datavorous/entrance-exam-dataset huggingface dataset:
An artificial pacemaker is implanted subcutaneously and connected to the heart in patients

Options:
A having $90 \%$ blockage of the three main coronary arteries
B having a very high blood pressure
C with irregularity in the heart rhythm
D suffering from arteriosclerosis.

The correct answer is:
with irregularity in the heart rhythm

Solution:
Artificial pacemaker is a medical device designed to regulate the beating of the heart. The purpose of an artificial pacemaker is to stimulate the heart when either the heart's native pacemaker is not fast enough or if there are blocks in the heart's electrical conducting system preventing the propagation of electrical impulses from the native pacemaker to the lower chambers of the heart, known as the ventricles. Generally, pacemakers do not treat fast rhythms of the heart.

Topic: Biology, Body Fluids and Circulation, AIIMS, AIIMS 2017
An artificial pacemaker is implanted subcutaneously and connected to the heart in patients

Options:
A having $90 \%$ blockage of the three main coronary arteries
B having a very high blood pressure
C with irregularity in the heart rhythm
D suffering from arteriosclerosis.

The correct answer is:
with irregularity in the heart rhythm

Solution:
A pacemaker (or "artificial pacemaker"), is a medical device designed to regulate the beating of the heart. The purpose of an artificial pacemaker is to stimulate the heart when either the heart's native pacemaker is not fast enough or if there are blocks in the heart's electrical conduction system preventing the propagation of electrical impulses from the native pacemaker to the lower chambers of the heart, known as the ventricles. Generally, pacemakers do not treat fast rhythms of the heart.

Topic: Biology, Body Fluids and Circulation, AIIMS, AIIMS 2004
An artificial satellite is made to move in circular orbits of different radii around the earth. The variations of its K.E., P.E. and total energy (E) in different orbits is shown in the figure by different curves.
Then for the satellites

Options:
A A represents the K.E., B & C the P.E. and C the total energy
B A represents the P.E., B the K.E. and C the total energy
C A represents the P.E., B the total energy and C the K.E.
D A represents the total energy, B the K.E. and C the P.E.

The correct answer is:
A represents the P.E., B the total energy and C the K.E.

Solution:
The K.E. is always positive.
∴ The curve C represents the K.E. Both P.E. and total energy are positive but E<P.E.
As P.E. = 2 (Total energy)
∴ B represents the total energy and A represents the P.E.

Topic: Physics, Gravitation, JEE Main
An artificial satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of earth is (where R is the radius of the Earth)

Options:
A R
B 2R
C 3R
D 4R

The correct answer is:
R

Solution:
122GMR=GMR + h
14Ã 2GMR= GMR + h
2R = R + h
h = R

Topic: Physics, Gravitation, JEE Main
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. The height of satellite above the surface of earth is -(where R=radius of earth)

Options:
A R
B 2 R
C 3 R
D 4 R

The correct answer is:
R

Solution:
escape velocity V s = 2Gm R
orbital velocity V 0 = Gm r
given V 0 = 1 2 V e
⇒ GmR+h =122GmR
⇒R+h=2R or h=R

Topic: Physics, Gravitation, JEE Main
An artillery piece of mass M1 fires a shell of mass M2 horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :

Options:
A M1(M1+M2)
B M2M1
C M2(M1+M2)
D M1M2

The correct answer is:
M2M1

Solution:
As linear momentum will be conserved, p⇀1=p⇀2

Now, kinetic energy can be written as KE=p22M.

Also p is the same for both masses. We can write,

KE∝1m

⇒KE1KE2=p22M1p22M2=M2M1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (31 Jan Shift 1)
An artillery piece which consistently shoots its shells with the same muzzle speed has a maximum range R. To hit a target which is R2 from the gun and on the same level, the angle of elevation of the gun should be

Options:
A 15°
B 45°
C 30°
D 60°

The correct answer is:
15°

Solution:
R max = R = u 2 g (at an angle of 45Â°
u 2 = Rg
Using Range = u 2 sin 2 Î¸ g
then R 2 = Rg sin 2 Î¸ g
sin 2 Î¸ = 1 2
2 Î¸ = 3 0 o â Î¸ = 1 5 o

Topic: Physics, Motion In Two Dimensions, JEE Main
An ‘Assertion’ and a ‘Reason’ is given below. Choose the correct answer from the following options:
Assertion A: Vinyl halides do not undergo nucleophilic substitution easily.
Reason R: Even though the intermediate carbocation is stabilized by loosely held π- electrons, the cleavage is difficult because of the strong bonding.

