{~ms}^{-2}$ downwards
B $4 \mathrm{~ms}^{-2}$ upwards
C $4 \mathrm{~ms}^{-2}$ downwards
D $14 \mathrm{~ms}^{-2}$ upwards

The correct answer is:
$4 \mathrm{~ms}^{-2}$ upwards

Solution:
Key Idea Apparent weight $>$ actual weight, then the lift is accelerating upward.
Here, lift is accelerating upward at the rate of $a$ Hence, equation of motion is written as
$$
\begin{gathered}
\mathrm{R}-\mathrm{mg}=\mathrm{ma} \\
28000-20000=2000 \mathrm{a} \\
\Rightarrow \mathrm{a}=\frac{8000}{2000}=4 \mathrm{~ms}^{-2} \text { upwards }
\end{gathered}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1) needed to be dissolved in 114 g octane to reduce its vapour pressure by 75 %, is:

Options:
A 37.5Â g
B 75Â g
C 150 g
D 50 g

The correct answer is:
150 g

Solution:
Molar mass of octane = 114 g/mol.

From the lowering of vapour pressure, we have,
∆PP=W2M2W2M2+W1M1
Where W2 and M2 are mass and molar mass of solute
and W1 and M1 are mass and molar mass of octane.
75100=W250 g/molW250 g/mol+114 g114 g/mol
0.75=W250W250+1
W250+1=W250×0.75
W2=150g

Topic: Chemistry, Solutions, JEE Main, JEE Main 2018 (16 Apr Online)
The mass of a non-volatile solute of molar mass $60 \mathrm{~g} \mathrm{~mol}^{-1}$ that should be dissolved in $126 \mathrm{~g}$ of water to reduce its vapour pressure to $99 \%$ will be

Options:
A $2.8 \mathrm{~g}$
B $5.6 \mathrm{~g}$
C $8.4 \mathrm{~g}$
D $4.2 \mathrm{~g}$

The correct answer is:
$4.2 \mathrm{~g}$

Solution:
Given, molar mass of solute, $\mu_{2}=60 \mathrm{~g} / \mathrm{mol}$
Mass of water $\left(W_{1}\right)=126 \mathrm{~g}^{\circ}$
Molar mass of water $\left(M_{1}\right)=18 \mathrm{~g} / \mathrm{mol}$
$p_{\text {Solution }}\left(p_{s}\right)=0.99 p^{\circ}$
Relative lowering of vapour pressure
$\quad \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{W_{2}}{M_{2}} \frac{M_{1}}{W_{1}}$
$\quad \frac{p^{\circ}-0.99 p^{\circ}}{W_{2} \times 18}=\frac{0.1 p^{\circ}}{60 \times 126}=\frac{W_{2} \times 18}{p^{\circ}} 60 \times 126$
$\Rightarrow \quad \frac{60 \times 126}{18 \times 100}=4.2 \mathrm{~g}$

Topic: Chemistry, Solutions, JEE Main
The mass of a pendulum bob is m = 2 kg and the length of the string is l. It is displaced through 90Â° from the vertical and released. What minimum tension (in N) should the string should be able to withstand so that it doesn't break during motion?

Options:

The correct answer is:
60

Solution:
Velocity at the lowest point, v=2gl
At the lowest point, the tension in the string
T=mg+mv2l=mg+ml2gl=3mg = 60 N

Topic: Physics, Laws of Motion, JEE Main
The mass of a planet and its diameter are three times those of earthâs. Then the acceleration due to gravity on the surface of the planet is: g=9.8 m s-2

Options:
A 3.3 m s-2
B 4.9 m s-2
C 19.6 m s-2
D 29.4 m s-2

The correct answer is:
3.3 m s-2

Solution:
gplanet=G3Me3Re2
g=GMeRe2
gplanet=9.83≈3.3 m s-2

Topic: Physics, Gravitation, JEE Main
The mass of a spaceship is $1000 \mathrm{~kg}$. It is to be launched from the earth's surface out into free space. The value of ' $\mathrm{g}$ ' and ' $R$ ' (radius of earth) are $10 \mathrm{~m} / \mathrm{s}^2$ and $6400 \mathrm{~km}$ respectively. The required energy for this work will be ;

