datavorous/entrance-exam-dataset huggingface dataset:
A ball with a speed of 9 m s-1 collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30° with the original direction. If the ratio of velocities of the balls after the collision is x:y, then what is the value of x?

Options:

The correct answer is:
1

Solution:
From conservation of momentum along y-axis.

P→iy=P→fy

0+0=mv1sin30°j^+mv2sin30°-j^

mv2sin30°=mv1sin30°

v2=v1 or v1v2=1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A ballet dancer is rotating about his own vertical axis on smooth horizontal floor with a time period 0.5 sec. The dancer folds himself close to his axis of rotation due to which his radius of gyration decreases by 20%, then his time period is

Options:
A 0.1 sec
B 0.25 sec
C 0.32 sec
D 0.4 sec

The correct answer is:
0.32 sec

Solution:
I1ω1=I2ω2
MK122πT1=MK222πT2
T 2 T 1 = ( K 2 K 1 ) 2
T2 = 0.64 × 0.5
= 0.32 sec

Topic: Physics, Rotational Motion, JEE Main
A ballet dancer suddenly folds her outstretched arms. Her angular velocity

Options:
A increases
B decreases
C remains the same
D may increase or decrease

The correct answer is:
increases

Solution:
According to law of conservation of angular momentum,
\(I \omega=\text { constant }\)...(i)
where, \(I=\) moment of inertia and \(\omega=\) angular velocity.
When a ballet dancer suddenly folds her outstretched arms, then her moment of inertia decreases, hence from Eq. (i), we observe that her angular velocity will increases.

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
A balloon carries a total load of 185 kg at normal pressure and temperature of 27°C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is -7°C. Assuming the volume constant?

Options:
A 214.15 kg
B 123.54 kg
C 219.07 kg
D 181.46 kg

The correct answer is:
123.54 kg

Solution:
ρair=gv=mg

ρair=V=185 kg

ρairgv=m'g

ρairP'P×TT'V=m'

1854576×300266=m'

m’ =123.54 kg

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A balloon filled with helium 32°C and 1.7atm bursts. Immediately after wards the expansion of helium can be considered as :

Options:
A irreversible isothermal
B irreversible adiabatic
C reversible adiabatic
D reversible isothermal

The correct answer is:
irreversible adiabatic

Solution:
A thermodynamic process that happens without transmitting heat or mass between the thermodynamic system and its surroundings is known as an adiabatic process.

An irreversible process is one in which the system and its surroundings are unable to revert to their original states.

Bursting of a balloon is an irreversible adiabatic process, because energy can not be restored in this process.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A balloon filled with oxygen is placed in a tank full of hydrogen gas at the same pressure, is pricked with a sharp pointed needle. The volume of balloon just after the pricking would be

Options:
A Shrunk
B Enlarge
C Completely collapsed
D remains unchanged in size

The correct answer is:
Enlarge

Solution:
According to graham's law of rate of diffusion, the rate of diffusion depends on pressure and molecular weight of the gas.
Rate of diffusion â partial pressure
â 1M
So According to graham's law of diffusion the rate of diffusion will be higher than the rate of diffusion of oxygen gas, So hydrogen gas will diffuse with a faster rate inside the balloon than oxygen gas will leave the balloon so balloon will enlarge in size.

Topic: Chemistry, States of Matter, JEE Main
A balloon has a volume 400 ml at 0°C . The balloon is distended to 4/5th of its maximum stretching capacity.
(i) Will burst if it is brought into a room at 40°C ?
(ii) What is the maximum temperature above which it will burst?

Options:
A Balloon will not burst at 40° C, balloon will burst above 40 ° C
B Balloon will burst at 40° C, balloon will not burst above 68.25 ° C
C Balloon will not burst at 40° C, balloon will burst above 68.25 ° C
D None of these

The correct answer is:
Balloon will not burst at 40° C, balloon will burst above 68.25 ° C

Solution:
(i) Maximum stretching capacity =400×54=500 ml
V1T1=V2T2
or 400273=V2313
V2=458.61 ml
If the volume becomes more than 500 ml it will burst. The volume is below 500 ml, hence it will not burst at 40 °C .
(ii) V1T1=V2T2
or 400273=500T2
T2=341.25 K = 68.25 °C
The temperature at which the volume becomes 500 ml the balloon will burst. So balloon burst above 68.26 °C.

