datavorous/entrance-exam-dataset huggingface dataset:
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the center of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of coil will be

Options:
A nB
B n2B
C 3nB
D n3B

The correct answer is:
n2B

Solution:
Suppose the length of wire is l
Then, for the circle of n turns
The magnetic field, B= Î¼ 0 In 2R , where R is the radius of the circle
Then Bs=Î¼0 I2R =B......(i)
Here, 2ÏR=l .......(ii)
Now, if the same wire is turned into a coil of n turns of radius a then,
2ÏaÃn=l .......(iii)
Thus, 2ÏR=2ÏaÃn
â a=Rn
Now, the magnetic field for the coil of n turns, is
Bn=Î¼0I2aÃn
=Î¼0I2RÃn2 âµ a=Rn
=n2B

Topic: Physics, Magnetic Effects of Current, JEE Main
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular loop of $n$ turns. The magnetic field at the centre of the coil will be

Options:
A
$\mathrm{nB}$

B
$\mathrm{n}^2 \mathrm{~B}$

C
$2 \mathrm{nB}$

D
$2 n^2 B$

The correct answer is:
$\mathrm{n}^2 \mathrm{~B}$

Solution:
$B^{\prime}=\frac{\pi \mu_0 I}{2 r^{\prime}}=n^2 \frac{\mu_0 \pi \pi}{\ell}=n^2 B$.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2004
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is \( B \). It is then bent into a circular coil of \( n \) turns. The magnetic field at the centre of this coil of \( n \) turns will be

Options:
A \( n B \)
B \( n^{2} B \)
C \( 2 n B \)
D \( 2 n^{2} B \)

The correct answer is:
\( n^{2} B \)

Solution:
B=μ0I2r.

When n turns are made, radius becomes r′.

n×2πr′=2πr⇒r′=rn.

Now, B′=μ0nI2r′=n2μ0I2r=n2B.

Topic: Physics, Magnetic Effects of Current, JEE Main
A longitudinal wave is represented by y=10sin2πnt-xλ cm. The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to

Options:
A 2π
B 5π
C π
D 5π2

The correct answer is:
5π

Solution:
As we know particle velocity, vp=dydt=10×2πn×cos2πnt-xλ

Therefore, vpmax=20πn

Now as we know, v=λn

As given in the question,

vpmax=4v⇒20πn=4λn⇒λ=5π

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A loop ABCDEFA of straight edges has six corner points \( A(0,0,0), B(5,0,0), C(5,5,0), D(0,5,0), E(0,5,5) \) and
\( F(0,0,5) \). The magnetic field in this region is \( \vec{B}=(3 \hat{i}+4 \hat{k}) T \). The quantity of flux through the loop ABCDEFA (in
Wb ) is

Options:

The correct answer is:
175

Solution:
ϕ=B→.A→=3i^+4k^.25i^+25k^
ϕ=2×25+4×25=175weber

Topic: Physics, Electromagnetic Induction, JEE Main
A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field. Identify the effect of the field on the wire.

Options:
A Loop assumes circular shape with its plane normal to the field.
B Loop assumes circular shape with its plane parallel to the field.
C Wire gets stretched to become straight.
D Shape of the loop remains unchanged.

The correct answer is:
Loop assumes circular shape with its plane normal to the field.

Solution:
Every part (dl ) of the wire is pulled by the force i(dl)B acting perpendicular to the current & magnetic field giving it the shape of a circle.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A loop PQR carries a current i=2 A as shown in the figure. A uniform magnetic field B=2 T exists in space parallel to the plane of the loop. The magnetic torque on the loop is

Options:
A 16  N m
B 8ââNâm
C zero
D 4ââNâm

The correct answer is:
8  N m

Solution:
Given,
i=2 A, B=2 T
Area of the loop is A=12×Base×Height=2 m2

τ=iABsin⁡θ [Here, θ=90°]
τ=2×2×2×sin⁡90°=2 N m

Topic: Physics, Magnetic Effects of Current, JEE Main
(A)Major ; Product A is

Options:
A
B
C
D

The correct answer is:

Solution:
This conversion occurs as follows

Topic: Chemistry, Amines, JEE Main
A machine gun fires a bullet of mass 40 g with a velocity 1200 m s-1. The man holding it can exert a maximum force of 144 N on the gun. The maximum number of bullets that the man can fire per second is

Options:

The correct answer is:
3

Solution:
Suppose he can fire n bullets per second
∴ Force = Change in momentum per second
144=n⁡×(401000)×(1200)
n⁡=144×100040×1200
n⁡=3

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A machine gun fires a bullet of mass $40 \mathrm{~g}$ with a velocity $1200 \mathrm{~ms}^{-1}$. The man holding it can exert a maximum force of $144 \mathrm{~N}$ on the gun. How many bullets can he fire per second at the most?