Options:
A Both A and R are correct statements but R is not the correct explanation of A.
B Both (A) and (R) are wrong statements.
C Both A and R are wrong statements and (R) is the correct explanation of A.
D A is a correct statement but R is a wrong statement.

The correct answer is:
A is a correct statement but R is a wrong statement.

Solution:
CH2=CH-Cl does not undergo unimolecular nucleophilic substitution (SN1) reaction due to the formation of highly unstable cation CH2=CH⊕, which can not be delocalized by π- electrons. It does not give SN2 because π e- cloud is exposed, so, hindrance is there.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2019 (12 Apr Shift 2)
An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the centre of the earth, it has a speed of 12 km s-1. Neglecting the effect of earth's atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km s-1) ? Give your answer to the nearest integer in km s-1__________.

Options:

The correct answer is:
16

Solution:
KEi+PEi=KEf+PEf
12mu02+-GMm10R=12mv2+-GMmR
v2=u02+2GMR1-110
v=u02+95GMR

=122+9511.222
=144+0.911.22=156.896

=16.028 km s-1

≃16 km s-1

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (08 Jan Shift 2)
An asteroid of mass m is approaching earth, initially at a distance 10Re with speed vi. It hite oarth with a speed vf , Re and Me are radius and mass of earth), then

Options:
A vf2=vi2+2GmRe1+110
B vf2=vi2+2GMeRe1+110
C vf2=vi2+2GMeRθ1-110
D vf2=vi2+2GmRe1-110

The correct answer is:
vf2=vi2+2GMeRθ1-110

Solution:
Initial energy of the asteroid is
Ei=Ki+Ui=12mvi2-GMθm10Re

Final energy of the asteroid,
Ef=12mvf2-GMθmRθ

According to law of conservation of energy, Ei=Ef

12mvi2-GMθm10Rθ=12mvf2-GMθmRθ

vf2-2GMθRθ=vi2-2GMθ10Rθ

⇒ vf2=vi2+2GMθRθ1-110

Topic: Physics, Gravitation, JEE Main
An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth's surface. The radius of the earth is R, while its mass is M. The gravitational pull FG on the astronaut is

Options:
A Zero since astronaut feels weightless
B GMmR+h2<FG<GMmR2
C FG=GMmR+h2
D 0<FG<GMmR2

The correct answer is:
FG=GMmR+h2

Solution:
As distance of the astronaut from the centre of the Earth is =R+h

Therefore, gravitational force =Gm1m2R+h2=GMmR+h2

Topic: Physics, Gravitation, JEE Main, JEE Main 2016 (10 Apr Online)
An astronaut on a strange planet finds that acceleration due to gravity is twice as that on the surface of earth. Which of the following could explain this?

Options:
A Both the mass and radius of the planet are half as that of earth
B Radius of the planet is half as that of earth, but the mass is the same as that of earth
C Both the mass and radius of the planet are twice as that of earth
D Mass of the planet is half as that of earth, but radius is same as that of earth

The correct answer is:
Both the mass and radius of the planet are half as that of earth

Solution:
Acceleration due to gravity on the surface of the planet is $g_{p}=\frac{G M_{p}^{2}}{R_{p}^{2}}$ Given, $\quad \mathrm{M}_{\mathrm{p}}=\frac{\mathrm{M}_{\mathrm{e}}}{2}$ and $\mathrm{R}_{\mathrm{p}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}$
$$
\therefore \quad g_{p}=\frac{G(M e / 2)}{(\operatorname{Re} / 2)^{2}}=\frac{2 G M_{e}}{R_{e}}=2 g_{e}
$$

