datavorous/entrance-exam-dataset huggingface dataset:
A dihaloalkane ' $X$, having formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Cl}_{2}$, on hydrolysis gives a compound, that can reduce Tollen's reagent. The compound $X^{\prime}$ is

Options:
A 1,2 -dichloropropane
B 1 , 1 -dichloropropane
C 1,3 -dichloropropane
D 2, 2-dichloropropane

The correct answer is:
1 , 1 -dichloropropane

Solution:
Since, the obtained compound reduces Tollen's reagent, it must be an aldehyde. Thus, it is obvious that both the $-\mathrm{Cl}$ atoms are present at $\mathrm{C}_{1} . \quad$ Hence, $\quad$ the $\quad$ compound $\quad{X}^{\prime}$ is 1, 1-dichloropropane and the reactions are as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is cm3. (Nearest integer) [Given : Faraday constant F=96500 Cmol-1 at STP, molar volume of an ideal gas is 22.7 Lmol-1]

Options:

The correct answer is:
127

Solution:
At anode

2H2O→O2g+4H++4e-

At cathode

2H++2e-→H2g

Now number of gm eq. =i×t96500

=0.1×2×60×6096500

=0.00746

VO2=0.007464×22.7=0.0423

VH2=0.007462×22.7=0.0846

VTotal≈127ml or cc

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved by approximately placing

Options:
A A convex mirror of suitable focal length
B A concave mirror of suitable focal length
C A convex lens of focal length less than 0.25 m
D A concave lens of suitable focal length

The correct answer is:
A convex lens of focal length less than 0.25 m

Solution:
Image can be formed on the screen if it is real. Real image of reduced size can be formed by a concave mirror or a convex lens
Let u=2f+x, then
1u+1v=1f
⇒12f+x+1v=1f
⇒1v=1f-12f+x=f+xf(2f+x)
⇒v=f(2f+x)f+x
It is given that u+v=1.0 m
2f+x+f(2f+x)f+x=2f+x1+ff+x<1.0m
Or 2f+x2f+x<1.0m
Or 2f+x2<(f+x)
This will be true only when f<0.25 m

Topic: Physics, Ray Optics, JEE Main
A diminished virtual image can be formed only in

Options:
A Plane mirror
B A concave mirror
C A convex mirror
D Concave-parabolic mirror

The correct answer is:
A convex mirror

Solution:
A convex mirror

Topic: Physics, Ray Optics, JEE Main
A diode AM detector with the output circuit consisting of R=1 kW and C=1 Î¼F would be more suitable for detecting a carrier signal of

Options:
A 0.1 kHz
B 0.5 kHz
C 1 kHz
D 10 kHz

The correct answer is:
10 kHz

Solution:
Given R=1 kΩ or R=1×103 Ω and C=1 μF or C=1×10-6 F.
In this condition, frequency of carrier signal,
1RC<<fc
11×103×10-6<<fc
∴fc>>1 kHz
Since, frequency must be greater than 1 kHz the possible answer will be fc=10 kHz.

Topic: Physics, Semiconductors, JEE Main
A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance of 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

Options:
A 5.31 MHz
B 5.31 kHz
C 10.62 MHz
D 10.62 kHz

The correct answer is:
10.62 kHz

Solution:
τ = RC = 1 0 0 × 1 0 3 × 2 5 0 × 1 0 - 1 2 sec
= 2.5 × 1 0 7 × 1 0 -1 2 sec

= 2.5 × 1 0 - 5 sec
The higher frequency which can be detected with tolerable distortion is
f = 1 2 π m a RC = 1 2 π × 0.6 × 2.5 × 1 0 - 5 Hz
=100×10425×1.2π Hz
= 4 1.2 π × 1 0 4 Hz
= 10.61 kHz
This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less then the rate of decay modulated signal voltage for proper detection of modulated signal.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2013 (07 Apr)
A diode having potential difference $0.5 V$ across its junction which does not depend on current, is connected in series with resistance of $20 \Omega$ across source. If 0.1. A passes through resistance then what is the voltage of the source

Options:
A $1.5 \mathrm{~V}$
B $2.0 \mathrm{~V}$
C $2.5 \mathrm{~V}$
D $5 V$

The correct answer is:
$2.5 \mathrm{~V}$

Solution:
$\begin{aligned} & V^{\prime}=V+I R \\ & =0.5+0.1 \times 20 \\ & =2.5 \mathrm{~V}\end{aligned}$

Topic: Physics, Semiconductors, JEE Main
A diol with following structure having 18 O is

Options:
A
B
C Both A and B
D

The correct answer is:

Solution:
In acidic medium H 2 O attack where +δ charge on carbon is more stable.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
A dip needle, which is free to move in a vertical plane perpendicular to magnetic meridian, will remain

Options:
A horizontal
B vertical
C somewhere in between horizontal and vertical
D none of these

The correct answer is:
vertical

Solution:
In the plane perpendicular to magnetic meridian, horizontal component of field is zero.
BH′=BHcos⁡90°=0
BV′=BV
∴ Only vertical component remains, so the dip needle will remain vertical.

