MCQ 511 Mark
A bar magnet has a magnetic moment equal to $5 \times 10^{-5}$ weber $\times m$. It is suspended in a magnetic field which has a magnetic induction $(B)$ equal to $8 \pi \times 10^{-4}$ tesla The magnet vibrates with a period of vibration equal to $15 \mathrm{sec}$. The moment of inertia of the magnet is
- A
$22.5 \mathrm{~kg} \times \mathrm{m}^2$
- B
$11.25 \times \mathrm{kg} \times \mathrm{m}^2$
- C
$5.62 \times \mathrm{kg} \times \mathrm{m}^2$
- ✓
$7.16 \times 10^{-7} \mathrm{~kg}-\mathrm{m}^2$
AnswerCorrect option: D. $7.16 \times 10^{-7} \mathrm{~kg}-\mathrm{m}^2$
Time period of a magnet $T=2 \pi \sqrt{\frac{I}{M B}}$
$ \text { or } I=\frac{T^2 M B}{4 \pi^2}=\frac{225 \times 5 \times 10^{-5} \times 8 \pi \times 10^{-4}}{4 \pi^2}$
$ \therefore I=7.16 \times 10^{-7} \mathrm{~kg}-\mathrm{m}^2$
View full question & answer→MCQ 521 Mark
A tangent galvanometer has a coil with 50 turns and radius equal to $4 \mathrm{~cm}$. A current of $0.1 \mathrm{~A}$ is passing through it. The plane of the coil is set parallel to the earth's magnetic meridian. If the value of the earth's horizontal component of the magnetic field is $7 \times 10^{-5}$ tesla and $\mu_0=4 \pi \times 10^{-7}$ weber / amp $\times m$, then the deflection in the galvanometer needle will be
- A
$45$
- ✓
$48.2$
- C
$50.7$
- D
$52.7$
AnswerCorrect option: B. $48.2$
For tangent galvanometer $I=\frac{2 r B}{\mu_0 n} \tan \theta$
$ \therefore \tan \theta=\frac{I \mu_0 n}{2 r B}=\frac{0.1 \times 4 \pi \times 10^{-7} \times 50}{0.04 \times 7 \times 10^{-5} \times 2}=1.12 $
$ \text { or } \theta=\tan ^{-1}(1.12)=48.2^{\circ}$
View full question & answer→MCQ 531 Mark
A bar magnet of length $10 \mathrm{~cm}$ and having the pole strength equal to $10.$ weber is kept in a magnetic field having magnetic induction $(B)$ equal to $4 \pi \times 10^{-3}$ Tesla. It makes an angle of $30$ with the direction of magnetic induction. The value of the torque acting on the magnet is $\left(\mu_0=4 \pi \times 10^{-7}\right.$ weber / amp $\left.\times \mathrm{m}\right)$
- ✓
$2 \pi \times 10^{-7} \mathrm{~N} \times \mathrm{m}$
- B
$2 \pi \times 10^{-5} \mathrm{~N} \times \mathrm{m}$
- C
$0.5 \mathrm{~N} \times \mathrm{m}$
- D
$0.5 \times 10^2 \mathrm{~N} \times \mathrm{m}$
AnswerCorrect option: A. $2 \pi \times 10^{-7} \mathrm{~N} \times \mathrm{m}$
Torque $\tau=M B_H \sin \theta$
$ =0.1 \times 10^{-3} \times 4 \pi \times 10^{-3} \times \sin 30^{\circ}=10^{-7} \times 4 \pi \times \frac{1}{2}$
$ =2 \pi \times 10^{-7} N \times m$
View full question & answer→MCQ 541 Mark
A small bar magnet $A$ oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is 60 . When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be