Options:
A
one

B
four

C
two

D
three

The correct answer is:
three

Solution:
Change in momentum for each bullet fired is
$$
=\frac{40}{1000} \times 1200=48 \mathrm{~N}
$$
If a bullet fired exerts a force of $48 \mathrm{~N}$ on man's hand so $\rho$ man can exert maximum force of $144 \mathrm{~N}$, number of bullets that can be fired $=144 / 48=3$ bullets.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2004
A machine gun has a mass of 5 kg. It fires 50 g bullets at the rate of 30 bullets per minute at a speed of 400 m s-1. The magnitude of force required to keep the gun in position (in N) is

Options:

The correct answer is:
10

Solution:
Force required=change in momentumtime taken
= ( 50Ã 10 â3 Ã30 )Ã(400â0) 60 =10 N

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A machine gun is mounted on a $2000 \mathrm{~kg}$ car on a horizontal frictionless surface. At some instant the gun fires bullets of mass $10 \mathrm{gm}$ with a velocity of $500 \mathrm{~m} / \mathrm{sec}$ with respect to the car. The number of bullets fired per second is ten. The average thrust on the system is

Options:
A $550 \mathrm{~N}$
B $50 \mathrm{~N}$
C $250 \mathrm{~N}$
D 250 dyne

The correct answer is:
$50 \mathrm{~N}$

Solution:
$\mathrm{F}=\mathrm{mnV}$ where $\mathrm{n}$ is the no of bullets fired per sec
$\therefore \mathrm{F}=\left(10 \times 10^{-3}\right) \times 10 \times 500=50 \mathrm{~N}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of 100 m s–1 each. The recoil velocity of the gun is :

Options:
A 0.02 m s-1
B 2.5 m s-1
C 1.5 m s-1
D 0.6 m s-1

The correct answer is:
0.6 m s-1

Solution:
Number of bullets fired per second 18060=3.

Applying momentum conservation, we get

320×10-3×100=10v

⇒v=0.6 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A machine has three parts, A, B and $\mathrm{C}$, whose chances of being defective are $0.02,0.10$ and $0.05$ respectively. The machine stops working if any one of the parts becomes defective. What is the probability that the machine will not stop working?

Options:
A $0.06$
B $0.16$
C $0.84$
D $0.94$

The correct answer is:
$0.84$

Solution:
Probability that machine stops working $=P(A \cup B \cup C)$
$\Rightarrow P(A \cup B \cup C)=P(A)+P(B)+P(C)$
$-P(A \cap B)-P(A \cap C)-P(B \cap C)$
$+P(A \cap B \cap C)$
$\begin{aligned} \Rightarrow P(A \cup B \cup C)=& P(A)+P(B)+P(C) \\ &-P(A) P(B)-P(A) P(C) \\ &-P(B) P(C)+P(A) P(B) P(C) \end{aligned}$
$(\because A, B \& C$ are independent events $)$ $\Rightarrow P(A \cup B \cup C)=0.02+0.1+0.05-(0.02 \times 0.1)$
$-(0.02 \times 0.05)-(0.1 \times 0.05)$
$+(0.02 \times 0.05 \times 0.1)$
$\Rightarrow P(A \cup B \cup C)=0.16$
$\therefore$ Probability that the machine will not stop working $=1-P(A \cup B \cup C)=1-0.16=0.84$

Topic: Mathematics, Probability, JEE Main
A machine which is 75 percent efficient, uses 12 joules of energy in lifting up a $1 \mathrm{~kg}$ mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in $\mathrm{ms}^{-1}$)