Topic: Physics, Gravitation, JEE Main
An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5 \mathrm{~km}$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}$. The value of $x$ is (take $\left.\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km}\right)$ :

Options:
A 10
B 12
C 9
D 11

The correct answer is:
11

Solution:
$\begin{aligned} & \mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right) \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{i}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}} \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\mathrm{h}\right)}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)} \\ & \Rightarrow \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}\end{aligned}$
Change in total mechanical energy
$\begin{aligned} & =\mathrm{TE}_{\mathrm{f}}-\mathrm{TE}_{\mathrm{i}} \\ & =\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\operatorname{Re}}\left[1-\frac{10}{21}\right]=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \operatorname{Re}}\end{aligned}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (09 Apr Shift 1)
An astronaut whose height is 1.50 m floats "feet down" in an orbiting space shuttle at a distance r=6.673Ã106 m away from the centre of Earth. The magnitude of difference between the gravitational acceleration at her feet and at her head is found to be NÃ10-6 m s-2. What is the value of N?
[ME=6Ã1024 kg and G=6.67Ã10-11 N m2 kg-2]

Options:

The correct answer is:
1.8

Solution:
The gravitational acceleration at any distance r from the center of Earth is ag=GMEr2.
On differentiating, we get dag=-2GMEr3dr, where dag is infinitesimal change in a due to differential change dr in r.
⇒dag=26.67×10-116×10246.673×1063×1.50
=1.8×10-6 m s-2
∴ N=1.8

Topic: Physics, Gravitation, JEE Main
An astronomical telescope has a large aperture to

Options:
A
reduce spherical aberration

B
have high resolution

C
increase span of observation

D
have low dispersion

The correct answer is:
have high resolution

Solution:
Large aperture increases the amount of light gathered by the telescope increasing the resolution.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2002
An astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, at what distance the lenses must be separated (in cm)?

Options:

The correct answer is:
54

Solution:
1V-1-200=140
1V=55140-1200
=5200-1200
1V=4200=150
V=50
â´d=50+4=54 cm

Topic: Physics, Ray Optics, JEE Main
An athlete completes one round of a circular track of radius $R$ in $40 \mathrm{sec}$. What will be his displacement at the end of $2 \mathrm{~min} .20 \mathrm{sec}$

Options:
A Zero
B 2R
C $2 \pi R$
D $7 \pi R$

The correct answer is:
2R

Solution:
Total time of motion is $2 \mathrm{~min} 20 \mathrm{sec}=140 \mathrm{sec}$.
As time period of circular motion is $40 \mathrm{sec}$ so in $140 \mathrm{sec}$. athlete will complete 3.5 revolution i.e., He will be at diametrically opposite point i.e., Displacement $=2 R$.

Topic: Physics, Motion In One Dimension, JEE Main
An athlete in the olympic games covers a distance of $100 \mathrm{~m}$ in $10 \mathrm{~s}$. His kinetic energy can be estimated to be in the range

Options:
A
$200 \mathrm{~J}-500 \mathrm{~J}$

B
$2 \times 10^5 \mathrm{~J}-3 \times 10^5 \mathrm{~J}$

C
$20,000 \mathrm{~J}-50,000 \mathrm{~J}$

D
$2,000 \mathrm{~J}-5,000 \mathrm{~J}$

The correct answer is:
$2,000 \mathrm{~J}-5,000 \mathrm{~J}$

Solution:
Approximate mass $=60 \mathrm{~kg}$
Approximate velocity $=10 \mathrm{~m} / \mathrm{s}$
Approximate $\mathrm{KE}=\frac{1}{2} \times 60 \times 100=3000 \mathrm{~J}$
$\mathrm{KE}$ range $\Rightarrow 2000$ to 5000 joule

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2008
An athlete, initially at rest, takes 2 s to reach his maximum speed of 36 km h-1. The magnitude of his average acceleration (in m s-2) is

Options:

The correct answer is:
5

Solution:
The final speed of the athlete, v=36 km h-1=36Ã518=10 m s-1

initial speed, u=0

time taken t=2 s

for average acceleration, a

v=u+at

a=v-ut=10-02=5 m s-2

Topic: Physics, Motion In One Dimension, JEE Main
An athlete is given 100 g of glucose C6H12O6 for energy. This is equivalent to 1800 kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is _____g (Nearest integer) Assume that there is no other way of consuming stored energy.