Topic: Chemistry, States of Matter, JEE Main
A balloon has mass of 10 g in air. The air escapes from the balloon at a uniform rate with velocity 4.5 cm s-1. If the balloon shrinks in 5 s completely. Then, the average force acting on that balloon will be (in dyne).

Options:
A 3
B 9
C 12
D 18

The correct answer is:
9

Solution:
Given that the mass of the air m=10 g and it escapes in time t=5 s with velocity v=4.5 cm s-1.

The average force on the balloon can be given as,

F=vdmdt

⇒F=4.5×105=9 dyne

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (28 Jul Shift 1)
A balloon is moving up in air vertically above a point A on the ground. When it is a height h1, a girl standing at a distance d (point B) from A (see figure) sees it at an angle 45° with respect to the vertical. When the balloon climbs up a further height h2, it is seen at an angle 60° with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height h2 is (given tan30°=0.5774):

Options:
A 1.464 d
B 0.732 d
C 0.464 d
D d

The correct answer is:
d

Solution:
t45=h1d

h1=d .......(i)

h1+h2d+2.464d=tan30°=13=0.5774

h1+h2=(3.464)d×0.5774

h1+h2=2d .......(ii)

∴ h2=d

Topic: Physics, Motion In One Dimension, JEE Main
A balloon is moving up in air vertically above a point A on the ground. When it is a height h1, a girl standing at a distance d (point B) from A (see figure) sees it at an angle 45° with respect to the vertical. When the balloon climbs up a further height h2, it is seen at an angle 60° with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height h2 is (given tan30°=0.5774):

Options:
A 1.464 d
B 0.732 d
C 0.464 d
D d

The correct answer is:
d

Solution:
t45=h1d

h1=d .......(i)

h1+h2d+2.464d=tan30°=13=0.5774

h1+h2=(3.464)d×0.5774

h1+h2=2d .......(ii)

∴ h2=d

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A balloon is pumped at the rate of $4 \mathrm{~cm}^{3}$ per second. What is the rate at which its surface area increases and radius is $4 \mathrm{~cm} ?$

Options:
A $1 \mathrm{~cm}^{2} / \mathrm{s}$ ?
B $2 \mathrm{~cm}^{2} / \mathrm{s}$
C $3 \mathrm{~cm}^{2} / \mathrm{s}$
D $4 \mathrm{~cm}^{2} / \mathrm{s}$

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

Solution:
Let $r$ be the radius of balloon.
Balloon is like a sphere and volume of sphere $=\frac{4}{3} \pi r^{3}$
$\therefore \quad V=\frac{4}{3} \pi r^{3}$
Differentiate both side w.r.t ' $\mathrm{t}^{\prime}$
$\Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}$
$\Rightarrow 4=\frac{4}{3} \pi .3(4)^{2} \frac{d r}{d t} \quad\left(\because \frac{d V}{d t}=4 \mathrm{~cm}^{3} / \mathrm{s}\right)$
$\Rightarrow \frac{d r}{d t}=\frac{1}{16 \pi}$
Now, surface area of balloon $=S=4 \pi r^{2}$
$$
\begin{aligned}
\frac{d S}{d t} &=4 \pi \cdot 2 r \frac{d r}{d t} \\
&=4 \pi \cdot 2 \times 4 \frac{1}{16 \pi}(\text { from }(1))=2 \mathrm{~cm}^{2} / \mathrm{s}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
A balloon is rising vertically up with a velocity of $29 \mathrm{~ms}^{-1}$. A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it, was $\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)$

Options:
A $100 \mathrm{~m}$
B $200 \mathrm{~m}$
C $400 \mathrm{~m}$
D $150 \mathrm{~m}$

The correct answer is:
$200 \mathrm{~m}$

Solution:
For stone to be dropped from rising balloon of velocity $29 \mathrm{~m} / \mathrm{s}, \mathrm{u}=-29 \mathrm{~m} / \mathrm{s}, \mathrm{t}=10 \mathrm{sec} .$

$\begin{array}{l}

\therefore \mathrm{h}=-29 \times 10+\frac{1}{2} \times 9.8 \times 100 \\

=-290+490=200 \mathrm{~m} .