Options:
A $\sqrt{24}$
B $\sqrt{32}$
C $\sqrt{18}$
D $\sqrt{9}$

The correct answer is:
$\sqrt{18}$

Solution:
Potential energy of a
$\begin{aligned}
& \text { body }=75 \% \text { of } 12 J \\
& \text { mgh }=9 J \Rightarrow h=\frac{9}{1 \times 10}=0.9 \mathrm{~m}
\end{aligned}$
Now when this mass allow to fall then it acquire velocity
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 0.9}=\sqrt{18} \mathrm{~m} / \mathrm{s} .$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A magnet freely suspended in a vibration magnetometer makes 40 oscillations per minute at place $A$ and 20 oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \mathrm{~T}$, then its value at $B$ is

Options:
A $30 \times 10^{-6} \mathrm{~T}$
B $9 \times 10^{-6} \mathrm{~T}$
C $144 \times 10^{-6} \mathrm{~T}$
D $288 \times 10^{-6} \mathrm{~T}$

The correct answer is:
$9 \times 10^{-6} \mathrm{~T}$

Solution:
Time period of magnet, $T=2 \pi \sqrt{\frac{I}{M H}}$
In both the condition magnet is same.
So,
$\begin{aligned} T & \propto \frac{1}{\sqrt{H}} \\ \frac{T_A}{T_s} & =\frac{\sqrt{H_n}}{H_A}\end{aligned}$
$\begin{gathered}\frac{60 / 40}{60 / 20}=\sqrt{\frac{H_n}{36 \times 10^{-4}}} \\ \frac{1}{2}=\frac{\sqrt{H_n}}{36 \times 10^{-6}}\end{gathered}$
Squaring both sides
$\frac{1}{4}=\frac{H_n}{36 \times 10^{-6}} \Rightarrow H_n=9 \times 10^{-4} \mathrm{~T}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnet is cut in four equal parts by cutting it parallel to its length. The time period of each part, if the time period of the original magnet in the same field is T0, will be

Options:
A T0
B T02
C T04
D 4T0

The correct answer is:
T0

Solution:
For each part, I1=I4 and M1=M4.
∴ T=2πIMB and T1=2πI1M1B
=2πI×44M×B
∴ T1T=1
∴ T1=T=T0

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnet makes 40 oscillations per minute at a place having magnetic field induction of 0.1Ã10â5 T. At another place, it takes 2.5âs to complete one vibration. The value of earth's horizontal field at that place is

Options:
A 0.25×10–6 T
B 0.36Ã10â6 T
C 0.66Ã10â8 T
D 1.2×10–6 T

The correct answer is:
0.36Ã10â6 T

Solution:
T=2ÏIMÃBHâTâ1BH
âT1T2=(BH)2(BH)1â60/402.5=(BH)20.1Ã10-5
â(BH)2=0.36Ã10-6 T

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnet of length $10 \mathrm{~cm}$ and magnetic moment $1 \mathrm{Am}^2$ is placed along side $A B$ of an equilateral triangle $A B C$. If the length of the side $A B$ is $10 \mathrm{~cm}$. The magnetic induction at the point $C$ is $\left(\mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right)$

Options:
A $10^{-9} \mathrm{~T}$
B $10^{-7} \mathrm{~T}$
C $10^{-5} \mathrm{~T}$
D $10^{-4} \mathrm{~T}$

The correct answer is:
$10^{-4} \mathrm{~T}$

Solution:
$l=10 \mathrm{~cm}, M=1 \mathrm{Am}^2$
Magnetic induction at point ${ }^{+} \mathrm{C}$
(In equatorial position),
$\begin{aligned} B & =\frac{\mu_a}{4 \pi} \frac{M}{(C O)^2}=10^{-7} \times \frac{1}{\left(\sqrt{10^2-5^2}\right)^3} \\ & =\frac{10^{-7} \times 1}{(\sqrt{75})^3}=\frac{10^{-7}}{\left(5 \sqrt{3} \times 10^{-2}\right)^3} \\ & =\frac{10^{-7}}{125 \times 3 \sqrt{3} \times 10^{-6}}=\frac{10^{-1}}{375 \times 1.732} \\ & =0.000154=1.54 \times 10^{-4} \mathrm{~T} \\ & =10^{-4} \mathrm{~T}\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnet of total magnetic moment 10-2 i^ A m2 is placed in a time varying magnetic field, Bi^cosωt where B=1 Tesla and ω=0.125 rad s-1. The work done for reversing the direction of the magnetic moment at t=1 second, is:

Options:
A 0.007 J
B 0.02 J
C 0.014 J
D 0.01 J

The correct answer is:
0.02 J

Solution:
As we know that, work done in rotating a magnetic dipole in magnetic field is,

W=MB(cosθ1-cosθ2) ...(1)

Now in the above given question we have, Magnetic Moment (M)=10-2 i^ A m2,Magnetic Field=Bi^cosωt with B=1 T , ω=0.125 rad s-1 and t=1 s

Let initial angle (θ1)=0∘ so on reversing the direction of magnetic moment final angle (θ2)=180∘

Now, substituting all the values in equation (1) we get,

⇒W=10-2×(1)cos0.125×1cos0∘-cos180∘

⇒W=10-2cos0.1251--1 (∵cos0∘=1 and cos180∘=-1)

⇒W=2×10-2×0.99

⇒W≈0.02 J

Therefore, the work done in reversing the direction of the magnetic moment is 0.02 J.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45Â° and 40 times per minute where the dip is 30Â°. If B1 and B2 are the net magnetic fields due to the earth at the two places respectively, then the ratio B1/B2 is approximately equal to

Options:
A 3.6
B 1.8
C 1.2
D 0.7

The correct answer is:
0.7

Solution:
Let f1 and f2 be the respective oscillation frequencies respectively for 45° and 30° dip. Then,

f1=12πMB1cos45°I

f2=12πMB2cos30°I

f1f2=B1cos45°B2cos30°

⇒302402=B1cos45°B2cos30°

⇒B1B2=900×cos30°1600×cos45°

⇒B1B2=0.688≈0.7

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45Â° and 40 times per minute where the dip is 30Â°. If B1 and B2 are the net magnetic fields due to the earth at the two places respectively, then the ratio B1/B2 is approximately equal to

Options:
A 3.6
B 1.8
C 1.2
D 0.7

The correct answer is:
0.7

Solution:
Let f1 and f2 be the respective oscillation frequencies respectively for 45° and 30° dip. Then,

f1=12πMB1cos45°I

f2=12πMB2cos30°I

f1f2=B1cos45°B2cos30°

⇒302402=B1cos45°B2cos30°

⇒B1B2=900×cos30°1600×cos45°

⇒B1B2=0.688≈0.7

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnetic dipole in a uniform magnetic field has: (Take zero potential energy when magnetic dipole is perpendicular to magnetic field)

Options:
A Maximum potential energy when the torque is maximum
B Zero potential energy when the torque is maximum
C Zero potential energy when the torque is minimum
D Minimum potential energy when the torque is maximum

The correct answer is:
Zero potential energy when the torque is maximum

Solution:
Torque, τ=MBsinθ
and potential energy, U=-MBcosθ
τ is maximum when θ=90o
U=-MBcos90o=0.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2017 (08 Apr Online)
A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75o. One of the fields has a magnitude of 15 mT. The dipole attains stable equilibrium at an angle of 30o with this field. The magnitude of the other field (in mT) is close to

Options:
A 1
B 11
C 36
D 1060

The correct answer is:
11

Solution:
Î¸ 1 +Î¸= 75 o
M B 1 sin Î¸ 1 =M B 2 sinÎ¸
â B 1 2 = 15 2
â B 1 = 15 2 â11 mT

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2016 (09 Apr Online)
A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75o. One of the fields has a magnitude of 15 mT. The dipole attains stable equilibrium at an angle of 30o with this field. The magnitude of the other field (in mT) is close to

Options:
A 1
B 11
C 36
D 1060

The correct answer is:
11

Solution:
Î¸ 1 +Î¸= 75 o
M B 1 sin Î¸ 1 =M B 2 sinÎ¸
â B 1 2 = 15 2
â B 1 = 15 2 â11 mT

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnetic dipole is placed in a uniform magnetic field of intensity B, oriented along the direction of the field. If the magnetic dipole moment is M, then the maximum work an external agent can perform in rotating the dipole will be

Options:
A 12MB
B 4MB
C 2MB
D MB

The correct answer is:
2MB

Solution:
The maximum work done by an external agent is

W=U180°-U0°

W=MB--MB

W=2MB

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnetic dipole is under the influence of two orthogonal magnetic fields, $B_1=0.5 \times 10^{-3} \mathrm{~T}$ and $B_2=0.866 \times 10^{-3} \mathrm{~T}$. If the dipole comes to stable equilibrium at an angle $\theta$ with respect to $B_2$ field, then the value of $\theta$ is