Options:
A
$6.4 \times 10^{11}$ Joules

B
$6.4 \times 10^8$ Joules

C
$6.4 \times 10^9$ Joules

D
$6.4 \times 10^{10}$ Joules

The correct answer is:
$6.4 \times 10^{10}$ Joules

Solution:
To launch the spaceship out into free space, from energy conservation,
$\frac{-\mathrm{GMm}}{\mathrm{R}}+\mathrm{E}=0$
$\mathrm{E}=\frac{\mathrm{GMm}}{\mathrm{R}}=\left(\frac{\mathrm{GM}}{\mathrm{R}^2}\right) \mathrm{mR}=\mathrm{mgR}$
$=6.4 \times 10^{10} \mathrm{~J}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2012 (Offline)
The mass of ammonia in grams produced when 2.8 kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is _____

Options:

The correct answer is:
3400

Solution:
Mole of N2=280028=100 & H2=10002=500

Initial  moleFinal moleN2g1000   +    3H2g500500−300  →  2NH3gLimiting reagent is N2200

Mass of NH3 formed =200×17=3400 gram

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The mass of $\mathrm{BaCO}_3$ produced when excess $\mathrm{CO}_2$ is bubbled through a solution of $0.205 \mathrm{~mol}$ $\mathrm{Ba}(\mathrm{OH})_2$ is

Options:
A $81 \mathrm{~g}$
B $40.5 \mathrm{~g}$
C $20.25 \mathrm{~g}$
D $162 \mathrm{~g}$

The correct answer is:
$40.5 \mathrm{~g}$

Solution:
$\mathrm{Ba}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{BaCO}_3+\mathrm{H}_2 \mathrm{O}$
Atomic wt. of $\mathrm{BaCO}_3=137+12+16 \times 3=197$
No. of mole $=\frac{\text { wt. of substance }}{\text { mol wt. }}$
囚 1 mole of $\mathrm{Ba}(\mathrm{OH})_2$ gives 1 mole of $\mathrm{BaCO}_3$
$\therefore 205$ mole of $\mathrm{Ba}(\mathrm{OH})_2$ will give 205 mole of $\mathrm{BaCO}_3$
$\therefore$ wt. of 0.205 mole of $\mathrm{BaCO}_3$ will be $.205 \times 197=40.385 \mathrm{gm} \approx 40.5 \mathrm{gm}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The mass of CaO obtained by heating 100 kg of 95% pure limestone CaCO3 is -

Options:
A 56 kg
B 28 kg
C 53.2 kg
D 50 kg

The correct answer is:
53.2 kg

Solution:
CaCO3(s)âCaO(s)+CO2(g)
mass =105Ã95100Ã100Ã56 gm
=53.2 kg

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The mass of carbon anode consumed (giving only carbon dioxide) in the production of $270 \mathrm{~kg}$ of a aluminium metal from bauxite by the Hall process is:
(Atomic mass : A1 = 27)

Options:
A $270 \mathrm{~kg}$
B $540 \mathrm{~kg}$
C $90 \mathrm{~kg}$
D $180 \mathrm{~kg}$

The correct answer is:
$90 \mathrm{~kg}$

Solution:
$3 \mathrm{C}+2 \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow 4 \mathrm{Al}+3 \mathrm{CO}_2$
(from bauxite)
4 moles of $\mathrm{Al}$ is produced by 3 moles of $\mathrm{C}$
1 mole of $\mathrm{Al}$ is produced by $\frac{3}{4}$ mole of $\mathrm{C}$
$10^4$ moles of $\mathrm{Al}$ is produced by $\frac{3}{4} \times 10^4$ moles of $\mathrm{C}$
Amount of carbon used
$\begin{aligned}
& =\frac{3}{4} \times 10^4 \times 12 \mathrm{~g} \\
& =90000 \mathrm{~g}=90 \mathrm{~kg}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The mass of gas adsorbed, x, per unit mass of adsorbate, m, was measured at various pressures, p. A graph between log xm and log p gives a straight line with slope equal to 2 and the intercept equal to 0.4771. The value of xm at a pressure of 4 atm is :