\end{array}$

Topic: Physics, Motion In One Dimension, JEE Main
A balloon is rising vertically upwards. At an instant, an observer on the ground, whose distance from the balloon is 100 meters, sees the balloon at an angle of elevation of 30o. If the balloon rises further vertically to a point where the angle of elevation as seen by the observer is 45o, then its height (in meters) from the ground is
(Take 3=1.73)

Options:

The correct answer is:
36.5

Solution:
Let A be the observer, P be the initial position of the balloon and it is at Q when the angle of elevation is 45 o

From triangle ABP and triangle ABQ,
âAB=503=BQ
and BP=50
So, PQ=BQ-BP=503-1

=500.73=36.5 meters

Topic: Mathematics, Heights and Distances, JEE Main
A balloon moving in a straight line passes vertically above two points A and B on a horizontal plane 10ft apart. When above A the balloon has an angle of elevation of 60o as seen from B. When above B it has an angle of elevation of 45o as seen from A. The distance of B from the point C where it will touch the plane is

Options:
A 53+1 ft
B 15 ft
C 53+3 ft
D None of these

The correct answer is:
53+1 ft

Solution:
In ∆ABE,
tan⁡45o=h10⇒h=10ft………..i
And in ∆ABD,
tan⁡60o=H10⇒H=103ft……ii
Also, H10+x=hBC (Similar triangle property)
Using equation i and ii, we get,

10310+x=10x
⇒3x=10+x
⇒x=103-1=53+1 ft

Topic: Mathematics, Heights and Distances, JEE Main
A balloon moving in a straight line passes vertically above two points A and B on a horizontal plane 300 ft apart. When above A it has an altitude of 45o as seen from B. When above B it has an altitude of 30o as seen from A. The distance of B from the point C where it will touch the plane is

Options:
A 1503+1ft
B 150ft
C 1503+3ft
D 3003+1ft

The correct answer is:
1503+1ft

Solution:
In ∆ABE,
tan⁡30o=h300⇒h=1003ft … 1
and in ΔABD,
tan⁡45o=H300⇒H=300ft … 2
Also, HAC=hBC (Similar triangles)
⇒300300+x=1003x [using Eq. 1 and 2 ]
⇒3x=300+x
⇒x=3003-1=1503+1 ft.

Topic: Mathematics, Heights and Distances, JEE Main
A balloon rises from rest with a constant acceleration of g8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is nhg seconds. Then what is the value of n?

Options:

The correct answer is:
2

Solution:
From equation v2=u2+2as
Given, u=0,a=g8
∴ We have, v=2g8h
When the stone released from this balloon. It will go upward with velocity v= gh
In this condition, time taken by stone to reach the ground,
−h=vt− 1 2 g t 2
t=vg1+1+2ghv2
= gh 2 g 1+1+2ghgh4
=2ghg=2hg

Topic: Physics, Motion In One Dimension, JEE Main
A balloon was moving upwards with a uniform velocity of 10 m s-1. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around: (takes the value of g as 10 m s-2)

Options:
A 300 m
B 200 m
C 125 m
D 250 m

The correct answer is:
125 m

Solution:
Object is projected as shown so as per motion under gravity,

S=ut+12at2

-75=+10t+12(-10)t2⇒t=5 s

Object takes t=5 s to fall on ground height of balloon from ground,

H=75+ut

=75+10×5=125 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}(2 x+1)$. Find the rate of change of its volume with respect to $x$.

Options:

None

Solution:
Dia of sphere $=\frac{3}{2}(2 x+1) ; \therefore$ Radius $=\frac{3}{4}(2 x+1)$ Vol. $=\frac{4}{3} \pi \cdot \frac{27}{64}(2 x+1)^3 \Rightarrow V=\frac{9 \pi}{16}(2 x+1)^3$
$\therefore=\frac{\mathrm{dV}}{\mathrm{dx}}=\frac{27 \pi}{8}(2 \mathrm{x}+1)^2$

Topic: Mathematics, Application of Derivatives
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10 \mathrm{~cm}$.

Options:

None

Solution:
Let $\mathrm{r}$ be the variable radius of the balloon which is in the form of sphere Vol. of the sphere
$\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 ; \frac{\mathrm{dv}}{\mathrm{dr}}=\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{4}{3} \pi \mathrm{r}^3\right)=4 \pi \mathrm{r}^2$
When $r=10, \frac{\mathrm{dv}}{\mathrm{dr}}=4 \pi \times 10 \times 10=400 \pi \mathrm{cm}^3 / \mathrm{sec}$.