- ✓
$\frac{T}{\sqrt{2}}$
- B
$T$
- C
$\sqrt{2} T$
- D
$2 T$
AnswerCorrect option: A. $\frac{T}{\sqrt{2}}$
$ T=2 \pi \sqrt{\frac{I}{M B}} \Rightarrow \frac{T}{T^{\prime}}=\sqrt{\frac{B^{\prime}}{B}}=\sqrt{\frac{B}{B_H}}$
$ \Rightarrow \frac{T}{T^{\prime}}=\sqrt{\frac{1}{\cos \phi}}=\sqrt{\frac{1}{\cos 60^{\circ}}}=\sqrt{2} $
$\Rightarrow T^{\prime}=\frac{T}{\sqrt{2}}$
View full question & answer→MCQ 551 Mark
The time period of oscillation of a magnet in a vibration magnetometer is $1.5$ seconds. The time period of oscillation of another magnet similar in size, shape and mass but having onefourth magnetic moment than that of first magnet, oscillating at same place will be
- A
$0.75 \mathrm{sec}$
- B
$1.5 \mathrm{sec}$
- ✓
$3 \mathrm{sec}$
- D
$6 \mathrm{sec}$
AnswerCorrect option: C. $3 \mathrm{sec}$
$T \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{M_2}{M_1}} $
$\Rightarrow \frac{1.5}{T_2}=\sqrt{\frac{M_1 / 4}{M_1}}=\frac{1}{2}$
$\Rightarrow T_2=3 \mathrm{sec}$
View full question & answer→MCQ 561 Mark
A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes $20$ oscillations per minute at a place where dip angle is $30$ and $15$ oscillations per minute at a place where dip angle is $60.$ The ratio of total earth's magnetic field at the two places is
- A
$3 \sqrt{3}: 8$
- ✓
$16: 9 \sqrt{3}$
- C
$4: 9$
- D
$2 \sqrt{3}: 9$
AnswerCorrect option: B. $16: 9 \sqrt{3}$
Given $v_1=\frac{20}{60}=\frac{1}{3} \sec ^{-1}$ and $v_2=\frac{15}{60}=\frac{1}{4} \mathrm{sec}^{-1}$
$ \text { Now } v=\frac{1}{2 \pi} \sqrt{\frac{M B_H}{I}}=\frac{1}{2 \pi} \sqrt{\frac{M B \cos \phi}{I}}\left(\because B_H=B \cos \phi\right) $
$ \therefore \frac{v_1}{v_2}=\sqrt{\frac{B_1 \cos \phi_1}{B_2 \cos \phi_2}} $
$\Rightarrow \frac{B_1}{B_2}=\left(\frac{v_1}{v_2}\right)^2\left(\frac{\cos \phi_2}{\cos \phi_1}\right)^2 $
$ \Rightarrow \frac{B_1}{B_2}=\left(\frac{1 / 3}{1 / 4}\right)^2 \frac{\cos 60^{\circ}}{\cos 30^{\circ}}=\frac{16}{9} \times \frac{1 / 2}{\sqrt{3} / 2}=\frac{16}{9 \sqrt{3}} .$
View full question & answer→MCQ 571 Mark
In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth's magnetic field is $2 \mathrm{sec}$. When a magnet is brought near and parallel to it, the time period reduces to $1 \mathrm{sec}$. The ratio $H / F$ of the horizontal component $H$ and the field $F$ due to magnet will be
- A
- B
(b) $1 / 3$
- C
(c) $\sqrt{3}$
- D
(d) $1 / \sqrt{3}$
View full question & answer→MCQ 581 Mark
The magnetic field at a point $x$ on the axis of a small bar magnet is equal to the field at a point $y$ on the equator of the same magnet. The ratio of the distances of $x$ and $y$ from the centre of the magnet is