Topic: Physics, Magnetic Properties of Matter, NEET
A dipole of dipole moment p→=pi^ lies along the x-axis in a non-uniform electric field E→=cxi^. The force acting on the dipole is

Options:
A zero
B -pcx2i^
C pcx2i^
D -pc2x2i^

The correct answer is:
-pcx2i^

Solution:
The force acting on a dipole is

F→=p→∂Ex∂x

F→=p→ddxcx=-pcx2i^

Topic: Physics, Electrostatics, JEE Main
A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of 3 Ω and 2 Ω respectively. The ratio of heat produced in the two resistances in same interval of time will be :

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

The correct answer is:
3:1

Solution:
Heat produced in resistor when direct current flows is H1=i2R1t

H1=423t=163t

Heat produced in resistor when alternating current flows is H2=irms2R2t irms=i02

H2=i022R2t

H2=422×2t=16t

Thus, the ratio is H1:H2=3:1

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A direct current of $5 \mathrm{~A}$ is superposed on an alternating current $I=10 \sin \omega t$ flowing through the wire. The effective value of the resulting current will be

Options:
A $(15 / 2) \mathrm{A}$
B $5 \sqrt{3} \mathrm{~A}$
C $5 \sqrt{5} \mathrm{~A}$
D $15 \mathrm{~A}$

The correct answer is:
$5 \sqrt{3} \mathrm{~A}$

Solution:
Total carrent, $1=(5+10 \sin \omega \mathrm{t})$

$$

\begin{aligned}

\Rightarrow I_{\text {eff }} &=\left[\frac{\int_{0}^{\mathrm{T}} \mathrm{I}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}\right]^{1 / 2} \\

&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(5+10 \sin \omega \mathrm{t})^{2} \mathrm{dt}\right]^{1 / 2} \\

&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}\left(25+100 \sin \omega \mathrm{t}+100 \sin ^{2} \omega \mathrm{t}\right)\right]^{1 / 2} \\

\text { But, } \quad \frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin \omega \mathrm{t} \mathrm{dt}=0

\end{aligned}

$$

and

$$

\begin{array}{l}

\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin ^{2} \omega \mathrm{t} \mathrm{dt}=\frac{1}{2} \\

\text { So, } \mathrm{I}_{\mathrm{eff}}=\left[25+\frac{1}{2} \times 100\right]^{1 / 2}=5 \sqrt{3} \mathrm{~A}

\end{array}

$$

Topic: Physics, Alternating Current, JEE Main
A disc is rotating with angular speed $\omega$. If a child sits on it, what is conserved?

Options:
A Linear momentum
B Angular momentum
C Kinetic energy
D Potential energy

The correct answer is:
Angular momentum

Solution:
$\tau=\frac{d L}{d t}$, so $=0 \Rightarrow L=\frac{d L}{d l}$ constant
Law of conservation of angular momentum.

Topic: Physics, Rotational Motion, JEE Main
A disc of mass 2 kg and radius 0.2 m is rotating with an angular velocity 30 rad s-1, about an axis passing through its centre and perpendicular to its plane. If a mass of 0.25 kg is stuck on the periphery of the disc, then its angular velocity will become

Options:
A 24 rad s-1
B 36 rad s-1
C 15 rad s-1
D 26 rad s-1

The correct answer is:
24 rad s-1

Solution:
When there is no external torque then the angular momentum remains constant.
i.e. τ=0⇒dLdt=0
∴ I1ω1=I2ω2 ... (i)
Here M=2 kg, m=0.25 kg, r=0.2 m,
ω1=30 rad s-1
Hence, putting the value of ω1 in equation (i), we get
⇒ 12×2×0.22×30=12×2+2×0.250.22×ω2
⇒ 1.2=0.05ω2
⇒ ω2=24 rad s-1