Topic: Mathematics, Application of Derivatives
A balloon which always remains spherical is being inflated by pumping in \( 10 \) cubic centimeters
of gas per second. Find the rate at which the radius of the balloon is increasing when the
radius is \( 15 \mathrm{cms} \).

Options:
A \( \frac{1}{90 \Pi} \mathrm{cm} s^{-1} \)
B \( \frac{1}{9 \Pi} \mathrm{cm} s^{-1} \)
C \( \frac{1}{30 \Pi} \mathrm{cm} s^{-1} \)
D \( \frac{1}{\Pi} \mathrm{cm} s^{-1} \)

The correct answer is:
\( \frac{1}{90 \Pi} \mathrm{cm} s^{-1} \)

Solution:
As we know that, the volume of sphere is given by
\[
V=\frac{4}{3} \Pi r^{3}
\]
\[
\begin{array}{l}
\text { So, } \frac{d V}{d t}=4 \Pi r^{2} \frac{d r}{d t} \\
\text { Here } \frac{d V}{d t}=10 \mathrm{~cm}^{3} \mathrm{~s}^{-1} \text { and } r=15 \mathrm{~cm} \text { we get } \\
10=4 \Pi(15)^{2} \frac{d r}{d t} \\
\Rightarrow \frac{d r}{d t}=\frac{10}{4 \Pi\left(15^{2}\right)}=\frac{1}{90 \Pi}
\end{array}
\]

Topic: Mathematics, Statistics, KCET, KCET 2015
A balloon, which always remains spherical on inflation, is being inflated by pumping in $\mathbf{9 0 0}$ cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15 \mathrm{~cm}$.

Options:

None

Solution:
Volume of the spherical balloon $=\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3$ $\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{4}{3} \pi \mathrm{r}^3\right) \cdot \frac{\mathrm{dr}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \cdot \frac{\mathrm{dr}}{\mathrm{dt}}$ $\because \frac{\mathrm{dv}}{\mathrm{dt}}=900 \mathrm{~cm}^3 / \mathrm{sec} . \quad \therefore 900=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}$ When radius is $15 \mathrm{~cm}$, putting $r=15$ $\therefore 900=4 \times \frac{22}{7} \times 15 \times 15 \times \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{7}{22}$ Hence, the radius of the balloon is increasing at the rate of $\frac{7}{22} \mathrm{~cm} / \mathrm{sec}$. i.e. $\frac{1}{\pi} \mathrm{cm} / \mathrm{sec}$

Topic: Mathematics, Application of Derivatives
A balloon with mass $m$ is descending down with an acceleration $a$ (where, $a < g$ ). How much mass should be removed from it so that it starts moving up with an acceleration $a$ ?

Options:
A $\frac{2 m a}{g+a}$
B $\frac{2 m a}{g-a}$
C $\frac{m a}{g+a}$
D $\frac{m a}{g-a}$

The correct answer is:
$\frac{2 m a}{g+a}$

Solution:
When the balloon is descending down with an acceleration $a$,
where, $B=$ buoyant force.
Here, we should assume that while removing some mass the volume of balloon and hence buoyant force will not change.
Let the new mass of the balloon is $m^{\prime}$, so
On solving Eqs. (i) and (ii), we get
$\begin{aligned}
& m g-m^{\prime} g=m a+m^{\prime} a \\
\Rightarrow \quad & m(g-a)=m^{\prime}(g+a) \Rightarrow m^{\prime}=\frac{m(g-a)}{g+a}
\end{aligned}$
So, mass removed, $\Delta m=m-m^{\prime}$
$=m\left[1-\frac{(g-a)}{(g+a)}\right]=m\left[\frac{g+a-g+a}{g+a}\right]=\frac{2 m a}{g+a}$

Topic: Physics, Laws of Motion, JEE Main
A band playing music at a frequency f is moving towards a wall at a speed vb. A motorist is following the band with a speed vm. If v is the speed of sound, the expression for the beat frequency heard by the motorist is