Given : The enthalpy of evaporation of water is 45 kJ mol-1

Molar mass of C,H&O are $12.1$ and 16 g mol-1.

Options:

The correct answer is:
360

Solution:
Overall reaction of metabolism of glucose in the body:
C6H12O6( s)+6O2→6CO2( g)+6H2O(l)

Extra energy used to convert H2O(l) into H2O(g), energy left

=18002=900 kJ
Moles of water vaporised from enthalpy of evaporation:

⇒900=nH2O×45

nH2O=90045=20 mole

WH2O=20×18=360 g

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (25 Jan Shift 1)
An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is n×10-4eV. The value of n is [Assume the atom to be stationary during the absorption and emission process] (Take h=6.6×10-34J s and c=3×108 m s-1 ).

Options:

The correct answer is:
4125

Solution:
The energy of a photon is given by E=hcλ.

It is given that λ1=500 nm, λ2=600 nm.

The net energy absorbed is

ΔE=hcλ1-hcλ2=hc10-9(1500-1600)

=6.6×10-34×3×108×100500×600×10-9

=6.6×330×10-19 J

1 eV=1.6×10-19 J

=6.6×330×1.6eV

=4125×10-4 eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (13 Apr Shift 2)
An atom of an element has electronic configuration 2, 8, 1. Which of the following statement is correct

Options:
A The element's valency is 7
B The element exists as a diatomic molecule
C The element is of non-metallic nature
D The element forms a basic oxide

The correct answer is:
The element forms a basic oxide

Solution:
The element is sodium, which is metallic in nature and will form basic oxide.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
An audio signal consists of two distinct sounds: one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal band width required to send both the signals together to the AM signal band width required to send just the human speech is:

Options:
A 3
B 5
C 6
D 2

The correct answer is:
3

Solution:
Band width for Music signal =15200â10200 =5000 Hz
Band width for speech signal =2700â200=2500Hz
Band with for Both = 2500+5000=7500 Hz
Band with for Human speech =2500 Hz
Required Ratio =75002500=31

Topic: Physics, Communication System, JEE Main, JEE Main 2016 (09 Apr Online)
An audio signal of \( 1 \mathrm{kHz} \) is used to modulate a carrier of \( 500 \mathrm{kHz} \). The bandwidth required is

Options:
A \( 499 \mathrm{kHz} \)
B \( 501 \mathrm{kHz} \)
C \( (500 \pm 1) \mathrm{kHz} \)
D \( 2 \mathrm{kHz} \)

The correct answer is:
\( 2 \mathrm{kHz} \)

Solution:
Given,
Modulating signal is, m=1 kHz.
Carrier signal is, C=500 kHz.
Lower side band of the signal is, LSB=C-m=500-1=499 Hz.
Upper side band of the signal is, USB=C+m=500+1=501 Hz.
So bandwidth of the signal is, BW=USB-LSB=501-499=2 kHZ.

Topic: Physics, Communication System, JEE Main
An audio signal vm=20sin2π1500t amplitude modulates a carrier vc=80sin2π100,000t. The value of percent modulation is

Options:

The correct answer is:
25

Solution:
% modulation=AmAc×100

% modulation=2080×100

% modulation=25%

=25

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (24 Feb Shift 1)
An automobile driver travels from plane to a hill station $120 \mathrm{~km}$ distant at an average speed of $30 \mathrm{~km}$ per hour. He then makes the return trip at an average speed of $25 \mathrm{~km}$ per hour. He covers another $120 \mathrm{~km}$ distance on plane at an average speed of $50 \mathrm{~km}$ per hour. His average speed over the entire distance of $360 \mathrm{~km}$ will be