- A
$2^{-3}$
- B
$2^{-1 / 3}$
- C
$2^3$
- ✓
$2^{1 / 3}$
AnswerCorrect option: D. $2^{1 / 3}$
$B_1=\frac{2 M}{x^3}$ and $B_2=\frac{M}{y^3}$As $B_1=B_2$
Hence $\frac{2 M}{x^3}=\frac{M}{y^3}$ or $\frac{x^3}{y^3}=2$ or $\frac{x}{y}=2^{1 / 3}$
View full question & answer→MCQ 591 Mark
At a certain place a magnet makes $30$ oscillations per minute. At another place where the magnetic field is double, its time period will be
AnswerCorrect option: D. $\sqrt{2} \mathrm{sec}$
$T=2 \pi \sqrt{\frac{I}{M B_H}} ; $
$\therefore \frac{T_1}{T_2}=\sqrt{\frac{\left(B_H\right)_2}{\left(B_H\right)_1}} $
$\Rightarrow T_2=T_1\sqrt{\frac{\left(B_H\right)_1}{\left(B_H\right)_2}}$
Here $n=30$ oscillation $/ \mathrm{min}=\frac{1}{2}$ oscillation $/ \mathrm{sec}$
$ \therefore T_1=\frac{1}{n_1}=2 \mathrm{sec} $
$ \therefore T_2=2 \sqrt{\frac{B_H}{2 B_H}}=2 \times \frac{1}{\sqrt{2}}=\sqrt{2} \mathrm{sec}$
View full question & answer→MCQ 601 Mark
In sum and difference method in vibration magnetometer, the time period is more if
- A
Similar poles of both magnets are on same sides
- ✓
Opposite poles of both magnets are on same sides
- C
Both magnets are perpendicular to each other
- D
AnswerCorrect option: B. Opposite poles of both magnets are on same sides
In sum position : $T_S=2 \pi \sqrt{\frac{I_s}{\left(M_1+M_2\right) B_H}}$In difference position : $T_d=2 \pi \sqrt{\frac{I_d}{\left(M_1-M_2\right) B_H}}$lt is clear that $T_d>T_s$
View full question & answer→MCQ 611 Mark
A magnetic needle is kept in a non-uniform magnetic field. lt experiences
- ✓
- B
- C
- D
Neither a torque nor a force
View full question & answer→MCQ 621 Mark
At a place, the horizontal and vertical intensities of earth's magnetic field is $0.30$ Gauss and $0.173$ Gauss respectively. The angle of dip at this place is
- ✓
$30^{\circ}$
- B
$90^{\circ}$
- C
$60^{\circ}$
- D
$45^{\circ}$
AnswerCorrect option: A. $30^{\circ}$
$\tan \phi=\frac{B_V}{B_H}=\frac{0.173}{0.30}=\frac{1.73}{3.0}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} \Rightarrow \phi=30^{\circ}$
View full question & answer→MCQ 631 Mark
A magnet of magnetic moment $M$ is situated with its axis along the direction of a magnetic field of strength B. The work done in rotating it by an angle of 180 will be
- A
$-M B$
- B
$+M B$
- C
$0$
- ✓
$+2 M B$
AnswerCorrect option: D. $+2 M B$
Work done $M B\left(\cos \theta_1-\cos \theta_2\right)$$\begin{aligned}& \theta_1=0^{\circ} \text { and } \theta_2=180^{\circ} \\& \Rightarrow W=M B(\cos 0-\cos 180)=2 M B\end{aligned}$
View full question & answer→MCQ 641 Mark
If magnetic lines of force are drawn by keeping magnet vertical, then number of neutral points will be