Options:
A $45^{\circ}$
B $30^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
From tangent law, we have, $m B_2 \sin \theta=m B_2 \sin \left(90^{\circ}-\theta\right)$
$$
\begin{aligned}
\Rightarrow \quad \tan \theta=\frac{B_2}{B_2} & =\frac{0.5 \times 10^{-3}}{0.866 \times 10^{-3}} \\
=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} & =\frac{1}{\sqrt{3}} \\
\tan \theta & =\tan 30^{\circ} \\
\theta \quad \theta & =30^{\circ}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A magnetic dipole of magnetic moment \( 6 \times 10^{-2} \mathrm{Am}^{2} \) and moment of inertia
\( 12 \times 10^{-6} \mathrm{~kg} \mathrm{~m}^{2} \) performs oscillations in a magnetic field of \( 2 \times 10^{-2} \mathrm{~T} \). The time taken by
the dipole to complete \( 20 \) oscillations is \( (\pi \simeq 3) \)

Options:
A \( 36 s \)
B \( 06 s \)
C \( 12 s \)
D \( 18 s \)

The correct answer is:
\( 12 s \)

Solution:
Given, magnetic moment, \( M=6 \times 10^{-2} \mathrm{Am}^{2} \), moment of inertia, \( I=12 \times 10^{-6} \mathrm{kgm}^{2} \); magnetic field,
\( B=2 \times 10^{-2} T \); number of oscillations \( =20 \).
Now, time period of one oscillation, \( t=2 \pi \sqrt{\frac{l}{B \times M}} \)
\( t=2 \pi \sqrt{\frac{12 \times 10^{-6}}{2 \times 10^{-2} \times 6 \times 10^{-2}}}=2 \Pi \times 10^{-1} \)
Therefore, time taken by dipole to complete \( 20 \) oscillations is
\( T=20 \times t=20 \times 2 \Pi \times 10^{-1}=20 \times 2 \times 3 \times \frac{1}{10}=12 s \)

Topic: Physics, Oscillations, JEE Main
A magnetic field can be produced by

Options:
A A moving charge
B A changing electric field
C None of these
D Both of these

The correct answer is:
Both of these

Solution:
A moving charge and changing electric field both produces magnetic field.

Topic: Physics, Magnetic Effects of Current, JEE Main
A magnetic field set up using Helmholtz coils (described in question 4.16) is uniform in a small region and has magnitude of $0.75 \mathrm{~T}$. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{-5} \mathrm{~V} \mathrm{~m}^{-1}$, make a simple guess as to what the beam contains. Why is the answer not unique?

Options:

None

Solution:
Narrow beam of charged particles remains undeflected and is perpendicular to both electric field and magnetic fields which are mutually perpendicular. So, the electric force is balanced by magnetic force.
$q E=q v \mathrm{~B}$
speed of charged particles
$v=E / B=\frac{9 \times 10^{-5}}{0.75}=12 \times 10^{-5} \mathrm{~ms}^{-1}$
Because the beam is accelerated through $15 \mathrm{kV}$, if charge is $q$, then kinetic energy gained by charged particles
$\frac{1}{2} m v^2=q V \Rightarrow \frac{m}{q}=\frac{2 V}{v^2}$
$$
=\frac{2 \times 15 \times 10^3}{\left(12 \times 10^{-5}\right)^2}=20.8 \times 10^1
$$
Here, we can only obtain charge to mass ratio and same ratio can be in Deuterium ions, $\mathrm{He}^{++}, \mathrm{Li}^{++}$, so the beam can contain any of these charged particles.

Topic: Physics, Magnetic Effects of Current
A magnetic moment of $1.73 \mathrm{BM}$ will be shown by one among the following

Options:
A $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
B $\left[\mathrm{NiCN}_4\right]^{\ell-}$
C $\mathrm{TiCl}_4$
D $\left[\mathrm{CoCl}_6\right]^{4-}$