Topic: Physics, Rotational Motion, JEE Main
A disc of mass 2 kg and radius 0.2 m is rotating with an angular velocity of 30 rad s-1. The angular velocity of the disc, if a mass of 0.25 kg is put on the periphery is

Options:
A 24 rad s-1
B 36 rad s-1
C 15 rad s-1
D 26 rad s-1

The correct answer is:
24 rad s-1

Solution:
If no external torque acts on a system of particle then angular momentum of the system remains constant, that is,
τ=0
⇒ dLdt=0
⇒ L=Iω=constant
⇒ I1ω1=I2ω2
∴ 12Mr2ω1=12M+2mr2ω2 ...(i)
Here, M=2 kg, m=0.25 kg and r=0.2 m
ω1=30 rad s-1
Hence, we get after putting the given values in Eq. (i)
12×2×0.22×30=12×2+2×0.250.22×ω2
⇒ 60=2.5ω2
∴ ω2=24 rad s-1

Topic: Physics, Rotational Motion, JEE Main
A disc of mass m and radius R is attached to a rectangular plate of the same mass, breadth R and length 2R as shown in figure. The moment of inertia of the system about the axis AB passing through the centre of the disc and along the plane is Î=1Î±313mR2, where value of Î± is

Options:

The correct answer is:
4

Solution:
ΙAB=Ιdisc+Ιplate+M3R22 (From Parallel Axis Theorem)
=14MR2+MR212+M3R22
=3112MR2

Topic: Physics, Rotational Motion, JEE Main
A disc of mass \( M \) and radius \( R \) is rolling with angular speed \( \omega \) on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin \( O \) is

Options:
A \( \left(\frac{1}{2}\right) M R^{2} \omega \)
B \( M R^{2} \omega \)
C \( \left(\frac{3}{2}\right) M R^{2} \omega \)
D \( 2 M R^{2} \omega \)

The correct answer is:
\( \left(\frac{3}{2}\right) M R^{2} \omega \)

Solution:
From the theorem
L → = L cm → +M( r → × v → )
Angular momentum about O = Angular momentum about CM + Angular momentum of CM about origin
L0 = Iω + MRv
= M R 2 ω 2 +M R 2 ω
=( 3 2 )M R 2 ω

Topic: Physics, Rotational Motion, JEE Main
A disc of moment of inertia \( I_{1} \) is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed \( \omega_{1} \). Another disc of moment of inertia \( I_{2} \) having zero angular speed is placed coaxially on a rotating disc. Now both the discs are rotating with constant angular speed \( \omega_{2} \). The energy lost by the initial rotating disc is

Options:
A \( \frac{1}{2}\left[\frac{I_{1}+I_{2}}{1_{1}}\right] \omega_{1}^{2} \)
B \( \frac{1}{2}\left[\frac{\mathrm{I} \mathrm{I}_{1}}{\mathrm{I}_{1}-\mathrm{I}}\right] \omega_{1}^{2} \)
C \( \frac{1}{2}\left[\frac{I_{1}-I_{2}}{I_{I_{2}}}\right] \omega_{1}^{2} \)
D \( \frac{1}{2}\left[\frac{\mathrm{I}, \mathrm{I}_{2}}{\mathrm{I}_{1}+\mathrm{I}_{2}}\right] \omega_{1}^{2} \)

The correct answer is:
\( \frac{1}{2}\left[\frac{\mathrm{I}, \mathrm{I}_{2}}{\mathrm{I}_{1}+\mathrm{I}_{2}}\right] \omega_{1}^{2} \)

Solution:
Here, angular momentum is conserved, so the initial momentum is equal to final momentum.

Ji=Jf

I1ω1=I1+I2ω2

⇒ω2=I1ω1I1+I2

Here, initial kinetic energy, K.E.i=12I1ω12

Thus, final kinetic energy, K.E.f=12I1+I2ω22,

now put the value, 12I1+I2I1ω1I1+I22=12I1ω12I1+I2

∆K.E.=K.E.i-K.E.f=12I1ω12-12I1ω12I1+I2

∆K.E.=12I1ω121-I1I1+I2=12ω12I2I1I1+I2

Topic: Physics, Rotational Motion, JEE Main
A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m s-1. It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is g=10 m s-2

Options:
A 2.4 m
B 0.9 m
C 2.7 m
D 1.8 m

The correct answer is:
1.8 m

Solution:
Let m be the mass of the disc. Then translational kinetic energy of the disc is:
KT=12mv2 ....(1)
When it ascends on a smooth track its rotational kinetic energy will remain same while translational kinetic energy will go on decreasing. At highest point Translational kinetic energy is transferred to Potential Energy.
KT=mgh
or 12mv2=mgh
or h=v22g=622Ã10=1.8 m