MCQ 651 Mark
At two places $A$ and $B$ using vibration magnetometer, a magnet vibrates in a horizontal plane and its respective periodic time are $2$ sec and $3 \mathrm{sec}$ and at these places the earth's horizontal components are $H$ and $H$ respectively. Then the ratio between $H$ and $H$ will be
- ✓
$9: 4$
- B
$3: 2$
- C
$4: 9$
- D
$2: 3$
AnswerCorrect option: A. $9: 4$
$T=2 \pi \sqrt{\frac{I}{M H}} \Rightarrow T \propto \frac{1}{\sqrt{H}} \Rightarrow \frac{T_A}{T_B}=\sqrt{\frac{H_B}{H_A}}$
$ \Rightarrow \frac{H_A}{H_B}=\left(\frac{T_B}{T_A}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}$
View full question & answer→MCQ 661 Mark
At a place the earth's horizontal component of magnetic field is $0.36 \times 10^{-4}$ weber $/ \mathrm{m}^2$. If the angle of dip at that place is 60 , then the vertical component of earth's field at that place in weber $/ m$ will be approximately
- A
$0.12 \times 10^{-4}$
- B
$0.24 \times 10^{-4}$
- C
$0.40 \times 10^{-4}$
- ✓
$0.62 \times 10^{-4}$
AnswerCorrect option: D. $0.62 \times 10^{-4}$
From the relation $B_H=B \cos \phi$ and $B_V=B \sin \phi$
$ \frac{B_V}{B_H}=\tan \phi \text { or } B_V=B_H \tan \phi$
$=0.36 \times 10^{-4} \times \tan 60^{\circ}=0.623 \times 10^{-4} \mathrm{~Wb} /\mathrm{m}^2$
View full question & answer→MCQ 671 Mark
At a certain place, the horizontal component of earth's magnetic field is $\sqrt{3}$ times the vertical component. The angle of dip at that place is
- A
$60^{\circ}$
- B
$45^{\circ}$
- C
$90^{\circ}$
- ✓
$30^{\circ}$
AnswerCorrect option: D. $30^{\circ}$
$B_H=\sqrt{3} B_V$, also $\tan \theta=\frac{B_V}{B_H}=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}$
View full question & answer→MCQ 681 Mark
Vibration magnetometer is used for comparing
MCQ 691 Mark
Magnetic moments of two bar magnets may be compared with the help of
MCQ 701 Mark
The radius of the coil of a Tangent galvanometer. which has 10 turns is $0.1 \mathrm{~m}$. The current required to produce a deflection of $60^{\circ}$ $\left(B_H=4 \times 10^{-5} T\right)$ is
- A
$3 \mathrm{~A}$
- ✓
$1.1 \mathrm{~A}$
- C
$2.1 \mathrm{~A}$
- D
$1.5 \mathrm{~A}$
AnswerCorrect option: B. $1.1 \mathrm{~A}$
$ B=B_H \tan \theta \Rightarrow \frac{\mu_0 n i}{2 r}=B_H \tan \theta$
$ \Rightarrow i=\frac{2 r \cdot B_H \tan \theta}{\mu_0 n}=\frac{2 \times 0.1 \times 4 \times 10^{-5}}{10 \times 4 \pi \times 10^{-7}}=1.1 \mathrm{~A}$
View full question & answer→MCQ 711 Mark
A magnet of length $0.1 \mathrm{~m}$ and pole strength $10^{-4} \mathrm{~A} . \mathrm{m}$. is kept in a magnetic field of $30 \mathrm{~Wb} / \mathrm{m}^2$ at an angle $30^{\circ}$. The couple acting on it is $\times 10^{-4} \mathrm{Nm}$.