Options:
A $\frac{30+25+50}{3} \mathrm{~km} / \mathrm{hr}$
B $(30,25,50)^{\frac{1}{3}}$
C $\frac{3}{\frac{1}{30}+\frac{1}{25}+\frac{1}{50}} \mathrm{~km} / \mathrm{hr}$
D None of these

The correct answer is:
$\frac{3}{\frac{1}{30}+\frac{1}{25}+\frac{1}{50}} \mathrm{~km} / \mathrm{hr}$

Solution:
Average speed $=\frac{120+120+120}{\frac{120}{30}+\frac{120}{25}+\frac{120}{50}}=\frac{\text{Total distance}}{\text{Total time}}$
$=\frac{3}{\frac{1}{30}+\frac{1}{25}+\frac{1}{50}} \mathrm{~km} / \mathrm{hr}$.

Topic: Mathematics, Statistics, JEE Main
An automobile moves on a road with a speed of 54 km h-1 . The radius of its wheels is 0.45âm and the moment of inertia of the wheel about its axis of rotation is 3 kg m2 . If the vehicle is brought to rest in 15 s , the magnitude of average torque transmitted by its brakes to the wheel is:

Options:
A 8.58 kg m2 s-2
B 10.86Â kgÂ m2Â s-2
C 2.86Â kgÂ m2Â s-2
D 6.66 kg m2 s-2

The correct answer is:
6.66 kg m2 s-2

Solution:
Ïi=150.45=1003 Ïf=0
Ïf=Ïi+Î±t
0=1003+-Î±(15)
Î±=10045
Ï=IÎ±=3Ã10045=6.66 N.M

Topic: Physics, Rotational Motion, JEE Main
An automobile of mass m accelerates starting from the origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by:

Options:
A 9P8 m12t32
B 8P9 m12t23
C 9m8P12t32
D 8P9 m12t32

The correct answer is:
8P9 m12t32

Solution:
P=const

P=FV=mv2 dvdx

∫0xPmdx=∫0vv2 dv

Pxm=v33

3Pxm13=v=dxdt

3Pm13∫0tdt=∫0xx-13dx

⇒x=8P9 m12t32

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (27 Jul Shift 2)
An automobile, traveling at 40 km h-1, can be stopped at a distance of 40 m by applying brakes. If the same automobile is traveling at 80 km h-1, the minimum stopping distance in metres is (Assume no skidding):

Options:
A 150 m
B 100 m
C 75 m
D 160 m

The correct answer is:
160 m

Solution:
S=u22a
S1S2=u12u22⇒S2=u2u12S1=22 40=160 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2018 (15 Apr)
An automobile, travelling at $40 \mathrm{~km} / \mathrm{h}$, can be stopped at a distance of $40 \mathrm{~m}$ by applying brakes. If the same automobile is travelling at $80 \mathrm{~km} / \mathrm{h}$, the minimum stopping distance, in metres, is (assume no skidding)

Options:
A
$75 \mathrm{~m}$

B
$160 \mathrm{~m}$

C
$100 \mathrm{~m}$

D
$150 \mathrm{~m}$

The correct answer is:
$160 \mathrm{~m}$

Solution:
According to question, $u_1=40 \mathrm{~km} / \mathrm{h}, v_1=0$ and $s_1=40 \mathrm{~m}$ using $v^2-u^2=2 a s ; 0^2-40^2=2 a \times 40$..(i) Again, $0^2-80^2=2$ as From eqn. (i) and (ii)
Stopping distance, $s=160 \mathrm{~m}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
An automobile travelling at 50 km h-1, can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 90 km h-1, all other conditions remaining the same and assuming no skidding, the minimum stopping distance in metre is

Options:
A 72
B 92.5
C 102.6
D 129.6

The correct answer is:
129.6

Solution:
According to kinematic equation,
v 2 = u 2 +2as
Therefore, the stopping distance is,
s= v 2 â u 2 2a
Therefore,
sâ u 2 as v=0 and a=constant
â s 2 s 1 = ( u 2 u 1 ) 2
Therefore, new stopping distance is,
s 2 = s 1 ( u 2 u 1 ) 2
s 2 =( 40 ) ( 90 50 ) 2
s 2 =129.6âm