(Given log 3=0.4771)

Options:

The correct answer is:
48

Solution:
xm=k(P)1n

logxm=logk+1nlogP

Slope =1n=2

so n=12

Intercept , logk = 0.477 So k = Antilog (0.477) = 3

so xm=k(P)1n

=3[4]2=48

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The mass of glucose that should be dissolved in $50 \mathrm{~g}$ of water in order to produce the same lowering of vapour pressure as is produced by dissolving lg of urea in the same quantity of water is

Options:
A $1 \mathrm{~g}$
B $3 \mathrm{~g}$
C $6 \mathrm{~g}$
D $18 \mathrm{~g}$

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

Solution:
Lowering in vapour pressure is directly proportional to moles of solute particles
$$
\begin{aligned}
(\Delta p)_{\text {glucose }} &=(\Delta p)_{\text {urea }} \\
\left(\chi_{B}\right)_{\text {glucose }} &=\left(\chi_{B}\right)_{\text {urea }} \\
\frac{W_{B}}{50} \times \frac{18}{180} &=\frac{1 \times 18}{50 \times 60} \\
W_{B} &=3 \mathrm{~g}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
The mass of lift is $2000 \mathrm{~kg}$. When the tension in the supporting cable is $28000 \mathrm{~N}$, then its acceleration is :

Options:
A $14 \mathrm{~ms}^{-2}$ upwards
B $30 \mathrm{~ms}^{-2}$ downwards
C $4 \mathrm{~ms}^{-2}$ upwards
D $4 \mathrm{~ms}^{-2}$ downwards

The correct answer is:
$4 \mathrm{~ms}^{-2}$ upwards

Solution:
$$
\begin{aligned}
& \Sigma F=m a \\
& \Rightarrow \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\
& \mathrm{a}=\frac{\mathrm{T}-\mathrm{mg}}{\mathrm{m}}=4 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main
The mass of lift is $2000 \mathrm{~kg}$. When the tension in the supporting cable is $28000 \mathrm{~N}$, then its acceleration is :

Options:
A $14 \mathrm{~ms}^{-2}$ upwards
B $30 \mathrm{~ms}^{-2}$ downwards
C $4 \mathrm{~ms}^{-2}$ upwards
D $4 \mathrm{~ms}^{-2}$ downwards

The correct answer is:
$4 \mathrm{~ms}^{-2}$ upwards

Solution:
$$
\begin{aligned}
& \Sigma F=m a \\
& \Rightarrow \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\
& \mathrm{a}=\frac{\mathrm{T}-\mathrm{mg}}{\mathrm{m}}=4 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2009 (Mains)
The mass of NH3 produced when 131.8 kg of cyclohexane carbaldehyde undergoes Tollen’s test is _____ kg. (Nearest Integer)

Molar mass of C=12 g/mol

N=14 g/mol

O=16 g/mol

Options:

The correct answer is:
60

Solution:
RCHO+2AgNH32OH→RCOONH4+2Ag+3NH3+H2O

The RCHO is cyclohexane carbaldehyde (C6H11CHO). The molar mass is112g.