- A
$7.5$
- B
$3.0$
- ✓
$1.5$
- D
$ 6.0$
Answer$\tau =M B \sin \theta=m \times(2 l) \times B \sin \theta $
$=10^{-4} \times 0.1 \times 30 \sin 30^{\circ}=1.5 \times 10^{-4} \mathrm{Nm}$
View full question & answer→MCQ 721 Mark
Which one of the following is a non-magnetic substance
MCQ 731 Mark
In a tangent galvanometer a current of $0.1 \mathrm{~A}$ produces a deflection of $30.$ The current required to produce a deflection of $60$ is
- A
$0.2 \mathrm{~A}$
- ✓
$0.3 \mathrm{~A}$
- C
$0.4 \mathrm{~A}$
- D
$0.5 \mathrm{~A}$
AnswerCorrect option: B. $0.3 \mathrm{~A}$
$ i \propto \tan \phi \Rightarrow \frac{i_1}{i_2}=\frac{\tan \phi_1}{\tan \phi_2}$
$ \Rightarrow \frac{0.1}{i_2}=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}=\frac{1}{3} \Rightarrow i_2=0.3 \mathrm{~A}$
View full question & answer→MCQ 741 Mark
A small bar magnet has a magnetic moment $1.2 \mathrm{~A}-\mathrm{m}$. The magnetic field at a distance $0.1 \mathrm{~m}$ on its axis will be : $(\mu=4 \pi \times 10^{-7}T-m / A)$
- A
$1.2 \times 10^5 T$
- ✓
$2.4 \times 10^{-4}T$
- C
$2.4 \times 10 . T$
- D
$1.2 \times 10^4 T$
AnswerCorrect option: B. $2.4 \times 10^{-4}T$
$B=\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{d^3} $
$\Rightarrow B=10^{-7} \times \frac{2 \times 1.2}{(0.1)^3}=2.4 \times 10^{-4} T$
View full question & answer→MCQ 751 Mark
The ultimate individual unit of magnetism in any magnet is called
MCQ 761 Mark
If the angular momentum of an electron is $\vec{J}$ then the magnitude of the magnetic moment will be
- A
$\frac{e J}{m}$
- ✓
$\frac{e J}{2 m}$
- C
el $2 m$
- D
$\frac{2 m}{e J}$
AnswerCorrect option: B. $\frac{e J}{2 m}$
As we know for circulating electron magnetic moment$M=\frac{1}{2} e v r$and angular momentum $J=m v r$
From equation (i) and (ii) $M=\frac{e J}{2 m}$
View full question & answer→MCQ 771 Mark
The error in measuring the current with a tangent galvanometer is minimum when the deflection is about
- A
$0^\circ$
- B
$30^\circ$
- ✓
$45^\circ$
- D
$60^\circ$
AnswerCorrect option: C. $45^\circ$
In case of tangent galvanometer as$i=k \tan \phi$Differentiating both side w.r.t. $\phi$
$ \frac{d i}{d \phi}=k \sec ^2 \phi \Rightarrow d i=k \sec ^2 d \phi $
$ \Rightarrow \frac{d i}{i}=\frac{d \phi}{\sin \phi \cos \phi}=\frac{2 d \phi}{\sin 2 \phi}$
Hence the error in the measurement will be least when
$\sin 2 \phi=\max =1 \Rightarrow 2 \phi=90^{\circ} \Rightarrow \phi=45^{\circ}$
View full question & answer→MCQ 781 Mark
The value of angle of dip is zero at the magnetic equator because on it
AnswerCorrect option: C. The value of $V$ is zero
View full question & answer→MCQ 791 Mark
At a certain place the horizontal component of the earth's magnetic field is $B$ and the angle of dip is $45.$ The total intensity of the field at that place will be
- A
$B$
- ✓
$\sqrt{2} B_0$
- C
$2 B$
- D
$B_0^2$
AnswerCorrect option: B. $\sqrt{2} B_0$
$B_H=B \sin \phi \Rightarrow B=\frac{B_H}{\sin \phi} \Rightarrow B=\frac{B_o}{\sin 45^{\circ}}=\sqrt{2} B_0$
View full question & answer→MCQ 801 Mark
A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S. unit is $M$, the length of wire will be
AnswerCorrect option: B. $\sqrt{\frac{4 \pi M}{i}}$
Magnetic moment of circular loop carrying current
$M=I A=I\left(\pi R^2\right)=I \pi\left(\frac{L}{2 \pi}\right)^2=\frac{I L^2}{4 \pi} \Rightarrow L=\sqrt{\frac{4 \pi M}{I}}$
View full question & answer→MCQ 811 Mark
The incorrect statement regarding the lines of force of the magnetic field $B$ is
- A
Magnetic intensity is a measure of lines of force passing through unit area held normal to it
- B
Magnetic lines of force form a close curve
- ✓
Inside a magnet, its magnetic lines of force move from north pole of a magnet towards its south pole
- D
Due to a magnet magnetic lines of force never cut each other
AnswerCorrect option: C. Inside a magnet, its magnetic lines of force move from north pole of a magnet towards its south pole
Inside a magnet, magnetic lines of force move form south pole to north pole.