Options:
A (v+vm)fv+vb
B v+vmfv-vb
C 2vb(v+vm)fv2-vb2
D 2vm(v+vb)fv2-vb2

The correct answer is:
2vb(v+vm)fv2-vb2

Solution:
The motorist receives two sound waves, direct one from the band and that reflected from the wall, figure. For direct sound waves, apparent frequency
f′=(v+vm)fv+vb
For reflected sound waves.
Frequency of sound wave reflected from the wall
f′′=v×fv-vb
Frequency of reflected waves as received by the moving motorist,
f′=v+vmf′′v=(v+vm)fv-vb
∴ Beat frequency =f′′-f′
=(v+vm)fv-vb-(v+vm)fv+vb=2vb(v+vm)fv2-vb2

Topic: Physics, Waves and Sound, JEE Main
A bandwidth of 10 MHz is available for AM transmission. If the maximum audio signal frequency used for modulating the carrier is not to exceed 5 kHz, how many stations can be broadcasted within this band simultaneously without interfering with each other?

Options:

The correct answer is:
1000

Solution:
Bandwidth required for each station
=2× maximum frequency of audio signal
2×5=10kHz
Number of stations which can be broadcast within bandwidth of 10MHz
=10×10610×103=1000

Topic: Physics, Communication System, JEE Main
A bar magnet has a magnetic moment equal to 5Ã10-5 Wb m. It is suspended in a magnetic field which has a magnetic induction B=8ÏÃ10â4âT. The magnet vibrates with a period of vibration equal to 15 s. The moment of inertia of magnet is

Options:
A 4.54Ã104 kg m2
B 4.54Ã 10 â5 kg m 2
C 4.54Ã 10 â4 kg m 2
D 4.54Ã 10 5 kgm 2

The correct answer is:
4.54Ã 10 5 kgm 2

Solution:
Here, magnetic moment is given in weber-meter, which is the unit of Î¼0 M.
â´ Î¼0 M=5Ã10-5 Wb-m
M=5Ã10-5Î¼0Am2
Also, B=8ÏÃ10-4=Î¼0H
â´ H=8ÏÃ10-4Î¼0
T=2ÏIMH
I=MHT24Ï2=5Ã10-5Ã8ÏÃ10-4Ã1524Ï2Î¼02
=5Ã10-5Ã8ÏÃ10-4Ã2254Ï24ÏÃ10-72
I=2250Ã10-9Ï(16Ï2)Ã10-14=4.54Ã105 kg m2

Topic: Physics, Magnetic Properties of Matter, JEE Main
A bar magnet has a magnetic moment of $200 \mathrm{Am}^{2}$. The magnet is suspended in a magnetic field of $0.30 \mathrm{NA}^{-1} \mathrm{m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ},$ will be

Options:
A $30 \mathrm{N}$ -m
B $30 \sqrt{3} \mathrm{N}-\mathrm{m}$
C $60 \mathrm{N}$ -m
D $60 \sqrt{3} \mathrm{N} \cdot \mathrm{m}$

The correct answer is:
$30 \mathrm{N}$ -m

Solution:
Given, $\mathrm{M}=200 \mathrm{A} \mathrm{m}^{2}$
$\mathrm{B}=0.30 \mathrm{NA}^{-} \mathrm{M}^{-1}$
and $\theta=30^{\circ}$
We know that the Torque,
$$
\begin{aligned}
\tau &=M \times B \\
\Rightarrow \quad|t| &=M B \sin \theta=200 \times 0.3 \times \frac{1}{2} \\
&=100 \times 0.3=30 \mathrm{N} \mathrm{m}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A bar magnet having a magnetic moment of 2.0×105 J T-1, is placed along the direction of uniform magnetic field of magnitude B=14×10-5 T. The work done in rotating the magnet slowly through 60° from the direction of field is

Options:
A 14 J
B 8 J
C 4 J
D 1.4 J

The correct answer is:
14 J

Solution:
Work done by the external force is equal to the change in the potential energy,

Wext=ΔU=Uf-Ui

⇒Wext=-MBcosθ2+MBcosθ1

⇒Wext=MBcosθ1-cosθ2

⇒Wext=14×10-5×2×105cos0°-cos60°

⇒Wext=14×2×12=14 J

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A bar magnet having a magnetic movement of $2 \times 10^4 \mathrm{JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=6 \times 10^{-4}$ $T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is :

Options:
A $2 \mathrm{~J}$
B $0.6 \mathrm{~J}$
C $12 \mathrm{~J}$
D $6 \mathrm{~J}$