The correct answer is:
$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

Solution:
Magneitc moment, $\mu$ is related with number of unpaired electrons as
$$
\begin{aligned}
\mu & =\sqrt{n(n+2)} \mathrm{BM} \\
(1.73)^2 & =n(n+2)
\end{aligned}
$$
On solving $n=1$
Thus, the complex/compound having one unpaired electron exhibit a magnetic moment of 1.73 $\mathrm{BM}$.
(a) $$
\begin{aligned}
& \ln \left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \\
& \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 d^9
\end{aligned}
$$
(Although in the presence of strong field ligand $\mathrm{NH}_3$, the unpaired electron gets excited to higher energy level but it still remains unpaired).
(b) $$
\begin{aligned}
& \ln \left[\mathrm{Nl}(\mathrm{CN})_4\right]^{2-} \\
& \mathrm{Ni}^{2+}=[\mathrm{Ar}] 30^8
\end{aligned}
$$
But $\mathrm{CN}^{-}$being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons $=0$.
(c) $\ln \left[\mathrm{TiCl}_4\right]$
$\mathrm{Ti}^{4+}=[\mathrm{Ar}]$ No unpaired electron.
(D) $$
\begin{aligned}
& \ln \left[\mathrm{CoCl}_6\right]^{4-} \\
& \qquad \mathrm{Co}^{2+}=[\mathrm{Ar}] 3 d^7
\end{aligned}
$$
It contains three unpaired electrons.
Thus, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ is the complex that exhibits a magnetic moment 1.73 BM.

Topic: Chemistry, Structure of Atom, NEET, NEET 2013 (All India)
A magnetic needle free to rotate in a horizontal plane is placed at the centre of a circular current-carrying coil whose axis is perpendicular to the magnetic meridian at that place. It is also known that the magnetic declination at this place is zero and in this condition, the magnetic needle is pointing towards the north-west. Now, if we reverse the direction of current in the coil, then the magnetic needle will

Options:
A point north-west
B point north-east
C point south-east
D point south-west

The correct answer is:
point north-east

Solution:
If there is no current in the coil, then the needle would point north but due to the magnetic field of the coil, it is pointing north-west. This means the magnetic field due to the coil is towards west and it is equal to the horizontal component of earth's magnetic field. So when we reverse the direction of current the magnetic field due to the coil will be towards east and the magnetic needle will point north-east.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnetic needle is kept in a non-uniform magnetic field. It experiences

Options:
A
a torque but not a force

B
neither a force nor a torque

C
a force and a torque.

D
a force but not a torque.

The correct answer is:
a force and a torque.

Solution:
In non uniform magnetic field, dipole experiences both force and torque

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2005
A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

Options:
A $2 \sqrt{3} w$
B $\sqrt{3} w$
C $\frac{\sqrt{3}}{2} w$
D $\frac{\sqrt{3}}{4} \mathrm{W}$

The correct answer is:
$\sqrt{3} w$

Solution:
Given, work done $=W$ and $\theta=60^{\circ}$ We know that
Hence,
$$
\begin{array}{l}
W=M B(1-\cos \theta) \\
W=M B\left(1-\cos 60^{\circ}\right) \\
W=\frac{M B}{2} \\
|\tau|=M B \sin 60^{\circ}=\sqrt{3} W
\end{array}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

Options:
A $\sqrt{3} W$
B $\frac{\sqrt{3}}{2} W$
C $\frac{W}{2}$
D $2 W$

The correct answer is:
$\sqrt{3} W$

Solution:
Work done
$\begin{aligned} W & =M B\left(1-\cos 60^{\circ}\right) \\ & =\frac{M B}{2}\end{aligned}$
The torque required to maintain the magnetic needle
$\begin{aligned} \tau & =M B \sin \theta \\ & =M B \sin 60^{\circ} \\ & =M B \frac{\sqrt{3}}{2} \\ \tau & =\sqrt{3} \mathrm{~W}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2012
A magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

Options:
A $\sqrt{3} W$
B $\frac{\sqrt{3}}{2} W$
C $\frac{W}{2}$
D $2 W$

The correct answer is:
$\sqrt{3} W$

Solution:
Work done
$\begin{aligned} W & =M B\left(1-\cos 60^{\circ}\right) \\ & =\frac{M B}{2}\end{aligned}$
The torque required to maintain the magnetic needle
$\begin{aligned} \tau & =M B \sin \theta \\ & =M B \sin 60^{\circ} \\ & =M B \frac{\sqrt{3}}{2} \\ \tau & =\sqrt{3} \mathrm{~W}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, TS EAMCET, TS EAMCET 2012