Topic: Physics, Work Power Energy, NEET
A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m s-1. It then ascends a smooth continuous track as shown in the figure. The height (in m) up to which it will ascend is : ( g=10 m s-2 )

Options:

The correct answer is:
2.7

Solution:
12Iw2+12mv2=mgh
12mr22.v2R2+12mv2=mgh
v24+v22=gh
v2+2v24=gh
34Ã36=10Ãh
2710=hâ2.7

Topic: Physics, Work Power Energy, JEE Main
A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

Options:
A 54R
B 23R
C 34R
D 32R

The correct answer is:
32R

Solution:
Ï=-mgRsinâ¡Î¸=-mgRÎ¸ âµsinâ¡Î¸âÎ¸
IÎ± =12mR2+mR2Î±=32mR2Î±
But Ï= IÎ±
32mR2Î±=-mgRÎ¸
Î±=-2g3RÎ¸=Ï2Î¸
Time period T=2Ï3R2g
For simple pendulum, T=2Ïlg
l=3R2

Topic: Physics, Oscillations, JEE Main
A disc of radius R and mass M is rolling horizontally without slipping with speed v. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:

Options:
A v2g
B 34v2g
C 12v2g
D 23v2 g

The correct answer is:
12v2g

Solution:
Only the translational kinetic energy of disc changes into gravitational potential energy and rotational KE remains unchanged as there is no friction.

12mv2=mgh

⇒h=v22g

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A disc of radius R rolls without slipping at a speed v along the positive x-axis. The velocity of a point P at the instant shown in the figure is

Options:
A v→P=v+v r sinθRi^+v r cosθRj^
B v→P=v+v r sinθRi^-v r cosθRj^
C v→P=v r sinθRi^+v r cosθRj^
D v→P=v r sinθRi^-v r cosθRj^

The correct answer is:
v→P=v+v r sinθRi^-v r cosθRj^

Solution:
We have Ï=vR for pure rolling.
The velocity of point P is resultant of two velocity vectors shown as arrows in the figure. Hence,
v â P = v x i ^ + v y j ^
or v â P = v + v r sin Î¸ R i ^ - v r cos Î¸ R j ^

Topic: Physics, Rotational Motion, JEE Main
A disc of radius R rotates with a constant angular velocity Ï about its own axis. The surface charge density of this disc varies as Ï=ar2 where r is the distance from the centre of the disc. Determine the magnetic field intensity at the centre of the disc.

Options:
A μ0aωR3
B μ0aωR36
C μ0aωR38
D μ0aωR33

The correct answer is:
μ0aωR36

Solution:
dA=2πr·dr
dq=σ·dA=2πar3·dr
i=dqf⁡=dqω2π=aωr3·dr
dB=μ0i2r=aμ0ωr2dr2
∴B=∫0RdB=μ0aωR36

Topic: Physics, Magnetic Effects of Current, JEE Main
A disc of the radius R is spun to an angular speed Ï0 about its axis and then imparted a horizontal velocity of magnitude (at t=0) with its plane remaining vertical. The coefficient of friction between the disc and the plane is Î¼. The sense of rotation and direction of its linear speed are shown in the figure. Choose the correct statement.

Options:
A disc will start rolling without slipping in the direction of V0
B slipping will never be ceased
C disc will return to initial point
D none of these

The correct answer is:
disc will return to initial point

Solution:
Angular acceleration α=-μmgrmr22=-2μgr
Time taken to stop rotation t1=ω0α=ω0r2μg
Linear acceleration a=-μg
Time taken to stop translation t2=V0μg=ω0r4μg
t2<t1
Hence disc will return to initial point

Topic: Physics, Laws of Motion, NEET
A disc rotates about its axis of symmetry in a hoizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distance of $1.25 \mathrm{~cm}$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A
$0.5$

B
$0.7$

C
$0.3$

D
$0.6$

The correct answer is:
$0.6$

Solution:
Using, $\mu m g=\frac{m v^2}{r}=m r \omega^2$ $\omega=2 \pi n=2 \pi \times 3.5=7 \pi \mathrm{rad} / \mathrm{sec}$ Radius, $r=1.25 \mathrm{~cm}=1.25 \times 10^{-2} \mathrm{~m}$ Coefficient of friction, $\mu=$ ?
$$
\mu m g=\frac{m(r \omega)^2}{r}(\because v=r \omega)
$$
$$
\begin{gathered}
\Rightarrow \quad \mu=\frac{r \omega^2}{g}=\frac{1.25 \times 10^{-2} \times\left(7 \times \frac{22}{7}\right)^2}{10} \\
=\frac{1.25 \times 10^{-2} \times 22^2}{10}=0.6
\end{gathered}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25 cm from the axis of rotation remains at rest on the disc. The coeï¬cient of friction between the coin and the disc is g=10 m s-2