The number of moles of cyclohexane carbaldehyde=131.8×103112=1.176×103 mol

1 mole of aldehyde produces 3 moles of NH3

1.176×103 mol aldehyde will produce

=3×1.176×103 moles of NH3

Mass of NH3 produced =3×1.176×103×17

=59.97×103 g

≈ 60 kg

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The mass of oxygen gas occupying a volume of 11.2 litres at a temperature $27^{\circ} \mathrm{C}$ and a pressure of $760 \mathrm{~nm}$ of mercury in kilograms is $[$ Molecular weight of oxygen $=32]$

Options:
A $0.001456$
B $0.01456$
C $0.1456$
D $1.1456$

The correct answer is:
$0.01456$

Solution:
At STP $(273 \mathrm{~K}$ and at $760 \mathrm{~mm}$ of $\mathrm{Hg})$ the mass of oxygen $=32 \mathrm{~g}$
From
$$
\frac{P_1 V_1}{m_1 T_1}=\frac{P_2 V_2}{m_1 T_2}
$$
$$
\begin{aligned}
& P_1=760 \mathrm{~mm}, V_1=2241 \text { (at STP) } \\
& T_1=273 \mathrm{~K} \\
& P_2=760 \mathrm{~mm} \text { (given), } V_2=11.2 \text { it (given) } \\
& T_2=27+273=300 \mathrm{~K} \\
& \therefore \frac{760 \times 224}{32 \mathrm{~g} \times 273}=\frac{760 \times 11.2}{m_2 \times 300} \\
& m_2=\frac{32 \times 273}{300 \times 2}=14.56 \mathrm{~g} \\
& =0.01456 \mathrm{~kg} \\
&
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
The mass of product liberated on anode in an electrochemical cell depends on
(where t is the time period, for which the current is passed)

Options:
A
$(\mathrm{It})^{1 / 2}$

B
IT

C
I/t

D
$\mathrm{I}^{2} \mathrm{t}$

The correct answer is:
IT

Solution:
From Faradays law of electrolysis, $\mathrm{m} \propto$ it.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2002
The mass of proton, neutron and helium nucleus are respectively 1.0073 u, 1.0087 u and 4.0015u. The binding energy of helium nucleus is:

Options:
A 14.2 MeV
B 28.4 MeV
C 56.8 MeV
D 7.1 MeV

The correct answer is:
28.4 MeV

Solution:
Given here, mHe=4.0015 u, mp=1.0073 u and mn=1.0087 u.

Mass defect

∆m = 2mp+2mn-mHe=2×1.0073+2×1.0087-4.0015=0.0305 u

Binding energy

E = ∆m×931 MeV=0.0305×931=28.4 MeV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The mass of silver (Molar mass of Ag:108gmol-1 ) displaced by a quantity of electricity which displaces 5600 mL of O2 at S.T.P. will be _____ g.

Options:

The correct answer is:
108

Solution:
Given Molar mass of Ag:108 gmol-1

Equal equivalents of substances deposited or liberated when some quantity of electricity passed through circuite in the electrolysis cell.

Equivalents of silver deposited= Equivalents of O2 liberated.
Let x gm silver displaced,
(Molar volume of gas at STP =22.4L )

Number equivalents of a substance = The number of moles of substance × Valence factor of that substance
x×1108=5.622.4×4,x=108 gm.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The mass of sodium acetate CH3COONa required to prepare 250 mL of 0.35M aqueous solution is _____ g. Molar mass of CH3COONa is 82.02 g mol-1) Round off to the nearest integer.

Options:

The correct answer is:
7

Solution:
Given, Molarity= 0.35 M

Moles = Molarity × Volume in litres

=0.35×0.25

No. of moles = Mass/ Molar mass

Mass = moles ×molar mass

=0.35×0.25×82.02=7.18 g

Ans. 7

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The mass of the earth is \( 6.0 \times 10^{24} \mathrm{~kg} \). The potential energy of a body of mass \( 50 \mathrm{~kg} \) at a distance of \( 6.3 \times 10^{9} \mathrm{~m} \) from the centre of the earth is

Options:
A \( 3.23 \times 10^{9} \mathrm{~J} \)
B \( 3.19 \times 10^{6} \mathrm{~J} \)
C \( 2.5 \times 10^{6} \mathrm{~J} \)
D \( 4.0 \times 10^{11} \mathrm{~J} \)