View full question & answer→MCQ 821 Mark
To measure which of the following, is a tangent galvanometer used
MCQ 831 Mark
The S.l. unit of magnetic permeability is
- A
$A m^{-1}$
- B
$\mathrm{Am}$
- ✓
$\mathrm{Henrym}^{-1}$
- D
No unit, it is a dimensionless number
AnswerCorrect option: C. $\mathrm{Henrym}^{-1}$
View full question & answer→MCQ 841 Mark
The unit of magnetic moment is
AnswerCorrect option: D. $A . m^2$
View full question & answer→MCQ 851 Mark
- A
- ✓
Attracts only magnetic substances
- C
Attracts magnetic substances and repels all non-magnetic substances
- D
Attracts non-magnetic substances and repels magnetic substances
AnswerCorrect option: B. Attracts only magnetic substances
View full question & answer→MCQ 861 Mark
The field due to a magnet at a distance $R$ from the centre of the magnet is proportional to
- A
$R^2$
- B
$R^3$
- C
$1 / R^2$
- ✓
$1 / R^3$
AnswerCorrect option: D. $1 / R^3$
$1 / R^3$
View full question & answer→MCQ 871 Mark
The value of the horizontal component of the earth's magnetic field and angle of dip are $1.8 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$ and $30^{\circ}$ respectively at some place. The total intensity of earth's magnetic field at that place will be
- ✓
$2.08 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$
- B
$3.67 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$
- C
$3.18 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$
- D
$5.0 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$
AnswerCorrect option: A. $2.08 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$
Horizontal component $B_H=B \cos \phi$ Total intensity of earth magnetic field
$B=\frac{B_H}{\cos \phi}$
$=\frac{1.8 \times 10^5}{\cos 30^{\circ}}=\frac{1.8 \times 10^{-5}}{\sqrt{3} / 2}=2.08 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$
View full question & answer→MCQ 881 Mark
Two tangent galvanometers having coils of the same radius are connected in series. A current flowing in them produces deflections of $60^{\circ}$ and $45^{\circ}$ respectively. The ratio of the number of turns in the coils is
AnswerCorrect option: D. $\sqrt{3} / 1$
In the first galvanometer$i_1=K_1 \tan \theta_1=K_1 \tan 60^{\circ}=K_1 \sqrt{3}$In the second galvanometer
$ i_2=K_2 \tan \theta_2=K_2 \tan 45^{\circ}=K_2 $
$ \text { In series } i=i \Rightarrow K_1 \sqrt{3}=K_2 \Rightarrow \frac{K_1}{K_2}=\frac{1}{\sqrt{3}}$
$\text { But } K \propto \frac{1}{n} \Rightarrow \frac{K_1}{K_2}=\frac{n_2}{n_1} \quad \therefore \frac{n_1}{n_2}=\frac{\sqrt{3}}{1} .$
View full question & answer→MCQ 891 Mark
A ferromagnetic material is heated above its curie temperature. Which one is a correct statement
- A
Ferromagnetic domains are perfectly arranged
- ✓
Ferromagnetic domains becomes random
- C
Ferromagnetic domains are not influenced
- D
Ferromagnetic material changes itself into diamagnetic material
AnswerCorrect option: B. Ferromagnetic domains becomes random
On heating, different domains have net magnetisation in them which are randomly distributes. Thus the net magnetisastion of the substance due to various domains decreases to minimum.
View full question & answer→MCQ 901 Mark
A bar magnet of magnetic moment $10 \% / T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10 \quad T$ to a direction $60^{\circ}$ from the field will be
AnswerCorrect option: A. $0.2 \mathrm{~J}$
Magnetic moment of bar $M=10^4 \mathrm{~J} / T$$B=4 \times 10^{-5} T$
Hence work done $W=\vec{M} \cdot \vec{B}$$=10^4 \times 4 \times 10^{-5} \times \cos 60^{\circ}=0.2 \mathrm{~J}$
View full question & answer→MCQ 911 Mark
Intensity of magnetic field due to earth at a point inside a hollow steel box is
AnswerNo magnetic lines of force passes through the steel box.