Options:
A 0.5
B 0.7
C 0.3
D 0.6

The correct answer is:
0.6

Solution:
3.5 rev s-1
1 rev→2π rad
3.5 rev→2π×3.5 rad
⇒ω=7π rad s-1
μmg=mv21.25
μmg=mrω2r
μmg=mrω2
μ=rω2g=1.25×10-2×7×227210
=1.25×10-2×22210
=0.6

Topic: Physics, Laws of Motion, NEET
A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity ω. The coefficient of static friction between the bottom of the beaker and the surface of the disc is μ. The beaker will revolve with the disc if :

Options:
A R≤μg2ω2
B R≤μgω2
C R≥μg2ω2
D R≥μgω2

The correct answer is:
R≤μgω2

Solution:
As the disc is revolving in circular motion about an axis passing through centre and perpendicular to its plane, thus centrifugal force must act on it. The forces acting on disc is shown below.

From FBD

For circular motion, frictional force must balance the required centripetal force.

So, f=mω2R

⇒μN≥mω2R

⇒μmg≥mω2R

⇒R≤μgω2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is 4 3 , and the diver's eyes are 15 cm below the surface of the water. Then the radius of the circle is :

Options:
A 15×73 cm
B 15×37 cm
C 15×37 cm
D 15×3×5 cm

The correct answer is:
15×37 cm

Solution:
∴ sin θ c = 1 μ
= 3 4
∴ r h 2 + r 2 = 3 4
∴ r 2 h 2 + r 2 = 9 1 6
∴ 1 6 r 2 = 9 h 2 + 9 r 2
7 r 2 = 9 h 2
r = h 9 7
= 3 h 7
= 3 7 × 1 5 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (09 Apr Online)
A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is:

Options:
A Real and at a distance of 6 cm from the convergent lens
B Real and at a distance of 40 cm from convergent lens
C Virtual and at a distance of 40 cm from convergent lens
D Real and at a distance of 40 cm from the divergent lens

The correct answer is:
Real and at a distance of 40 cm from convergent lens

Solution:
Diverging lens forms the image at its focus. The image formed by diverging lens is the object for converging lens.
∴ Distance of the object from converging lens is 40 cm i.e., object is at double focal point.
∴ Image forms at double focal point on the other side of the lens.
1v-14=1f
1v-1-40=120
1v+140=120
1v=120-140=140
v=40

Topic: Physics, Ray Optics, JEE Main, JEE Main 2017 (02 Apr)
A DNA molecule having labelled thymidine in one of its strand is allowed to replicate in a radioactive medium. Calculate the number of strands with labelled thymine in the DNA after three successive divisions.

Options:
A 15
B 13
C 3
D 16

The correct answer is:
15

Solution:
DNA replication is the process of making a new double stranded nucleic acid. This process is semi-conservative where one is a new strand synthesized and the other is the old strand from the parent DNA used as template for replication. The given dsDNA- one strand was labelled with labelled thymidine and other was non-labelled. This dsDNA was replicated in a medium of labelled nucleotide. In subsequent 3 generations of replication, only one strand (parent dsDNA) will remain unlabeled and rest all will be labelled as the new nucleotides used for making new strand are from the labelled medium. Therefore, after 3 divisions there will 16 strands from 8 dsDNA but 15 strands will be labelled.

Topic: Biology, Molecular Basis of Inheritance, NEET
A doctor prescribed the drug Equanil to a patient. The patient was likely to have symptoms of which disease?