The correct answer is:
\( 3.19 \times 10^{6} \mathrm{~J} \)

Solution:
Mass of the earth (M) = 6.0 x 1024 kg;
Mass of the body (m) = 50 kg and distance (r) = 6.3 x 109 m Potential energy
P.E = GMm r = 6 . 7 × 1 0 - 1 1 × 6 . 0 × 1 0 2 4 × 5 0 6 . 3 × 1 0 9
= 3.19 x 106 J

Topic: Physics, Gravitation, JEE Main
The mass of the moon is 1144 times the mass of a planet and its diameter 116 times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:

Options:
A v3
B v4
C v12
D v6

The correct answer is:
v3

Solution:
As we know, escape velocity is given by, vescape=2GMR

Now, vplanet=2GMR=v

vMoon=2GM×16144×R=132GMR=v3

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The mass of the moon is $7.34 \times 10^{22} \mathrm{~kg}$ and the radius is $1.74 \times 10^6 \mathrm{~m}$. The value of gravitation force will be

Options:
A $1.45 \mathrm{~N} / \mathrm{kg}$
B $1.55 \mathrm{~N} / \mathrm{kg}$
C $1.75 \mathrm{~N} / \mathrm{kg}$
D $1.62 \mathrm{~N} / \mathrm{kg}$

The correct answer is:
$1.62 \mathrm{~N} / \mathrm{kg}$

Solution:
$g=\frac{G M}{R^2}=\frac{6.67 \times 10^{-11} \times 7.34 \times 10^{22}}{\left(1.74 \times 10^6\right)^2}=1.62 \mathrm{~N} / \mathrm{kg}$

Topic: Physics, Gravitation, JEE Main
The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to

Options:
A relative equivalent weight
B gram equivalent weight
C specific equivalent weight
D None of the above

The correct answer is:
gram equivalent weight

Solution:
The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to gram equivalent weight.

Topic: Chemistry, Electrochemistry, JEE Main
The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is ____×10-4 g. (Atomic mass of zinc =65.4amu)

Options:

The correct answer is:
46

Solution:
Given, t = 15 min

Zn+2+2e-⟶Zn

Quantity of charge passed (Q) =Current in amperes ×Time in seconds.
=(0.015A)×(15×60)s=13.5C

Gram equivalent mass of zinc = 65.42=32.7g
Now, 96500 C of charge deposit zinc =32.7 g
13.5 C charge will deposit zinc = (32.7g)(96500C)×(13.5C)=4.57×10−3 g

=45.75×10-4gm

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The mass per unit length of a uniform wire is 0.135 g cm-1. A transverse wave of the form y=-0.21sinx+30t is produced in it, where x is in meter and t is in second. Then, the expected value of tension in the wire is x×10-2 N. Value of x is

(Round-off to the nearest integer)

Options:

The correct answer is:
1215

Solution:
μ=0.135 gm cm-1=0.0135 kg m-1

y=-0.21sinx+30t

(x in meter & t in sec)

v=ωk=301=30 m s-1

v=Tμ⇒T=v2μ=3020.0135

=12.15

=x×10-2 N

⇒x=1215

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The mass percentage of nitrogen in histamine is ________

Options:

The correct answer is:
37.8

Solution:
Structure of Histamine is
Molecular formula of Histamine is C5HgN3
Molecular mass of Histamine is 111
Percentage nitrogen by mass in Histamine =42111×100=37.84% .

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The mass percentage of nitrogen in Histamine is?

Options:
A 18.92
B 57.2
C 37.84
D 34.64

The correct answer is:
37.84

Solution:
The structure of Histamine is

Molecular formula =C5H9N3
Molecular mass =111
% of N =wM×100
% of N=42111×100=37.84%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The masses and radii of the earth and moon are M1, R1 and M2, R2 respectively. Their centres are at a distance r apart. Find the minimum escape velocity for a particle of mass m to be projected from the middle of these two masses :

Options:
A ve=4GM1+M2r
B ve=122GM1+M2r
C ve=124GM1+M2r
D ve=2GM1+M2r

The correct answer is:
ve=4GM1+M2r

Solution:
On conserving mechanical energy. The initial total energy is equal to final total energy, ⇒-GM1r2+GM2r2m+12mve2=0
⇒ve=4GM1+M2r.