Topic: Physics, Motion In One Dimension, NEET
An automobile travelling with speed of $60 \mathrm{~km} / \mathrm{h}$, can brake to stop within a distance of $20 \mathrm{~cm}$. If the car is going twice as fast, i.e $120 \mathrm{~km} / \mathrm{h}$, the stopping distance will be

Options:
A
$20 \mathrm{~m}$

B
$40 \mathrm{~m}$

C
$60 \mathrm{~m}$

D
$80 \mathrm{~m}$

The correct answer is:
$80 \mathrm{~m}$

Solution:
If the initial speed is doubled, the stopping distance becomes four times, i.e. $80 \mathrm{~m}$.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2004
An average force of 125 N is applied on a machine gun firing bullets each of mass 10 g at the speed of 250 m s-1 to keep it in position. The number of bullets fired per second by the machine gun is :

Options:
A 50
B 25
C 100
D 5

The correct answer is:
50

Solution:
The formula to calculate the net momentum of the gun and bullet system can be written as

F=nmv ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the required number of bullets.

125=n×101000×(250)⇒n=50

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (11 Apr Shift 1)
An average induced e.m.f. of \( 1 \mathrm{~V} \) appears in a coil when the current in it is changed from \( 10 \mathrm{~A} \) in one direction to \( 10 \mathrm{~A} \) in opposite direction in \( 0.5 \mathrm{~s} \). Self-inductance of the coil is

Options:
A \( 25 \mathrm{mH} \)
B \( 50 \mathrm{mH} \)
C \( 75 \mathrm{mH} \)
D \( 100 \mathrm{mH} \)

The correct answer is:
\( 25 \mathrm{mH} \)

Solution:
From Faraday's law of electromagnetic induction, magnitude of average induced emf is given by, ε=-L∆i∆t, where L, i & t represents coefficient of self inductance, current and time respectively.
On substituting corresponding parameters in above relation, we have
⇒1=-L10-(-10)0.5
⇒L=1×0.520=140=0.025 H=25 mH

Topic: Physics, Electromagnetic Induction, JEE Main
An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass =180.0 g mol-1) needed to meet this energy requirement is ______g.
(Use : ΔCH (glucose)=-2700 kJ mol-1)

Options:

The correct answer is:
667

Solution:
1 mole glucose give 2700 kJ energy so mole of glucose needed for 104 kJ energy
=100002700=3.7037 moles
wt. of glucose =3.7037×180

=666.666
≈667gm

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (20 Jul Shift 1)
An azeotropic mixture of two liquids has boiling point lower than either of them when it:

Options:
A Shows a negative deviation from Raoult's law
B Shows no deviation form Raoult's law
C Shows positive deviation from Raoult's law
D Is saturated

The correct answer is:
Shows positive deviation from Raoult's law

Solution:
For solution having positive deviation from Raoult's law

When an azeotropic mixture shows positive deviation, then vapour pressure of mixture is greater than that of either liquids at that temperature. Hence, mixture quickly boils i.e., its boiling point becomes less.

Topic: Chemistry, Solutions, JEE Main
An carbonyl compounds gives nuclrophilic addition reaction with
(1) $\mathrm{HCN}$
(2) $\mathrm{NaSHO}_3$
(3) $R-M g X$.
(4) $\mathrm{NH} 2-\mathrm{NH} 2$

Options:
A 1,2
B 2,3
C $1,2,4$
D $1,2,3$

The correct answer is:
$1,2,3$

Solution:
The correct answer is (4): $1,2,3$.
Explanation: Carbonyl compounds typically undergo nucleophilic addition reactions with:
1. HCN (Hydrogen cyanide) - Adds across the carbonyl group to form cyanohydrins.
2. NaHSO3 (Sodium bisulfite) - Forms bisulfite adducts with aldehydes and ketones.
3. $\mathrm{RMgX}$ (Grignard reagents) - Adds to carbonyl compounds to form alcohols after hydrolysis.
The options in (D) correspond to these three reagents that react with carbonyl compounds via nucleophilic addition.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main