Options:
A Stomach ulcers
B Hyperacidity
C Anxiety and stress
D Depression and hypertension

The correct answer is:
Depression and hypertension

Solution:
Tranquilizers are neurologically active drugs used to treat stress and severe or mild mental disorders.
Equanil is an example of a tranquilizer that helps in controlling depression and hypertension.
This medication is used short term to treat symptoms of anxiety and nervousness. It acts on certain centres of the brain to help calm the nervous system.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A domain in a ferromagnetic substance is in the form of a cube of side $1 \mu \mathrm{m}$. If it
contains $8 \times 10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24}$
$\mathrm{A} / \mathrm{m}^{2}$, then the magnetisation of the domain is

Options:
A $7 \cdot 2 \times 10^{9} \mathrm{Am}^{-1}$
B $7 \cdot 2 \times 10^{5} \mathrm{Am}^{-1}$
C $7 \cdot 2 \times 10^{12} \mathrm{Am}^{-1}$
D $7 \cdot 2 \times 10^{3} \mathrm{Am}^{-1}$

The correct answer is:
$7 \cdot 2 \times 10^{5} \mathrm{Am}^{-1}$

Solution:
(A)
Volume of cube $=\ell^{3}=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$
Total magnetic moment $\mathrm{M}=8 \times 10^{10} \times 9 \times 10^{-24}$
$=72 \times 10^{-14} \mathrm{~A}-\mathrm{m}^{2}$
Magnitization $=\frac{\mathrm{M}}{\mathrm{V}}=\frac{72 \times 10^{-14}}{10^{-18}}=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$

Topic: Physics, Magnetic Properties of Matter, MHT CET, MHT CET 2020 (19 Oct Shift 2)
A door 1.6 m wide requires a minimum force of 1 N to be applied at the free end to open or close it. The minimum force that is required at a point 0.4 m away from the hinges for opening or closing the door is

Options:
A 1.2 N
B 2.4 N
C 3.6 N
D 4 N

The correct answer is:
4 N

Solution:
It is given in the question, torque Ï=1.6Ã1=1.6 N m
So, when d=0.4 m,
F=Ïd=1.60.4=4 N.

Topic: Physics, Rotational Motion, JEE Main
A door 1.2 mwide requires a force of $1 \mathrm{~N}$ to be applied perpendicular at the free end to open or close it. The perpendicular force required at a point 0.2 mdistant from the hinges for opening or closing the door is

Options:
A $2.4 \mathrm{~N}$
B $3.6 \mathrm{~N}$
C $6.0 \mathrm{~N}$
D $1.2 \mathrm{~N}$

The correct answer is:
$6.0 \mathrm{~N}$

Solution:
$\tau$ torque required to open/close the door will remain constant.
$\begin{aligned} & \tau=r_1 \times F_1=r_2 \times F_2 \\ & 1.2 \times 1=0.2 \times F_2 \\ & \Rightarrow F_2=6 \mathrm{~N}\end{aligned}$

Topic: Physics, Rotational Motion, MHT CET, MHT CET 2022 (10 Aug Shift 2)
A door \( 1.6 \mathrm{~m} \) wide requires a minimum force of \( 1 \mathrm{~N} \) to be applied at the free end to open or close it. The minimum force that is required at a point \( 0.4 \mathrm{~m} \) away from the hinged edge for opening or closing the door is:

Options:
A \( 1.2 \mathrm{~N} \)
B \( 2.4 \mathrm{~N} \)
C \( 3.6 \mathrm{~N} \)
D \( 4 \mathrm{~N} \)

The correct answer is:
\( 4 \mathrm{~N} \)

Solution:
Torque required in initial case is,

τ=1.6×1=1.6 N m

Now when the point of application of force is changed to, d=0.4 m

F=τd=1.60.4=4 N

Topic: Physics, Rotational Motion, JEE Main
A double convex lens has focal length $25 \mathrm{~cm}$. The radius of curvature of one of the surfaces is doubled of the other. Find the radii, if the refractive index of the material of the lens is 1.5.

Options:
A $100 \mathrm{~cm}, 50 \mathrm{~cm}$
B $25 \mathrm{~cm}, 50 \mathrm{~cm}$
C $18.75 \mathrm{~cm}, 37.5 \mathrm{~cm}$
D $50 \mathrm{~cm}, 100 \mathrm{~cm}$

The correct answer is:
$18.75 \mathrm{~cm}, 37.5 \mathrm{~cm}$

Solution:
\(\begin{aligned} & \frac{1}{\mathrm{f}}=(\mu-1)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & \frac{1}{25}=0.5\left[\frac{1}{\mathrm{R}}-\left(\frac{1}{2 R}\right)\right] \\ & \frac{1}{25}=\frac{0.5}{10}\left[\frac{3}{\mathrm{R}}\right] \\ & \mathrm{R}=\frac{75}{4}=18.75 \\ & \mathrm{R}_1=18.75 \mathrm{~cm}, \mathrm{R}_2=37.5 \mathrm{~cm}\end{aligned}\)

Topic: Physics, Ray Optics, JEE Main