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The masses and radii of the earth and the moon are M1, R1 and M2, R2 respectively. Their centres are at a distance d apart. What should be the minimum speed with which a body of mass m should be projected from a point midway between their centres, so as to escape to infinity?

Options:
A GM1+M2d
B 2GM1+M2d
C GdM1+M2
D M1+M2Gd

The correct answer is:
2GM1+M2d

Solution:
The P.E. of the body at C is given by
U=GM1md2-GM2md2=-2GmdM1+M2
Initially it is at rest, hence its K.E. = 0
∴ Total energy of the body =-2GmdM1+M2 and Binding energy =2GmdM1+M2
Let ve be the escape velocity that should be given to the body and its K.E. = Binding energy
∴12mve2=2GmdM1+M2
∴ve=2GM1+M2d

Topic: Physics, Gravitation, JEE Main
The masses M1, M2 and M3 are 5,â2 and 3âkg respectively. These have been joined using massless, inextensible pieces of strings as shown in the figure. If the whole system is going upward with an acceleration of 2 m s-2, then the value of tensions T1,T2 and T3 respectively are [take g=9.8 m s-2]

Options:
A 29.4âN, 98 N, 98 N
B 98âN, 49 N, 29.4 N
C 118âN, 59 N, 35.4 N
D 35.4âN, 118 N, 59 N

The correct answer is:
118âN, 59 N, 35.4 N

Solution:
Given T1=M1+M2+M3(g+a)
=5+2+39.8+2=118 N
T2=M2+M3g+a=2+39.8+2=59 N
T3=M3g+a=3(9.8+2)=35.4 N

Topic: Physics, Laws of Motion, JEE Main
The masses of five balls at rest and lying at equal distances in a straight line are in geometrical progression with ratio 2 and their coefficients of restitution are each 23. If the first ball be started towards the second with velocity u, then the velocity communicated to 5th ball is

Options:
A 59u
B 592u
C 593u
D 594u

The correct answer is:
594u

Solution:
We know that velocity of 2nd ball after collision is given by
v2=u11+em1(m1+m2)+u2(m2-m1e)(m1+m2)
In present problem u2=0,m2=2m1 and , hence
v2=u1+23m1(m1+2m1)=59u
As four exactly similar type of collisions are taking place successively, hence velocity communicated to fifth ball
v5=594u

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The masses of neutron, proton and deutron in amu are 1.00893, 1.00813 and 2.01473, respectively. The packing fraction of the deutron in amu is

Options:
A $11.65 \times 10^{-4}$
B $23.5 \times 10^{-4}$
C $33.5 \times 10^{-4}$
D $47.15 \times 10^{-4}$

The correct answer is:
$11.65 \times 10^{-4}$

Solution:
$m_n=1.00893, m_p=1.00813, m_d=201473$
Deuteron $\left.{ }_1 \mathrm{H}^2\right)$ nucleus consist of one proton and one neutron.
$\begin{aligned}
\therefore \text { Mass defect } & \Delta m=\left(m_n+m_p\right)-m_d \\
& =(1.00893+1.00813)-201473 \\
& =0.00233
\end{aligned}$
$\begin{aligned} \text { Packing fraction } & =\frac{\Delta m}{A} \\ & =\frac{0.00233}{2}=11.65 \times 10^{-4}\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
The masses of three copper wires are in the ratio 2:3:5 and their lengths are in the ratio 5:3:2. Then, the ratio of their electrical resistances is