View full question & answer→MCQ 921 Mark
Rate of change of torque $\tau$ with deflection $\theta$ is maximum for a magnet suspended freely in a uniform magnetic field of induction $B$, when
- ✓
$\theta=0^{\circ}$
- B
$\theta=45^{\circ}$
- C
$\theta=60^{\circ}$
- D
$\theta=90^{\circ}$
AnswerCorrect option: A. $\theta=0^{\circ}$
$\tau=M B_H \sin \theta$ or $\frac{d \tau}{d \theta}=M B_H \cos \theta$
This will be maximum. when $\theta=0^{\circ}$.
View full question & answer→MCQ 931 Mark
- A
- ✓
- C
Tend to crowd far away from the poles of magnet
- D
Do not pass through vacuum
View full question & answer→MCQ 941 Mark
The angle of dip at a place on the earth gives
- A
The horizontal component of the earth's magnetic field
- B
The location of the geographic meridian
- C
The vertical component of the earth's field
- ✓
The direction of the earth's magnetic field
AnswerCorrect option: D. The direction of the earth's magnetic field
View full question & answer→MCQ 951 Mark
Tangent galvanometer is used to measure
- ✓
- B
- C
Magnetic moments of bar magnets
- D
View full question & answer→MCQ 961 Mark
Vibration magnetometer works on the principle of
- ✓
Torque acting on the bar magnet
- B
Force acting on the bar magnet
- C
Both the force and the torque acting on the bar magnet
- D
AnswerCorrect option: A. Torque acting on the bar magnet
View full question & answer→MCQ 971 Mark
Magnetic field intensity is defined as
- A
Magnetic moment per unit volume
- ✓
Magnetic induction force acting on a unit magnetic pole
- C
Number of lines of force crossing per unit area
- D
Number of lines of force crossing per unit volume
AnswerCorrect option: B. Magnetic induction force acting on a unit magnetic pole
Number of lines of force passing through per unit area normally is intensity of magnetic field, hence option (c) is incorrect. The correct option is (b).
View full question & answer→MCQ 981 Mark
The time period of a bar magnet suspended horizontally in the earth's magnetic field and allowed to oscillate
- A
Is directly proportional to the square root of its mass
- B
Is directly proportional to its pole strength
- C
Is inversely proportional to its magnetic moment
- ✓
Decreases if the length increases but pole strength remains same
AnswerCorrect option: D. Decreases if the length increases but pole strength remains same
Decreases if the length increases but pole strength remains same
View full question & answer→MCQ 991 Mark
A magnet of magnetic moment $20$ C.G.S. units is freely suspended in a uniform magnetic field of intensity $0.3$ C.G.S. units. The amount of work done in deflecting it by an angle of $30$ in C.G.S. units is
- A
$6$
- B
$3 \sqrt{3}$
- ✓
$3(2-\sqrt{3)}$
- D
$3$
AnswerCorrect option: C. $3(2-\sqrt{3)}$
Work done $W=M B_H(1-\cos \theta)$
$=20 \times 0.3\left(1-\cos 30^{\circ}\right)=6\left(1-\frac{\sqrt{3}}{2}\right)=3(2-\sqrt{3})$
View full question & answer→MCQ 1001 Mark
Two magnets $A$ and $B$ are identical in mass, length and breadth but have different magnetic moments. In a vibration magnetometer, if the time period of $B$ is twice the time period of $A$. The ratio of the magnetic moments $M_A / M_B$ of the magnets will be
Answer$T=2 \pi \sqrt{\frac{I}{M B_H}} \Rightarrow T \propto \frac{1}{\sqrt{M}}$
$ \Rightarrow \frac{M_A}{M_B}=\left(\frac{T_B}{T_A}\right)^2=\frac{4}{1}$
View full question & answer→