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

Solution:
$$
\begin{aligned}
\mathrm{W} & =\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& =2 \times 10^4 \times 6 \times 10^{-4}\left(\cos \theta-\cos 60^{\circ}\right) \\
& =12 \times \frac{1}{2}=6 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A bar magnet having a magnetic movement of $2 \times 10^4 \mathrm{JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=6 \times 10^{-4}$ $T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is :

Options:
A $2 \mathrm{~J}$
B $0.6 \mathrm{~J}$
C $12 \mathrm{~J}$
D $6 \mathrm{~J}$

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

Solution:
$$
\begin{aligned}
\mathrm{W} & =\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& =2 \times 10^4 \times 6 \times 10^{-4}\left(\cos \theta-\cos 60^{\circ}\right) \\
& =12 \times \frac{1}{2}=6 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2009 (Mains)
A bar magnet having magnetic moment $M$ is bent as an arc. Its magnetic moment

Options:
A decreases
B increases
C does not change
D may increase or decrease

The correct answer is:
decreases

Solution:
Magnetic moment of a bar magnet,
$\mathbf{M}=m(2 \mathbf{l})$
$=$ pole strength $\times$ distance between poles
When bar magnet is bent as an arc distance between poles decreases whereas pole strength remains same.
So, its magnetic moment also, decreases.

Topic: Physics, Magnetic Effects of Current, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 2)
A bar magnet is $10 \mathrm{~cm}$ long is kept with its north $(N)$-pole pointing north. A neutral point is formed at a distance of $15 \mathrm{~cm}$ from each pole. Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet is

Options:
A 9 A-m
B 6.75 A-m
C 27 A-m
D 1.35 A-m

The correct answer is:
6.75 A-m

Solution:
$\begin{aligned} & \text { Length of magnet }=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}, \\ & \qquad r=15 \times 10^{-2} \mathrm{~m}\end{aligned}$
$O P=\sqrt{225-25}=\sqrt{200} \mathrm{~cm}$
Since, at the neutral point, magnetic field due to the magnet is equal to $B_H$
$\begin{gathered}
B_H=\frac{\mu_0}{4 \pi} \cdot \frac{M}{\left(O P^2+A O^2\right)^{3 / 2}} \\
0.4 \times 10^{-4}=10^{-7} \times \frac{M}{\left(200 \times 10^{-4}+25 \times 10^{-4}\right)^{3 / 2}} \\
\frac{0.4 \times 10^{-4}}{10^{-7}} \times\left(225 \times 10^{-4}\right)^{3 / 2}=M \\
0.4 \times 10^3 \times 10^{-6}(225)^{3 / 2}=M \\
M=1.35 \mathrm{~A}-\mathrm{m}
\end{gathered}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A bar magnet is cut into two equal halves by a plane parallel to the magnetic axis. The following physical quantities, the one which remains unchanged is

Options:
A Pole strengths
B Magnetic moment
C Intensity of magnetization
D Moment of inertia

The correct answer is:
Intensity of magnetization

Solution:
For each half pole strength m becomes half
M=mÃ2l becomes half and volume V=aÃ2l also becomes half therefore, magnetic intensity remains constant.

Topic: Physics, Magnetic Properties of Matter, NEET
A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The coercivity of the bar magnet is:

Options:
A 285 A/m
B 2600 A/m
C 520 A/m
D 1200 A/m

The correct answer is:
2600 A/m

Solution:
Coercivity =nI=1000.2×5.2 A/m
=2600 A/m

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

Options:
A motion remains simple harmonic with time period $=\frac{T}{2}$
B motion remains S.H.M with time period $=2 T$
C motion remains S.H.M. with time period $=4 T$
D motion remains S.H.M and period nearly constant

The correct answer is:
motion remains S.H.M with time period $=2 T$

Solution:
Time period of bar magnet $T$
$$
=2 \pi \sqrt{\frac{I}{14 H}}
$$
Where $\mathrm{I}=$ M.I. $=\frac{m \mathrm{~L}^2}{12}$
$m=$ mass of magnet
$\therefore T=2 \pi \sqrt{\frac{m L^2}{12 \mathrm{MH}}}$
$\Rightarrow T \propto \sqrt{m}$
$\therefore$ If mass is quadruplet then $T$ becomes twice.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2003