Options:
A 1:9:15
B 2:3:5
C 5:3:2
D 125:30:8

The correct answer is:
125:30:8

Solution:
As we know that R=ρlA.
So, R1:R2:R3=l1A1:l2A2:l3A3
=l12V1:l22V2:l32V3
=l12m1d:l22m2d:l32m3d
=l12m1:l22m2:l32m3
=522:323:225
=252:3:45=125:30:8

Topic: Physics, Current Electricity, JEE Main
The masses of two radioactive substances are same and their half-lives are 1 year and 2 years, respectively. The ratio of their activities after 6 years will be

Options:
A $1: 4$
B $1: 2$
C $\mathbf{l}: 3$
D $1: 6$

The correct answer is:
$1: 4$

Solution:
Let initial masses of two radioactive substances be $N_{0}$.
If $T_{1}$ and $T_{2}$ be the half-lives of two radioactive substances,
Then, $\quad T_{1}=1$ yr
Hence, remaining mass of the first substances after
$$
\begin{aligned}
t &=6 \mathrm{yrs} \\
N_{1} &=N_{0}\left(\frac{1}{2}\right)^{\frac{t}{T}}=N_{0}\left(\frac{1}{2}\right)^{6 / 1} \\
&=\frac{N_{0}}{2^{6}} \\
\Rightarrow \quad N_{1} &=\frac{N_{0}}{64}
\end{aligned}
$$
Similarly, remaining mass of the second substance after $t=6 y r \mathrm{~s}$
$N_{2}=N_{0}\left(\frac{1}{2}\right)^{\frac{t}{T_{2}}}=N_{0}\left(\frac{1}{2}\right)^{6 / 2}=\frac{N_{0}}{8}$
$\Rightarrow \quad N_{2}=\frac{N_{0}}{8}$
$\therefore \quad \frac{N_{1}}{N_{2}}=\frac{\frac{N_{0}}{\frac{64}{N_{0}}}}{8}=\frac{1}{8} \Rightarrow \frac{N_{1}}{N_{2}}=\frac{1}{8}$
Since, activity, $A=\lambda N=\frac{0.693}{T} N$
$\Rightarrow \quad A \propto \frac{N}{T}$
$\therefore \quad \frac{A_{1}}{A_{2}}=\frac{N_{1} T_{2}}{N_{2} T_{1}}=\frac{1}{8} \times \frac{2}{1}=\frac{1}{4}$
$\Rightarrow \quad A_{1}: A_{2}=1: 4$

Topic: Physics, Nuclear Physics, COMEDK, COMEDK 2018
The material filled between the plates of a parallel plate capacitor has resistivity 200 Ω m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permittivity of material is 50)

Options:
A 0.9 mA
B 9.0 mA
C 9.0 μA
D 0.9 μA

The correct answer is:
0.9 mA

Solution:
ρ=200 Ωm

K=50

C=ϵ0KAd=2×10−12 F

R=ρdA

I=VR=AVρd=Vρ⋅Cϵ0K=40200×2×10−128.85×10−12×50=0.9mA

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The material that has the highest energy gap at room temperature is:

Options:
A Cds
B Ge
C GaAs
D $\mathrm{Si}$

The correct answer is:
GaAs

Solution:
Energy bandgap
- An energy band gap is the distance between the valence band of electrons and the conduction band.
- The band gap represents the minimum energy that is required to excite an electron to move from the valence band to the conduction band where it can participate in the conduction process.
Among the following options, Cds have the highest energy gap at room temperature.
Cadium sulphide (Cds) has a band gap of $2.42 \mathrm{eV}$.
Germanium (Ge) has a band gap of 0.7 eV.
Gallium arsenide (GaAs) has a band gap of $1.42 \mathrm{eV}$.
Silicon (Si) has a band gap of $1.1 \mathrm{eV}$.

Topic: Physics, Semiconductors, JEE Main
The materials suitable for making electromagnets should have

Options:
A
high retentivity and high coercivity

B
low retentivity and low coercivity

C
high retentivity and low coercivity

D
low retentivity and high coercivity

The correct answer is:
low retentivity and low coercivity

Solution:
No Solution Available

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2004